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KCL Problem with Dependant sources

  1. Jan 24, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the voltage drop V1 of the following circuit, assuming Vin=1v, and for the life of me can't seem to get the right answer.
    QWnesTY.jpg

    2. Relevant equations
    KCL, Ohm's law

    3. The attempt at a solution
    I basically tried to use KCL at the top node,
    Code (Text):
    Vin/2+2V1 = V1/3 + Vin/6, so 1/2=-2V1+V1/3+V1/6, (1/2)/(1.5)=V1=1/3, but apparently that's the incorrect answer
    Thanks for any help
     
  2. jcsd
  3. Jan 24, 2017 #2

    gneill

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    Staff: Mentor

    Hi bran_1, Welcome to Physics Forums.

    I don't understand the terms of your node equation. For example, the first term of the LHS (left hand side) is Vin /2. How do you arrive at that? And on the RHS you have Vin/6, but Vin is not the node voltage. Can you explain your reasoning?
     
  4. Jan 24, 2017 #3
    Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node. As for the Vin over 6, since the 3ohm and 6ohm resistors are in parallel, they share the same voltage, do they not? This is just ohm's law again, except for the current leaving the node through this branch.


    Edit: I didn't get it correct though, so my logic is flawed somewhere...
     
    Last edited: Jan 24, 2017
  5. Jan 24, 2017 #4

    cnh1995

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    Homework Helper

    You need not assume any value for Vin.
    You should combine the 3 ohm and 6 ohm resistors in parallel and use their equivalent resistance (for simplicity). You'll still have the two nodes, and three currents instead of four.

    Write the KCL (node voltage) equation again and simplify.
     
    Last edited: Jan 25, 2017
  6. Jan 24, 2017 #5
    I tried that also, it didn't seem to make a difference, as I got the same answer.
    Vin/2 + 2V1 = V1/2
    Vin = -3V1
    V1=-(1/3)V, which is incorrect.

    Is there some rule with dependent sources that I'm unaware of?
     
  7. Jan 25, 2017 #6

    gneill

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    Staff: Mentor

    Vin is not the potential difference between the two ends of the 2 Ω resistor. Only the left side of that resistor is at potential Vin. What is the potential at its other end? So then what is the potential difference?
     
  8. Jan 25, 2017 #7
    Wouldn't it be VA (of the node) on the other side? I'm not sure how to find the node voltage though
     
  9. Jan 25, 2017 #8

    gneill

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    Staff: Mentor

    It's an unknown at this point. You'll solve for it after writing your node equation.
     
  10. Jan 25, 2017 #9
    Alright, so I'd have (Vin-Va)/2+2V1=V1/2, right? Could I call the bottom node node B, and say it's V=0? So then V1=Va-Vb? -> V1=Va?
     
  11. Jan 25, 2017 #10

    gneill

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    Staff: Mentor

    Better.

    Note that you can see that Va = V1 just by inspecting the circuit diagram. You don't need to introduce a different variable. Why not just use V1?

    Yes, the bottom node makes an excellent choice for the reference node.

    upload_2017-1-25_9-13-34.png
     
  12. Jan 25, 2017 #11
    So using that, I have
    Vin-V1=-3V1+V1 -> V1=-0.5V
     
  13. Jan 25, 2017 #12

    gneill

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    Staff: Mentor

    Looks good.
     
  14. Jan 25, 2017 #13
    So then the power delivered by the dependent source would be:

    P=VI , -0.5 * (2*-0.5) =1/2 watts
     
  15. Jan 25, 2017 #14

    gneill

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    Staff: Mentor

    Yes, that looks right.
     
  16. Jan 25, 2017 #15
    Thanks for all your help!
     
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