What Is Electric Power?

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:


Forces that work is done and  forces not doing work.

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.

The potential difference between any two points in an electric circuit, which gives rise to a voltage and when connected causes electron to move and current to flow. This is one of a good example in which forces causing motion, thus causing work to be done.

Talking about work in electric circuit, there is also a electric power which is the time rate of doing work done of moving electrons from point to point. It is represented by the symbol P, and the unit of power is watt, which is usually represented by the symbol W. Watt 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.

The Useful Power Formula

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 power.

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 power consumed in a resistor is determined by the voltage measured across it, multiplied by the current flowing through it. Then it becomes,

Power = Voltage x Current
Watts  = Volts x Amperes

P = E x I  or P = EI ------> formula no.1

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 Ohm's Law. 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:

P = EI
P = (IR)I or P = I²R ------------> formula no.2

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.

P = EI
P = E(E/R) or P = E²R ---------> formula no. 3

For guidance regarding expressing of units of power are the following:
a. Quantities of power greater than 1,000 watts are generally expressed in (kW).
b. Quantities greater than 1,000,000 watts are generally expressed as megawatts (MW).
c.  Quantities less than 1 watt are generally expressed in (mW).

The Power Rating of Equipment

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.

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. 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.

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.

Take a look at the sizes of carbon resistors 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.


Carbon resistors with comparative sizes of different wattage ratings
of 1/4 watt, 1/2 watt,1 and 2 watts

If the resistors greater then 2 watts rating are needed, wire-wound resistors are used. They are ranges between 5 and 200 watts, with special types being used for power in excess of 200 watts.



Use wire wound resistors if higher than 2 watts are needed

Fuses

We all know that when current passes through the resistors, the electrical energy is transformed into heat which raises the temperature of the resistors. 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.

Fuses are resistors using special metals with very low resistance value and a low melting point. When the power 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.



This is the good fuse




This is the blown fuse

 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 ohmmeter.

There are two types of fuses in use today - conventional fuses, which blow immediately when the circuit is overloaded. The slow-blowing (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.

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.

Various types of fuses are made for various equipments

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.

Later we will be study circuit breaker which is another protective devices for over current protection.


Electrical Power in Series, Parallel and Complex Circuits

The principle of getting the total power of the circuit is just simple. There is no need to elaborate this topic.

The total power consumed by the circuit is the sum of all power consumed in each resistance.

Therefore, we just only sum up all power consumed in each resistance whether it a series, parallel or a complex circuits. Thus,

P_t= P₁+P₂+P₃+P_n watts  ---------->formula no. 4

From the problem in my previous post about complex circuit, try to calculate each power of the resistance and the total power as well. Constant practice always makes you perfect!

Cheers!

Applications: A Few Tips in Solving DC Series Circuit Problems

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. Learn Electrical Engineering for Beginners will provide you a technique on how you will overcome those scenarios.

The first problem that you will encounter is somewhat an application of a simple transmission lines. 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.

Let's begin...

Problem 1: 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.

This is how you will going to solve it...

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:

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 2 x 425 ft = 850 ft length. Please take note on this because there are many beginners who are getting mistakes when solving this type of problem.

Moving on...

The voltage drop would not exceed 5 % of the generator emf therefore,
Line Voltage Drop = 240 x 0.05 = 12 volts

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,
Load Voltage = Generator Emf - Line Voltage Drop = 240 - 12 = 228 volts

You have to consider the line voltage drop when dealing with transmission line.

The Line Current across the 4.1 ohm load resistor would be,
Line Current = 228 / 4.1 = 55.6 amperes

and the Line Resistance would also be,
Line Resistance, Rl = 12 / 55.6 = 0.216 ohms

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,

Resistance (1000 ft)/ 1000 ft = Line Resistance / length of transmission wire
Resistance per (1000 ft) = (1,000 ft /850 ft) x 0.216 ohm = 0.254 ohm

a.) Consulting the AWG table below it shows that the standard wire size is No. 4; 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.

b.) Getting the actual line resistance would be,
Actual Line Resistance / 850 ft = 0.253 ohm / 1000 ft
Actual Line Resistance = 0.85 x 0.253 = 0.21505 ohm

We need to get the following data in order to get line power loss.
Total Series-Circuit Resistance = 4.1 + 0.21505= 4.31505 ohms
Total Current I = 240 / 4.31505 = 55.6193 amperes
Line Loss Power = (55.6193)^2 x 0.21505 = 665 watts - answer

c.) Getting the efficiency of transmission line can be derived in terms of load power and total power. This can be expressed as:

Efficiency = Load Power / Total Power = [(55.6193)^2 x 4.1 / 240 x 55.6193 ] x 100 %
Efficiency = 95% - answer

This is not not formal discussion about transmission line. 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.

