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Homework Help: Finding the Potential (Kirchoff's Law)

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    ƩV = 0

    3. The attempt at a solution
    Starting behind the left EMF source I sum the potential changes using Kirchoff's Voltage Law
    150V - R2I - 50V - R1I = 0
    Solving for I, I get 20Amps

    However I know this can't be right because the potential at point Q is 97V
    and after traversing the circuit, the potential according ot the calculated I is 110 Volts..

    Can someone perhaps show me step-by-step to do problems like these?
  2. jcsd
  3. Feb 24, 2012 #2


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    Homework Helper

    Note that there is no Earth connected to this circuit, so no point on the circuit is necessarily at 0 V.

    If the current flows anti-clockwise, the potential goes up 150 as you pass through the left battery, drop 40 or 60 through the bottom resistor [having trouble which resistor is which - one you start to reply the picture of the circuit is gone], drop another 50 going "backwards" through the ideal battery on the right, then another 60 or 40 through the top resistor.
    So I think P should be at -13 or 7 depending which resistor was the 2Ω and which was the 3Ω
  4. Feb 24, 2012 #3
    Okay I understand everything but the very end. What calculations did you do to get those two numbers?
  5. Feb 24, 2012 #4
    Using loop laws you can get the above answer ! join PQ by arbitrary line and can calculate the voltage difference between points P and Q using Loop law.
  6. Feb 24, 2012 #5


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    Homework Helper

    As I said, as soon as I start answering the post, I no longer have access to the circuit diagram. I remembered that the 150V battery was on the left and the 50V battery was on the right. I didn't remember which resistor was on the top and which was on the bottom. I do know there is a 60v drop across the 3Ω resistor and 40V drop across the 2Ω. Depending which one was top/bottom determines whether Q is at -13 [110 below], or 7 [90 below] point Q.
    I figured you would be able to work out which one - as you ahve the original circuit, and a greater need to solve the problem.
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