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Trying to apply superposition method to circuit, not working

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data

    A double loop circuit is shown in the figure below.

    FVTMCtE.png
    The Emf mimetex.gif x = 6.00 V and the Emf mimetex.gif w = 19.00 V. Both are shown on the diagram. Calculate the power dissipated in the 6 Ω resistor located on the extreme right in the circuit.


    2. Relevant equations

    V=IR, P=VI

    3. The attempt at a solution

    I tried using superposition, where I replaced one battery with a short and solved for the voltage and current through the 6 ohm resistor, and the did the same by replacing the other battery with a short. The total current and voltage is the sum of the individual values. I plugged those final values into P=VI and got the wrong answer. Should this method work and I just did the work wrong or is there something I'm not aware of?
     
  2. jcsd
  3. Oct 19, 2015 #2

    phinds

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    2016 Award

    Sounds like you did it right but without your posting your work it's impossible for us to see where you might have gone wrong.
     
  4. Oct 20, 2015 #3

    rude man

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    You cannot add power by superposition. You can compute current I by superposition:
    I = I1 + I2.
    Then P = I^2 R.
    Or you could have computed V1 and V2 by superposition, then P = VI = (V1 + V2)(V1 + V2)/R.
     
  5. Oct 20, 2015 #4

    phinds

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    DOH ! I assumed that's what he was doing but on rereading I see I just glossed over what he really said. Good catch.
     
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