# Fast Circuit Analysis

1. Apr 22, 2013

### EngPF

Hello guys!

I'm studying Electrical Circuits right now and come to this particular circuit where I need to find the current I, take a look:

I'm able to find it, it is not hard at all. But what I would like to know are these two things:

1. What do you think is the fastest way to do it? (I need to do it as fast as possible).
2. Is it true that removing the 6 ohm resistor doesn't change the circuit? I have been told it, but I cant see it. Could anyone explain this to me?

Sorry if these are too simple for you! :) Thanks

2. Apr 22, 2013

### phyzguy

(1) I think you just need to write a series of equations for the node voltages and currents and solve them. I don't see a quicker way.

(2) Since the 6 ohm resistor is connected across the two supplies, its presence or absence doesn't change the other voltages and currents. So you can remove it, analyze the rest of the circuit, and then put it back in. However, there is current flowing through the 6 ohm resistor, so it needs to be included in the final current analysis.

3. Apr 22, 2013

### EngPF

Thanks. After removing the 6 ohm resistor I solve the problem using thevenin. Since I was interested only in the current I, I believe that is the fastest way. Vth could be found really fast, just some voltage divider. Do you guys think there's a faster way in doing it?

Well, thanks for the answer about the 6 ohm resistor. That helped. Also when you are calculating Rth and you substitute the voltage sources with a short the 6 ohm is short circuited to ground and is out of the circuit. Is that a sign that the resistor could be removed?

Do you guys have any other examples of circuits that would become way easier to solve if one could see that there's some components that could be removed without changing the analysis you are trying to do?

4. Apr 22, 2013

### rude man

Removing the 6 ohm resistor does not change the current i as depicted in your image.

The other, perhaps faster, approach is to use superposition:
let the right-hand source be a short (V = 0) & compute i = i1.
Then restore that source but let the left-hand source be a short & compute i = i2.
The final current i = i1 + i2.

This works provided the sources are not dependent on another in any way.

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