tag:blogger.com,1999:blog-14205339509046064522024-03-14T04:02:47.468+08:00Learn Electrical Engineering for BeginnersAnonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.comBlogger30125tag:blogger.com,1999:blog-1420533950904606452.post-87219266542181011452013-06-15T15:15:00.000+08:002013-06-15T15:15:01.231+08:00Generation of a Sine Wave of Voltage <div dir="ltr" style="text-align: left;" trbidi="on">
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There are two facts that the <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Voltage">voltage</a> developed in a coil of a generator changes; the first one is<i> it changes in magnitude from instant to instant as varying values of flux are cut per second</i> and the other one is <i>it changes in direction as coil side change positions under north and south poles</i>, implies that alternating emf is generated. This means that the voltage is maximum as mentioned in our last topic<a href="http://electricalengineeringforbeginners.blogspot.com/2013/06/generation-of-alternating-emfs_14.html"> here</a> when the position of the coil is just like shown in the figure below:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-vcQ-nY3j7gA/UbvfGITs7kI/AAAAAAAAAhc/hywPwJN3lCM/s1600/two+pole+ac+generator.jpg" imageanchor="1"><img border="0" height="175" src="http://3.bp.blogspot.com/-vcQ-nY3j7gA/UbvfGITs7kI/AAAAAAAAAhc/hywPwJN3lCM/s1600/two+pole+ac+generator.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Initial position of the coil</td></tr>
</tbody></table>
and will diminish to zero as the coil rotates clockwise toward the position as shown below:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-S3RuYDjh4V0/UbvgLXJeYXI/AAAAAAAAAhs/sZmgih0B14w/s1600/AC+Generator+one+quarter.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="198" src="http://4.bp.blogspot.com/-S3RuYDjh4V0/UbvgLXJeYXI/AAAAAAAAAhs/sZmgih0B14w/s320/AC+Generator+one+quarter.png" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">As the coil rotates clockwise</td></tr>
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Then, as the coil continues to rotate clockwise, the polarities will change. Assuming uniform flux distribution between north and south poles, the generated voltage in a coil located from the vertical will be:<br />
<br />
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<br />
e = Em sin α<br />
<br />
Consider the figure below for us to analyze why this relationship mentioned above happened.<br />
<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-dNqXi1Jx2NE/Ubv9VZxHMzI/AAAAAAAAAiM/IFHqtKpxIS0/s1600/instanteneous+voltage.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="279" src="http://1.bp.blogspot.com/-dNqXi1Jx2NE/Ubv9VZxHMzI/AAAAAAAAAiM/IFHqtKpxIS0/s640/instanteneous+voltage.jpg" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Illustrating the generated voltage is proportional to sin alpha </td></tr>
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It was come up to the relationship between instantaneous voltage <b>e</b> and maximum voltage <b>Em</b> is that a coil side such as<b> a</b>, moving tangentially to a circle as indicated, cut lines of force in proportion to its vertical component of the motion. If the vector length <b>ay </b>in the figure above represents a constant rotating velocity, it should be obvious that vector <b>xy </b>is, its vertical component; the vector length <b>ax </b>is the horizontal component and it emphasize that motion in this direction involves no flux- cutting action. Since the velocity ratio xy/ay=sinα is also a measure of the voltage in coil side <b>a </b>with respect to the maximum voltage (when the coil is located horizontally) it follows that sinα is a varying proportionality factor that equates <b>e</b> to<b> Em</b>.<br />
<br />
The equation above may be used to determine a succession of generated voltage values in a coil as it rotates through a complete revolution. This is just by computing with its selected angular displacements.<br />
<br />
A more convenient way of representing the instantaneous voltage equation mentioned above is to draw a graph to illustrate a smooth variation of voltage with respect to the angular position of the coil, this graph is called a <i>sine wave</i>. The wave repeats itself and it is called a <i>periodic, </i>then each complete succession of values is called a <i>cycle, </i>while each positive or negative half of the cycle is called <i>alternation</i>.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-hn39Dy3_j3g/UbwKg3C98II/AAAAAAAAAic/0uXTMbndT5w/s1600/sinusoidal+voltage.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="238" src="http://4.bp.blogspot.com/-hn39Dy3_j3g/UbwKg3C98II/AAAAAAAAAic/0uXTMbndT5w/s320/sinusoidal+voltage.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Sinusoidal Voltage Wave</td></tr>
</tbody></table>
Now, we can say that an alternating voltage as an emf that varies in magnitude and direction periodically. Then, when the emfs are proportional to the trigonometric sine function, it is referred to a <i>sinusoidal alternating voltage</i>. However, there are also some periodic waves which do not follow this shape and they are called <i>non sinusoidal waves. </i>This topic will be covered when we reached more complicated analysis is AC Circuits.<br />
<br />
Lets have a practical example of a problem using the equation above just for you to appreciate the presented formula above:<br />
<br />
Problem : The voltage in an ac circuit varies harmonically with time with a maximum of 170V. What is the instantaneous voltage when it has reached 45 degree in its cycle?<br />
<br />
Using, e = Em sin α = 170V x sin (45 degree) = 170V x 0.71 = 120 V.<br />
<br />
In the common 60 cycle ac circuit, there are 60 complete cycle each second; i.e. the time interval of 1 cycle is 1/60 sec. It should be noted that this corresponds to a reversal in a direction of the current every 1/120 sec. (since the direction reverses twice during each cycle). </div>
Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-20335131028946052632013-06-14T14:38:00.000+08:002013-06-15T10:59:57.365+08:00Generation of Alternating EMF's<div dir="ltr" style="text-align: left;" trbidi="on">
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A voltage can be developed in a coil of wire in one of the three ways:<br />
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<br />
1. By changing the flux through the coil.<br />
2. By moving the coil through the magnetic field.<br />
3. By altering the direction of the flux with respect to the coil.<br />
<br />
The first one is that voltage is said to be induced emf and in accordance with Faraday's law, its magnitude at any instant of time is given by the formula as shown below:<br />
<br />
e = N(dΦ/dt) x 10 <sup>-8</sup> volts<br />
<br />
where N is the number turns in a coil<br />
dΦ/dt = rate at which the flux in maxwells changes through the coil<br />
<br />
Please take note that in this method of developing an emf, there is no physical motion of coil or magnet; the current through the exciting coil that is responsible for the magnetism is altered to change the flux through the coil in which the voltage is induced. For the second and third method mentioned above, there is actual physical motion of coil or magnet, and in altered positions of coil or magnet flux through the coil changes. A voltage developed on these ways is called a generated emf and is given by the equation:<br />
<br />
e = Blv x 10<sup>-8</sup> volts<br />
<br />
where B is the flux density in lines per square inch<br />
l is the length of the wire, in., that is moved relative to the flux<br />
v is the velocity of the wire, in.per sec., with respect to the flux<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-RRtHAi86m_o/UbqgQj-XvAI/AAAAAAAAAgs/B_F2oR1QQN8/s1600/two+pole+ac+generator.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="175" src="http://4.bp.blogspot.com/-RRtHAi86m_o/UbqgQj-XvAI/AAAAAAAAAgs/B_F2oR1QQN8/s320/two+pole+ac+generator.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Two-pole single AC Generator</td></tr>
</tbody></table>
The figure above illustrates an elementary a-c generator. The single turn coil may be moved through the magnetic field created by two magnet poles N and S. As you can see, the ends of the coil are connected to two collectors upon which two stationary brushes rest on it. For the clockwise rotation as shown, the side of the coil on north pole N is moving vertically upward to cut the maximum flux under north pole N, while the other side of the coil on south pole S is moving vertically downward to cut the maximum flux under south pole S. After the coil is rotated one quarter of a revolution to the position as shown below:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-L0v6EELyJls/UbquTYlVtKI/AAAAAAAAAg8/twFSYoS-uOA/s1600/AC+Generator+one+quarter.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="198" src="http://4.bp.blogspot.com/-L0v6EELyJls/UbquTYlVtKI/AAAAAAAAAg8/twFSYoS-uOA/s320/AC+Generator+one+quarter.png" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Rotated 90 degree</td></tr>
</tbody></table>
the coil sides have no flux to be cut and no voltage is generated. As the coil proceeds to rotate, the side of the coil on south pole S will cut the maximum flux on north pole N. Then, the side of the coil previously on north pole N will cut the maximum flux on south pole S. With this change in the polarity that are cut by the conductors, reversal in brush potential will occur. There are two important points that would like to emphasize in connection with the rotation of the coil of wire through a fixed magnetic field:<br />
<br />
1. The voltage changes from instant to instant.<br />
2. The electrical polarity (+) and minus (-) changes with alternating positions under north and south poles.<br />
<br />
In actual, ac generator rotate a set of poles that is placed concentrically within a cylindrical core containing many coils of wires. However, a moving coil inside a pair of stationary poles applies equally well to the rotating poles construction; in both arrangements there is a relative motion of one element with respect to the other.</div>
Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-650582647813021242013-03-30T11:17:00.001+08:002013-06-12T07:28:05.308+08:00Introduction to Alternating Current<div dir="ltr" style="text-align: left;" trbidi="on">
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Last time, we studied the first part of Learn Electrical Engineering for Beginners and this is all about <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/DC%20Circuits">DC Circuits</a>. Today, we will be dealing with our Part 2 of our module and this is all about <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/AC%20Circuits">Alternating Current Circuits</a>.
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<br />
So, you may now start to learn what this ac is and how it behaves. Alternating Current does not flow through a conductor in the same direction as what dc does. Instead, it flows back and forth in the conductor at the regular interval, continually reversing its direction of flow and can do so very quickly. It is measured in amperes, just as dc is measured too. Remember, one couloumb of electrons is passing a given point in a conductor in one second. This definition also applies when ac is flowing- only now some of the electrons during that 1 second flow past the given point going in one direction, and the rest flow past it going in the opposite directions.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-k2N3ynDA4vc/UVZPBQb9FaI/AAAAAAAAAfs/mRXoN2b3CyU/s1600/Alternating+Current.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="132" src="http://4.bp.blogspot.com/-k2N3ynDA4vc/UVZPBQb9FaI/AAAAAAAAAfs/mRXoN2b3CyU/s320/Alternating+Current.png" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Difference between DC and AC</td></tr>
</tbody></table>
<br />
<br />
The industrial applications of alternating current are widespread. These include the many types of induction motor, ranging in size employed in wind tunnels and reclamation projects, transformer equipment used in connection with welders and many kinds of control devices, communication systems, and many others.<br />
<br />
The advantages of ac generation are, however, apparent when it is recognized that it can be accomplished economically in large power plants where fuel and water are abundant. But nowadays, solar power is becoming popular as power plants through solar panels. Moreover, generators and associated equipment may be large, an important matter in so far as cost per kilowatt is concerned; also transmission over networks of high-voltage lines to distant load centers is entirely practicable.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-TDD-mUYlXE0/UVZVKs5zKrI/AAAAAAAAAf8/wOe4KznQQDg/s1600/Power+Plants+AC.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="268" src="http://4.bp.blogspot.com/-TDD-mUYlXE0/UVZVKs5zKrI/AAAAAAAAAf8/wOe4KznQQDg/s640/Power+Plants+AC.png" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Transmission Lines to distant load centers</td></tr>
</tbody></table>
<br />
In Part 2 of Learn Electrical Engineering for Beginners, you will study the nature, behavior and uses of time-varying or alternating current. You will study the for the first time two components - the inductor and the capacitor which are frequently used to control direct as well as alternating current and voltage. The resistors, in which we all know acted in such a way as to restrict the flow of current directly. In other words, the bigger the resistor you put in, the more you restrict the current flow. The inductor and the capacitor, on the other hand, act to control the current and voltage in different ways, and you will see that they do depends on how often the current is reversed. These three components - the resistor, inductor and the capacitor are basic elements of electric and electronic circuits.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-5eeXNpZKOwc/UVZWhmbsuCI/AAAAAAAAAgE/OlHfVJLYaas/s1600/RLC.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="379" src="http://1.bp.blogspot.com/-5eeXNpZKOwc/UVZWhmbsuCI/AAAAAAAAAgE/OlHfVJLYaas/s640/RLC.gif" width="640" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Resistor, Capacitor and Inductor Behavior in AC Circuits</td></tr>
</tbody></table>
<br />
As of now, you will not understand the meaning of the behavior of the given diagram shown above. But as we started the first topic of AC Circuits on my next post, you will appreciate and understand gradually what really mean by AC Circuits.</div>
Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-36463638444493069202013-03-17T13:51:00.000+08:002013-03-30T14:03:00.954+08:00Simple Example of Network Analysis<div dir="ltr" style="text-align: left;" trbidi="on">
Last time, I have already imparted to you about the theoretical aspects of <a href="http://electricalengineeringforbeginners.blogspot.com/2011/06/network-analysis-for-electric-circuits.html">Network Analysis</a> here in my simple <a href="http://electricalengineeringforbeginners.blogspot.com/">Electrical Engineering</a> educational site. Today, I will just give you some simple example for you to appreciate the last topic I had posted over a year ago. :)<br />
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<br />
Take a look at the simple electric circuits below. If you have a voltage divider with an external resistance, you could do this by using <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Ohm%27s%20Law">Ohm' Law</a> and calculating the parallel resistance of R2 and R(Load) and then the voltage divider itself. The simpler method that you can use is the <a href="http://electricalengineeringforbeginners.blogspot.com/2011/06/network-analysis-for-electric-circuits.html">Thevenin's theorem</a> which enables you to calculate quickly the effect of any load.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-k7-Qp3MpmIM/UUVIuyN1TRI/AAAAAAAAAfE/bU9uhMl0p4c/s1600/Thevenin's+Theorem+1.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="242" src="http://1.bp.blogspot.com/-k7-Qp3MpmIM/UUVIuyN1TRI/AAAAAAAAAfE/bU9uhMl0p4c/s320/Thevenin's+Theorem+1.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Simple Circuit using Thevenin's Theorem</td></tr>
</tbody></table>
<br />
Considering you have R(Load) equivalent to 40 ohms is in open circuit condition. We can now calculate the equivalent Thevenin resistance.<br />
<br />
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<a href="http://3.bp.blogspot.com/-JgaMTjGingc/UUU53n_YltI/AAAAAAAAAe0/lfcIj20Zjd0/s1600/Thevenin's+Theorem+3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-JgaMTjGingc/UUU53n_YltI/AAAAAAAAAe0/lfcIj20Zjd0/s1600/Thevenin's+Theorem+3.jpg" /></a></div>
<br />
Therefore,<br />
R (Thevenin) = R1R2 / R1+R2 = 20X40/ 20+40 = 13.33 ohms.<br />
<br />
Also, you can calculate the voltage across R2 at no load using the voltage division method:<br />
E(Thevenin) = Ein R2 / R1+R2 = 20x40 /20+40 = 13.33 Volts<br />
<br />
When the resistance load or R(Load) of 40 ohms is added as shown above, by using Ohms Law for simple series circuits you can now obtain the ouput voltage:<br />
<br />
E(Load) = E(Thevenin) x R (Load) / R(Thevenin) + R(Load) = 13.33 x 40/ 13.33+40 = <b>10 volts - Ans</b>.<br />
<br />
Using other method like <a href="http://electricalengineeringforbeginners.blogspot.com/2011/06/network-analysis-for-electric-circuits.html">Norton's Theorem</a> is also a good method to used in this given problem. Now, instead of open circuit condition, let's make it short circuit condition at the load side R(Load).<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-Ml19cX-m1p4/UUVJEQFhcOI/AAAAAAAAAfM/7klivEk0jwE/s1600/Norton's+Theorem.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="242" src="http://4.bp.blogspot.com/-Ml19cX-m1p4/UUVJEQFhcOI/AAAAAAAAAfM/7klivEk0jwE/s320/Norton's+Theorem.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Simple Circuit Using Norton's Theorem</td></tr>
</tbody></table>
In this condition, we can calculate the current when R2 is shorted. Let's called it I (Norton):<br />
I (Norton) = Ein / R1 = 20V / 20 Ohms = 1 Ampere<br />
<br />
Therefore, the equivalent circuit will comes out like this:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-SajHcply26U/UUVQYSVDGMI/AAAAAAAAAfc/x6WblpFaCjI/s1600/Norton+Theorem+1.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="170" src="http://4.bp.blogspot.com/-SajHcply26U/UUVQYSVDGMI/AAAAAAAAAfc/x6WblpFaCjI/s320/Norton+Theorem+1.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Norton's Equivalent Circuit</td></tr>
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From Thevenin's equivalent circuit above for R1 and R2, the parallel combination was computed as 13.33 ohms. We know that the voltage across R(Norton) and R(Load) are the same when R(Load) is connected but the total current is still 1 Ampere.<br />
<br />
Let's make an analysis now:<br />
<br />
We can get load voltage E (out) in two ways:<br />
1---> E(out) = I1 x R(Norton) = I1 x 13.33<br />
2---> E(out) = I2 x R(Load) = I2 x 40<br />
<br />
Then, we can equate the two equations above:<br />
3--->I1 x 13.33=I2 x 40<br />
<br />
Using <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Kirchhoff%27s%20Law">Kirchoff's current law</a>,<br />
I(total) = I1 + I2 = 1 Ampere<br />
<br />
We can now solve I1 by substitution method: (<b>Remember, you should used you math technique sometimes to solve particular problems is Electric Circuits</b>)<br />
<br />
We can substitute I2 value: I2 = I(total)-I1 or 1- I1 to equation 3.<br />
I1 x 13.33 = (1-I1) 40<br />
I1 = 0.75 Ampere<br />
<br />
Since, I(total) is 1 Ampere, therefore I2 = 1-0.75 = 0.25 Ampere<br />
Finally, we can calculate E(out) in either equation 1 or 2 above:<br />
<br />
E(out) = I1 x R(Norton) or I2 x R(Load) = 0.25 x 40 = <b>10 volts - Ans</b>.<br />
<br />
The two methods used either Thevenin's or Norton's still have same results obtained. This is the magic of Network Analysis!</div>
Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-84221892688506903302011-06-23T15:58:00.001+08:002013-03-30T14:00:41.205+08:00Network Analysis for Electric Circuits<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="font-family: Georgia, "Times New Roman", serif;">Network Analysis for <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Electric%20Circuits">electric circuits</a> are the different useful techniques related to several <a href="http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html">currents</a>, <a href="http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html">emfs</a>, and <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Resistance">resistance</a> voltages in such circuit. This is somewhat the collection of techniques of finding the voltages and currents in every component of the network. Some of those techniques are already mentioned in this online tutorial of Electrical Engineering. </span><br />
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<br/>
<ul>
<li><span style="font-family: Georgia, "Times New Roman", serif;"><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/dc-series-circuit-part-1.html">Kirchhoff''s Voltage Law or KVL</a></span></li>
<li><span style="font-family: Georgia;"><a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-2.html">Kirchhoff''s Current Law or KCL</a></span></li>
</ul>
<span style="font-family: Georgia;">There are six remaining useful techniques that we are going to learn. The practical example of each analysis will be given in my next post. This is for you to comprehend first what each theory is all about. So, let's begin the first useful technique in analyzing network. </span><br />
<br />
<span style="font-family: Georgia;"><strong>Thevenin's Theorem</strong></span><br />
<br />
<span style="font-family: Georgia;">Consider the figure below which schematically represents the two-terminal network of constant emf's and resistances; a high-resistance voltmeter, connected to the accessible terminals, will indicate the so called <em>open circuit voltage v<sub>oc</sub></em>. If an extremely low-resistance ammeter is next connected to the same terminals, as in fig.(b), which is so called the <em>short-circuit current i<sub>sc</sub> </em>will be measured. </span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-CXJsUnoMvPQ/TgKn3G1nCYI/AAAAAAAAARU/hRzkQWFWNIU/s1600/thevenins.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" i="" src="http://4.bp.blogspot.com/-CXJsUnoMvPQ/TgKn3G1nCYI/AAAAAAAAARU/hRzkQWFWNIU/s1600/thevenins.jpg" true="" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Test circuits for Thevenin's Theorem</em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">Now the two quantities determined above may be used to represent <em>an equivalent simple network </em>consisting of the single resistance R<sub>TH</sub>, which is equal to <em>v<sub>oc</sub>/i<sub>sc</sub></em>. If the resistor R<sub>L</sub> is connected to the two terminals, the load current of the circuit will be</span><br />
<br />
<span style="font-family: Georgia;">I<sub>L</sub> = v<sub>oc</sub> / R<sub>TH</sub>+R<sub>L</sub><strong>---------------> equation no.1</strong> </span><br />
<br />
<span style="font-family: Georgia;">The analysis leading to the equation no.1 above was first proposed by M.L. Thevenin the latter part of the nineteenth century, and has been recognized as an important principle in electric circuit theory. His theory was stated as follows: <em>In any two-terminal network of fixed resistances and constant sources of emf, the current in the load resistor connected to the output terminals is equal to the current that would exist in the same resistor if it were connected in series with (a) a simple emf whose voltage is measured at the open-circuited network terminals and (b) a simple resistance whose magnitude is that of the network looking back from the two terminals into the network with all sources of emf replaced by their internal resistances.</em></span><br />
<br />
<span style="font-family: Georgia;">Thevenin's Theorem has been applied to many network solutions which considerably simplify the calculations as well as reduce the number of computations.</span><br />
<br />
<span style="font-family: Georgia;"><strong>Norton's Theorem</strong></span><br />
<br />
<span style="font-family: Georgia;">From the previous topic above, it was learned that a somewhat modified approach of Thevenin was formulated. This modified approach is to convert the original network into a simple circuit in which a parallel combination of <em>constant-current source and looking-back resistance "feeds" the load resistor. </em>Take a look on the figure below</span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-iYhS6LFS294/TgK33x1IMjI/AAAAAAAAARY/rQOs3p3uCAs/s1600/norton.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="108" i="" src="http://2.bp.blogspot.com/-iYhS6LFS294/TgK33x1IMjI/AAAAAAAAARY/rQOs3p3uCAs/s400/norton.jpg" true="" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Norton's equivalent circuit </em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">Take note that Norton's theory also make use of the resistance looking back into the network from the load resistance terminals, with all potential sources replaced by the zero-resistance conductors. It also employs a fictitious source which delivers a <em>constant current, </em>which is equal to the current that would pass into a short circuit connected across the output terminals of the original circuit. </span><br />
<br />
<span style="font-family: Georgia;">From the fig (b) above of Norton's equivalent circuit, the load current would be</span><br />
<br />
<span style="font-family: Georgia;">I<sub>L</sub> = I<sub>N</sub> R<sub>N</sub> / R<sub>N</sub>+R<sub>L</sub></span> <strong><span style="font-family: Georgia;">---------------> equation no.2</span></strong><br />
<br />
<strong><span style="font-family: Georgia;">Superposition Theorem</span></strong><br />
<br />
<span style="font-family: Georgia;">The theorem states like this: <em>In the network of resistors that is energized by two or more sources of emf, (a) the current in any resistor or (b) the voltage across any resistor is equal to: (a) the algebraic sum of the separate currents in the resistor or (b) the voltages across the resistor, assuming that each source of emf, acting independently of the others, is applied separately in turn while the others are replaced by their respective internal values of resistance. </em></span><br />
<br />
<span style="font-family: Georgia;">This theorem is illustrated in the given circuit below:</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-lZJbzjjl4nI/TgLLRvm9dpI/AAAAAAAAARc/_Lmb64YLRwQ/s1600/superposition.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="132" i="" src="http://1.bp.blogspot.com/-lZJbzjjl4nI/TgLLRvm9dpI/AAAAAAAAARc/_Lmb64YLRwQ/s400/superposition.gif" true="" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Illustration of Superposition Theorem</em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">The original circuit above ( left part ) have one emf source and a current source. If you like to obtain the current I which is equal to the sum of I' + I"using the superposition theorem, we need to do the following steps:</span><br />
<br />
<span style="font-family: Georgia;">a. Replace the current source I<sub>o </sub>by an open circuit. Therefore, an emf source v<sub>o</sub> will act independently having a current I' as the first value obtained when the circuit computed.</span><br />
<br />
<span style="font-family: Georgia;">b. Replace emf source v<sub>o </sub>by a short circuit. This time I<sub>o</sub> will act independently and I" now will be obtained when the circuit computed.</span><br />
<br />
<span style="font-family: Georgia;">c. The two values obtained ( I' and I") with emf and current source acting independently will be added to get I = I' + I"</span><br />
<br />
<span style="font-family: Georgia;"><strong>Maxwell's Loop (Mesh) Analysis</strong></span><br />
<br />
<span style="font-family: Georgia;">The method involves the set of independent loop currents assigned to as many meshes as exists in the circuit, and these currents are employed in connection with appropriate resistances when Kirchhoff voltage law equations are written. Take a look on the given circuit below. </span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-HlaE57wq5Wc/TgLT-7Ja0cI/AAAAAAAAARg/ejCcWRFg_7I/s1600/mesh.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="200" i="" src="http://4.bp.blogspot.com/-HlaE57wq5Wc/TgLT-7Ja0cI/AAAAAAAAARg/ejCcWRFg_7I/s400/mesh.jpg" true="" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Given circuit can be analyze using mesh method </em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">The given circuit above have two voltage sources V1 and V2 are connected with a five-resistor network in which there are two loop currents i1 and i2. Observe that they are shown directed <em>clockwise, </em>a convention that is generally adopted for convenience. The following Kirchhoff's voltage-law may now be written as:</span><br />
