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Power dissipated by a resistor in a circuit

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data

    Screen Shot 2015-02-05 at 11.23.06 AM.png

    2. Relevant equations

    P = V^2 / R
    1/Rt = sum(1/Ri) (parallel)

    3. The attempt at a solution

    Since all the answer choices express the solution in terms of V^2 / R, I tried to find the voltage drop across resistor #2 to find the power. I collapsed the two parallel resistors 2 and 3 into one and found their effective resistance to be R/2 (parallel formula). Using Kirchhoff's voltage law,

    V = IR + I(R/2) = 3IR / 2 where I is the total current in the circuit

    Since V = 3IR/2, I can rewrite the voltage drop across the collapsed resistors as (⅓)V. Therefore the voltage drops across resistors #2 and #3 are both (⅓)V since they are in parallel. Plugging into the power formula,

    P = (1/3V)^2 / R = V^2 / 9R, answer choice D. But the answer key says the solution is B. Why?
     
  2. jcsd
  3. Feb 5, 2015 #2

    TSny

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    Your reasoning looks correct to me. I agree that D is the correct answer. Apparently the answer key is wrong.
     
  4. Feb 5, 2015 #3

    CWatters

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    I agree.
     
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