Which circuit schematics violate KVL/KCL?

[Marcin H]

Homework Statement

Homework Equations

V=IR
KVL/KCL

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.

 

[NascentOxygen]

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.

 

[Marcin H]

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.

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?

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.

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.

 

[NascentOxygen]

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?

Yes. Though you need to keep in mind it has polarity, too.

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.

-2V is right.

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

Is (d) a valid arrangement?

 

[Marcin H]

Yes. Though you need to pay attention to polarity, too.

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?

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!

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.

Is (d) a valid arrangement?

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?

 

[NascentOxygen]

If you draw a voltage arrow pointing to the left across the current source in (d) and label it V_s 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.

 

[Marcin H]

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.

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?

 

[NascentOxygen]

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

 

[CWatters]

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.

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?

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.

 

[Marcin H]

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

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.

Ok cool. Thanks!

 

[alej27]

Circuit (a) violates KVL because 2 V (the left voltage source) is not equal to 3 V (the right voltage source).

Circuit (b) doesn’t violate any law. The current through the circuit is 4 A. Thus the voltage across the resistor is 4 A • 1 Ω = 4 V, with the positive reference terminal at the top. Hence the voltage across the current source is +2 V - 4 V = -2 V, with the positive reference terminal to the left.

Circuit (c) violates KVL because 3 V (the voltage sources) is not equal to 0 V (the ideal short-circuit).

Circuit (d) violates KCL because 2 A (the left current source) is not equal to -3 A (the right current source).

Circuit (e) doesn’t violate any law. The voltage across the 1-Ω resistor is 6 V, and the current through the 2-Ω resistor is 2 A. Therefore the current through the 1-Ω resistor is (6 V)/(1 Ω) = 6 A, with the positive reference direction downwards; and the voltage across the 2-Ω resistor is 2 A • 2 Ω = 4 V, with the positive reference terminal at the top. So the current through the voltage source is + 6 A + 2 A = 8 A, with the reference direction upwards; and the voltage across the current source is +6 V - 4 V = 2 V, with the positive reference terminal to the left.

 

[berkeman]

Circuit (a) violates KVL because 2 V (the left voltage source) is not equal to 3 V (the right voltage source).

Circuit (b) doesn’t violate any law. The current through the circuit is 4 A. Thus the voltage across the resistor is 4 A • 1 Ω = 4 V, with the positive reference terminal at the top. Hence the voltage across the current source is +2 V - 4 V = -2 V, with the positive reference terminal to the left.

Circuit (c) violates KVL because 3 V (the voltage sources) is not equal to 0 V (the ideal short-circuit).

Circuit (d) violates KCL because 2 A (the left current source) is not equal to -3 A (the right current source).

Circuit (e) doesn’t violate any law. The voltage across the 1-Ω resistor is 6 V, and the current through the 2-Ω resistor is 2 A. Therefore the current through the 1-Ω resistor is (6 V)/(1 Ω) = 6 A, with the positive reference direction downwards; and the voltage across the 2-Ω resistor is 2 A • 2 Ω = 4 V, with the positive reference terminal at the top. So the current through the voltage source is + 6 A + 2 A = 8 A, with the reference direction upwards; and the voltage across the current source is +6 V - 4 V = 2 V, with the positive reference terminal to the left.

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