<br />
<span style="font-family: Georgia;"> -V1 + i1R1 + R3(i1-i2) + i1R2 = 0 (loop 1)</span><br />
<span style="font-family: Georgia;"> -V2 + i2R5 +R3(i2-i1) + i2R4 = 0 (loop 2)</span><br />
<br />
<span style="font-family: Georgia;">You may simplify the equations by using the simple algebra. This will be well explained on my next post for more practical examples of Network Analysis.</span><br />
<br />
<span style="font-family: Georgia;"><strong>Nodal Analysis</strong></span><br />
<br />
<span style="font-family: Georgia;">For this analysis, every junction in the network that represents a connection of three or more branches is regarded as a <em>node</em>. Considering one of the nodes as a reference or zero-potential point, current equations are then written for the remaining junctions, thus a solution is possible with <em>n-1</em> equations, where <em>n </em>is the number of nodes. </span><br />
<br />
<span style="font-family: Georgia;">There are three basics steps to follow when using nodal analysis</span><br />
<br />
<span style="font-family: Georgia;">a. Label the node voltages with respect to ground.</span><br />
<span style="font-family: Georgia;">b. Apply KCL to each of the nodes in terms of the node voltages.</span><br />
<span style="font-family: Georgia;">c. Determine the unknown node voltages by solving simultaneous equations from step b.</span><br />
<br />
<span style="font-family: Georgia;">Take a look on the snapshot on how nodal analysis is being done. This is illustration by Stephen Mendez. Don't worry I will give you the technique on my next post on how to solve nodal analysis. </span><br />
<br />
<span style="font-family: Georgia;">Note: In the snapshot below, he used <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html">conductance</a> which is G = 1/R. </span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-xB0Tik-3C1o/TgLqG5noP1I/AAAAAAAAARk/fYRdlUMaJ5k/s1600/nodal.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="640" i="" src="http://3.bp.blogspot.com/-xB0Tik-3C1o/TgLqG5noP1I/AAAAAAAAARk/fYRdlUMaJ5k/s640/nodal.gif" true="" width="561" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>The Nodal Analysis Snapshot</em></td></tr>
</tbody></table>
<br />
<span style="font-family: Georgia;"><strong>Millman's Theorem</strong></span><br />
<br />
<span style="font-family: Georgia;">Any combination of parallel-connected voltage sources can be represented as a single equivalent source using Thevenin's and Norton theorems appropriately. This can be illustrated as :</span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-IZyCEtxVanI/TgLtgf7f9XI/AAAAAAAAARo/RpQUQA7c3mo/s1600/millman.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="170" i="" src="http://1.bp.blogspot.com/-IZyCEtxVanI/TgLtgf7f9XI/AAAAAAAAARo/RpQUQA7c3mo/s400/millman.png" true="" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>This is Millman's Theorem </em></td></tr>
</tbody></table>
<span style="font-family: Georgia;">The formula above can be written as:</span><br />
<br />
V<sub>L</sub> = V<sub>1</sub>/R<sub>1</sub> + V<sub>2</sub>/R<sub>2 </sub>+ .....Vn / Rn <br />
-------------------------------<br />
1/R<sub>1</sub> + 1/R<sub>2 </sub>+ ......1/Rn + 1/R<sub>L</sub><br />
<br />
<span style="font-family: Georgia;">where:</span><br />
<br />
<span style="font-family: Georgia;">V1, V2, V3... Vn are the voltages of the individual voltage sources.</span><br />
<span style="font-family: Georgia;">R1, R2, R3... Rn are the internal resistances of the individual voltage sources.</span><br />
<br />
<span style="font-family: Georgia;">Vout or V<sub>L</sub>= load voltage</span><br />
<span style="font-family: Georgia;">RL = load resistor</span><br />
<br />
<span style="font-family: Georgia;">I think this is enough for today. Don't forget to read my next post for more practical examples regarding Network Analysis/Theorems.</span><br />
<br />
<span style="font-family: Georgia;">Cheers!</span></div>
Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com13tag:blogger.com,1999:blog-1420533950904606452.post-23923361748606040852011-06-22T14:41:00.004+08:002013-03-30T14:05:12.876+08:00What Is Electric Power?<div dir="ltr" style="text-align: left;" trbidi="on"><span style="font-family: Georgia, "Times New Roman", serif;">If we are going to recall our Physics subject, it is said that whenever a force is applied that causes motion the work is said to be done. Take a look on the illustration below:</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-DFrMaJDs_IU/TgAQo_jGAQI/AAAAAAAAAQ8/1qf_mqogWL4/s1600/work.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="192" i$="true" src="http://1.bp.blogspot.com/-DFrMaJDs_IU/TgAQo_jGAQI/AAAAAAAAAQ8/1qf_mqogWL4/s400/work.gif" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Forces that work is done and forces not doing work.</em></td></tr>
</tbody></table><span style="font-family: Georgia, "Times New Roman", serif;">The first figure shown above are combination of forces which work is done and forces which work is not done. (a)The picture in which the shelf is held under tension does not cause motion, thus work is not done. (b) The second picture in which the woman pushes the cart causes motion, thus the work is done. (c) The man applied tension in the string is not working since as there is no movement in the direction of the force. (d) The track applied horizontal force on the log is doing work. </span><br />
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<br />
<span style="font-family: Georgia;">The potential difference between any two points in an <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Electric%20Circuits">electric circuit</a>, which gives rise to a <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Voltage">voltage</a> and when connected causes electron to move and <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Current">current</a> to flow. This is one of a good example in which forces causing motion, thus causing work to be done. </span><br />
<br />
<span style="font-family: Georgia;">Talking about work in electric circuit, there is also a electric <a href="http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html">power</a> which is the time rate of doing work done of moving electrons from point to point. It is represented by the symbol <strong>P</strong>, and the unit of power is <strong>watt</strong>, which is usually represented by the symbol <strong>W</strong>. <strong>Watt</strong> is practically defined as the rate at which work is being done in a circuit in which the current of 1 ampere is flowing when the voltage applied is 1 volt.</span><br />
<br />
<span style="font-family: Georgia;"><strong>The Useful Power Formula</strong></span><br />
<br />
<span style="font-family: Georgia;">Electric Power can be transmitted from place to place and can be converted into other forms of energy. One typical energy conversion of electrical energy are heat, light or mechanical energy. Energy conversion is what the engineers really mean for the word <strong>power</strong>.</span><br />
<br />
<span style="font-family: Georgia;">The power or the rate of work done in moving electrons through a resistor in electric circuit depends on how many electrons are there to moved. It only means that, the <em>power consumed in a resistor is determined by the voltage measured across it, multiplied by the current flowing through it</em>. Then it becomes,</span><br />
<br />
<span style="font-family: Georgia;">Power = Voltage x Current</span><br />
<span style="font-family: Georgia;">Watts = Volts x Amperes</span><br />
<br />
<span style="font-family: Georgia;">P = E x I or <strong>P = EI ------> formula no.1</strong></span><br />
<br />
<span style="font-family: Georgia;">The power formula above can be derived alternatively in other ways in terms of resistance and current or voltage and resistance using our concept of <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/basic-concept-of-ohms-law.html">Ohm's Law</a>. Since E=IR in Ohm's Law, the E in the power formula above can be replaced by IR if the voltage is unknown. Therefore, it would be:</span><br />
<br />
<span style="font-family: Georgia;">P = EI</span><br />
<span style="font-family: Georgia;">P = (IR)I or <strong>P = I<sup>2</sup>R</strong> <strong>------------> formula no.2</strong></span><br />
<br />
<span style="font-family: Georgia;">Alternatively if I = E/R in Ohm'Law, we can also substitute it to E in the power formula which is terms of voltage if the resistance is unknown. </span><br />
<br />
<span style="font-family: Georgia;">P = EI</span><br />
<span style="font-family: Georgia;">P = E(E/R) or <strong>P = E<sup>2</sup>R ---------> formula no. 3</strong></span><br />
<br />
<span style="font-family: Georgia;">For guidance regarding expressing of units of power are the following:</span><br />
<span style="font-family: Georgia;">a. Quantities of power greater than 1,000 watts are generally expressed in (kW).</span><br />
<span style="font-family: Georgia;">b. Quantities greater than 1,000,000 watts are generally expressed as megawatts (MW).</span><br />
<span style="font-family: Georgia;">c. Quantities less than 1 watt are generally expressed in (mW).</span><br />
<br />
<span style="font-family: Georgia;"><strong>The Power Rating of Equipment</strong></span><br />
<br />
<span style="font-family: Georgia;">Most of the electrical equipment are rated in terms of voltage and power - volts and watts. For example, electrical lamps rated as 120 volts which are for use in 120 volts line are also expressed in watts but mostly expressed in watts rather than voltage. Probably you would wonder what wattage rating all about.</span><br />
<br />
<span style="font-family: Georgia;"><em>The wattage rating of an electrical lamps or other electrical equipment indicates the rate at which electrical energy is changed into another form of energy, such as heat or light</em>. It only means the greater the wattage of an electrical lamp for example, the faster the lamp changes electrical energy to light and the brighter the lamp will be.</span><br />
<br />
<span style="font-family: Georgia;">The principle above also applies to other electrical equipment like electric soldering irons, electrical motors and resistors in which their wattage ratings are designed to change electrical energy into some forms of energy. You will learn more about other units like horsepower used for motors when we study motors. </span><br />
<br />
<span style="font-family: Georgia;">Take a look at the sizes of <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/resistors-part-1-use-and-properties.html">carbon resistors</a> below. Their sizes are depends on their wattage rating. They are available with same resistance value with different wattage value. When power is used in a material having resistance, electrical energy is changed into heat. When more power are used, the rate at which electrical energy changed into heat increases, thus temperature of the material rises. If the temperature of the material rises too high, the material may change it composition: expand, contract or even burn. In connection to this reason, all types of electrical equipment are rated for a maximum wattage. </span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-rj3vVcsNOS4/TgFpkDg3XyI/AAAAAAAAARA/jDWiL5Ln2rM/s1600/CarbonFilmResistors.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="269" i$="true" src="http://1.bp.blogspot.com/-rj3vVcsNOS4/TgFpkDg3XyI/AAAAAAAAARA/jDWiL5Ln2rM/s320/CarbonFilmResistors.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Carbon resistors with comparative sizes of different wattage ratings</em><br />
<em>of 1/4 watt, 1/2 watt,1 and 2 watts</em></td></tr>
</tbody></table><span style="font-family: Georgia, "Times New Roman", serif;">If the resistors greater then 2 watts rating are needed, <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/resistors-part-1-use-and-properties.html">wire-wound resistors</a> are used. They are ranges between 5 and 200 watts, with special types being used for power in excess of 200 watts.</span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-JtibWXx_x2w/TgFwacdL0TI/AAAAAAAAARE/g3O-ZskEoXc/s1600/wire_wound_resistors.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" i$="true" src="http://4.bp.blogspot.com/-JtibWXx_x2w/TgFwacdL0TI/AAAAAAAAARE/g3O-ZskEoXc/s320/wire_wound_resistors.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Use wire wound resistors if higher than 2 watts are needed</em></td></tr>
</tbody></table><br />
<span style="font-family: Georgia, "Times New Roman", serif;"><strong>Fuses</strong></span><br />
<br />
<span style="font-family: Georgia;">We all know that when current passes through the resistors, the electrical energy is transformed into heat which raises the <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html">temperature of the resistors</a>. If the temperature rises too high, the resistor may be damaged thereby opening the circuit and interrupting the current flow. One answer for this is to install the fuse. </span><br />
<br />
<span style="font-family: Georgia;">Fuses are resistors using special metals with very low resistance value and a low melting point. When the <strong>power</strong> consumed by the fuses raises the temperature of the metal too high, the metal melts and the fuse blows thus open the circuit when the current exceeds the fuse's rated value. What is the identification of blown fuse? Take a look on the picture below.</span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-cKk419b9p3U/TgF_eyu5aBI/AAAAAAAAARI/Pw4EqsJt8rk/s1600/goodfuse.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="158" i$="true" src="http://1.bp.blogspot.com/-cKk419b9p3U/TgF_eyu5aBI/AAAAAAAAARI/Pw4EqsJt8rk/s320/goodfuse.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>This is the good fuse</em></td></tr>
</tbody></table> <br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-yIpg4oRG268/TgF_rl1JRgI/AAAAAAAAARM/uH2JMLN1IfQ/s1600/blownfuse.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="153" i$="true" src="http://1.bp.blogspot.com/-yIpg4oRG268/TgF_rl1JRgI/AAAAAAAAARM/uH2JMLN1IfQ/s320/blownfuse.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>This is the blown fuse</em></td></tr>
</tbody></table> <span style="font-family: Georgia, "Times New Roman", serif;">In other words, blown fuses can be identified by broken filament and darkened glass. You can also check it by removing the fuse and using the <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/resistors-part-2-color-code-and-how.html">ohmmeter</a>.</span><br />
<br />
<span style="font-family: Georgia;">There are two types of fuses in use today - <strong>conventional fuses</strong>, which blow immediately when the circuit is overloaded. The <strong>slow-blowing</strong> (slo-blo) fuses accepts momentary overloads without blowing, but if the overload continues, it will open the circuit. This slo-blo fuses usually used on motors and other appliances with a circuit that have a sudden rush of high currents when turned on.</span><br />
<br />
<span style="font-family: Georgia;">Fuses are rated in terms of current. Since various types of equipments use different currents, fuses are also made with different sizes, shapes and current ratings.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-xLOkdyrj0g0/TgGGyUIIGpI/AAAAAAAAARQ/8sGiXhM7VaY/s1600/fuses.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="157" i$="true" src="http://2.bp.blogspot.com/-xLOkdyrj0g0/TgGGyUIIGpI/AAAAAAAAARQ/8sGiXhM7VaY/s320/fuses.gif" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Various types of fuses are made for various equipments</em></td></tr>
</tbody></table><span style="font-family: Georgia, "Times New Roman", serif;">Proper rating of fuse is needed and very important. It should be slightly higher than the greatest current you expect in the circuit because too low current rating of fuse will result to unnecessary blowouts while too high may result to dangerously high current to pass.</span><br />
<br />
<span style="font-family: Georgia;">Later we will be study circuit breaker which is another protective devices for over current protection.</span><br />
<br />
<br />
<span style="font-family: Georgia;"><strong>Electrical Power in Series, Parallel and Complex Circuits</strong></span><br />
<br />
<span style="font-family: Georgia;">The principle of getting the total power of the circuit is just simple. There is no need to elaborate this topic. </span><br />
<br />
<em><span style="font-family: Georgia;">The total power consumed by the circuit is the sum of all power consumed in each resistance.</span></em><br />
<br />
<span style="font-family: Georgia;">Therefore, we just only sum up all power consumed in each resistance whether it a series, parallel or a complex circuits. Thus,</span><br />
<br />
<span style="font-family: Georgia;">P<sub>t</sub>= P<sub>1</sub>+P<sub>2</sub>+P<sub>3</sub>+P<sub>n</sub> <strong>watts ---------->formula no. 4</strong></span><br />
<br />
<span style="font-family: Georgia;">From the problem in my previous post about <a href="http://electricalengineeringforbeginners.blogspot.com/2011/06/ohms-law-series-parallel-circuits.html">complex circuit</a>, try to calculate each power of the resistance and the total power as well. Constant practice always makes you perfect!</span><br />
<br />
<span style="font-family: Georgia;">Cheers!</span></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com1tag:blogger.com,1999:blog-1420533950904606452.post-60413822876507687222011-06-16T13:32:00.002+08:002013-03-30T13:47:21.653+08:00Ohm's Law Series-Parallel Circuits Calculation<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="font-family: Georgia, "Times New Roman", serif;">To end up the discussion of <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/DC%20Series-Parallel%20Circuits">Series-Parallel Circuits</a>, I would like to post this last one remaining topic which is about <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Ohm%27s%20Law">Ohm's Law</a> of Series-Parallel Circuits for <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Current">currents and voltages</a>. I did not even mentioned in my previous topics on how to deal with its currents and voltages regarding this type of circuit connection. </span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-X9jOtRe4klk/TflzG_GvVHI/AAAAAAAAAQo/fLEoBzHEHHs/s1600/SeriesParallel.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="244" i="" src="http://2.bp.blogspot.com/-X9jOtRe4klk/TflzG_GvVHI/AAAAAAAAAQo/fLEoBzHEHHs/s320/SeriesParallel.jpg" true="" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Ohms Law in Series-Parallel Circuits </em></td></tr>
</tbody></table>
<br />
<span style="font-family: Georgia;"><strong>Ohm's Law in Series-Parallel Circuits - Current</strong></span><br />
<br />
<span style="font-family: Georgia;">The total current of the series-parallel circuits depends on the total resistance offered by the circuit when connected across the voltage source. The current flow in the entire circuit and it will divide to flow through parallel branches. In case of parallel branch, the current is inversely proportional to the resistance of the branch - that is the greater current flows through the least resistance and vice-versa. Then, the current will then sum up again after flowing in different circuit branch which is the same as the current source or total current. </span><br />
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<br/>
<span style="font-family: Georgia;">The total circuit current is the same at each end of a series-parallel circuit, and is equal to the current flow through the voltage source.</span><br />
<br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;"><strong>Ohm's Law in Series-Parallel Circuits - Voltage</strong></span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">The voltage drop across a series-parallel circuits also occur the same way as in series and parallel circuits. In series parts of the circuit, the voltage drop depends on the individual values of the resistors. In parallel parts of the circuit, the voltage across each branch are the same and carries a current depends on the individual values of the resistors. </span><br />
<br />
<span style="font-family: Georgia;">If in case of circuit below, the voltage of the series resistance forming a branch of the parallel circuit will divide the voltage across the parallel circuit. If in case of the single resistance in a parallel branch, the voltage across is the same as the sum of the voltages of the series resistances.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-xFFLzUGsKO0/TfmDRlVELoI/AAAAAAAAAQw/0ucn0ow76Eo/s1600/circuit2.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" i="" src="http://4.bp.blogspot.com/-xFFLzUGsKO0/TfmDRlVELoI/AAAAAAAAAQw/0ucn0ow76Eo/s1600/circuit2.png" true="" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>The sum of the voltage across R3 and R4 is the same </em><br />
<em>as the voltage across R2.</em></td></tr>
</tbody></table>
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<span style="font-family: Georgia;">Finally, the sum of the voltage drop across each paths between the two terminal of the series-parallel circuit is the same as the total voltage applied to the circuit.</span><br />
<br />
<span style="font-family: Georgia;">Let's have a very simple example of this calculation for this topic. Considering the circuit below with its given values, lets calculate the total current, current and voltage drop across each resistances. </span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-5mxcunFvzsk/TfmNKVvie5I/AAAAAAAAAQ0/aoc46wG2KmI/s1600/series-parallelcal.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" i="" src="http://4.bp.blogspot.com/-5mxcunFvzsk/TfmNKVvie5I/AAAAAAAAAQ0/aoc46wG2KmI/s1600/series-parallelcal.png" true="" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>What is the total current, current and voltage across each resistances</em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;"> Here is the simple calculation of the circuit above:</span><br />
<br />
<span style="font-family: Georgia;">a. Calculate first the total resistance of the circuit:</span><br />
<br />
<span style="font-family: Georgia;">The equivalent resistance for R2 and R3 is:</span><br />
<br />
<span style="font-family: Georgia;"><strong>R2-3</strong> = 25X50/ 25+50 = 16.67 ohms</span><br />
<br />
<span style="font-family: Georgia;"><strong>R total</strong> = 30 ohms + 16.67 ohms = 46.67 ohms</span><br />
<br />
<span style="font-family: Georgia;">b. Calculate the Total Current using Ohm's Law:</span><br />
<br />
<span style="font-family: Georgia;"><strong>I1</strong> = 120V / 46.67 Ohms = <strong>2.57 Amp</strong>. Since R1 is in series connection, the total current is the same for that path. </span><br />
<br />
<span style="font-family: Georgia;">c. Calculating the voltage drop for R1:</span><br />
<br />
<span style="font-family: Georgia;"><strong>VR1</strong> = 2.57 Amp x 30 ohms = <strong>77.1 volts</strong></span><br />
<br />
<span style="font-family: Georgia;">d. Calculate the voltage drop across R2 and R3.</span><br />
<br />
<span style="font-family: Georgia;">Since the equivalent resistance for R2 and R3 as calculated above is 16.67 ohms, we can now calculate the voltage across each branch.</span><br />
<br />
<span style="font-family: Georgia;"><strong>VR2 = VR3</strong> = 2.57 Amp x 16.67 ohms = <strong>42. 84 volts</strong></span><br />
<br />
<span style="font-family: Georgia;">e. Finally, we can now calculate the individual current for R2 and R3:</span><br />
<br />
<span style="font-family: Georgia;"><strong>I2</strong> = VR2 / R2 = 42.84 volts / 25 ohms = <strong>1.71 Amp</strong>.</span><br />
<span style="font-family: Georgia;"><strong>I3</strong> = VR3 / R3 = 42.84 volts / 50 ohms = <strong>0.86 Amp</strong>.</span><br />
<br />
<span style="font-family: Georgia;">You may also check if the current in each path of the parallel branch are correct by adding its currents:</span><br />
<br />
<span style="font-family: Georgia;">I1 = I2 + I3 = 1.71 Amp + 0.86 Amp = <strong>2.57 Amp</strong>. which is the same as calculated above. Therefore, we can say that our answer is correct. </span><br />
<br />
<br />
<span style="font-family: Georgia;">Cheers!</span></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-39986116962752233982011-06-15T14:01:00.001+08:002013-03-30T17:44:11.263+08:00The Bridge Resistor Circuit<div dir="ltr" style="text-align: left;" trbidi="on"><span style="font-family: Georgia,"Times New Roman",serif;">This is already the Part-3 lessons for <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/DC%20Series-Parallel%20Circuits">Series-Parallel Circuits</a>. Today we will be dealing with another type of complex circuit which you do not know yet - particularly for the beginners.</span><br />
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<span style="font-family: Georgia;">Suppose you have a type of simple circuit below. You will notice that there is an extra <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Resistors">resistor</a> of R3 connecting to the two parallel branches of the <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/DC%20Parallel%20Circuits">parallel circuit</a> connection and in such way it was interrupted to the leads of the new resistor. This new resistor (R3) is called a <b>bridge</b>. </span><br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-jehPxcGkMYs/TfgZWoCbTII/AAAAAAAAAQY/4eEQrl6bCcQ/s1600/bridge+resistor.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="280" src="http://4.bp.blogspot.com/-jehPxcGkMYs/TfgZWoCbTII/AAAAAAAAAQY/4eEQrl6bCcQ/s400/bridge+resistor.gif" t8="true" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>R3 is called the Bridge Resistor</i></td></tr>
</tbody></table> <span style="font-family: Georgia,"Times New Roman",serif;">Take a look at the circuit above. If you look at the upper part of R3 resistor, wherein R1, R2 and R3 are all connected together. You will notice a new arrangement of connection. This arrangement from its similarity to the shape of the Greek letter D (delta), is said to be <b>delta connected</b>. Here is the diagram below to see clearly what I'm talking about.</span><br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-mSKEDclWflk/Tfgek-1skVI/AAAAAAAAAQc/z7nlzLNTVKA/s1600/delta.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-mSKEDclWflk/Tfgek-1skVI/AAAAAAAAAQc/z7nlzLNTVKA/s1600/delta.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>This is the illustrative diagram for delta connection</i></td></tr>
</tbody></table> <br />
<div style="text-align: left;"></div> <br />
<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-8UEGWmpwSEI/TfghLV6zVYI/AAAAAAAAAQg/i_w9VQ7A_Qw/s1600/wye-delta.png" imageanchor="1" style="clear: left; cssfloat: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-8UEGWmpwSEI/TfghLV6zVYI/AAAAAAAAAQg/i_w9VQ7A_Qw/s1600/wye-delta.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>The equivalent connection of left diagram is </i><br />
<i>called the Y connection </i></td></tr>
</tbody></table> <span style="font-family: Georgia,"Times New Roman",serif;">Take a look at the diagram at the left side. If you will devise a circuit shown in the delta connection Ra, Rb and Rc to shaped like a <b>Y (wye)</b>. This Y circuit would fit onto the rest of the original circuit in such a way that you could solve its values without difficulty. Look at the diagram (at the left) in Y connection. </span><br />
<br />
<span style="font-family: Georgia;">The resistors connected in Y are R1, R2 and R3. Take note that their values must be such that the terminal resistances at N1 and N3 are exactly where they were in the original circuit. The problem now is how would you able to solve the values of R1, R2 and R3 (said to be unknown values) in terms of Ra, Rb and Rc whose values are known. </span><br />
<br />
<span style="font-family: Georgia;"><b>How to Solve Bridge Resistor Circuit</b></span><br />
<br />
<span style="font-family: Georgia;">Lets use again this previous diagram. Then, take note that both circuits must give exactly the same values of resistance across every corresponding pair of terminals. This operation that we'll set is called <b>delta-Y conversion or transformation</b>.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-8UEGWmpwSEI/TfghLV6zVYI/AAAAAAAAAQg/i_w9VQ7A_Qw/s1600/wye-delta.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-8UEGWmpwSEI/TfghLV6zVYI/AAAAAAAAAQg/i_w9VQ7A_Qw/s1600/wye-delta.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>delta -Y conversion</i></td></tr>
