Discover the Fastest Way to Analyze Electrical Circuits

[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 can't see it. Could anyone explain this to me?

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

 

[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.

 

[EngPF]

(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.

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?

 

[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.

 

[rezeew]

I would suggest using Kirchhoff's laws to analyze this electrical circuit. Kirchhoff's laws are a set of rules that help determine the current and voltage in a circuit. They can be used to quickly and accurately analyze complex circuits, making them the fastest way to analyze electrical circuits.

To answer your second question, removing the 6 ohm resistor does not change the circuit because it is in parallel with the 8 ohm resistor. In parallel circuits, the total resistance is always less than the smallest individual resistor. In this case, removing the 6 ohm resistor would result in a total resistance of 4 ohms, which is still in parallel with the 8 ohm resistor. Therefore, the circuit remains the same.

I hope this helps and good luck with your studies! Don't worry about asking simple questions, we all start somewhere and it's important to have a strong understanding of the basics in order to tackle more complex problems. Keep up the good work!