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Finding the voltage in a non- series/parallel circuit

  1. Jul 14, 2012 #1

    JJBladester

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    1. The problem statement, all variables and given/known data

    Find VAB and the power supplied by the source.

    circuit_problem.JPG

    2. Relevant equations

    [itex]\sum V=0[/itex] around a closed loop
    [tex]V_x=\left (\frac{R_x}{R_T} \right )E[/tex]
    [tex]I_x=\left (\frac{R_T}{R_x} \right )I[/tex]

    3. The attempt at a solution

    problem_5.jpg

    I end up getting the correct equivalent resistance (40.64 Ω), source current (1.23 A), and source power (61.5 W). However, my result for VAB is .096 V but according to my circuit simulation software it should be 29.73 V. Where did I go wrong?
     
  2. jcsd
  3. Jul 14, 2012 #2

    lewando

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    Why are you shorting out A and B? In the circuit you have shown, R4 is just dangling out there, not dissipating any power.
     
  4. Jul 14, 2012 #3

    CWatters

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    I haven't checked your attempt but did you notice that when calculating Vab you can:

    1) ignore R1 (eg remove it because it's parallel with V1).
    2) You can also short R4 because the current in R4 = 0

    Redraw the circuit and you end up with a v. simple two resistor circuit with just R2 and R3 as a potential divider..

    Vab = V1 * R3/(R2+R3)

    = 50 * 220/(150+220)

    = 29.73 V
     
  5. Jul 14, 2012 #4

    CWatters

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    To calculate the power apply KCL to the battery positive node.

    Hint: Calculate the current flowing in R3 from Vab/R3.
     
  6. Jul 14, 2012 #5
    No current flows through R4 so Vab is the same as the V across R3.
    R2 and R3 form a potential divider
     
  7. Jul 14, 2012 #6

    JJBladester

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    There was no reason for me to short A and B. It was my mistake. It appears that R4 is a red herring in this problem.

    I think I have gotten to the solution. After thinking about what you both said, it was a much easier route than what I was attempting to do. The big aha moment was the realization that R4 was not contributing as it was "dangling" on the original circuit.

    problem_5_revised.jpg
     
  8. Jul 14, 2012 #7

    CWatters

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    I would do it this way..

    Apply KCL at the battery positive node..

    Ibat + IR1+IR2 = 0
    or
    - Ibat = IR1+IR2

    Note that with R4 open circuit then..
    IR2=IR3=VR3/R3=29.73/220= 135mA

    As R1 is in parallel with the 50V supply then..
    IR1 = 50/47= 1.064A

    Substitute these two values into the above

    -Ibat = 1.064 + 0.135
    = 1.199A call it -1.2A

    Power = V * I
    = 50 * -1.2
    = -60W
     
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