</tbody></table> <span style="font-family: Georgia,"Times New Roman",serif;">If you consider the sum of the resistances between N3 and N1, then assume N2 is to be disconnected. In delta combination you will see that between these two points there is a series combination of Rc and Rb in parallel across Ra. You can now express the resistance N3 and N1 applying the knowledge of <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/DC%20Parallel%20Circuits">parallel circuits</a> we have:</span><br />
<br />
<div style="text-align: center;"><span style="font-family: Georgia;"><b>Ra (Rb + Rc) / Ra + Rb + Rc</b></span></div><div style="text-align: center;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia,"Times New Roman",serif;">Considering the Y circuit connection above, the total resistance between N3 and N1 is R1 + R3. Then, since we all know that these resistances must be equal, you can now write down the first equation as:</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R1 + R3 = Ra (Rb + Rc)/ Ra + Rb + Rc</b> ---------> EQUATION no. 1</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">For the remaining terminals, you can exactly do the same way for total resistances between N3 - N2 and between N1 - N2 in terms of Ra, Rb, Rc and R1, R2 and R3. Then, you will get the two remaining equations: </span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R2 + R3 = Rc (Ra + Rb) / Ra + Rb + Rc</b> ---------> EQUATION no. 2</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R1 + R2 = Rb (Ra + Rc) / Ra + Rb + Rc </b>---------> EQUATION no. 3</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">Then, do a little algebra from the three equations above to obtain the values in terms of R1, R2 and R3. Finally, we can have the following formula:</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R1 = Ra x Rb / Ra + Rb + Rc</b> ------------> Formula no. 1</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R2 = Rb x Rc / Ra + Rb + Rc </b>------------> Formula no. 2</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R3 = Ra x Rc / Ra + Rb + Rc </b>------------> Formula no. 3</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">The problem below was given in the board exam way back 1997 - two years before my <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Electrical%20Engineering%20Board%20Exams">EE board examination</a>.</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">(EE April'97) <i>A circuit consisting of three resistors rated : 10 ohms, 15 ohms and 20 ohms are connected in DELTA. What would be the resistances of the equivalent WYE connected load?</i></span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>Solution:</b> </span></div><div style="text-align: left;"><span style="font-family: Georgia;">Just get the pattern of the above formula, this would give us the following:</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">R1 = 10 X 15 / 10 + 15 + 20 = <b>3. 33 ohms - answer</b></span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">R2 = 15 x 20 / 10 + 15 + 20 = <b>6. 67 ohms</b> <b>- answer</b></span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">R3 = 10 X 20 / 10 + 15 + 20 = <b>4.44 ohms</b> <b>- answer</b></span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><b><span style="font-family: Georgia;">Y to Delta Conversion</span></b></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">For the reverse conversion which is Y to delta conversion considering the given circuit.</span></div> <br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-wLjG-9-k8EE/TfhDd0K8jLI/AAAAAAAAAQk/c_3DCkMg3I8/s1600/star+delta.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="165" src="http://1.bp.blogspot.com/-wLjG-9-k8EE/TfhDd0K8jLI/AAAAAAAAAQk/c_3DCkMg3I8/s400/star+delta.jpg" t8="true" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>Y to delta conversion or transformation</i></td></tr>
</tbody></table> <br />
<div style="text-align: left;"><span style="font-family: Georgia,"Times New Roman",serif;">The general idea here is to compute the resistance in the delta circuit by:</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>R- delta = Rp / R opposite </b></span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">where: <b>Rp</b> is the sum of the product of all pairs of resistances in the Y circuit and <b>R opposite </b>is the resistance of the node in the Y circuit which is opposite the edge with <b>R- delta. </b></span><span style="font-family: Georgia,"Times New Roman",serif;">You will have the following formula for you to get the equivalent delta load in terms of Ra, Rb and Rc. </span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>Ra = R1R2 + R2R3 + R3R1 / R2</b> ---------> Formula no. 4</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>Rb = R1R2 + R2R3 + R3R1 / R3</b> ---------> Formula no. 5</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;"><b>Rc = R1R2 + R2R3 + R3R1 / R1</b> ---------> Formula no. 6</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">These are the formula that you'll going to use from our future topics since this is already the part of Network Theorems. Don't ever forget it...</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">Cheers!</span></div></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com2tag:blogger.com,1999:blog-1420533950904606452.post-68533836696367454102011-06-09T12:10:00.001+08:002013-03-30T17:46:34.243+08:00Series-Parallel Circuits- Part 2<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="font-family: Georgia, "Times New Roman", serif;">This is just the continuation of my post yesterday about <a href="http://electricalengineeringforbeginners.blogspot.com/2011/06/series-parallel-circuits-part-1.html">Series-Parallel Circuits- Part 1</a>. I've already provided you the steps on how to simplify a simple series-parallel connections. Today, I will give you an example on how to solve that circuit using that steps mentioned before.</span><br />
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<br />
<span style="font-family: Georgia;">The practical example that I will show you below is how to break down a complex circuits to find the total resistance. Refer to figure below:</span><br />
<br />
<div class="separator" style="clear: both; text-align: center;">
</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-uTLxGEC9O_o/TfAMwE4M_QI/AAAAAAAAAP8/oN4Q9AoNFO4/s1600/series-parallel.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-uTLxGEC9O_o/TfAMwE4M_QI/AAAAAAAAAP8/oN4Q9AoNFO4/s1600/series-parallel.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-family: Georgia, "Times New Roman", serif;"><em>Circuit Problem for Series-Parallel</em></span></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">Let's say: R1= 7 ohms, R2= 10 ohms, R3= 6 ohms and R4= 4 ohms. We are required to get the total resistance of the circuit.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Using the steps that previously discussed <a href="http://electricalengineeringforbeginners.blogspot.com/2011/06/series-parallel-circuits-part-1.html">here</a>. We can redraw an equivalent circuit in a way that we can understand it well. The figure below is the redrawn circuit for the given problem above.</span><br />
<br />
<span style="font-family: Georgia;">1. Redraw the circuit.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-kG3AFPAFIvs/TfAV14La6aI/AAAAAAAAAQA/jXJ1jJuiqZk/s1600/series-parallelredrawn.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-kG3AFPAFIvs/TfAV14La6aI/AAAAAAAAAQA/jXJ1jJuiqZk/s1600/series-parallelredrawn.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Redrawn Series-Parallel</em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">From the redrawn circuit above. we can now simplify R3 and R4. Lets name it R3-4 = 6+4 = 10 ohms.</span><br />
<br />
<span style="font-family: Georgia;">2. The next step is by getting the resistance between R2 and R3-4 connected in parallel. The circuit now will be simplify as shown below:</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-N7FceTFDATk/TfAhkgyAwxI/AAAAAAAAAQE/B36e5H2wzzo/s1600/simplifyseries-parallel.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-N7FceTFDATk/TfAhkgyAwxI/AAAAAAAAAQE/B36e5H2wzzo/s1600/simplifyseries-parallel.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Simplified R3-4 to be combine with R2</em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">Next, we will simplify the R2 and R3-4 using the formula of <a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-2.html">two resistances connected in parallel</a>. Let's name it Ra= R2 X R3-4 / R2 + R3-4 = 10 x 10 / 10 +10 = 5 ohms.</span><br />
<br />
<span style="font-family: Georgia;">3. Now, take a look on the next circuit figure below. The circuit was already simplified into series circuit and we can already get the total resistance of the circuit. </span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-Rn2P_0_MQmU/TfAxd-pE3eI/AAAAAAAAAQI/njdtq-hZxw4/s1600/series-equivalent.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-Rn2P_0_MQmU/TfAxd-pE3eI/AAAAAAAAAQI/njdtq-hZxw4/s1600/series-equivalent.png" t8="true" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Equivalent Simplified Circuit </em></td></tr>
</tbody></table>
<span style="font-family: Georgia;">The total resistance of the given circuit based on the simplified circuit above would be: Rt= R1 + Ra = 7 + 5 = <strong>12 ohms</strong>. Very easy right? </span><br />
<br />
<span style="font-family: Georgia;">It only means that a complex circuit can be broken down into simplified circuit to get the total resistance Rt = 12 ohms.</span><br />
<br />
<span style="font-family: Georgia;">I will leave the next circuit as your exercise. This is a bit complicated than above circuit problem. The application is still the same. Given that R1= 1 ohm, R2= 2 ohms, R3= 3 ohms, R4= 4 ohms, R5= 5 ohms, R6= 6 ohms, R7= 7 ohms, R8= 8 ohms and R9= 9 ohms. What will be the total resistance of the circuit below?</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-PHmbaB-rMOk/TfBAK0wkJrI/AAAAAAAAAQQ/tLdxp2Zsunk/s1600/exercise.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="136" src="http://2.bp.blogspot.com/-PHmbaB-rMOk/TfBAK0wkJrI/AAAAAAAAAQQ/tLdxp2Zsunk/s320/exercise.gif" t8="true" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><em>Exercise: What is the total resistance of the given circuit? </em></td></tr>
</tbody></table>
<span style="font-family: Georgia, "Times New Roman", serif;">After further simplication, I found out that the total resistance of the given circuit above is <strong>11.10 ohms</strong>. Did you get the same answer? If not, you may leave your comment below and let's discuss...</span><br />
<br />
<span style="font-family: Georgia;">Since you've already understand the concept of series-parallel circuits, it is now time to move fast on another topic on my next post. This is already the Part 3 and would be dealing with bridge resistor circuits. </span><br />
<br />
<span style="font-family: Georgia;">Stay tune! </span><br />
<br />
<span style="font-family: Georgia;">Cheers!</span></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com2tag:blogger.com,1999:blog-1420533950904606452.post-28714548246422770262011-06-08T13:02:00.002+08:002011-06-09T06:35:53.238+08:00Series-Parallel Circuits- Part 1<div dir="ltr" style="text-align: left;" trbidi="on"><span style="font-family: Georgia, "Times New Roman", serif;">It's been a long time ago when I posted my last topic about <a href="http://electricalengineeringforbeginners.blogspot.com/search/label/Electric%20Circuits">Electric Circuits</a>. Though its very difficult to have time to write a topic for this blog, this site will always be alive for you. I would like to thank first those who have subscribed to this blog.</span><br />
<br />
<span style="font-family: Georgia;">Well, let's talk about another basic topic about Basic <strong>Electrical Engineering</strong>. This is about Series-Parallel Circuits. For those who are just new with this site, you can surely catch up with my previous post at <a href="http://electricalengineeringforbeginners.blogspot.com/p/electrical-engineering-syllabus.html">Electrical Engineering Syllabus</a> that I've provided last time.</span><br />
<br />
<span style="font-family: Georgia;">Circuits can be connected into complex circuits consisting of three or more resistors. One part of the circuit is in series and the other part could be connected in parallel. This connection is called the <strong>Series-Parallel Circuits</strong>.</span><br />
<br />
<span style="font-family: Georgia;">There are two types of series-parallel connections: the first one is the resistance in series with a parallel combination and the other one is the series in which the parallel combination have a series of resistances. Let's see the figure below for better understanding of this theory.</span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">a. This is a series resistances with a parallel combination.</span><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-XNzlczGuBX4/Te7pfvNx0RI/AAAAAAAAAPs/scWAm1_G69E/s1600/seriesparallelfig1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-XNzlczGuBX4/Te7pfvNx0RI/AAAAAAAAAPs/scWAm1_G69E/s1600/seriesparallelfig1.jpg" t8="true" /></a></div><span style="font-family: Georgia;">b. This is a series resistance and series of resistances in a parallel combination.</span><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-HydUpPHiBsI/Te7q9Wqvl7I/AAAAAAAAAPw/5o2gMtVRGtk/s1600/seriesparallelfig2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://2.bp.blogspot.com/-HydUpPHiBsI/Te7q9Wqvl7I/AAAAAAAAAPw/5o2gMtVRGtk/s320/seriesparallelfig2.jpg" t8="true" width="220" /></a></div><span style="font-family: Georgia;">Take for example you have three lamps to be connected in a source of a battery. There are two ways that you can connect it. The first one is that: connecting the first lamp connected in series to the parallel combination of the second and third lamp. The second one is that: first and second lamp is connected is series then connect it parallel to the third lamp.</span><br />
<span style="font-family: Georgia;"></span><br />
<br />
<strong><span style="font-family: Georgia, "Times New Roman", serif;">How to Simplify a Series-Parallel Circuit Connection?</span></strong><br />
<br />
<span style="font-family: Georgia;">In dealing with series-parallel connection, there's nothing something new formula to be use here except for concept of Ohm's Law. </span><br />
<br />
<span style="font-family: Georgia;">In terms of simplifying a circuit, all you need to do is to start first with the most complex part before you get the overall resistances of the entire circuit. Take the following steps below as your guide. This is what I've always follow when I was still a student. </span><br />
<br />
<span style="font-family: Georgia;">1. First, redraw the circuit in a most compre</span><span style="font-family: Georgia;">hensive way if necessary. Some circuits looks like complicated at first glance, but if you will redraw it equivalent to the original circuit, you could easily deal with it.</span><br />
<span style="font-family: Georgia;">2. Start to simplify the circuit in the complex part. In the parallel combination with branches consisting of two or more resistors in series, start to simplify them by adding its value. </span><br />
<div style="text-align: left;"><span style="font-family: Georgia;">3. Then using the formula of the parallel resistances, get the value of resistances of parallel parts of the circuit.</span></div><div style="text-align: left;"><span style="font-family: Georgia;">4. Then, combined the resistances of the entire circuit.</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">Is it clear? Ok let's proceed...</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">Let's take a sample figure 2 below:</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-D3v8bHO-Zdw/Te79McBbTvI/AAAAAAAAAP0/N_gpMrBOWDM/s1600/seriesparallelfig2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://4.bp.blogspot.com/-D3v8bHO-Zdw/Te79McBbTvI/AAAAAAAAAP0/N_gpMrBOWDM/s320/seriesparallelfig2.jpg" t8="true" width="220" /></a></div><div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Using the steps above:</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">1. You don't need to redraw the circuit above, since it is obviously where to start simplifying the circuit.</span></div><div style="text-align: left;"><span style="font-family: Georgia;">2. Simplify the resistance of D and E first using the equivalent <a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-1.html">resistances in parallel</a> formula. Then, add the combined resistances of D and E to C using the <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/dc-series-circuit-part-1.html">resistances in series</a> formula.</span></div><div style="text-align: left;"><span style="font-family: Georgia;">3. Since you already get the value of combined resistances for D and E to C. Then, you may now get the combined resistances of B to D, E and C using resistances in parallel.</span></div><div style="text-align: left;"><span style="font-family: Georgia;">4. Get the overall resistance Rt = Ra + R(combined resistances of B, C, D and E).</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">Now you get the clear understanding of series-parallel circuits. On my next post, I will show you more illustrative examples which were already given in previous <strong>Electrical Engineering Board </strong>Exams.</span></div><div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia;">Cheers!</span></div></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com2tag:blogger.com,1999:blog-1420533950904606452.post-6951986381026377882010-12-24T12:16:00.001+08:002013-03-30T17:48:08.488+08:00Electric Shock - Save Yourselves!<div dir="ltr" style="text-align: left;" trbidi="on">
Its been a long time when I post my last topic here in <a href="http://electricalengineeringforbeginners.blogspot.com/"><strong>Learn Electrical Engineering for Beginners</strong></a>. Although the author of this blog is very busy with his life, he's trying to make a post in order to feed his readers with good information. <br />
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<br />
It's Christmas time! We're expecting that everyone is busy preparing to celebrate this holiday season. In connection to this, accidents are also very prone this holiday season. There are fires, car accidents and a lot more. One of the topic that I would like to stress now is related to my blog which is Electric/Electrical Shocks which is one of the common accident this Christmas Season.<br />
<br />
Electric/Electrical Shock is the injury caused by an electrical current passing through the body. The electricity maybe atmospheric (lightning) or man-made (high-voltage transmission and low-voltage lines). Possible injuries include burns and physiological disturbances, which may range from a minor burn to death in severe cases.<br />
<br />
<strong>What are the common causes?</strong><br />
<ul>
<li>Touching high-tension wires that fall during the storm.</li>
<li>Touching an electric socket or worn out cord.</li>
<li>Mixing water and electricity.</li>
<li>Being struck by lightning. A bolt of lightning carries as many as 30 million volts.</li>
</ul>
<strong>What are the symptoms?</strong><br />
<ul>
<li>Shocking sensations. Numbness or tingling. A change in vision, speech, or in any sensation.</li>
<li>Burns or open wounds. These occur where the electricity enters and exits the body.</li>
<li>Muscle spasms or contractions.</li>
<li>Sudden immobility or fractures. A body part may looked deformed.</li>
<li>Interrupted breathing. Irregular heartbeat or chest pain.</li>
<li>Seizures.</li>
<li>Unconsciousness.</li>
</ul>
<strong>What are the steps to save the victim from electric/electrical shocks?</strong><br />
<ol>
<li>The victim usually gets stuck to the source of the electricity, and it important that you first separate him from electrical source.</li>
<li>Turn off the power supply switch and disconnect the plug. It's best to simply turn off the main power supply or pull out the fuse. Often, simply turning off the switch may not stop the flow of electricity. </li>
<li>DO NOT touch the victim with your bare hand, or the electric current will pass through you as well.</li>
<li>Water with electrolyte, fruit/vegetable juice should be given to the victim.</li>
<li>If you are barefoot, stand of some clothes or any hand non-conductive material like wood or paper. Make sure that you are not standing on anything that is wet.</li>
<li>Throw a blanket over the victim and try to separate him from the source. Make sure you don't touch him though. You could also use dry, non-conductive material such as wooden broom handle or a chair to separate the victim from the live current, whatever is handy.</li>
<li>Once the victim has been separated, check to see if he is breathing. If breathing has stopped or seems slow, administer CPR immediately. </li>
<li>Cover the victim with a blanket.</li>
<li>If the victim has a burn, remove the clothing from the burned area (unless it's stuck to the skin) and rinse it in cool running water. Cover the burn with a dressing.</li>
<li>Don't apply ice or any ointment or cotton dressing to the burn.</li>
</ol>
Video related to electric shock first aid:<br />
<br />
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<embed src="http://www.youtube.com/v/2sAPsz4_2E0?fs=1&hl=en_US" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"></embed></object><br />
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I hope I had given you another learnings here in Learn Electrical Engineering for Beginners where you are always safety and have knowledge power!<br />
<br />
Have a Merry Christmas to everyone! Stay tune for more post.</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-61174417725563098192010-08-22T14:09:00.006+08:002013-03-30T17:49:55.854+08:00What Jobs Awaits for Electrical Engineering Graduates Today?<span style="font-family: Georgia, "Times New Roman", serif;">Let me pause for awhile in posting lectures about <strong>Electrical Engineering</strong>. Let's have a short talk about the jobs and opportunity awaits for all graduates of Electrical Engineering. Details are fact based on the present situation nowadays and also based on my own personal experienced.</span><br />
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<span style="font-family: Georgia;">According to National Statistics Coordination Board in the Philippines, unemployment rise from 7.5% of April 2009 to 8.0% of April 2010. The most common rise up of this unemployment rate is because of the newly graduates and one of these are the Electrical Engineering graduates. There are approximately about 1,611 newly graduates in the Philippines (Registered for Electrical Engineering Board Exams)which are expected to be unemployed nowadays. Still grabbing for a new job and opportunity that may suit them. </span><br />
<br />
<span style="font-family: Georgia;">In U.S. according to Department of Labor's Bereau of Labor Statistics employment grew for Electrical Engineers by 7.8% from the fourth quarter of 2009 to the first quarter of 2010 and now 16.1% above compared to its very low employment last first quarter of 2009. This rise of employment for Electrical Engineering field was due to positive recovery in the U.S. economy. </span><br />
<br />
<span style="font-family: Georgia;">What's the reason of giving this short comparison to you? </span><br />
<br />
<span style="font-family: Georgia;">Sad to say, there are only limited of opportunities and jobs for Electrical Engineering here in the Philippines even before during my time way back 1999 after passing my licensure exams. If there is an opportunity or jobs available for Electrical Engineering, the salary or the compensation is very low. Besides, majority of them will required you a minimum two years experience for the position even in the Junior/Cadet Engineering position. That's why other Electrical Engineering graduates like me were all shifted to semiconductor field with attractive compensation and benefits and got their luck related to their field and some aren't got any but dive into last resort which is in the semiconductor operations/processes. Others with no luck in this field grabbed their fate in the call centers- poor engineers...</span><br />
<br />
<span style="font-family: Georgia;">In connection to this, many Electrical Engineering graduates and experts in the Philippines try their luck to get their job somewhere in US, Australia, Taiwan, Japan, North Korea, Saudi Arabia, Abu dhabi, Kuwait, Dubai, Dammam, Doha, Qatar and sometimes in Canada but quiet limited unlike Saudi Arabia. This means that the Philippine Government should work more on jobs availability most especially to our professionals and experts. </span><br />
<br />
<span style="font-family: Georgia;">Take a look on this snapshot in JobStreet: Click on the picture to enlarge.</span><br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/THCaiyoHhKI/AAAAAAAAAOU/L2g8nvt4sTc/s1600/EE+Jobs.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img alt="Electrical Engineering Jobs" border="0" height="187" ox="true" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/THCaiyoHhKI/AAAAAAAAAOU/L2g8nvt4sTc/s320/EE+Jobs.PNG" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-family: Georgia, "Times New Roman", serif;">If you click "ALL Locations" will appear only limited opportunities in the Phillipines where most of the jobs are in Saudi Arabia in Electrical Engineering.</span></td></tr>
</tbody></table><br />
<span style="font-family: Georgia, "Times New Roman", serif;">Take a look on the National Capital Region- Philippines. There are only 97 jobs available below considering this region is the center for opportunities and jobs in any field. Would you think newly graduates of Electrical Engineering could accomodate them all? If you read the qualifications almost none of them offers for newly graduates. If there is, only FEW. </span><br />
<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/THCd_nG0MhI/AAAAAAAAAOc/_RSZnIOW2v0/s1600/EE+Jobs+2.PNG" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img alt="Electrical Engineering Jobs" border="0" height="187" ox="true" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/THCd_nG0MhI/AAAAAAAAAOc/_RSZnIOW2v0/s320/EE+Jobs+2.PNG" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-family: Georgia, "Times New Roman", serif;">If you click the National Capital Region, only 97 jobs available with almost no one affers jobs for Electrical Engineering newly graduates.</span></td></tr>
</tbody></table><br />
<span style="font-family: Georgia, "Times New Roman", serif;">In my information above, I've already given you a brief idea where most of the jobs and opportunities are coming in for Electrical Engineering field. </span><br />
<br />
<span style="font-family: Georgia;">So, what are the recommended job positions for a newly graduate of Electrical Engineering?</span><br />
<br />
<span style="font-family: Georgia;">If you are a newly graduate of Electrical Engineering, it is recommended that you start with a Junior position for your job. Some company are required you to train with their traning program if you are a newly graduate but sad to say, NOT ALL. That's why before you accept their offer, be sure they have some training for you before you start with your current job. Some company have higher expectations with their applicants to have a certain number of years of experience for the job position although you are only a newly graduate. Make sure you won't missed this step. </span><br />
<br />
<span style="font-family: Georgia;">Some of the known positions for Junior position are Assistant Project Engineer, Junior Design Engineer, Estimates, Electrical Inspectors, Electrical Engineer (Junior Position) and many others related to Electrical Engineering jobs. There are some company that accepts newly Electrical Engineering graduates in the middle position considering if you have an attractive qualifications like:</span><br />
<ul><li><span style="font-family: Georgia;">Board Topnochers</span></li>
<li><span style="font-family: Georgia;">With above average rating in academic performance or an Honor Student.</span></li>
<li><span style="font-family: Georgia;">With certain special skills for their requirements: i.e autocad, excellent communication skills, programming, etc...</span></li>
<li><span style="font-family: Georgia;">Experience in certain field they required. Majority one year experience. </span></li>
</ul><span style="font-family: Georgia;">Not all newly graduates have these qualifications. If you did not fall under these qualifications. Don't loose HOPE. </span><br />
<br />
