# Homework Help: Which circuit schematics violate KVL/KCL?

1. Aug 30, 2016

### Marcin H

1. The problem statement, all variables and given/known data

2. Relevant equations
V=IR
KVL/KCL

3. The attempt at a solution
I am stuck on part E of this problem. This looks like a valid circuit, but I am struggling with KVL/KCL to prove it.

First can I say that the voltage across the 1 ohm resistor is 6V because it is in parallel with the 6V battery? But wouldn't that mean that the voltage across the 2 ohm is also 6V? I'm not too sure how the current source there would affect that. Can I combine the current source with the 2 ohm resistor to make it into a battery and then use kvl? Right now I am trying loops for the left side and the right side and the whole thing and I am not getting anywhere. Can I assign a ground at the bottom of the middle node and try using node voltage method?

Left loop: 6V = i(1) --> i=6A? But that doesn't seem right.

The other loops I'm not too sure about because I am not sure if my logic with the voltage is correct. If it is then I would use that to try finding the KCL for the other loops.

2. Aug 30, 2016

### Staff: Mentor

The voltage across the 2Ω with 2A through it is 4V according to Ohm's Law.

Voltage across the (current source + 2Ω) combo = 6V because the voltage source dictates it. Remember, an ideal current source can carry any voltage across its terminals while maintaining a fixed current; you used this fact earlier to show part (b) is valid.

3. Aug 30, 2016

### Marcin H

So does that mean the voltage across the current source is 2V then? Then you can use KVL to show whether or not it works?

I'm not sure why B is valid. I had it as invalid originally. If I do the same thing here as the problem e then the voltage across the 1ohm resistor would be 4V right? Does that mean the voltage across the 4A source is -2V? How can I prove that is the correct method? This way of doing it feels like you will always get a right answer.

4. Aug 30, 2016

### Staff: Mentor

Yes. Though you need to keep in mind it has polarity, too.
-2V is right.

Whilever Kirchoff's Laws are not broken, then the analysis is valid!

Is (d) a valid arrangement?

5. Aug 30, 2016

### Marcin H

Would it be (+ to -) from left to right? Then Using KVL the big loop outside will give you the sum of the voltages is 0 meaning that KVL works. Right? And that is all you have to show to prove the the circuit is valid?

so would the polarities for the current source be (+ t -) from left to right with a voltage of -2V? If I flip the arrow, making it point left, then can I change the voltage to 2V? I would have to change the polarities too then.

For D I just looked at the bottom left node and noticed that the current in would not equal the current out, so I said that the circuit would be invalid. But now I feel like this is incorrect. Do I have to find the voltages across all the sources and resistors first and then use KVL?

6. Aug 30, 2016

### Staff: Mentor

If you draw a voltage arrow pointing to the left across the current source in (d) and label it Vs then you also label that arrow +2V because the left is more +ve than the right.

In (d) one source wants to force 2A through the second, while the second wants to force 3A through the first. There can not be a winner here. It is not a valid arrangement.

7. Aug 30, 2016

### Marcin H

Right, but is there a way to prove this using KVL or KCL? Or is my way of looking at the node a valid explanation?

8. Aug 31, 2016

### Staff: Mentor

What you said is right: any point in a circuit must have as much current entering as leaving.

9. Aug 31, 2016

### CWatters

KCL can be applied to components as well as nodes. For example the net current going into a resistor is always zero. The net current coming out of a transistor is always zero. The net current coming out of a battery is always zero.

10. Aug 31, 2016

### Marcin H

Ok cool. Thanks!!!