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Circuit with 2 Batteries and 6 Resistors

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 56 Ω, R2 = R6 = 157 Ω R3 = 118 Ω, and R4 = 97 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is V4?

    2. Relevant equations

    V=IR

    3. The attempt at a solution

    I have been trying to apply Kirchoff's laws for junctions and loops.

    I picked 2 loops and got the following equations-

    I4R4 +I6R6 = V2

    R5I5 + R4I4 + R6I6 + R2I2 + R3I3 = V1

    From the junction rule, I got that I1 + I3 = I5 = I2

    I'm not sure if this is right though- I5 = I2 = I4 = I6

    I don't know what to do next. I think I need another equation.
     

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  3. Feb 13, 2012 #2

    gneill

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    I think you need to step back from the analysis for a moment and ask yourself, "what is determining the voltage across resistors R4 and R5?"
     
  4. Feb 14, 2012 #3

    vela

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    Your picture is so small it's illegible. Can you post a bigger version?
     
  5. Feb 14, 2012 #4
    If you click on it, I think it gets bigger.
     
  6. Feb 14, 2012 #5
    Okay, I got that part. Now, to find I3, can I use my equations and Kirchoff's laws?
     
  7. Feb 14, 2012 #6
    Never mind! Got it! The problem also asks for the potential difference between V(a) and V(b).

    Is this just V=IR where R= R6 and I=I2?
     
  8. Feb 14, 2012 #7

    vela

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    Yes, that's right.
     
  9. Feb 28, 2012 #8
    How'd you figure out I3?
     
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