<span style="font-family: Georgia;">There are also some other ways you can explore your skills in Electrical Engineering if this is only your line of interest or the job that you want. You should not depend on what the company could offer job for you. Life must go on. You can also start on your own. For example, building your own repair shop related to electrical machineries, electrical layout job for residential, job for electrical installations etc. </span><br />
<br />
<span style="font-family: Georgia;">Examples above are the acquired skills during your college days and you can explore these things out. If you think you can shift to other line of interest, you may do so. It really depends where you can find your luck in line with Electrical Engineering jobs. Your own decision is still a MUST. </span><br />
<br />
<span style="font-family: Georgia;">In my personal experience, I have no luck in finding jobs for Electrical Engineering here in the Philippines because of the limited opportunities. Since, I have other skills that I can explore, I shifted in the field of Semiconductor operations and now in the Supervisory position. This is my decision and choice. But still hoping I can find my luck going abroad for Electrical Engineering jobs if there is any opportunity for me. Skills that I have in Electrical Engineering is still there and I believe I can find my luck anytime if I would pursue. </span><br />
<br />
<span style="font-family: Georgia;">Before I close this post, here are some of the sites where you can browse for jobs in Electrical Engineering:</span><br />
<ul><li><span style="font-family: Georgia;">Jobstreet.com - <a href="http://www.jobstreet.com/" rel="nofollow" target="_blank">http://www.jobstreet.com/</a></span></li>
<li><span style="font-family: Georgia;">Engineering Central - <a href="http://www.engcen.com/electrical.asp" rel="nofollow" target="_blank">http://www.engcen.com/electrical.asp</a></span></li>
<li><span style="font-family: Georgia;">Engineer.net- <a href="http://www.engineer.net/electrical.php" rel="nofollow" target="_blank">http://www.engineer.net/electrical.php</a></span></li>
<li><span style="font-family: Georgia, "Times New Roman", serif;">engineering4professionals.com- <a href="http://www.engineering4professionals.co.uk/" rel="nofollow" target="_blank">http://www.engineering4professionals.co.uk/</a></span></li>
<li><span style="font-family: Georgia;">careerbuilder.com- <a href="http://www.engineering.careerbuider.com/" rel="nofollow" target="_blank">http://www.engineering.careerbuider.com/</a></span></li>
<li><span style="font-family: Georgia;">engineerjobs.com- <a href="http://www.engineerjobs.com/jobs/electrical-engineering" rel="nofollow" target="_blank">http://www.engineerjobs.com/jobs/electrical-engineering</a></span></li>
<li><span style="font-family: Georgia;">Yuvajobs.com - <a href="http://www.yuvajobs.com/" rel="nofollow" target="_blank">http://www.yuvajobs.com/</a></span></li>
<li><span style="font-family: Georgia;">JustEngineers - <a href="http://www.justengineers.net/" rel="nofollow" target="_blank">http://www.justengineers.net/</a></span></li>
<li><span style="font-family: Georgia;">Click India - <a href="http://www.clickindia.com/" rel="nofollow" target="_blank">http://www.clickindia.com/</a></span></li>
<li><span style="font-family: Georgia;">Career Overview - <a href="http://www.careeroverview.com/" rel="nofollow" target="_blank">http://www.careeroverview.com/</a></span></li>
</ul><span style="font-family: Georgia;"><strong>P.S.</strong> These are the top ten sites I have chosen for you as your reference for finding jobs in Electrical Engineering. These are based on the Google search engine. My mission is to give my readers a high quality content readings for this site especially for beginners.</span><br />
<br />
<span style="font-family: Georgia;">The list of sites above are subject for change as long as I find other sites that offers good job for electrical engineering. So much for this...</span><br />
<br />
<span style="font-family: Georgia;">Cheers!</span><br />
<span style="font-family: Georgia;">Donny</span>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com2tag:blogger.com,1999:blog-1420533950904606452.post-17911300447578615782010-08-20T16:18:00.008+08:002013-03-30T17:53:47.738+08:00Electrical Engineering Tutorial Still Alive?<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="font-family: Georgia, "Times New Roman", serif;">It's been nine months ago when I stopped making a post here at <strong>Electrical Engineeering</strong> for Beginners. I had received more comments for some other topics I published. I just want to apologized for my very late response. As you know, I'm not a full time writer of this blog related to Electrical Engineering.</span><br />
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<span style="font-family: Georgia;">For this 2010 onwards, I will be making my own website about Electrical Engineering. But that will depend on my time since there are no other support I'm receiving to complete this course outline. As of now I'm busy learning an HTML or the Hyper Text Markup Language and CSS or the Cascading Style Sheets to make a website of my own. Besides, I'm a full time worker in one of the semiconductor company where all my time are being spend only to support my family. For all of you readers out there, good news! this Online Electrical Engineering tutorial will resume on September 2010. </span><br />
<br />
<span style="font-family: Georgia;">I will continue making a post about Electrical Engineering topics from where I stopped. This blog is now ranking well in Google SERP's which makes me to inspire because it's now PR4 which beats other big sites out there. This will push me helping out our young wanna be Electrical Engineering students today. I would like to thank those who are continue subscribing to this site though it's not been updated for a long time. Now, its your turn to open your laptop and see my two to three times post per week. </span><br />
<br />
<span style="font-family: Georgia;">More things to clarify out....</span><br />
<br />
<span style="font-family: Georgia;">I also received an email about this blog was being nominated as Top Blogs conducted by other big sites out there. Ok, well then there are you again links that would like to enter my site in order to have a one way link. You are encouraging me to put your badge to my site containing your links. I checked your site and it's only PR0. Well I'm sorry, but I refused to put your badge to my site. I would like to have a quality links to my site. I just only link to higher PR sites related to Electrical Engineering.</span><br />
<br />
<span style="font-family: Georgia;">There are also some concerns that I would like to clarify to my readers out there. One time, I received an email requesting me to post about a topic related also to Electrical Engineering. Ok, I will just make a post for your request if there are more requesting for that particular topic.</span><br />
<br />
<span style="font-family: Georgia;">I also received an emails on how to pass Electrical Engineering board exam. They are also asking me on how to get some manuals or review materials. Well, my advice to you is to enroll yourselves to any known Electrical Engineering Review Centers. In the Philippines, the well known Review Center is the Powerline. This is where I took my review and refresher. One thing that you should put in your mind especially to our Electrical Engineers wanna be. It's not on the review center that matters most, its on how you really understand the principles behind Electrical Engineering. Review Centers will just only give you the technique. Your knowledge is still that matters most here and your confidence on how you will pass your Electrical Engineering Board. To those who are new to this site. You may read my post here:</span><br />
<br />
<ul>
<li><span style="font-family: Georgia;"><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/how-to-prepare-for-electrical.html">How to Prepare for Electrical Engineering Board Exam?</a></span></li>
</ul>
<span style="font-family: Georgia, "Times New Roman", serif;">So much for that, I also received an email on exchanged links to other Electrical Engineering sites. Today, my decision is to stop having an exchanged links because Google will detect my site as spam. I'll better do a high quality content to sustain my SERP. Please stop sending me an email for exchanged links. I only do for one way link. </span><br />
<br />
<span style="font-family: Georgia;">Now, how do I convinced you to come back to my site?</span><br />
<br />
<span style="font-family: Georgia;">First, Learning Electrical Engineering keyword have a Global Monthly Search of 3,600. Not too bad to build this site. Advertising competition is only medium. </span><br />
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<span style="font-family: Georgia;">See this: keyword for "<strong>Learning Electrical Engineering</strong>"</span><br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/TG4yqg0GZJI/AAAAAAAAAOE/xeyHwkglRHU/s1600/Google+page.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="187px" ox="true" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/TG4yqg0GZJI/AAAAAAAAAOE/xeyHwkglRHU/s320/Google+page.PNG" width="320px" /></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">In Google Search Engine, this site is still number one considering nine months without post or update. Take a look on the other side below:</span><br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/TG40rxecuLI/AAAAAAAAAOM/roAdZegsdH8/s1600/Yahoo+Page.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="187px" ox="true" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/TG40rxecuLI/AAAAAAAAAOM/roAdZegsdH8/s320/Yahoo+Page.PNG" width="320px" /></a></div>
<br />
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<span style="font-family: Georgia, "Times New Roman", serif;">In Yahoo, Learn Electrical Engineering is still No.3 first page for the mentioned keyword. Considering my site is only hosted free, it also rank well with those mentioned big search engines. You can still find this site easily. Ranking well in search engines means this site has a higher authority and with good quality contents over the other. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">For my readers out there, please help me to continue this mission of teaching online for Electrical Engineering. I want to optimized my site with a most competitive keyword "Electrical Engineering". I hope I can still make it to the top. </span><br />
<br />
<span style="font-family: Georgia;">Stay tune for more updates here in Learn <strong>Electrical Engineering</strong> for Beginners.</span><br />
<br />
<span style="font-family: Georgia;">Cheers!</span><br />
<span style="font-family: Georgia;">Donny</span></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-49202821613009548182009-11-11T13:00:00.013+08:002013-03-30T17:56:10.446+08:00Applications for DC Parallel Circuits<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: justify;">
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<span style="font-family: Georgia, "Times New Roman", serif;">It's been a month and a half ago when I did an update on this blog. I've been busy with some things which I cannot explained to you right now. The reason why is that this project which is initiated by me is really intended for 2010. Although I'm doing a little update on this blog about <span style="font-weight: bold;">Learning Electrical Engineering</span>, the project still alive until such time I completed the course outline of this site.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">I checked my stats awhile ago and I was surprised that my traffic still in tact with an average of 45 and above visitors a day. The number of subscribers is still increasing a bit compared when I started this blog. One thing you can expect from me, though I'm doing a little update on this blog, this site will still remain in search engines. I have now a PR2 from PR0. I would like to give thanks for those who supported and still visiting this site. Target keyword "<span style="font-weight: bold;"> learning electrical engineering</span> " now dominates my rank in Google and Yahoo. Now let's continue of what I have left before.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">The following illustrative problems below are just an applications of what I had presented on my post about DC Parallel Circuits.</span></div>
<ul>
<li><div style="text-align: left;">
<a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-1.html"><span style="font-family: Georgia, "Times New Roman", serif;">DC Parallel Circuits 1</span></a></div>
</li>
<li><div style="text-align: left;">
<a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-2.html"><span style="font-family: Georgia, "Times New Roman", serif;">DC Parallel Circuits 2</span></a></div>
</li>
</ul>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">From the previous post, I had illustrated to you the theory of the DC Parallel Circuits which consists of Part 1 and 2. Today, I will show you the application through problem solving.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Problem No. 1 : Let's begin solving parallel circuits in which I will show you how the current divides in each branch of circuits and how can we obtained the unknown values using Kirchhoff's Law. Suppose I have the circuit as shown in the diagram below. We need to find the unknown currents. We need to find the currents at I1, I4 , I6 and I7 . Click the image to enlarge.</span></div>
<br />
<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SvosdF6XmzI/AAAAAAAAANc/-yhVm3eRHkw/s1600-h/DC+PARALLEL+CIRCUITS.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5402679581299153714" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SvosdF6XmzI/AAAAAAAAANc/-yhVm3eRHkw/s400/DC+PARALLEL+CIRCUITS.jpg" style="cursor: pointer; display: block; height: 300px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Obviously, we can find the current at junction A by simply performing the KCL or the Kirchoff's Current Law. We can see that the current divides at junction A. Therefore, I1 = I2 + I3 = 7A + 3 A= <span style="font-weight: bold;">10 amperes.</span></span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Taking a look at junction C. We can see that the current entering this junction I2 divides into I4 and I5 which becomes a total current to I2. Therefore we can say, I2 = I4 + I5, which becomes 7 A = I4 + 5 A, and since we are looking for I4 = <span style="font-weight: bold;">2 Amperes</span>.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Now, we would like to solve for I6. Since, the flow of currents I3 and I4 are flowing toward junction B. We can say that, I6 = I3 + I4 = 3 A + 2 A = <span style="font-weight: bold;">5 Amperes</span>.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">The last requirement is I7. Since the currents I5 and I6 are flowing toward junction D. we can say that I7 = I5 + I6 = 5 A + 5A = <span style="font-weight: bold;">10 Amperes</span>.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">The example above I had shown you is how the currents divides in each branch. The next example will illustrate you the application of unequal resistors in parallel.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Problem No.2 : Three loads A, B and C are connected in parallel to a 230 volt source. Load A takes 9.2 KW, load B takes a current of 60 amperes and load C is a resistance of 4.6 ohms. Calculate (a) the resistance of loads A and B, (b) the total resistance of the three paralleled loads, (c) the total current, (d) the total power. Click the image to enlarge.</span></div>
<br />
<div style="text-align: left;">
<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/Svo26_Oq3kI/AAAAAAAAANk/dcUYP0ThWbI/s1600-h/PARALLEL+CIRCUITS+SAMPLE+PROBLEM.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5402691090017607234" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/Svo26_Oq3kI/AAAAAAAAANk/dcUYP0ThWbI/s400/PARALLEL+CIRCUITS+SAMPLE+PROBLEM.jpg" style="cursor: pointer; display: block; height: 300px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Since we have different parameters given for each load, We have to solve first some missing requirements that we need in our calculations.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">(a) Calculating Ra = Vt ^2 / Pa = 230 ^ 2 / 9,200 = <span style="font-weight: bold;">5.75 ohms </span></span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">for Rb = Vt / Ib = 230 / 60 = <span style="font-weight: bold;">3.83 ohms</span></span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Note : The solutions shown above are just an applications of ohms law that we previously discussed.</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">(b) Calculating the total equivalent resistance of the circuits will follow:</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Req = 1/ 1/5.75 + 1/3.83 + 1/4.6 = 1/ 0.174+ 0.261+0.217 = <span style="font-weight: bold;">1.53 ohms </span></span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Note : the formula used was just discussed on the previous post above </span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-2.html"><span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">DC PARALLEL CIRCUITS</span></a><span style="font-family: Georgia, "Times New Roman", serif;">.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">(c) Since we have already calculated the missing values, we can now solve for total current.</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">It = 230 / 1.53 = <span style="font-weight: bold;">150 Amperes</span></span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">(d) To get the total power, since we already know the values of Vt and It. It will be now,</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Pt = 230 x 150 = <span style="font-weight: bold;">34,500 watts or 34.5 KW</span>.</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">For the above illustrative problem ( No.2 ) , I had shown you how the Ohms Law was also applied in solving the unknown quantities. This will surely applied when one value is missing. This is the best and basic technique that you can applied anytime you encountered such problems like this. The mentioned technique will also be applied when we reached complex AC circuitry.</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">This is the end of our basic circuitry in parallel connections. The next post will deal about series-parallel circuits lecture.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">See you again on my next post.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Cheers!</span></div>
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</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com1tag:blogger.com,1999:blog-1420533950904606452.post-51459221663584376282009-09-25T20:10:00.008+08:002013-03-30T18:35:28.996+08:00DC Parallel Circuits Part 2<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-family: Georgia, "Times New Roman", serif;">Yes, let's continue of what we had left last time here in <span style="font-weight: bold;">Electrical Engineering for Beginners</span>. I was glad that you are still there and an increasing number of subscribers makes me feel more energetic in writing more in this <span style="font-weight: bold;">Electrical Engineering</span> course. But before you rolled your eyes over me, the coverage of this lesson for today is all about the unequal resistors, kirchoff's law and applying ohm's law in parallel circuits.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Last time, I had mentioned about solving the total resistance in parallel with equal resistors. I will tell you how it was derived when I reached the topic of solving unequal resistors in parallel within today. Let's begin to have a short introduction of unequal resistors in parallel then, I will insert Kirchhoff's first law before continue discussing unequal resistors in parallel. I did it that way because Kirchhoff's first law has something to do with the flow of current.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Moving on...</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">If the circuit contains resistors in parallel whose values are not equal or unequal, we have some difficulty in assessing the total resistance of the circuits. One easy way to get the total resistance in parallel is by using your ohmmeter to measure the total resistance. Suppose you have an R1 and R2 connected in parallel with 40 and 80 ohms respectively, you would obviously measure a total resistance of 27 ohms for that circuit.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Wondering how it was obtained?</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">In our previous lesson, </span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/09/dc-parallel-circuits-part-1.html"><span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">DC Parallel Circuits Part 1</span></a><span style="font-family: Georgia, "Times New Roman", serif;">. I had mentioned there that the current flowing in each branch of the parallel circuits are not equal if the resistances were also different from each other. More current will flow on the smaller resistance compared to that with bigger resistance value. All of them were mentioned in this post without some problem illustrations. I just only show you how the current divides parallel connections with varying values of resistances.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Since, it is not often possible to get the total resistance of the circuit by using an ohmmeter especially in this case our circuit connection is getting more complex, we ought to know how to get such values by using calculations. Previously, we had learned the useful concepts of Ohm's law by solving circuit values in series circuit connection. But in this case, there is another equation which you will need this time. It is what we have been waiting for. It was known as <span style="font-weight: bold;">Kirchhoff's First Law</span> - Second Law was already discussed</span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/dc-series-circuit-part-1.html"><span style="font-family: Georgia, "Times New Roman", serif;"> <span style="font-weight: bold;">here</span></span></a><span style="font-family: Georgia, "Times New Roman", serif;">.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">What was it all about?</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Kirchhoff's Law is true in every type of circuit. The concerns of this law is not the circuit as the whole but only individual junctions where currents combine within the circuit itself. It's law states that : </span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">The sum of the currents flowing toward a junction always equal the sum of all currents flowing away from that junction</span><span style="font-family: Georgia, 'Times New Roman', serif;">.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Or, other states like this...</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">The algebraic sum of the currents at any junction of an electrc circuit is zero</span><span style="font-family: Georgia, 'Times New Roman', serif;">. This statement has something to do with the algebraic signs of the currents coming and moving away of the node. In order for you to understand this principle, take a look on the illustration below:</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SqHp8fZ18jI/AAAAAAAAAMk/Ho6K-PMT0S4/s1600-h/KCL2.gif" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377836655488397874" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SqHp8fZ18jI/AAAAAAAAAMk/Ho6K-PMT0S4/s400/KCL2.gif" style="cursor: pointer; display: block; height: 320px; margin: 0px auto 10px; text-align: center; width: 270px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">The image above is the simple representation of a circuit junctions. Suppose you have a four junctions there and all that conductors are carrying a current in the direction shown above. If you look at the image above IA are delivering stream of electrons at its <span style="font-weight: bold;">node</span>. It is obviously, that when the currents leaves that node, the current divides into IB, IC and ID which is equivalent to IA. Thus, making it IA = IB+IC+ID.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">or, in other ways of expressing it...</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">IA- IB - IC -ID = 0, which also states on the above Kirchhoff's first law. In this case, it is important to know the direction of current. The current coming to the node is (+) positive while the current leaving, we'll assign a (-) negative sign for it.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">I will be giving a pure problem illustrations of this topic on my next post this coming first week of October 2009 for you to comprehend well this topic. I reduced the frequency of posting due to my busy schedule at work.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Uhmm....</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">Unequal Resistors in Parallel Circuits</span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">Here are some of the important rules to remember when dealing with unequal resistors in parallel:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">1. The same voltage is impressed across all resistors.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">2. The individual-resistor currents are inversely proportional to their respective magnitudes. You will understand this fully when I give you the sample problem on my next post.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">3. The total current for the circuit is : <span style="font-weight: bold;">It = I1 + I2 + I3+...</span></span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">4. The total equivalent resistance of the circuit is:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">Req = 1/ (1/R1) + (1/R2) + (1/R3) + ....</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Note : When two unequal resistors are connected in parallel their equivalent resistance is equal to their product divided by their sum.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">R xy = Rx x Ry / Rx + Ry</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">Ohms Law in Parallel Circuits</span><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">Just like series circuits, we were also need to apply Ohms Law when dealing with the parallel circuits. We will be using this law to calculate some other unknown quantities like current, voltage, and resistance in such circuits. This law would require less time and effort if you would have to know such quantities mentioned above.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Let's say you have a number of resistors connected in parallel but you like to measure the </span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">resistance </span><span style="font-family: Georgia, 'Times New Roman', serif;">of a particular resistor using your ohmmeter. Of course, you would first disconnect the resistor to be measured from the circuit otherwise, you will measure or the ohmmeter reads the total resistance of the circuits.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Another one, if you would like to know the current across the particular resistor of a combination of parallel resistors using an ammeter. Again, this time you would have to disconnect it and insert an ammeter to read only the current flow through that particular resistor.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Knowing the voltage requires no disconnection. But of course, Ohms Law is the very pratical use in knowing such quantities for </span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">electrical engineers</span><span style="font-family: Georgia, 'Times New Roman', serif;"> like us.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">These are just a short concepts for our Part 2 of DC Parallel Circuits. On my next post it would be a little bit lengthy for I will illustrate to solve problems related to this topic.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">I will come back on</span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;"> first week of October 2009.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Cheers!</span></div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com2tag:blogger.com,1999:blog-1420533950904606452.post-44655553545446763432009-09-04T10:23:00.033+08:002013-03-30T18:28:22.043+08:00DC Parallel Circuits Part 1<div dir="ltr" style="text-align: left;" trbidi="on">
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Before I proceed with my new post here on <span style="font-weight: bold;">Learn</span> <span style="font-weight: bold;">Electrical Engineering for Beginners</span>, I would like to thank those who sent their email for some questions. Keep those emails coming in. If you don't received an email from me that means the answer is already here on my blog or I will answer you on my future post. Please read below portion of this site for your guidance.<br />
