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Find I through one of three resistors in a 3-battery circuit

  1. Apr 5, 2016 #1
    1. The problem statement, all variables and given/known data
    tmp_12234-20160405_043842-639066610.jpg


    Sorry for the quality of the image. This is the best I can come up with with my cellphone.

    What is the current through the 10 ohms resistor?
    Is the current from left to right or right to left?

    2. Relevant equations
    Kirchhoff's loop law
    Kirchhoff's junction law



    3. The attempt at a solution

    1) I1 = I2 + I3 ==> I3 = I1 -I2

    2) -5(I1) + 12V - 10(I2) + 3V = 0
    15V = 10(I2) + 5(I1)
    I1 = (15V - 10(I2))/5

    3) -5(I3) + 9V + 3V + 10(I2) = 0
    -5(I1-I2) + 12V + 10 (I2) = 0
    12V = -15 (I2) + 5 (I1)
    = -15 (I2) + 15V - 10 (I2)
    -3V = -25 (I2) ===> I2 = 0.12 A, from right to left.

    This answer is incorrect. The correct answer is 0.36 A. Can someone please give me a hint as to what I'm doing wrong ?
     
  2. jcsd
  3. Apr 5, 2016 #2

    Simon Bridge

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    Check the signs of the voltages in (3).
     
  4. Apr 5, 2016 #3
    3) -5 (I3) + 9V - 3V + 10(I2) = 0
    -5 ( I1-I2 ) + 6V + 10 (I2) = 0
    -5 (I1) + 5 (I2) + 6V + 10(I2) = 0
    6V = 5 (I1) - 15 (I2)
    6V = 15V - 25 (I2)
    -9V/-25 ohms = I2 = 0.36A

    I got to the right answer by trial and error. I don't understand why the 3V has to be negative when the current ( I2) is flowing through it from its negative terminal to its positive one. ...Can someone please explain?
     
  5. Apr 5, 2016 #4

    gneill

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    The potential difference across an ideal voltage supply does not depend upon the current or the direction of the current flowing through it. When you do your "KVL walk" around the loop you add or subtract the supply voltage strictly depending upon the direction that you pass through the fixed potential difference.
     
  6. Apr 5, 2016 #5
    Ok. I get it. Thanks!
     
  7. Apr 5, 2016 #6

    Simon Bridge

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    What I usually do is draw potential difference arrows next to the components ... the arrow points from the low potential end to the higher potential end: so it points from negative to positive by a battery and the opposite direction to the (proposed) current by a resistor.
    When I traverse the loop, I add potential differences where I go in the direction of the arrow and subtract the ones where I go against the arrow.
     
  8. Apr 5, 2016 #7
    Thanks for the tip!
     
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