Applications for DC Parallel Circuits

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 Learning Electrical Engineering, the project still alive until such time I completed the course outline of this site.

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 " learning electrical engineering " now dominates my rank in Google and Yahoo. Now let's continue of what I have left before.

The following illustrative problems below are just an applications of what I had presented on my post about DC Parallel Circuits.

• DC Parallel Circuits 1
• DC Parallel Circuits 2

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.

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.

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= 10 amperes.

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 = 2 Amperes.

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 = 5 Amperes.

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 = 10 Amperes.

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.

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.

Since we have different parameters given for each load, We have to solve first some missing requirements that we need in our calculations.

(a) Calculating Ra = Vt ^2 / Pa = 230 ^ 2 / 9,200 = 5.75 ohms

for Rb = Vt / Ib = 230 / 60 = 3.83 ohms

Note : The solutions shown above are just an applications of ohms law that we previously discussed.

(b) Calculating the total equivalent resistance of the circuits will follow:

Req = 1/ 1/5.75 + 1/3.83 + 1/4.6 = 1/ 0.174+ 0.261+0.217 = 1.53 ohms

Note : the formula used was just discussed on the previous post above DC PARALLEL CIRCUITS.

(c) Since we have already calculated the missing values, we can now solve for total current.

It = 230 / 1.53 = 150 Amperes

(d) To get the total power, since we already know the values of Vt and It. It will be now,

Pt = 230 x 150 = 34,500 watts or 34.5 KW.

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.

This is the end of our basic circuitry in parallel connections. The next post will deal about series-parallel circuits lecture.

See you again on my next post.

Cheers!

Yes, let's continue of what we had left last time here in Electrical Engineering for Beginners. I was glad that you are still there and an increasing number of subscribers makes me feel more energetic in writing more in this Electrical Engineering 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.

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.

Moving on...

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.

Wondering how it was obtained?

In our previous lesson, DC Parallel Circuits Part 1. 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.

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 Kirchhoff's First Law - Second Law was already discussed here.

What was it all about?

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 : The sum of the currents flowing toward a junction always equal the sum of all currents flowing away from that junction.

Or, other states like this...

The algebraic sum of the currents at any junction of an electrc circuit is zero. 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:

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 node. 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.

or, in other ways of expressing it...

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.

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.

Uhmm....

Unequal Resistors in Parallel Circuits

Here are some of the important rules to remember when dealing with unequal resistors in parallel:

1. The same voltage is impressed across all resistors.

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.

3. The total current for the circuit is : It = I1 + I2 + I3+...

4. The total equivalent resistance of the circuit is:

Req = 1/ (1/R1) + (1/R2) + (1/R3) + ....

Note : When two unequal resistors are connected in parallel their equivalent resistance is equal to their product divided by their sum.

R xy = Rx x Ry / Rx + Ry

Ohms Law in Parallel Circuits

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.

Let's say you have a number of resistors connected in parallel but you like to measure the resistance 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.

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.

Knowing the voltage requires no disconnection. But of course, Ohms Law is the very pratical use in knowing such quantities for electrical engineers like us.

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.

I will come back on first week of October 2009.

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!