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I hope you are now learning with this site. We are now moving on with our topic and let's study the next part of Electric Circuits which is <span style="font-weight: bold;">DC Parallel Circuits</span>.<br />
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One objective of this lesson is for you to understand that you can solve any circuits because all circuits are made of combinations of series and /or parallel circuits.<br />
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Previously, in DC Series Circuits we defined that whether resistors, lamps or cells are connected <span style="font-weight: bold;">end-to-end. </span>Today, the scenario would be completely different. Instead of being connected <span style="font-weight: bold;">end-to end </span>as in series circuit, they are connected<span style="font-weight: bold;"> side by side </span>therefore it would create more than one path in which the current can flow. If this is so, we can say that resistors/resistances are said to be <span style="font-weight: bold;">parallel connected </span>or <span style="font-weight: bold;">connected in parallel</span>. The circuit would now be called a <span style="font-style: italic;">parallel circuit</span>. The image below is one example of the parallel circuit. I show you this illustration because the diagram already explains everything.<br /><br /><a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SqCIv4y9p_I/AAAAAAAAAL8/af_SU9J4i58/s1600-h/parallel_circuit.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377448311361742834" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SqCIv4y9p_I/AAAAAAAAAL8/af_SU9J4i58/s400/parallel_circuit.jpg" style="cursor: pointer; display: block; height: 341px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a>As what I mentioned in our earlier topic about electric circuits, we cannot say a complete circuit if we do not have a source of emf connected to them. For instance, we have two resistors connected their wire in parallel or parallel connected. When any two terminals are connected across the voltage source just as what had shown above, the whole arrangement- both resistors, the wires connecting them together and the voltage source - forms <span style="font-weight: bold;">complete parallel circuit.<br /></span>The circuit above shows that there are more than one path of current to flow. This means that these two resistors shares on the total current drawn from the battery. A part of the total current goes through the first resistor and at the same time a part of the total current also drawn for the second resistor. If you will intend to connect a device with polarity, for example a cell or batteries, you must connect the <span style="font-weight: bold;">positive terminals together </span>and the <span style="font-weight: bold;">negative terminals together</span>.<br /><br /><span style="font-weight: bold;"><span style="font-weight: bold;"></span>The Voltage in Parallel Circuit<br /><span style="font-weight: bold;"><br /></span></span>When the resistances are connected in parallel just like the diagram above and connected across the voltage source, the voltage across each resistors are always the same. Observe the diagram that the labels used were just the same. It was self-explained in the diagram that the voltage are just the same.<br /><br />Since, it is the fact that voltages across each resistances are just always the same. It has a practical consequence. What do I mean with this? This means that all components which are to be connected in parallel must have <span style="font-weight: bold;">the same voltage rating</span> if they are to work properly. Did you noticed that?<br /><br />The line voltage throughout the Philippines is 220 volts. In U.S. 120 volts. I think you are aware that some of our appliances are rated 120 volts or 220 volts work properly well. If in case you have a lamp or a bulb rated 20 volts. What the hell do you think will happen? The bulb will be burn immediately because the excess current will flow through it.<br /><br />Since, all appliances are connected across the same voltage source, the same voltage will also experienced across each load. Each load must be properly rated to handle this voltage.<br /><br /><span style="font-weight: bold;">How the Current Flow in Parallel Circuits<br /></span>In order to understand well the flow of current in the circuit for parallel connections. I made a little details on the diagram above. Take a look on the diagram below:<br /><br /><a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SqCfHznIlPI/AAAAAAAAAME/h0LWfXzM_xo/s1600-h/parallel_circuit.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377472911542621426" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SqCfHznIlPI/AAAAAAAAAME/h0LWfXzM_xo/s400/parallel_circuit.PNG" style="cursor: pointer; display: block; height: 341px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a>As I mentioned earlier, the flow of current in the parallel connections divides through each of the parallel paths. In the circuit diagram above, the two branches named them AB and CD connected in parallel. Observe that as the current flowing in each branches will divide and reunite them at node B returning to voltage source. You will wonder how the amount of current divides in each branches.<br /><br />The amount of current that will serve in each branches will depend on the resistance value. This will be the principle: <span style="font-weight: bold;">The current flowing through the several branches of a parallel circuit divides in inverse proportions, governed by the comparative resistance of the individual branches.<br /></span>And so? what does this mean?<br /><br />It only means that the lower resistance value in any branch circuit in proportion to the resistance of other branches in the same parallel circuit, the higher will be the current value or proportion which that branch will take.<br /><br />Simply...<br /><br /><span style="font-weight: bold;">In parallel circuit, branches having low resistance draw more current than other branches having high resistance</span>.<br /><br />Like voltage, the flow of current connected in parallel circuit is also of a great importance. For instance, like we know that every electrical appliances connected in parallel, the current will divide unequally to each branch since they have differing value of resistances- <span style="font-weight: bold;">the highest current flowing through the lowest resistance. </span>You will learn more about this when we reached the <span style="font-weight: bold;">protection against excessive flow of current</span>.<br /><br />Ohhh... I love to illustrate example again through problem solving. Let's take a sample problem for you to show that it really happens. Let's take again the circuit above and assign values for them.<br /><br /><span style="font-weight: bold;">Given that, I = 9 amperes ; I1 = 3 amperes ; I2 = 6 amperes</span><br /><br /><a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SqCqn0RtcGI/AAAAAAAAAMM/YCtJxZnqHGs/s1600-h/parallel_circuit.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377485556104917090" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SqCqn0RtcGI/AAAAAAAAAMM/YCtJxZnqHGs/s400/parallel_circuit.PNG" style="cursor: pointer; display: block; height: 341px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a>For instance, a total current I = 9 amperes is flowing through the parallel branch of R1 and R2. If you will observe, the value of R1 is twice the value of R2. We have mentioned earlier that <span style="font-weight: bold;">the current divides in inverse proportion to the values of the two resistors</span>. Therefore, only 3 amperes will flow on R1 and 6 amperes will flow on R2. If for example, R1 triple its resistance from 40 ohms to 120 ohms. The current flowing through R1 will be reduced from 3 amperes to 3/3 ampere or 1 ampere while I2 would remain unchanged. Thus the total current would be 6 amperes + 1 ampere = <span style="font-weight: bold;">7 amperes</span>.<br /><br />What does it implies?<br /><br />Since, in series circuit we said that all currents are the same throughout the circuit. In parallel circuit we just add it. This can be expressed mathematically as, <span style="font-weight: bold;">It = I1 + I2 + I3 +...</span><br /><br />What if the resistances are equal in parallel circuit? This will be the next topic.<br /><br /><span style="font-weight: bold;">Equal Resistors in Parallel Circuits<br /><br /><span style="font-weight: bold;"></span></span>Let's consider a water pipes connected in parallel as shown in the diagram below.<br /><br /><a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SqC1gmARdpI/AAAAAAAAAMU/Ph78aR28Jyk/s1600-h/pipes+parallel.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377497526642505362" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SqC1gmARdpI/AAAAAAAAAMU/Ph78aR28Jyk/s400/pipes+parallel.jpg" style="cursor: pointer; display: block; height: 260px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a>Let's say we have a constant water pressure incoming or the head as we learned in fluid dynamics. Assuming the same cross-sectional area of water pipes connected in parallel. The amount of water here would flow is equivalent to cross sectional area of pipe 1 + pipe 2.What do you think would happened to the amount of water flowing in the system if you would convert it to single pipe only? The amount of water will be less than that connected in parallel. Why? because you reduced cross sectional area in which the water would flow. The bigger the cross sectional area, the more water would flow on the system shown above.<br /><br />The same thing in resistor. The bigger the cross sectional of resistor, the more current would flow because the resistor value diminishes as the cross sectional area is getting bigger.<br /><br />The conclusion here is that :<span style="font-weight: bold;"> resistors or loads connected in parallel present a lower combined resistance or load than does any one of them individually.<br /></span>This means that if you have four 400 ohms resistors connected in parallel. The resistance of the combined load will be equally divided into 4 equal resistances thereby giving you a 100 ohms total resistance. In other words, 100 ohms is your combined resistances of four 400 ohms connected in parallel.<br /><br />If you don't get my point here, let's discuss it when we illustrate more problem solving in DC Parallel Circuits.<br /><br /><span style="font-weight: bold;">To be continued.....</span><br /><br />Cheers!<br /><span style="font-weight: bold;"></span></div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com8tag:blogger.com,1999:blog-1420533950904606452.post-71252786752477925962009-09-03T09:59:00.023+08:002013-03-30T18:31:20.608+08:00Applications: A Few Tips in Solving DC Series Circuit Problems<div dir="ltr" style="text-align: left;" trbidi="on">
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Before we proceed with to the Parallel Circuits, let's study first some other worded problems that you may be encountered during your board exam using the concept of DC Series Circuit. <span style="font-weight: bold;">Learn Electrical Engineering for Beginners</span> will provide you a technique on how you will overcome those scenarios.<br />
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The first problem that you will encounter is somewhat an application of a simple <span style="font-weight: bold;">transmission lines. </span>I just want to open this topic earlier because we will be dealing with this topic on my future post. I will just show you the snapshot on how those concepts that we studied in my previous post are being applied.<br />
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Let's begin...<br />
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<span style="font-weight: bold;">Problem 1</span>: The problem states that a load resistor of 4.1 ohms, 425 ft from 240-volt generator, is to be supplied with power through a pair of standard-size copper wires. If the voltage drop in the wires is not to exceed 5 percent of the generator emf, calculate (a) the proper AWG wire that must be used, (b) the power loss in the transmission line, (c) the transmission efficiency.<br />
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This is how you will going to solve it...<br />
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Let's have a simple visual of what is being said on the above problem. It could be shown just like the simple representation of transmission line below:<br />
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/Sp9MoR_gdaI/AAAAAAAAALU/s2aroD71cHk/s1600-h/simple+transmission+line.png" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377100735011911074" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/Sp9MoR_gdaI/AAAAAAAAALU/s2aroD71cHk/s400/simple+transmission+line.png" style="cursor: pointer; display: block; height: 228px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a>This simple transmission line can be considered as a circuit consisting of a 4.1 ohms resistor in series with 850 ft length of copper wire connected to a 240 volt generator. You would wonder why I say 850 ft length of wire while in the problem states that it is 425 ft from 240 volt generator. This is because in the given problem, it only mentioned the distance of the load resistor from the generator emf. It is not pertaining to the length of wire. Since, the load resistor has 2 ends connected to the wire across the generator emf, the actual length of wire should be <span style="font-weight: bold;">2 x 425 ft = 850 ft length</span>. Please take note on this because there are many beginners who are getting mistakes when solving this type of problem.<br />
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Moving on...<br />
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The voltage drop <span style="font-weight: bold;">would not exceed 5 %</span> of the generator emf therefore,<br />
<span style="font-weight: bold;">Line Voltage Drop</span> = 240 x 0.05 = 12 volts<br />
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Remember that in Ohm's Law in Series Circuit, the sum of all voltages across each resistances in series circuit is equivalent to the source emf therefore,<br />
<span style="font-weight: bold;">Load Voltage</span> = Generator Emf - Line Voltage Drop = 240 - 12 = 228 volts<br />
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You have to consider the line voltage drop when dealing with transmission line.<br />
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The Line Current across the 4.1 ohm load resistor would be,<br />
<span style="font-weight: bold;">Line Current</span> = 228 / 4.1 = 55.6 amperes<br />
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and the Line Resistance would also be,<br />
<span style="font-weight: bold;">Line Resistance</span>, Rl = 12 / 55.6 = 0.216 ohms<br />
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We are about to find the resistance per 1000 ft. Using the ratio and proportion to get the resistance per 1000 ft. It would be,<br />
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Resistance (1000 ft)/ 1000 ft = Line Resistance / length of transmission wire<br />
<span style="font-weight: bold;">Resistance per (1000 ft)</span> = (1,000 ft /850 ft) x 0.216 ohm = 0.254 ohm<br />
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a.) Consulting the AWG table below it shows that the standard wire size is <span style="font-weight: bold;">No. 4</span>; this wire has a resistance of 0.253 ohm per 1, 000 ft. This is based on the table below. Click the image to enlarge.<br />
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/Sp82y5kUzJI/AAAAAAAAALM/JloIdWb2Z34/s1600-h/awg-table.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377076728178199698" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/Sp82y5kUzJI/AAAAAAAAALM/JloIdWb2Z34/s400/awg-table.jpg" style="cursor: pointer; display: block; height: 400px; margin: 0px auto 10px; text-align: center; width: 248px;" /></a>b.) Getting the actual line resistance would be,<br />
Actual Line Resistance / 850 ft = 0.253 ohm / 1000 ft<br />
<span style="font-weight: bold;">Actual Line Resistance</span> = 0.85 x 0.253 = 0.21505 ohm<br />
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We need to get the following data in order to get line power loss.<br />
<span style="font-weight: bold;">Total Series-Circuit Resistance</span> = 4.1 + 0.21505= 4.31505 ohms<br />
<span style="font-weight: bold;">Total Current I</span> = 240 / 4.31505 = 55.6193 amperes<br />
<span style="font-weight: bold;">Line Loss Power</span> = (55.6193)^2 x 0.21505 = <span style="font-weight: bold;">665 watts</span> -<span style="font-weight: bold;"> answer</span><br />
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c.) Getting the efficiency of transmission line can be derived in terms of load power and total power. This can be expressed as:<br />
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Efficiency = Load Power / Total Power = [(55.6193)^2 x 4.1 / 240 x 55.6193 ] x 100 %<br />
<span style="font-weight: bold;">Efficiency </span>= <span style="font-weight: bold;">95% - answer</span><br />
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This is not not formal discussion about <span style="font-weight: bold;">transmission line</span>. We will touch it more in depth on my succeeding post. I'm just showing you how the concepts are being applied with these kind of application.<br />
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You would also encountered some tricky problems like what I will going to show you. This is just a simple one but you would used a simple ohm's law solving for unknown values. Let's have this example below:<br />
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<span style="font-weight: bold;">Problem 2</span>: A dc generator may be characterized by an ideal voltage source in series with a resistor. At the terminals of the generator, voltage and current measurements for two different operating conditions are Vt = 115 V at I = 10 A and Vt = 105 V at I = 15 A. Model the generator by a voltage source in series with a resistor.<br />
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This is already an application to dc generator. The problem required you to model the generator by a voltage source in series with a resistor. So, let's do what is being said. I have here the figure that shows the simple model mentioned in the problem in order to visualize it. I always love to draw the figure first when solving problems in <span style="font-weight: bold;">electrical engineering. </span>In the first place I'm a visual person. It's the easy way to understand what the problem is asking for.<br />
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/Sp9ThwdJ87I/AAAAAAAAALc/S0CyGdy29E8/s1600-h/series+wound+generator.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5377108319511638962" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/Sp9ThwdJ87I/AAAAAAAAALc/S0CyGdy29E8/s400/series+wound+generator.jpg" style="cursor: pointer; display: block; height: 229px; margin: 0px auto 10px; text-align: center; width: 336px;" /></a>With the circuit model of the generator shown above, I will defined the symbols as Vo for the dc generator voltage, I is the current, Rg is the series field connected to the dc generator and Vt is the terminal voltage or the generator output. Since there are two conditions mentioned above, let's expressed it in ohm's law for finding the dc generator voltage Vo which has no value given in the above problem and in terms of resistance which we also need to know here.<br />
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Therefore, we can state that: <span style="font-weight: bold;">Vo = Vt + IRg</span><br />
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Condition 1 : Given that Vt = 115 V and I = 10 A<br />
Vo = 115 V + (10 ) Rg, will served as <span style="font-weight: bold;">equation 1</span>.<br />
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Condition 2 : Given that Vt = 105 V and I= 15 A<br />
Vo = 105 V + (15) Rg will served as<span style="font-weight: bold;"> equation 2.</span><br />
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Let's equate 1 and 2 since the value of Vo in equation 1 and 2 are equal. We can therefore expressed it mathematically as,<br />
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115 V+ 10Rg = 105 V + 15 Rg , we can now solve for Rg.<br />
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<span style="font-weight: bold;">Rg = 2 ohms- answer</span><br />
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Then solving Vo will give, you may substitute the value of Rg from either equation 1 or 2 above will yield, I will choose equation 1.<br />
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<span style="font-weight: bold;">Vo = 115 V + (10)(2) = 135 V - answer</span><br />
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These are some of the illustrative problems that I could share with you. on DC series Circuit Just always remember that when solving problems like what I illustrated above:<br />
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1. Try to make an simple illustration of the problem to picture out and understand the scenario of the problem.<br />
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2. Always try collect first the given data before proceeding in solving the problem.<br />
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3. There are some problems that you need to solved first the missing data before solving what was being required. Problem number 1 and 2 above are the best examples of this one.<br />
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4. Know what is the subject matter of the given problem is also one that you should not forget.<br />
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5. Always review and finalize your answers.<br />
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My next post will continue our study of Circuits connected in Parallel.<br />
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Hope you learned some tricks for today here in <span style="font-weight: bold;">Learn </span><span style="font-weight: bold;">Electrical Engineering for Beginners</span>.<br />
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Cheers!</div>
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</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com1tag:blogger.com,1999:blog-1420533950904606452.post-65268281043280239522009-09-02T18:09:00.026+08:002013-03-30T18:10:24.649+08:00DC Series Circuit Part 2<div dir="ltr" style="text-align: left;" trbidi="on">
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Hello folks, I'm glad that you're still there hunting for my new post here in our study of basic electrical engineering which is very recommended for the beginners. Well, if you find this site useful for you, then tell your friends and let them subscribe to my articles.</div>
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It's a little bit hectic on my work schedule including posting my blog here. Because I have to double my effort just for you... How sweet...I will not make my introduction again get longer because I know that you're really want to learn more here. So, let's continue of what we've left last moment which is the <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/dc-series-circuit-part-1.html"><strong>DC Series Circuit Part 1</strong></a>. For those who missed it, you can still catch up with the lecture. </div>
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Let's study the continuation...</div>
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<strong>The Voltage Division in the Series Circuit</strong></div>
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In a series circuit, you would be able to find the voltage across at any point in the series circuit. The voltage which we called it the step-down voltage. </div>
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A circuit used for this is shown in the circuit diagram below. This is called a <strong>voltage divider</strong>. You may click the picture to enlarge.</div>
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/Sp5MiGzo6HI/AAAAAAAAAKE/LXhPmNRwxZ4/s1600-h/voltage+division.jpg"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5376819153953417330" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/Sp5MiGzo6HI/AAAAAAAAAKE/LXhPmNRwxZ4/s400/voltage+division.jpg" style="display: block; height: 216px; margin: 0px auto 10px; text-align: center; width: 216px;" /></a> Let's assume in the given circuit above that the applied voltage with the electrical notation of E or Ein is 100 volts and the values of R1 and R2 are 15 and 20 ohms respectively. You may want to know what is the applied voltage across R2 with the electrical notation of Vout or Eout which is written in another way. Please take note that the electrical notation of Vout or Eout could be an input voltage for another circuit. which of course will also become an Ein once more). That topic would be study on my succeeding post here in Electrical Engineering. <br />
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So, this is how you will going to do it... <br />
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We know that the total resistance in the circuit is 15 + 20 = 35 ohms. Given that the given circuit voltage is 100 volts. You may now use the Ohm's Law to find the circuit current. This is:<br />
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I = Ein / Rt = 100/35 = 2.86 amperes<br />
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Then, let's go to R2 in the given circuit. We all know that the resistance is 20 ohms and we have just calculated that the current is 2.86 amperes (from the conditions mentioned earlier in Ohm's Law Series Circuit that the current in the series circuit are just the same throughout). </div>
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Therefore, you will obtain that,<br />
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Eout = I x R2 = 2.86 x 20 =<strong> 57.2 volts - answer</strong><br />
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You may observe that with the appropriate choice of resistor values in voltage divider chain, an input voltage of 100 volts has been <strong>stepped down </strong>to an output voltage of 57.2 volts. By using ohm's law, you would be able to calculate it.<br />
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Since we have an illustrative example above, let's expressed the above example into equation. Given that the total resistance is R1 + R2 and ohm's law tells you that the circuit current would be: Ein/ R1+R2. This is also the current across R2. Using Ohm's Law again, we can calculate the Eout = I x R2 will give an equation just by substitution:<br />
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Eout = (Ein/R1+R2) x R2</div>
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Expressing it in a correct and understandable way would give you...</div>
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<strong>Eout = (R1/R1+R2) x Ein </strong><br />
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The equation above is the simplified formula that you can use in the given condition like this in voltage division chain. To put the equation into words. <strong>The voltage across any resistor in a voltage divider chain can be calculated by multiplying the value of that resistor by the input voltage dividedby the total resistance of the circuit.</strong><br />
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Norton's and Thevenin's Theorem are commonly used principles when solving such voltage divider problems. This will be another basic concept that we will study on my succeeding post. For the meantime, let's absorbed first what we have now.</div>
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<strong>The Variable Resistors</strong><br />
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You can also vary the resistance of the circuit as well. If you are not aware, you always done it by yourself by adjusting volume of your radio. This is what we called a variable resistor.<br />
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The resistor can be made variable in this way by means of sliding arm made of good conducting material to be arranged so that it can be moved along the length of the resistor. The resistor is then connected into the circuit with one of its end fastened to the sliding arm. By moving its sliding arm along the resistor, the value of the resistor can be varied at between maximum and minimum (zero). <br />
