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....

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

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!

## 2 comments:

I find this site very useful for beginners. I currently subcribed to your article.

Thank you...

Thanks, I have a new post tomorrow related to my recent post.

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