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Circuit With Three Batteries Two Resistors

  1. Mar 14, 2012 #1
    In the above circuit:
    V1 = 12.00 V V2 = 6.00 V V3 = 3.00 V
    R1 = 220.0 R3 = 270.0

    Because of the direction of the batteries, a student looking at the above circuit makes this guess for the direction of current across resistors R1 and R3. Is the student right, or wrong?

    Find the current across resistors R1 and R3. If the student's choice of direction is RIGHT, the sign of the respective current you enter should be POSITIVE. If the student's choice of direction is WRONG, the sign of that current should be NEGATIVE. (Don't discount the possibility that a current might be zero, either!)

    Solve for I3=
    I1=

    (If you have trouble, this hint: Note that the central battery polarity is OPPOSITE what you would expect to make a nice, neat loop. How do you think you should handle two batteries in a loop that are hooked up with OPPOSITE, not the same, polarities?)http://www.webassign.net/userimages/qid1025007.bmp?db=v4net&id=76445

    Attempt at a Solution: I know I use Kirchoff's Laws but I cannot figure out what to do with no resistor in the central wire. I know that that means infinite current but cannot determine the effect it would have on the other two.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 14, 2012 #2

    gneill

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    Staff: Mentor

    For infinite current there would have to be a path without any resistance between the terminals of a voltage source. Can you show that such a path exists for the middle battery?

    What have you learned about Kirchhoff's laws?
     
  4. Mar 14, 2012 #3
    We have learned how to apply Kirchoff's rules. I am sure that this problem is a lot easier than I am making it out to be but we have not done many examples in class.
     
  5. Mar 14, 2012 #4

    gneill

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    Staff: Mentor

    Why don't you try to apply Kirchhoff's voltage law around one of the loops? Can you write the resulting equation?
     
  6. Mar 14, 2012 #5
    Top Loop: starting from junction at lower left corner -I1R1+V1-V2=0
     
  7. Mar 14, 2012 #6
    Bottom Loop: Starting from bottom right junction -V2+V3-I3R3=0
    Is this correct? If so, I know how to go about solving from here.
     
  8. Mar 14, 2012 #7

    gneill

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    Staff: Mentor

    Both look fine. Solve for I1 and I3.
     
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