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If the variable resistor is used in this way, it is called the <strong>rheostat</strong>. It is used to control the current flow in the circuit. </div>
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/Sp5o-4kBfWI/AAAAAAAAAKU/vrt_YhjHpCU/s1600-h/pot_rheostat.gif"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5376850434671607138" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/Sp5o-4kBfWI/AAAAAAAAAKU/vrt_YhjHpCU/s320/pot_rheostat.gif" style="display: block; height: 207px; margin: 0px auto 10px; text-align: center; width: 320px;" /></a> The maximum value of resistance was obtained when the slider moves on the lower position as what had shown on the illustration above- left portion. (You may click the image to enlarge). Likewise, when the slider moves upward would obtain the minimum value of resistance. This is the simple function of a variable resistor. <br />
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A variable resistor may have either two or three circuit connections. The first picture that you see below is the example of the three terminal teminal connections variable resisitor which are commonly known as <strong>Potentiometer</strong>.<br />
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This is one typical sample for three terminal connections...<br />
<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/Sp52iMqRqzI/AAAAAAAAAK0/dSMhDKkUZLs/s1600-h/three+terminal+var.jpg"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5376865335013124914" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/Sp52iMqRqzI/AAAAAAAAAK0/dSMhDKkUZLs/s320/three+terminal+var.jpg" style="display: block; height: 284px; margin: 0px auto 10px; text-align: center; width: 320px;" /></a><strong></strong><br />
Typical potentiometer looks like this. <strong> </strong><br />
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<strong>The Potentiometer Connections</strong><br />
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/Sp5egP4mo5I/AAAAAAAAAKM/6KaGkUvejV4/s1600-h/volume-control-potentiometer-connections-schematic.JPG"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5376838913239720850" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/Sp5egP4mo5I/AAAAAAAAAKM/6KaGkUvejV4/s400/volume-control-potentiometer-connections-schematic.JPG" style="display: block; height: 221px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a> The circuit diagram of a potentiometer is really no more than that of a voltage divider chain. R1-R2 is a single resistor effectively divided by the sliding arm C, whose movement alters the relative values of R1 and R2. Please refer to circuit diagram above.<br />
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The output voltage can vary from zero (when C is lowered so that R2=0) to full circuit voltage (when C is moved up so that R1=0)<br />
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Variable resistors, like fixed resistors, can be made with resistance material of carbon or can be wired-wound, depending on the amount of current to be controlled - wire-wound for large currents and carbon for small currents.<br />
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Wire-wound variable resistors are constructed by winding resistance wire on a porcelain or bakelite circular form, with a contact arm which can be adjusted to any position on the circular form by means of a rotating shaft. A lead connected to this movable contact can then be used, with one or both of the end leads, to vary the resistance used.<br />
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For controlling small currents, carbon variable resistors are constructed by depositing a carbon compound on a fiber disk. A contact on a movable arm actsto vary resistance as the arm shaft is rotated.</div>
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On my next post, let's have some practical applications for DC Series Circuit here in <strong>Learn Electrical Engineering for Beginners.</strong></div>
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Cheers!</div>
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Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-79112098721071407222009-08-29T10:08:00.062+08:002013-03-30T18:52:08.467+08:00DC Series Circuit Part 1<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-family: Georgia, "Times New Roman", serif;">I just want clarify something before we start our new topic for today. I think you realized why I'm posting such very basic topic. This is not to insult your intelligence but the purpose of this is for you to comprehend well the basic first, because in reality some of you do not understand the basic. This is one of the common mistakes of the <span style="font-weight: bold;">beginners </span>especially taking up the course like<span style="font-weight: bold;"> Electrical Engineering</span>. If you would ask me if I have a plan of posting an advanced concept. Simply the answer is YES , PERIOD. But this is after posting up all the basic concepts.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Please bear with me for a sort of introduction. Just want to clarify something.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Moving on...</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">Resistance in Series Circuit</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Have you already read the definition of a series circuit in your Physics? Well, if you forgot it already I will define it for you again to refresh your mind. If you still remember our example of a light bulb connected across the battery source, that's already a typical example of a series circuit. But in this case, since we are dealing of resistive circuits, we will define it in this way.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">A<span style="font-weight: bold;"> series circuit </span>is formed when two or more resistors are connected from terminal to terminal or simply end-to-end in a circuit in such a way that there is only <span style="font-weight: bold;">one path</span> for current to flow.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Connecting the resistor in series so as to form series connection is much easier. You don't have to worry about connecting positive and negative terminal. Resistors (unlike cells) have no polarity.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">I just want to clear this things up for you. If you have for example a two lamps connected together from one terminal to another leaving each other terminals of each lamps unconnected, this is a<span style="font-weight: bold;"> series connected</span> but you would not have a so called<span style="font-weight: bold;"> series circuit. </span>In order to have a series circuit, you would have to connect the lamps across the voltage terminals such as a battery for example using the unconnected terminals to complete a </span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">series circuit.<br />
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<span style="font-weight: bold;"></span></span>Remember, any number of resistance, lamps or any devices connected together in a series would only be a series circuit provided their end to end terminals are connected across to a voltage source and would offer one path of current flow. Ohhh... I just can't proceed with the discussion without showing you the typical example of a series circuit. A very good representation of a series circuit was given below.</span></div>
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpicnkR2VXI/AAAAAAAAAIY/-KbNRM347vY/s1600-h/DC_CIRCUIT_%28SERIES%29.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><span style="font-family: Georgia, "Times New Roman", serif;"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5375218358834058610" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpicnkR2VXI/AAAAAAAAAIY/-KbNRM347vY/s320/DC_CIRCUIT_%28SERIES%29.jpg" style="cursor: pointer; display: block; height: 219px; margin: 0px auto 10px; text-align: center; width: 320px;" /> </span></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">One important thing that we need to remember here in series circuit is that their values are just added. The regular reader of this blog still remember the discussion about </span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html"><span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">factors upon which the resistance of conductor depends</span></a><span style="font-family: Georgia, "Times New Roman", serif;">. It was mentioned there that the resistance of conductor increases as the length of the conductor increases. In order for you to get my point here, imagine you have a 3 different length of wires in your hand with different resistances. Then, lets try to connect it together. After connecting them together, the resistance of the wire in full length would be equal to sum of resistances of wire 1 + wire 2 + wire 3.<br />
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Let's take it again into more detail... another example if I have a three lengths of wires. The first one having a resistance of 3 ohms, the second is 4 ohms and the third one is 5 ohms. If to try to connect them together, the total resistance of the end-to-end terminal would be 12 ohms. <span style="font-weight: bold;">The conclusion here is that any types of resistances connected in series, their total resistance would be equal to the sum of their individual resistances</span>.<br />
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One more thing that we need to consider when dealing with resistance connected in series is the proper identification. So, what is this all about? When dealing with a circuit, you would surely encounter different equivalent value of resistances just like what we have in our examples above. In order for you to distinguish same device with different resistances to another we have a so called <span style="font-style: italic;">subscript </span>for them to identify.</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/Spip80HcE5I/AAAAAAAAAIg/7YPxNpigsoc/s1600-h/series+circuit+resistor.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><span style="font-family: Georgia, "Times New Roman", serif;"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5375233017513776018" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/Spip80HcE5I/AAAAAAAAAIg/7YPxNpigsoc/s400/series+circuit+resistor.jpg" style="cursor: pointer; display: block; height: 288px; margin: 0px auto 10px; text-align: center; width: 400px;" /> </span></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">This subscript was now written in a different manner. Today it was written like R1 ( not offset) which is similar to R1 ( which is offset) like the above diagram. R1, R2, R2 identifies same resistor with different or same resistances. The same manner was also applied to voltage E1, E2, E3 or sometimes they use V1, V2, V3 just like given in the above example diagram. For uniformity, lets use Rt with the small subscript t for the total resistance. For example on the above diagram our total resistance is Rt = R1 + R2 + R3.<br />
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Remember to use the correct subscript, because when we deal with more complicated circuits, you would not be confused. In any aspect aside from this, it is very important to distinguished one equivalent value to another.</span><br />
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<strong><span style="font-family: Georgia;">The Flow of Current in Series Circuit</span></strong></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><br />
<span style="font-family: Georgia, "Times New Roman", serif;">Since we already know that the current will flow in one path in a series circuit. This only means that the current will flow in every component of the circuit. Since I'm a visual person, I would give you a practical example of this. Supposed you have a circuit just like what we have on the above example, let's put an ammeter across a resistor one at a time to get the current reading. <span style="font-weight: bold;">It would show that there are identical amount of current is flowing in every component in the circuit.</span><br />
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Take note that the current must be capable of passing in every component in the circuit without being damaged. What does this mean? Let's take gain another example. If you have a light bulb to be connected in circuit, it must be rated. If it is rated too low, the light of the bulb would be very bright and the tendency of this light bulb would be burn out because of excessive flow of current. The same thing would happen if in practical application, you mistakenly use different ratings would result sometimes to a serious trouble. The circuit would probably stop functioning or will not function properly. This is very very important.</span><br />
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">Voltages in Series Circuit - The Kirchhoff's Second Law</span><br />
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</span><br />
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/Spi7wslILrI/AAAAAAAAAIo/n75-g4cmIP4/s1600-h/kirchhoffs+second.png" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><span style="font-family: Georgia, "Times New Roman", serif;"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5375252600541687474" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/Spi7wslILrI/AAAAAAAAAIo/n75-g4cmIP4/s400/kirchhoffs+second.png" style="cursor: pointer; display: block; height: 400px; margin: 0px auto 10px; text-align: center; width: 330px;" /> </span></a></div>
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<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">We all know that the current are just identical in every component for a series circuit. Now, let's consider the voltage across each component in a series circuit. This is somewhat we called it a<span style="font-weight: bold;"> potential drop </span>or the <span style="font-weight: bold;">voltage drop</span>. Since they have same current, the energy expended in pushing this equal amount of current through the individual component must be also the same.<br />
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Let's consider the diagram given above. Supposed you have three resistors connected in series. The diagram shows that you have 45 V connected across the circuit. If we measure the voltage across R1, the reading would be 10 V, getting the voltage across R1 and R2 would be 30 V and lastly getting the voltage across R1, R2 and R3 would be obviously 45V. This is similar when we get the voltage across each resistor and sum it up would be also equivalent to 45V. See the diagram how it is clearly illustrated.<br />
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The fact above was expressed by the German Physicist Kirchhoff (1824- 1887) which is known as Kirchhoff s Second Law. The first law would be discuss on my next succeeding post. Please take note of what this law being state and it is very important too.</span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">Kirchhoff s Second Law states: The sum of the voltage drop across the resistances of a closed circuit equals the total voltage applied to the circuit.<br />
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Applying Ohm's Law in Series Circuit<br />
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<span style="font-weight: bold;"></span></span>Simplifying the facts that we have above about series circuit. We now know that:<br />
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1. The current in the series circuit is the same everywhere. This can be mathematically expressed as It = I1= I2= I3 and so on.<br />
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2. That the total resistance in a series circuit is equivalent to the sum of the individual resistances in the circuit. This can be expressed as Rt= R1 + R2 + R3 and so on.<br />
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3. That the voltage drop across each resistances when added is equal to the voltage source connected across the circuit in series. This can be expressed as, Et = E1 + E2 + E3, and so on.<br />
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Applying ohm's law in a series circuit is very helpful especially in terms of application. For instance, you do not know the value of the resistance connected across the circuit but you have some data to resolve that problem. Ohm's Law would be a great help for this.<br />
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Some useful application of Ohm's Law in the series circuit is the simplication process. Let's take an example diagram below.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span><a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpjKlDsjCtI/AAAAAAAAAIw/9wR5kWRrDIQ/s1600-h/Series+Circuit+simplified.JPG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><span style="font-family: Georgia, "Times New Roman", serif;"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5375268893262809810" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpjKlDsjCtI/AAAAAAAAAIw/9wR5kWRrDIQ/s400/Series+Circuit+simplified.JPG" style="cursor: pointer; display: block; height: 201px; margin: 0px auto 10px; text-align: center; width: 400px;" /> </span></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">You will in the example above that the right hand diagram is the simplified version of the left hand diagram. With this theory, you can find some missing factor that you want to know in your circuit. You should always try to simplify the series of resistances into single resistance equivalent circuit just like what had shown above. Let's simulate a good example of this.<br />
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A sample problem...<br />
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A circuit contains two resistors connected in series across 100 Volts. The circuit current flow is 2 Amperes. One of the resistor R1 have known value of 10 ohms. You wish to know the resistance of the entire circuit, the value of the second resistor R2 and the voltage drop across each of the two resistors.<br />
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First, in Ohm's Law Rule, lets draw the diagram just what had shown above. Since we already visualize on how we are going to simplify it, let's make a list of the given and unknown values.<br />
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<span style="font-weight: bold;">Data:</span><br />
Et = 100 Volts<br />
It = 2 Amperes<br />
Rt= (unknown)<br />
<br />
E1= (unknown)<br />
I1 = 2 Amperes since It = I1 = I2 = 2 Amperes<br />
R1= 10 ohms<br />
<br />
E2= (unknown)<br />
I2 = 2 Amperes<br />
R2 = (unknown)<br />
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Now let's solve it!<br />
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Since we are looking for the resistance of the entire circuit, from Ohm's Law we use the magic triangle. Put your thumb on R. Now we obtained that </span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">Rt = Et/ It = 100 / 2 = 50 ohms - answer</span>If you will see, you cannot obtain the value of the second resistance R2 without knowing first the voltage drop across R1 which is E1. After knowing the value of E1, we can now proceed in solving the voltage drop across R2 which is E2 therefore we can solve for the unknown R2. Let's do it...<br />
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From the magic triangle, put your thumb on E, therefore voltage drop across R1 is <span style="font-weight: bold;"><span style="font-weight: bold;"></span>E1 = I1 X R1 = 2 x 10 = 20 Volts -answer</span></span><span style="font-weight: bold;"><span style="font-weight: bold;"></span><br />
<span style="font-weight: bold;"><br />
</span></span><span style="font-family: Georgia, "Times New Roman", serif;">From our concept above, the sum of the voltage drop across each resistances is equivalent to a voltage source connected across them. - <span style="font-weight: bold;">Kirchhoff s Second Law</span><br />
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Et = E1 + E2 ; 100 Volts = 20 Volts + E2, solving for <span style="font-weight: bold;">E2 = 100 - 20 = 80 Volts - answer </span>(simple algebra dear)<br />
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Since we already know the value of voltage drop across the R2, we may now solve the value of R2. Putting your thumb on R of the magic triangle, we obtain that </span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">R2 = E2 / I2 = 80 / 2 = 40 ohms - answer</span>Try to practice more solving a series circuit using ohm's law. It is important that you should grasp the use of Ohm's Law in solving series circuit.<br />
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I will continue the discussion of DC Series Circuit here in <span style="font-weight: bold;">Electrical Engineering</span> course.<br />
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See you again.<br />
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<span style="font-weight: bold;">Note : I'll be back on September 3, 2009 evening (Philippines Time) for continuation of DC Series Circuit Part 2. </span><br />
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Cheers!</span></div>
</div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com1tag:blogger.com,1999:blog-1420533950904606452.post-15427136332050024112009-08-28T10:23:00.045+08:002013-03-30T18:58:50.740+08:00Resistors Part 2- Color Code and How Resistance is Measured<div dir="ltr" style="text-align: left;" trbidi="on"><div style="text-align: justify;"><div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Let's continue of what we have discussed yesterday before we go on the full discussion of a Series Circuit which will be our next topic that we will going to study here in <span style="font-weight: bold;">Learn Electrical Engineering for Beginners</span>.</span><br />
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</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Today, we will be dealing on how the resistor color coding is being used and how do we going to interpret it to obtain the reading. Then, afterward s we will touch a little bit on how the resistance is being measured. That's all will be discussed within this new post.</span></div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Let's begin now... timer start now!</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">The Resistor Color Code</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">We all know that we can find the resistance value of any resistor by using an ohmmeter. But what if we don't have an ohmmeter to use? Most of the case we can find the resistance value easier by interpreting its marking. Some resistors like wire-wound resistor have its printed value in ohms in their body. If they don't have the mark, you would require to use an ohmmeter. An example of a resistor which usually have all of the data printed directly on the resistor body with the information such as tolerance, temperature characteristics, and exact resistance value is the precision wire wound resistor. Other resistor like the carbon resistors usually do not have the data of characteristics directly marked on them, instead they have a so called <span style="font-weight: bold;">color code</span> by which they can be identified. You will wonder why it is being done this way for carbon resistors. The reason of using a color code for a carbon resistor is that they are small which is difficult to read the printed values especially when they are mounted.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Before we forgot something, there are two types of carbon resistors. The <span style="font-weight: bold;">radial </span>and an <span style="font-weight: bold;">axial</span>. They are only differ in the the way the leads are connected to the body of the resistor. Both employ the same color code but they are printed in the different manner. Radial lead resistors are not found in modern equipment. They are widely used in the past. I can't see any example of this now. Below is an example of an axial resistor.</span></div><br />
<div style="text-align: left;"><a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpdX75DFu5I/AAAAAAAAAHI/50bJtQ7Y9_U/s1600-h/axial+carbon+resistor.JPG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374861366727654290" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpdX75DFu5I/AAAAAAAAAHI/50bJtQ7Y9_U/s320/axial+carbon+resistor.JPG" style="cursor: pointer; display: block; height: 250px; margin: 0px auto 10px; text-align: center; width: 250px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">In the picture above this axial lead resistors have its leads molded into the ends of the carbon rod of the resistor body. If you will see, the leads extends straight out in line from the body of the resistor. The carbon rod is coated with a good insulator.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Moving on...</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Color coding system for resistors consists of three colors to indicate the resistance value in ohms of a certain resistor, sometimes the fourth color indicate the tolerance value of the resistor. By reading the color coded in correct order and substituting the correct value of each corresponding color coded as shown in the table below, you can immediately tell all you need to know about the resistor. The only thing that you will practice on how to use it and familiar yourselves for those value so that you can easily determine the value of the resistor color coded at a glance.</span></div><br />
<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpdtCu1bbEI/AAAAAAAAAHo/Zrku7R9NTbo/s1600-h/color+coded+resistor.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374884573989268546" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpdtCu1bbEI/AAAAAAAAAHo/Zrku7R9NTbo/s400/color+coded+resistor.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
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<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">This is how you will do it.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">The color of the first color band indicates the first digit of the resistance value or the first significant digit. Let's have an example below. Supposed that you have a given resistor below, the first color is yellow. If you would look at the table above it is equivalent to 4.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">The second color coded of the resistor given below is violet, so this is now your second digit which is equivalent to 7 as shown in the table above.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">The third color would served as your multiplier. In the case below since it is color red which is equivalent to 100 multiplier, or just simply add 2 zeros so this would look like this now:</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">47 ohm x 100 =<span style="font-weight: bold;"> 4, 700 ohms</span> or <span style="font-weight: bold;">4.7 kilohms</span></span></div><div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;"><br />
</span></div><div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Ooooppps! it seems that we are not done yet. The last color band or the fourth color band is gold which have 5% tolerance according to our table above. Therefore our final answer would be:</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">4.7 kilohms +/- 5%</span> - <span style="font-weight: bold;">answer</span></span></div><br />
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpdsKzlmHHI/AAAAAAAAAHY/w3rjl6tgq58/s1600-h/sample+color+coded+resistor.gif" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374883613192363122" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpdsKzlmHHI/AAAAAAAAAHY/w3rjl6tgq58/s400/sample+color+coded+resistor.gif" style="cursor: pointer; display: block; height: 234px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
<div style="text-align: left;"><br />
</div><div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">How to Measure the Resistance</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">We all know that voltmeter and ammeter are used for measuring the voltage and the current respectively. For the resistance, the meters that use to measure it is the <span style="font-weight: bold;">ohmmeter</span>. When using an ohmmeter, there should be no voltage present across the resistors except for the ohmmeter battery, otherwise your ohmmeter would be damaged. I can see two types of ohmmeter nowadays, the analog and the digital. Among the two ohmmeters, digital is widely used nowadays.</span><br />
</div><br />