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:

Problem 2: 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.

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 electrical engineering. In the first place I'm a visual person. It's the easy way to understand what the problem is asking for.

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.

Therefore, we can state that: Vo = Vt + IRg

Condition 1 : Given that Vt = 115 V and I = 10 A
Vo = 115 V + (10 ) Rg, will served as equation 1.

Condition 2 : Given that Vt = 105 V and I= 15 A
Vo = 105 V + (15) Rg will served as equation 2.

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,

115 V+ 10Rg = 105 V + 15 Rg , we can now solve for Rg.

Rg = 2 ohms- answer

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.

Vo = 115 V + (10)(2) = 135 V - answer

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:

1. Try to make an simple illustration of the problem to picture out and understand the scenario of the problem.

2. Always try collect first the given data before proceeding in solving the problem.

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.

4. Know what is the subject matter of the given problem is also one that you should not forget.

5. Always review and finalize your answers.

My next post will continue our study of Circuits connected in Parallel.

Hope you learned some tricks for today here in Learn Electrical Engineering for Beginners.

Cheers!

DC Series Circuit Part 2

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.

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 DC Series Circuit Part 1. For those who missed it, you can still catch up with the lecture.

Let's study the continuation...

The Voltage Division in the Series Circuit

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.

A circuit used for this is shown in the circuit diagram below. This is called a voltage divider. You may click the picture to enlarge.

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.

So, this is how you will going to do it...

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:

I = Ein / Rt = 100/35 = 2.86 amperes

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).

Therefore, you will obtain that,

Eout = I x R2 = 2.86 x 20 = 57.2 volts - answer

You may observe that with the appropriate choice of resistor values in voltage divider chain, an input voltage of 100 volts has been stepped down to an output voltage of 57.2 volts. By using ohm's law, you would be able to calculate it.

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:

Eout = (Ein/R1+R2) x R2

Expressing it in a correct and understandable way would give you...

Eout = (R1/R1+R2) x Ein

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. 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.

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.

The Variable Resistors

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.

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).

If the variable resistor is used in this way, it is called the rheostat. It is used to control the current flow in the circuit.

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.

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 Potentiometer.

This is one typical sample for three terminal connections...

Typical potentiometer looks like this.

The Potentiometer Connections

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.

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)

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.

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.

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.

On my next post, let's have some practical applications for DC Series Circuit here in Learn Electrical Engineering for Beginners.

Cheers!

DC Series Circuit Part 1

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 beginners especially taking up the course like Electrical Engineering. 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.

Please bear with me for a sort of introduction. Just want to clarify something.

Moving on...

Resistance in Series Circuit

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.

A series circuit 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 one path for current to flow.

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.

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 series connected but you would not have a so called series circuit. 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 series circuit.

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.

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 factors upon which the resistance of conductor depends. 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.

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. 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.

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 subscript for them to identify.

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.

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.

The Flow of Current in Series Circuit

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. It would show that there are identical amount of current is flowing in every component in the circuit.

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.

Voltages in Series Circuit - The Kirchhoff's Second Law

 

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 potential drop or the voltage drop. Since they have same current, the energy expended in pushing this equal amount of current through the individual component must be also the same.

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.

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.

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.


Applying Ohm's Law in Series Circuit

Simplifying the facts that we have above about series circuit. We now know that:

1. The current in the series circuit is the same everywhere. This can be mathematically expressed as It = I1= I2= I3 and so on.

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.

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.

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.

Some useful application of Ohm's Law in the series circuit is the simplication process. Let's take an example diagram below.

 

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.

A sample problem...

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.

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.

Data:
Et = 100 Volts
It = 2 Amperes
Rt= (unknown)

E1= (unknown)
I1 = 2 Amperes since It = I1 = I2 = 2 Amperes
R1= 10 ohms

E2= (unknown)
I2 = 2 Amperes
R2 = (unknown)

Now let's solve it!

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 Rt = Et/ It = 100 / 2 = 50 ohms - answerIf 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...

From the magic triangle, put your thumb on E, therefore voltage drop across R1 is E1 = I1 X R1 = 2 x 10 = 20 Volts -answer

From our concept above, the sum of the voltage drop across each resistances is equivalent to a voltage source connected across them. - Kirchhoff s Second Law

Et = E1 + E2 ; 100 Volts = 20 Volts + E2, solving for E2 = 100 - 20 = 80 Volts - answer (simple algebra dear)

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 R2 = E2 / I2 = 80 / 2 = 40 ohms - answerTry 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.

I will continue the discussion of DC Series Circuit here in Electrical Engineering course.

See you again.

Note : I'll be back on September 3, 2009 evening (Philippines Time) for continuation of DC Series Circuit Part 2.

Cheers!