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Kirchhoff's Laws with 3 voltage sources

  1. Jul 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the potential differences across the 5 ohm, 4 ohm, 3 ohm, and 2 ohm resistors.

    2. Relevant equations
    5I_1-4I_2+3-12=0 Left Loop
    3-18+5I_3-4I_2=0 Right Loop
    12-18+5I_3-5I_1=0 Whole loop

    3. The attempt at a solution
    I simplified the first diagram by combining the 3 ohm and 2 ohm resistors into one resistor. Every time I solve for one equation and plug into the next equation, my variable start cancelling each other out. I don't think that I have my current direction and polarities correct. I'm not sure what to do next with this new setup of 3 voltage sources, etc. I'm not exactly sure of the current flow because it seems like one resistor would have to be switched to maintain polarity when using the loop equations. What am I doing wrong? Please see the attachment...the second diagram down is the circuit with the combined 2 and 3 ohm resistors.


    Attached Files:

    • 5.jpg
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  2. jcsd
  3. Jul 29, 2008 #2


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    Hi ae4jm,

    Your attachment isn't available yet. However, it seems likely that your circuit only has two simple loops to it, so that when you wrote down the equations for the left loop and right loop, that covered the entire circuit. If so, then writing the equation for the outer loop will not give you any additional information (you can get it from just subtracting your left loop equation from your right loop equation).

    However, Kirchoff's laws covered more than just the loop equations. You can also write down relationships between the currents at the nodes. This should give you enough to solve the problem.
  4. Jul 29, 2008 #3
    I'll look over my notes and in my book about how I might do this. I think that I'll be doing, for example, I_1=I_2+I_3, something like this. I just checked the image and it is now available, but I had to log in...I guess that it is just available to me at the moment. Thanks for the help...this problem is tougher than we had for homework out of the book and in class.

    Thanks again
    Last edited: Jul 29, 2008
  5. Jul 29, 2008 #4


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    The attachment has been approved now, so looking at your circuit and the chosen current directions, I think you have some sign errors in your equations (including the current equation in your last post). It looks to me like the errors mainly come from the signs you have labeled on your resistors. Remember that the voltage decreases across a resistor when you go in the direction of your chosen currents.
  6. Jul 29, 2008 #5
    Thanks. So far I've figured out that my current at I_1 was headed the wrong direction and you showed me that I had my polarity of the resistor incorrect since the voltage decreases as the current flows through the resistor. I have found I_1 to be -0.323A or -323mA and V_5ohm to be 1.62V. So, I know that my current was correct because the answers that the prof. gave us matches for the voltage across the 5 ohm resistor. Thanks. I'm sure that I'm not finished with my questions yet, I've still got 3 other voltages and 2 currents to figure. I also setup the node equation to be I_1+I_2+I_3=0, since I'm thinking that the currents meet, not sure if this is possible but I've sure confused myself with the current direction and I was thinking that currents will not meet. Since all three branches of currents that I have labeled now seem to meet head on and do not flow into each other, would it be safe to say that I have more than one of the current directions reversed?

    I switched the polarities of R1, R2, and R3,4 of the second circuit in my attached image.
    Last edited: Jul 29, 2008
  7. Jul 29, 2008 #6
    I've gotten the correct answers, I do believe.

    left loop equation: -12-5I_1+4I_2+3=0
    right loop equation: -3-4I_2+5I_3+18=0
    node equation: I_1=-I_2-I_3

    I_5ohm= -0.323A
    I_2ohm= -1.846A
    I_3ohm= -1.52
    I_2ohm= -1.52

    V_5ohm= -1.62V
    V_2ohm= -3.05V
    V_3ohm= -4.57V
    V_4ohm= -7.38V

    So I know that all of my currents should have went in the opposite directions, which would have made all my values positive instead of negative. Does this sound right? The way that the circuit was put together really confused me on the currents and polarities. If you don't mind my asking, how would you have redrawn the circuit or labeled everything?

    Thanks for all of your help and I truely appreciate it!
  8. Jul 29, 2008 #7


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    This current (the one going through the middle branch) does not look right to me; I think you picked up a sign error somewhere. (And shouldn't it be called I_4ohm?) Notice that your current (node) equation is

    I_1 + I_2 + I_3 =0

    so the only way that equation is true is if either one current is positive and two are negative, or one is negative and two are positive. But you can't have all three be negative.
    The way you labeled the currents is fine; the only thing wrong with the diagram was the labeled polarities on the resistors.

    When I do Kirchoff's problems, I usually don't worry about trying to determine the "correct" current direction to label my diagram. To me, that is a great strength of Kirchoff's approach--you don't need to figure all that out. Just label the currents, and the answer will tell you which way the currents are really going. (Of course, being able to determine from the diagram which way the currents are going is a great way to check your answers. I've definitely found errors that way.)
  9. Jul 30, 2008 #8
    Thanks, Alphysicist.

    You were saying that the middle branch didn't look correct...so, I make I_4ohm flow towards the 3V source making the negative of the resistor closest to the 3V battery. Then,
  10. Jul 30, 2008 #9


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    No, go ahead and leave all three currents downwards since your equations are based on that choice. So the equations that you got were:

    left loop equation: -12-5I_1+4I_2+3=0
    right loop equation: -3-4I_2+5I_3+18=0
    node equation: I_1=-I_2-I_3

    I'm just saying that when you solve for I2, you don't get -1.846A, you get +1.846A.
  11. Jul 30, 2008 #10
    Okay, I understand now. I appreciate you pointing that out to me. I did get a positive 1.846A, but I entered a negative into my table and I typed from the table. Thanks for catching my mistake.

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