<div style="text-align: left;"><a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/Spd55OVAc5I/AAAAAAAAAHw/6Z2D4Q7Dnjs/s1600-h/Ohmmeter-+digital.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374898704295687058" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/Spd55OVAc5I/AAAAAAAAAHw/6Z2D4Q7Dnjs/s320/Ohmmeter-+digital.jpg" style="cursor: pointer; display: block; height: 320px; margin: 0px auto 10px; text-align: center; width: 240px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">The above ohmmeter usually used to measure the resistance of the resistors. Ohmmeter ranges usually vary from 0-1,000 ohms to 0 -10 megohms. There are some special ohmmeters called the MEGGERS. This ohmmeter was used to measure high resistance values which are over 10 megohms. Some meggers use high voltage batteries and other use special type of hand generator to obtain the necessary voltage. These megohmeters is used to measure and test the resistance of insulation. Picture below is the example of a megger.</span></div><br />
<div style="text-align: left;"><a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/Spd-wuXwV6I/AAAAAAAAAII/8gj919cA6jA/s1600-h/MEGGER_247001.JPG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374904055836465058" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/Spd-wuXwV6I/AAAAAAAAAII/8gj919cA6jA/s320/MEGGER_247001.JPG" style="cursor: pointer; display: block; height: 240px; margin: 0px auto 10px; text-align: center; width: 320px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Ohmmeter is very easy to use by following two steps. First, the voltage must be set to the proper value. This is done with the zero adjustment by shorting outor by connecting together the two leads from the ohmmeter and setting to zero ohms on the meter with the zero adjustment control. This should be always done whenever you changed the meter range selector switch to a different scale. Now, the meter is now calibrated for the given range, you will notice that when the leads shorted out, the meter reads zero ohm, but when it opens, the meter reads infinity which indicates an open circuit. Therefore, when these leads touches the resistors subject for measurement, it will directly read the resistance in the meter multiplying it with the range selector switch. The range selector switch is serves as the multiplier or the multiplying factor whenever you are measuring the resistance using ohmmeter. The range selector switch usually marked as R, RX 10, RX 100, RX1,000, etc... </span><br />
<span style="font-family: Georgia, "Times New Roman", serif;">For example if the ohmmeter is switch on to R X 1,000 meaning the value of the meter will be multiplied to 1,000 to get the actual value of the resistance being measured.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">That's it for today.</span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Tomorrow we'll continue dealing with circuits here in <span style="font-weight: bold;">Electrical Engineering.</span></span><br />
</div><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;"><span style="font-family: Georgia, "Times New Roman", serif;">Cheers!</span></div></div></div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-80062182761781051472009-08-27T10:41:00.039+08:002013-03-30T19:07:08.037+08:00Resistors Part 1 - Use and Properties<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Now that you have learned the basic concept of Ohm's Law, we can now proceed in discussing the use, properties, and construction of resistors. All you can learn it here in <span style="font-weight: bold;">Electrical Engineering</span>.</span></div>
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<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Before we continue our study of circuits, we need to know more about resistance and resistors though we have touched it a little on my previous post. But its just a review. Today, it would be a bit deeper.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, 'Times New Roman', serif;">We know that there is a certain amount of resistance in all electrical equipment which we use. Sometimes, this resistance is not enough to control the flow of current to the extent desired. If you did not get my point here, let's have a few example of this. I will going to give an example by illustration as what had shown below. The circuit shown below, a switch and a current limiting resistor are used to control the flow of current through the motor. When starting a motor, the switch is kept open and the resistance is thereby added into the circuit to control the flow of current. After the motor has started, the switch is then closed in order to bypass the current limiting resistor.</span></div>
<div style="text-align: justify;">
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<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpYAoqYZvqI/AAAAAAAAAFw/quVssLu3u0o/s1600-h/resistor.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374483903884672674" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpYAoqYZvqI/AAAAAAAAAFw/quVssLu3u0o/s400/resistor.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">There are wide variety of resistors, some of them have fixed values and some are variables.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">What resistor is made up of?</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Resistors are made up of special resistance wire, graphite (carbon) composition, or of metal film. Wire wound resistors are usually used to control large currents, while carbon resistors controls current which are relatively small.</span></div>
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpYJ7t1s3DI/AAAAAAAAAF4/jt_f6WBZTzw/s1600-h/Vitreous+Enamel+Res+DVR.JPG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374494126835031090" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpYJ7t1s3DI/AAAAAAAAAF4/jt_f6WBZTzw/s400/Vitreous+Enamel+Res+DVR.JPG" style="cursor: pointer; float: left; height: 178px; margin: 0pt 10px 10px 0pt; width: 387px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Vitreous enameled wire-would resistors are constructed by winding resistance wire on a porcelain base, attaching the wire ends to metal terminals, and coating the wire and base with powdered glass and bake enamel to protect the wire and conduct heat away from it.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Fixed wire wound resistors with a coating other than vitreous enamel are also used. The example below is the example of this one.</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpYL4aRIOsI/AAAAAAAAAGI/4i6vuuPfHW4/s1600-h/Fixed+wire+wound.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374496269065009858" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpYL4aRIOsI/AAAAAAAAAGI/4i6vuuPfHW4/s400/Fixed+wire+wound.jpg" style="cursor: pointer; display: block; height: 200px; margin: 0px auto 10px; text-align: center; width: 200px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Wire-wound resistors may have fixed taps, which can be also used to change the resistance value in steps, or sliders, which can be adjusted to change the resistance of any fraction of the total resistance. The picture below is the example of this one.</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpYNyYsY5rI/AAAAAAAAAGQ/SSvzgQbEWgI/s1600-h/wire+wound+fixed.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374498364586518194" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpYNyYsY5rI/AAAAAAAAAGQ/SSvzgQbEWgI/s400/wire+wound+fixed.jpg" style="cursor: pointer; display: block; height: 134px; margin: 0px auto 10px; text-align: center; width: 245px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Precision wound resistors of manganin wire (a special wire that does not change resistance very much with high temperature) are used where the resistance value must be very accurate, such as in test instruments. The picture below is the example of this one.</span></div>
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<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpYTP8MQuZI/AAAAAAAAAGg/rVut6iwgI_s/s1600-h/precision+wire+wound.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374504369889786258" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpYTP8MQuZI/AAAAAAAAAGg/rVut6iwgI_s/s400/precision+wire+wound.jpg" style="cursor: pointer; display: block; height: 182px; margin: 0px auto 10px; text-align: center; width: 217px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Carbon resistors are used for low current applications. They are made from a rod of compensated graphite (carbon) that is mixed with clay and binders. By varying the amount of each component, it is possible to vary the resistance values obtained over a very wide range. Two lead wires are called <span style="font-style: italic;">pigtails</span> are attached to the end of the resistance rod, and the rod is embedded in a ceramic or plastic covering, leaving the pigtails protruding from the ends. Take a look for a sample below.</span></div>
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpYVbEemEqI/AAAAAAAAAGo/gery-90LtJ0/s1600-h/carbon+resistor.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374506760115983010" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpYVbEemEqI/AAAAAAAAAGo/gery-90LtJ0/s400/carbon+resistor.jpg" style="cursor: pointer; display: block; height: 200px; margin: 0px auto 10px; text-align: center; width: 200px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">You will find other type of resistor called a deposited film resistor used for special applications. These resistors are made by depositing a thin film of resistance metal or carbon on a ceramic core and then coating the resistor with either a ceramic or enamel protective coating. In many cases you will find that these resistors have radial leads meaning the leads come off at right angles to the body of the resistor. In some cases the deposited film is laid down on the core as spiral, similar to winding a wire around the tube, in order to increase the length of the resistance element without making the resistor too long. The example of this one was shown below.</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpYa6sATv8I/AAAAAAAAAGw/rcdeCPxSOoA/s1600-h/film+resistor.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374512800860454850" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpYa6sATv8I/AAAAAAAAAGw/rcdeCPxSOoA/s400/film+resistor.jpg" style="cursor: pointer; display: block; height: 135px; margin: 0px auto 10px; text-align: center; width: 300px;" /></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">Resistor Tolerance and Values</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">Let's consider this topic before we go on the color code for resistors. You need to find out the something about resistor tolerances and something about the preferred values of resistance that you will find in the circuits. Special resistors may have tolerances of as little as 1% , 0.1% or even 0.01% but most resistors that you will see have much greater tolerances. Large wire-wound resistors usually have tolerances of 10% or 5%. Carbon resistors are available in 20%, 10% and 5% tolerances.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">So what are those tolerance mentioned above means? Let's take an example...</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">If you had a 10 kilohm resistor with a tolerance of 20%, the actual value of the resistor could anywhere from 10 kilohm - 10 kilohm (0.20) = </span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">8 kilohm </span><span style="font-family: Georgia, 'Times New Roman', serif;">to 10 kilohm + 10(0.20) = </span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">12 kilohm</span><span style="font-family: Georgia, 'Times New Roman', serif;">. That is how you will going to use the tolerance given for a specific resistor.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">You will wonder how many different resistance values you can get for a resistor. It depends on the</span><span style="font-family: Georgia, 'Times New Roman', serif; font-style: italic;"> tolerance</span><span style="font-family: Georgia, 'Times New Roman', serif;">. Considerations such as this have led to the establishment of a set of preferred values of resistance in each tolerance where the highest tolerance of one value is about equal to the lowest tolerance of the next highest value. The table of preferred resistance was shown below. Later, you will find that resistors are available in different power ratings as well.</span></div>
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<a href="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpYsDI-GifI/AAAAAAAAAG4/pQ-AGBHNkTM/s1600-h/tolerance.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374531637772454386" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SpYsDI-GifI/AAAAAAAAAG4/pQ-AGBHNkTM/s400/tolerance.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">The numbers on the chart above show only the first two digits. Thus, it means for example, 33 means that 3.3, 330, 3.3 kilohm, 330 kilohm and 3.3 megohm resistors are available.</span></div>
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<span style="font-family: Georgia, 'Times New Roman', serif;">On my next post I will show you how the resistor color code is being used for carbon resistors and how the resistance is being measured.</span><span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, 'Times New Roman', serif;">I will tee off shortly. Stay more here in </span><span style="font-family: Georgia, 'Times New Roman', serif; font-weight: bold;">Learn Electrical Engineering for Beginners</span><span style="font-family: Georgia, 'Times New Roman', serif;">.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Cheers!</span></div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0tag:blogger.com,1999:blog-1420533950904606452.post-1984635116530168732009-08-26T16:30:00.047+08:002013-03-30T19:08:36.189+08:00The Basic Concept of Ohm's Law<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-family: Georgia, "Times New Roman", serif;">I just can't wait for another morning without my new post here in<span style="font-weight: bold;"> Learn </span><span style="font-weight: bold;">Electrical Engineering for Beginners.</span> It's going interesting here because I've noticed that there are new another batch of subscriber who want to learn about <span style="font-weight: bold;">Electrical Engineering</span>. Since you've found this site, you're halfway to learning the Electrical Engineering. For those who are new in this site, you can still catch up with our previous 2 last topics here.</span></div>
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<a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/what-is-electric-circuit.html"><span style="font-family: Georgia, "Times New Roman", serif;">What is Electric Circuit - Part 1</span></a></div>
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<a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/what-is-electric-circuit-part-2.html"><span style="font-family: Georgia, "Times New Roman", serif;">What is Electric Circuit - Part 2</span></a></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Ok, let's begin with our new basic topic.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">The Relationship of Voltage, Current, and Resistance</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">This next topic, you will learn about Ohm's Law, this is one of the most basic and important that you will use throughout your career here in <span style="font-weight: bold;">Electrical Engineering</span>.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">The concept is just simple: given a <span style="font-style: italic;">constant resistance </span>in a circuit, the current flow increases as the voltage applied to the circuit increases. Given a <span style="font-style: italic;">constant voltage</span> (emf) applied to the circuit, current flow decreases as the resistance of the circuit increases. You can finalize these ideas or concepts as follows: <span style="font-weight: bold;">Current flow in the circuit increases as the voltage is increased, and decreases as the resistance is increased</span>.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">The relationship of voltage, current, and resistance was studied by German Physicist, George Simon Ohm. His statement of this relationship, called <span style="font-style: italic;">Ohm's Law</span>. Obviously, the title of the law was derived itself on his name. This is one of the fundamental laws of Physics.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">What does this vital laws state? This is simple as:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">With Constant Resistance </span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">lower</span> voltage gives<span style="font-weight: bold;"> small </span>current.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">higher</span> voltage gives <span style="font-weight: bold;">large</span> current.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">With Constant EMF</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">lower</span> resistance passes <span style="font-weight: bold;">large</span> current.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">higher</span> resistance passes <span style="font-weight: bold;">small</span> current.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">In a more technical expression, you can state it as:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Ohm's Law states that the current flowing in a circuit is <span style="font-weight: bold;">directly proportional </span>to the voltage applied (emf) and inversely proportional to the resistance.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">It is also possible to express Ohm's Law as a mathematical equation (relationship) as further indicated below:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">CURRENT = EMF / RESISTANCE</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">In electrical terms (notation), current is always represented by the letter "I", resistance by the letter "R" and the voltage by the letter "E" or you can used letter "V". You can therefore rewrite the mathematical given above in another way like what as shown below:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">I = E / R</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Using our simple algebra, you can also derived it in these ways:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-weight: bold;">E = I x R</span> or as <span style="font-weight: bold;">R = E / I</span></span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Which of the three ways ( formulas) of expressing Ohm's Law you might choose to employ depends on two things: 1. what facts you know and 2. what facts you need to know about the circuit you are considering.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Ohhh... I think you've got a headache on how to remember those formulas above. Let's have a little trick on how you can remember it better. Let's draw a triangle with a horizontal line across it half-way up from its base. Write letter<span style="font-weight: bold;"> E</span> in the small triangle, which has been formed above the line, and the letters <span style="font-weight: bold;">I</span> and <span style="font-weight: bold;">R </span>below the line, it will look like as what as shown below. This is what you called the <span style="font-weight: bold;">magic triangle</span>!</span></div>
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<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpUIcFgIckI/AAAAAAAAAFA/y7O2mXQDwPo/s1600-h/Ohms+Law.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374211008942862914" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpUIcFgIckI/AAAAAAAAAFA/y7O2mXQDwPo/s400/Ohms+Law.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">Ok! now consider a circuit in which you know the values of any two of the three factors - voltage, current, and resistance- and want to find out the third. The rule for working the magic triangle to give the correct formula is as follows:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif; font-style: italic;">Put your thumb over the letter in the triangle whose value you want to know- and the formula for calculating that value is given by the two remaining letters.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Here is how to do it:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">1. If you know the values of current and resistance in the circuit. But the value of voltage is unknown and you cannot measure it because of some reason ( no available voltmeter to use). Draw the magic triangle, put your thumb you want to calculate, which in this case the E- and you are left with the formula you need. ---- I X R.</span></div>
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/SpUQAQ_ghKI/AAAAAAAAAFI/3M7QsiSqp00/s1600-h/thumb+1.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374219327083938978" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/SpUQAQ_ghKI/AAAAAAAAAFI/3M7QsiSqp00/s400/thumb+1.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">2. You know the values of current and voltage, but in this case you have no ohmmeter to measure the resistance. Put your thumb over the letter R and you are left with the formula E / I. Substitute the known values for E and I, and your answer is R.</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpURjkAwjUI/AAAAAAAAAFQ/zuqOU0EhHZI/s1600-h/thumb+2.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374221032996506946" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SpURjkAwjUI/AAAAAAAAAFQ/zuqOU0EhHZI/s400/thumb+2.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">3. The voltage and the resistance of a circuit are known to you; but in this case the ammeter you need to measure the current is lost or broken. Put your thumb over the symbol I. and read off the formula you need: E/R.</span></div>
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/SpUVSaYcwYI/AAAAAAAAAFY/O2PHkE5zgt4/s1600-h/thumb+3.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374225136400253314" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/SpUVSaYcwYI/AAAAAAAAAFY/O2PHkE5zgt4/s400/thumb+3.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Ohm's law cannot work properly if you will not know on how to expressed its values in the correct units of measurement. The next topic will show you the Ohm's Law Rules.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">Ohm's Law Rules</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Ohm's Law will work for you and give you the correct answer in any problem situations which you may try to solve. In the Ohm's Law equation,<span style="font-weight: bold;"> the first rule</span> is that:</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">CURRENT is ALWAYS expressed in AMPERES.</span></div>
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">VOLTAGE is ALWAYS expressed in VOLTS.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">RESISTANCE is ALWAYS expressed in OHMS.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Let's have an example of the above rule.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Take you have a circuit in which you have to measure the resistance to be 100 ohms, and the current to be 300 milliamperes (mA). Obviously, if you use the Ohm's Law without knowing the correct rule mentioned above, you will arrived like this: E = I X R = 100 X 300 = 30,000 your answer will be definitely <span style="font-weight: bold;">wrong</span> by a factor of 1,000.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">You have to use the conversion tables as what you have learned in Physics, and you must rewrite all factors in the simple expression above in amperes, volts and ohms. When you did this, you will obtain:</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">E = I X R = 100 X 0.3 = </span><span style="font-family: Georgia, "Times New Roman", serif;">30 volts - <span style="font-weight: bold;">answer</span></span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">Thus, giving you the correct answer.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">There is a <span style="font-weight: bold;">second rule</span> in which you must apply whenever you are attempting to solve an Ohm's Law problem involving quantities and values in an electric circuit. The rule is: <span style="font-weight: bold;">Always sketch a rough diagram of the circuit you are considering, before you start making calculations based on the values in the circuit which are already known to you.</span> This rule is very useful especially when we already dealing with the more complex circuits.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Let's have an example of the above second rule:</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Supposed you have an unknown resistor connected across the battery, and you find by measurement that the voltage across it is 24 volts. You measure the current flowing as 6 amperes. You want to know the resistance of the resistor; but you have no ammeter.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">This is how you will solve it.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">First, draw the circuit diagram, and fill it in with the information you already have. Sketch out the magic triangle ( illustration given below) . The magic triangle tells you that R = E/ I. </span><br />
<br />
<span style="font-family: Georgia, "Times New Roman", serif;">Into this equation, you may substitute the known values and get R = 24 / 6 = 4 ohms - <span style="font-weight: bold;">answer</span></span></div>
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<div style="text-align: left;">
<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpUjcvz2EpI/AAAAAAAAAFg/YAj8e3gkz7o/s1600-h/second+rule.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5374240707113783954" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SpUjcvz2EpI/AAAAAAAAAFg/YAj8e3gkz7o/s400/second+rule.jpg" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><span style="font-family: Georgia, "Times New Roman", serif;">Since you know now already the basic concept of Ohm's Law, you are now ready to face the challenges here in<span style="font-weight: bold;"> Electrical Engineering</span>. On my next post let's discuss some applications about it.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">Hope you learn something new basic technique here in Learn Electrical Engineering for Beginners.</span></div>
<span style="font-family: Georgia, "Times New Roman", serif;"></span><br />
<div style="text-align: left;">
<span style="font-family: Georgia, "Times New Roman", serif;">Cheers to all and Good night!</span></div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com2tag:blogger.com,1999:blog-1420533950904606452.post-60934391538004376302009-08-20T10:31:00.019+08:002013-03-30T19:09:47.981+08:00What is Electric Circuit? - Part 2<div dir="ltr" style="text-align: left;" trbidi="on">
Before I proceed with our new topic for today, I just want to give thanks for those who are currently subscribed here in <a href="http://electricalengineeringforbeginners.blogspot.com/"><span style="font-weight: bold;">Learn Electrical Engineering for Beginners</span></a>. I hope this will be a helpful site for you which I will always give a full and detail information for every subject matter.<br />
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<br />
Well, it's enough for my short introduction because I'm already running out of time for the updates. Yesterday I had mentioned on my previous post about our topic on full definition of <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/what-is-electric-circuit.html"><span style="font-weight: bold;">Electric Circuits</span></a> that I will going to differentiate between DC and AC Circuits. Please keep in mind that after discussing the difference between the two, I will going to discuss first to you all DC Circuits related topics so that we would not be confused. The principles that we will be discussing here in DC Circuits will also be used again when we touch AC Circuits. Please keep that in mind and this is very important. I just want to keep my ideas and discussions organize here in <span style="font-weight: bold;">Electrical Engineering</span> site. I was also trying to catch up the attention of those <span style="font-weight: bold;">Electrical Engineering</span> who study <span style="font-weight: bold;">online</span> as well as those <span style="font-weight: bold;">non- electrical engineers</span>.<br />
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Moving on...<br />
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Still remember on our review on Physics on my previous post in <a href="http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html"><span style="font-weight: bold;">Voltage, Current, Power and Energy</span></a>. I already give the definition of DC and AC Current. But for those recent readers, here is the short definition and which is almost the same when dealing with circuits.<br />
