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Which circuit schematics violate KVL/KCL?

  1. Aug 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-08-28 at 6.16.15 PM.png

    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. jcsd
  3. Aug 30, 2016 #2

    NascentOxygen

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    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.
     
  4. Aug 30, 2016 #3
    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.
     
  5. Aug 30, 2016 #4

    NascentOxygen

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    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?
     
  6. Aug 30, 2016 #5
    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?
     
  7. Aug 30, 2016 #6

    NascentOxygen

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    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.
     
  8. Aug 30, 2016 #7
    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?
     
  9. Aug 31, 2016 #8

    NascentOxygen

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    Staff: Mentor

    What you said is right: any point in a circuit must have as much current entering as leaving.
     
  10. Aug 31, 2016 #9

    CWatters

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    Science Advisor
    Homework Helper

    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.
     
  11. Aug 31, 2016 #10
    Ok cool. Thanks!!!
     
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