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Magnitude of the current in a battery

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the magnitude of the current in the 14V cell.
    Answer in units of Amperes.

    2. Relevant equations
    Kirchoff's junction & loop laws:
    I1 = I2 + I3

    ƩΔV = 0

    V = IR

    Rseries = R1 + R2 + ...

    1/Rparallel = 1/R1 + 1/R2 + ...

    3. The attempt at a solution
    At first I assumed that the current inside an ideal battery is 0 amps. But it appears that is incorrect. I've been going over my notes and looking through my book for a while now, but I have had no luck in finding out how to find the current in a battery in a multi-loop circuit. I have also tried setting up 2 different loop equations but those were wrong too. I know that the value has to be positive because it wants the magnitude. I'm not going to give up on this but I do need help from a reliable source. Any help is welcome.
  2. jcsd
  3. Mar 3, 2012 #2


    User Avatar
    Gold Member

    How do you know your "2 different loop equations" were wrong?

    Also, in the context of this problem, I'm 98.3% positive that "1/Rparallel = 1/R1 + 1/R2 + ..." is irrelevant.
  4. Mar 3, 2012 #3
    What I had done was Loop 1 being the circuit containing the 14 V battery and the 29V battery. Setting up the equation 12I2 - 23I1 = -43 for the top loop and 12I1 - 27I2 = -66 for the bottom loop. Then I put together a system of equations. I multiplied the top loop equation by 9 and the bottom by 4 so that the I2 would cancel. This gave me the value of I1 to be about 4.094 amps. Knowing that the current in a series of resistors is equal to each other, I had assumed that the battery was some sort of resistor so I had put that answer down and the system showed to be wrong.

    Also, thanks for the quick response!
  5. Mar 3, 2012 #4


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    Staff: Mentor

    Ah. When you went around your loops and summed the voltage supplies, you didn't take into account the fact that you're passing through one going from positive to negative, and the other negative to positive.
  6. Mar 3, 2012 #5
    Gah, it is always the little mistakes! Thank you so much, I finally got it! :D
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