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In electricity, we deal both on direct current (DC) and alternating current (AC). In <span style="font-weight: bold;">DC Circuits</span>, the current always flows in the <span style="font-style: italic;">same </span>or one direction. In AC Circuits, the direction of current flow <span style="font-style: italic;">reverses periodically</span>- this means at one instant the current flow in one direction and in the next instant, in the opposite direction. Remember in our review? this flow reversal in AC current is usually done regularly. What does it mean? If we talk about 60Hz AC power, we mean that the direction of flow reverses 60 times ( or cycles) per second. Graphically, here are the difference for you to comprehend it well. ( You may click the image to enlarge)<br />
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/SozDap60sjI/AAAAAAAAAEo/W6DPRBm3qdE/s1600-h/ac+and+dc+circuit.png" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5371883318242751026" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/SozDap60sjI/AAAAAAAAAEo/W6DPRBm3qdE/s400/ac+and+dc+circuit.png" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
Ohhh... before I forgot there are other types of current such as Exponential Current and Dumped Sinusoidal Current. But these are somewhat on the deeper concepts that we might be able to touch on our future discussion. But to give you an idea here are the graphical difference between the two. ( You may click the image to enlarge)<br />
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<a href="http://1.bp.blogspot.com/_cQJDlnQI2zY/SozINIZGfTI/AAAAAAAAAEw/QmxQoyIiNr4/s1600-h/dumped+and+exponential.png" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5371888583462780210" src="http://1.bp.blogspot.com/_cQJDlnQI2zY/SozINIZGfTI/AAAAAAAAAEw/QmxQoyIiNr4/s400/dumped+and+exponential.png" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
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As what I had mentioned earlier, on the first part of the circuitry discussion we will deal with the function of direct current in circuits containing resistance (resistive circuits) and we will use Ohms Law and Kirchhoff's Laws as the tools for analysis and understanding the relationship between current, voltage and resistance. This will be your foundation for your future understanding of ac circuitry. Therefore, it is very important that you completely understand every concept I presented here in dc circuitry.<br />
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Continuing our understanding of Electric Circuit.<br />
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It may help you to grasp the concept of an electric current flowing through a closed circuit. Imagine that the electrons which make up the current form a moving stream which revolves through the complete circuit.<br />
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This moving streams of electrons maintains a constant density throughout its entire length. The number of electrons entering the positive terminal of a battery from a wire is always exactly balanced by the number of electrons which the battery forces to move its own negative terminal and out into the wire.<br />
<br />
Therefore, there is no way either the conductor wire or the battery possess either more or less electrons in a complete circuit. If the circuit loop is broken, the electron orbiting stream instantly stops revolving through the circuit; but both wire and battery will still hold exactly same number of electrons as they did when the circuit was made. The only difference is that the wire is now holding some of the electrons which was previously in the battery. Likewise, the battery had taken an equal number from the wire.<br />
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The number of electrons in the electron stream is depends on the strength of the voltage applied forcing the electrons to move. The lower the voltage, the weaker.<br />
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I will just tell you this in advance that when a resistance of any kind is inserted into the circuit loop, it also restrict the number of electrons flowing therefore reduces the current. You will notice that in some of our applications when we touch different circuitry laws. The flow of current is restricted by this resistance.<br />
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Also keep in mind a closed loop of wire is not always an electric circuit. Remember that in our definition of <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/what-is-electric-circuit.html"><span style="font-weight: bold;">Electric Circuits</span></a>, I had mentioned there the 2 conditions that makes up an electric circuit. Current, voltage and resistance are present in any electric circuit where electrons move around the close loop. The pathway for current flow is actually the circuit, and its resistance controls the amount of current flow around the circuit.<br />
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DC circuits consists of a source of DC voltage, such as batteries plus the combined resistance of the electrical load connected across this voltage. While working with DC circuits, you will find out how the total loads can be changed with various combinations of resistances, and how these combination of resistances control the circuit current and affect the voltage. This concept will also be applied in AC circuitry.<br />
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There will be two types of circuits that we will be dealing with: these are <span style="font-weight: bold;">series circuits</span> and <span style="font-weight: bold;">parallel circuits</span>. No matter how complex the circuit is, still it can be simplified down into series connection to or a parallel circuit connection.<br />
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One last thing.<br />
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<span style="font-weight: bold;">The Load</span><br />
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Previously on our last topic we had mentioned about the load in the electric circuit. So what the heck is the load? How does it works in the electric circuit?<br />
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In basic electric circuit the device that transforms the electrical energy from the source of power (emf) into some useful function -such as heat, light, mechanical power, etc.- is called the<span style="font-weight: bold;"> load</span>. The load aside from transforming and electrical energy into some useful purposes, can be utilized to changed or control the amount of energy being delivered from the source.<br />
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A load could be a motor, a telephone, a lamp, a heater or some other appliances -( name it ). the term <span style="font-weight: bold;">load</span> means the <span style="font-weight: bold;">electric power delivered by the source</span>. If you don't get it, I will give you an example. When it is stated that the load is increased or decreased, it means that the source is delivering less or more power. Remember a load can be: a device which utilizes the power from the source and the power that is taken from the source.<br />
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One more last thing...<br />
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<span style="font-weight: bold;">The Switches</span><br />
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I included this because this is one of the common part of electric circuitry either on DC or AC Circuitry. We have been using switches everyday and all our life. We could see it in our lamps, a radios, flahlights etc. It is a controlling device which open and close the circuit. There are many types of switches you will encounter in your study of <span style="font-weight: bold;">Electrical Engineering</span>. But this will be discussing separately when we touch practical applications in our course outline.<br />
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This is where it ends our topic discussing what an electric circuit is. On my succeeding post, we will now begin to study in detail the relationship between voltage, current and resistance.<br />
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Hope you appreciate this post today presenting it in my own little way. Learning is fun here in <span style="font-weight: bold;">Electrical Engineering for Beginners</span>.<br />
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Cheers!<br />
<br /></div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com1tag:blogger.com,1999:blog-1420533950904606452.post-63720953306465922562009-08-19T09:12:00.017+08:002013-03-30T19:11:14.094+08:00What is Electric Circuit?<div dir="ltr" style="text-align: left;" trbidi="on">
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Last time we have a little review of basic Physics. The reason why I included those topics here in <span style="font-weight: bold;">Electrical Engineering </span>as a beginning topic because those topics are mostly given in the board exam during our time. Please bear with me for three times posting for I have to do a little SEO for this blog. The reason why I'm doing this is to drive more traffic to this site. I was currently standing number 6 for a keyword <span style="font-weight: bold;">Electrical Engineering for Beginners</span> for which this site is really intended to.</div>
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Anyway, let's begin with our new topic for today. This is already the start of our major discussion where every detail should be understand by everyone.<br />
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<span style="font-weight: bold;">The Electric Circuits</span><br />
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Since we will be dealing with electric circuit starting today up to the rest of the topic, it is recommended that you should have an accurate picture of what electric circuit is and how the electric current behaves on it.<br />
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Just recall for a moment what we had reviewed on our previous topic about the current flow on <a href="http://electricalengineeringforbeginners.blogspot.com/2009/07/voltage-current-power-and-energy.html" style="font-weight: bold;">Voltage, Current, Power and Energy</a>. You have learned that if you connect a length of wire or a conductor across the positive and negative terminals of a source of electromotive force ( emf) let say, a battery, the potential difference (voltage) makes the current flow and also that electrical energy is needed to keep the current flowing.<br />
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Any combination of a conductor and a source of electricity connected together to permit the electrons to travel around a continuos stream is called <span style="font-style: italic;">electric circuits</span>. The figure shown below is a simple electric circuit that we are talking about ( you may click the image to enlarge)<br />
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<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SotuLc_1wSI/AAAAAAAAAEI/DFjjMqgBlTM/s1600-h/electric+circuit.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5371508123611021602" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SotuLc_1wSI/AAAAAAAAAEI/DFjjMqgBlTM/s400/electric+circuit.jpg" style="cursor: pointer; display: block; height: 306px; margin: 0px auto 10px; text-align: center; width: 400px;" /></a><br />
Many millions of free electrons that have already been separated from the outer orbits of their respective atoms by the heat of room temperature, and which have been wandering aimlessly in all directions through the wire, now come under a <span style="font-weight: bold;">common controlling force</span>. They repelled by the <span style="font-weight: bold;">more negative </span>( or less positive) charge which have been set-up at one end of the wire, and strongly attracted by the <span style="font-weight: bold;">less negative</span> (or more positive) charge which have been connected at the other end. This movement are converted into a disciplined current flow from more negative to more positive, and the electric current flows.<br />
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Take note that electrons are negative charges of electricity and have practically no weight at all. It only means that when a potential difference is applied to the wire, they respond to it <span style="font-weight: bold;">immediately</span>. Likewise, when the potential difference is removed, the electrons stop their disciplined flow in a single direction at once and resume their wanderings through the conductor material.<br />
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Now...</div>
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What are the conditions required in order to maintain the flow of an electric current in a circuit? Please take note of these conditions. They are simple but very important when it comes to actual application.</div>
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1. There must be a <span style="font-weight: bold;">source</span> of potential difference or voltage to provide the energy which forces electrons to move in a disciplined way in a specific direction.</div>
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2. There must be <span style="font-weight: bold;">continous </span>(complete) external path for the electrons to flow from negative terminal to the positive terminal of the source of voltage.</div>
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We have mentioned about <span style="font-weight: bold;">external path</span>. This is usually made up of two parts: the <span style="font-style: italic;">conductors </span>or the wires, and the <span style="font-style: italic;">load</span> to which the electric power is to be delivered for some useful effects. In the above illustration given, the<span style="font-weight: bold;"> lamp</span> or the small bulb is <span style="font-style: italic;">load</span> in the given sample.</div>
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An electric circuit is thus completed its electrical pathway, consisting of not only a conductor in which the current will flow (negative to positive), but also of a path through a source of potential difference from positive back to negative.<br />
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A small bulb connected across a dry cell is an example of electric circuit. Current flows from the negative (-) terminal of the cell, through the small bulb (the load), to the positive terminal. The action of the cell is that it provides a regenerative path for the flow of electrons to be maintained.<br />
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As long as this electrical pathway remains unbroken at any point, it is a <span style="font-style: italic;">closed</span> circuit and the current flows. But if the pathway is broken, it becomes an <span style="font-style: italic;">open </span>circuit and no current flows.</div>
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<span style="font-weight: bold;">An Open Circuit</span></div>
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<a href="http://4.bp.blogspot.com/_cQJDlnQI2zY/SouDl35T99I/AAAAAAAAAEQ/UdXxkgTd5_I/s1600-h/open_circuit.gif" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5371531667252180946" src="http://4.bp.blogspot.com/_cQJDlnQI2zY/SouDl35T99I/AAAAAAAAAEQ/UdXxkgTd5_I/s400/open_circuit.gif" style="cursor: pointer; display: block; height: 216px; margin: 0px auto 10px; text-align: center; width: 316px;" /></a><span style="font-weight: bold;">A Closed Circuit</span></div>
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<a href="http://2.bp.blogspot.com/_cQJDlnQI2zY/SouFMrXm-iI/AAAAAAAAAEY/-5s7B9n1Thc/s1600-h/closed+circuit.gif" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5371533433416120866" src="http://2.bp.blogspot.com/_cQJDlnQI2zY/SouFMrXm-iI/AAAAAAAAAEY/-5s7B9n1Thc/s400/closed+circuit.gif" style="cursor: pointer; display: block; height: 216px; margin: 0px auto 10px; text-align: center; width: 316px;" /></a>This is how the Electric Circuit is defined. On my next post, I will going to differentiate about DC and AC Circuits. We should be able to comprehend well the difference between the two. We will be discussing it separately on our course outline for they have completely separate ideas. I would like to organize my ideas and lessons presented in this <span style="font-weight: bold;">Electrical Engineering</span> site.<br />
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I hope you enjoy with this simple discussion today. I'll be back shortly.<br />
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Cheers!</div>
</div>
Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com4tag:blogger.com,1999:blog-1420533950904606452.post-52922423953504546092009-08-14T10:32:00.071+08:002013-03-30T19:12:59.230+08:00Solutions To Brain Teasers Number 2<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;">As I promised last time, I will going to show you the solutions to teasers on our previous topic about the </span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html"><span style="font-family: Georgia, "Times New Roman", serif; font-weight: bold;">Factors Affecting Resistance</span></a><span style="font-family: Georgia, "Times New Roman", serif;"> here in </span></span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%; font-weight: bold;">Electrical Engineering</span><span style="font-size: 100%;">. By the way, this topic is still covered on our review of basic Physics for this is very important one when we reached our major topics in this </span><span style="font-size: 100%; font-weight: bold;">Electrical Engineering</span><span style="font-size: 100%;"> course.</span></span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;">Last time, I leaved you four worded problems in order for you to analyze and understand the principles fully. But if you still running out of time to solve it, I will show to you it now.</span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">The first problem given was:</span></div>
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<strong><span style="font-family: Georgia, "Times New Roman", serif;">Problem 1 :</span></strong><span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;"> The resistance of a copper wire 2, 500 cm long and a 0.090 cm in diameter is 0.67 ohm at 20 degree celcius. What is the resistivity of copper at this temperature?<br />
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<span style="font-size: 100%;">The solution here is quiet simple. We just have to substitute the given values from the formula since we have uniform units:</span></span></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%; font-weight: bold;">p = RA/ l</span></span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%;"> = 0.67 ohm x [ TT ( 0.090 cm) ^2/ 4 ] / 2, 500 cm<br />
= </span><span style="font-size: 100%; font-weight: bold;">1.7 x 10 ^-6 ohm.cm</span><span style="font-size: 100%;"> -</span><span style="font-size: 100%; font-weight: bold;"> answer</span></span><br />
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<strong></strong><span style="font-size: 100%;"><span style="font-family: Georgia;">The unit of resistivity in the British engineering system of units differs from that just given in that different units of length and area are employed. The unit of area is the <span style="font-size: 100%; font-style: italic;">circular mil</span></span><span style="font-family: Georgia; font-size: 100%;">., the area of the circle 1 mil (0.001 in) in diameter, and the unit of length is the foot. Since the areas of two circles are proportional to the squares of their diameters, the area of a circle in circular mils is equal to the square of its diameter in mils. In this system of units the resistivity of a substance is numerically equal to the resistance of a sample of that substance 1 ft long and 1 circular mil in area, and is expressed in ohm-circular mils per foot.<br />
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The abbreviation CM is often used for circular mils. This should not be confused with the abbreviation used for centimeters (cm). We will use the more standard cmil.<br />
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Let's solve another problem using the principles above.</span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%; font-weight: bold;">Problem 2 :</span><span style="font-size: 100%;"> Find the resistance of 100 ft of copper wire whose diameter is 0.024 in and whose resistivity is 10.3 ohm.cmils/ft.</span></span></span><br />
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<span style="font-size: 100%;"></span></span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%;">Convert first the given diameter in mils. Since, 1 mil = 0.001 in as mentioned above.</span></span></span><br />
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<span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%;">Therefore, d= 0.024 in = 24 mils.<br />
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Then getting the area we have, A = d^2 = 24 ^2 cmils. Substituting the values:</span><br />
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<span style="font-size: 100%; font-weight: bold;">R = pl / A </span><span style="font-size: 100%;">= (10.3 ohm.cmils/ft)(100 ft) / 24 ^ 2 cmils.</span><br />
<span style="font-size: 100%; font-weight: bold;">R = 1.8 ohm - answer</span></span></span><br />
<span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;"><strong></strong><br />
<span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%; font-weight: bold;">Problem 3 : </span><span style="font-size: 100%;">A silver wire has a resistance of 1.25 ohm at 0 degree celcius and the temperature coefficient of resistance of 0.00375 per degree celcius. To what temperature must the wire be raised to double the resistance?</span><br />
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<span style="font-size: 100%;">Since we are asking for the temperature, just derive the formula of EQ . 1 in our <a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html">previous post</a> here in <span style="font-size: 100%; font-weight: bold;">Electrical Engineering </span><span style="font-size: 100%;">topics</span><span style="font-size: 100%;">. It will be:</span></span></span></span></span></span></span><br />
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<span style="font-size: 100%;"><strong></strong></span><span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%; font-weight: bold;"></span></span></span></span> </div>
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<span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%;"><strong>t = Rt - R0 / Ro oo </strong><span style="font-size: 100%;">= ( 2.50 -1.25 ) ohm / 1.25 ohm x 0.00375 /C</span></span><br />
<span style="font-size: 100%; font-weight: bold;">t = 266 degree celcius</span><span style="font-size: 100%;"> -</span><span style="font-size: 100%; font-weight: bold;"> answer</span><br />
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</span></span><span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;">It should be clearly understood that R0 in the above equation ordinarily refers to the resistance at 0 degree celcius and not to the resistance of any other temperature. A value of oo based upon the resistance at room temperature, for example, is appreciably different from the value based upon 0 degree celcius. This may be made clearer by the graphic analysis of the variation of resistance with temperature. ( Click the image to enlarge)<br />
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<img alt="" border="0" id="BLOGGER_PHOTO_ID_5369766928787825458" src="http://3.bp.blogspot.com/_cQJDlnQI2zY/SoU-kk0J4zI/AAAAAAAAAD4/tb0F16hNd64/s400/drop_shadow_109270+for+R+vs+T.png" style="cursor: pointer; display: block; height: 302px; margin: 0px auto 10px; text-align: center; width: 400px;" /></span></span></span></span></span></span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%;">In the above illustration, the resistance Rt of a conductor at any temperature t is plotted. For a pure metal, this curves gives a linear relation ( approximately). Note the fact that the curve does not pass through the origin; i.e. at 0 degree celcius the resistance is not zero. Hence, we cannot say that R oo t. The slope of the curve delta R/delta t is constant . Since,<br />
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oo = delta</span><span style="font-size: 100%; font-weight: bold;"> R</span><span style="font-size: 100%;"> / delta</span><span style="font-size: 100%; font-weight: bold;"> t </span></span><span style="font-family: Georgia, "Times New Roman", serif; font-size: 100%;">/Ro = slope / Ro<br />
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it is clear that the value of oo depends upon the base temperature chosen for Ro. In computations involving temperature variation of resistance, the value of Ro must be obtained by using the equation below:</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><strong>Rt = Ro + Ro oo t = Ro( 1 + oo t)</strong></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;">Understanding the principles of effect of temperature in resistance.</span></div>
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<span style="font-family: Georgia, "Times New Roman", serif;"><strong>Problem 4 :</strong></span><span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%;"> A tungsten filament has a resistance of 133 ohm at 150 degree celcius. If oo = 0.0045/C., what is the resistance of the filament at 500 degree celcius?<br />
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From the </span><span style="font-size: 100%; font-weight: bold;">EQ 1</span></span><span style="font-size: 100%;"><span style="font-family: Georgia, "Times New Roman", serif;">: ( </span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html"><span style="font-family: Georgia, "Times New Roman", serif;">of our previous post</span></a><span style="font-family: Georgia, "Times New Roman", serif;">)<br />
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Ro = Rt / 1 + oo t = 133 ohm / 1 + (0.00450/C ) x 150 degree celcius<br />
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Ro = 79.4 ohms<br />
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Getting the resistance at 500 degree celcius,<br />
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R500 = R0( 1 + oo t500)<br />
R500 = 79.4 ohms [ 1 + (0.00450/C) x 500 degree celcius]</span></span><br />
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<span style="font-family: Georgia, "Times New Roman", serif;"><span style="font-size: 100%; font-weight: bold;">R500 = 258 ohms - answer</span><br />
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Since it is the resistivity factor that changes with temperature. The equations of EQ 1 and EQ 2 from our</span><a href="http://electricalengineeringforbeginners.blogspot.com/2009/08/factors-upon-which-resistance-of.html"><span style="font-family: Georgia, "Times New Roman", serif;"> previous lecture</span></a><span style="font-family: Georgia, "Times New Roman", serif;"> may be written with<span style="font-size: 100%; font-weight: bold;"> p</span><span style="font-size: 100%;"> in place of </span><span style="font-size: 100%; font-weight: bold;">R</span><span style="font-size: 100%;">.</span><br />
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<strong>pt = p0( 1 + oot) ------------------------------------- EQ. 3</strong><br />
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<span style="font-size: 100%;"><span style="font-size: 100%;">I will not give the table for resistivities and temperature coefficients of resistivity of materials for it is always given in the problem. You don't have to memorize it.</span></span></span></div>
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<span style="font-size: 100%;"><span style="font-size: 100%;"><span style="font-size: 100%;"></span></span></span>I think this is now over for the basic review....<br />
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On my next post, we will now begin to discuss the real scope of <span style="font-size: 100%; font-weight: bold;">Electrical Engineering</span></span><span style="font-family: Georgia, "Times New Roman", serif; font-size: 100%;">.<br />
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Cheers!</span></div>
</div>Anonymoushttp://www.blogger.com/profile/12533189977281361236noreply@blogger.com0