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What happens to the total power dissipated in the circuit ?

  1. Apr 9, 2016 #1
    1. The problem statement, all variables and given/known data
    What happens to the total power dissipated in the circuit once the switch is closed?
    tmp_3989-20160409_2113301282769602.jpg


    2. Relevant equations
    Pr=Pbat=dEth/dt=(dq/dt)*Vr=I*Vr



    3. The attempt at a solution
    V total and I toral remain the same so the power dissipated by the circuit stays the same...?
     
  2. jcsd
  3. Apr 9, 2016 #2

    phyzguy

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    Why do you think Itotal remains the same? What is the resistance across the battery with the switch open and with it closed?
     
  4. Apr 9, 2016 #3

    gneill

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    Can you show some equations to back up your guess?
     
  5. Apr 9, 2016 #4
    Well....for the current, it's current conservation. Switch open: I total goes through A, B and C ( they're in series) and then goes back to the battery. Switch closed: I total goes through A then B then splits into let's say I1( through the middle resustless wire) and I2 ( through C) at the upper junction. Then it becomes I total again at the lower junction. I think the closed switch short-circuits C but I am not sure this means I2 is null....As for the voltages, I think the voltage across C has to decrease once the switch is closed ( V=IR, I decreases; V decreases). But the total voltage has to stay the same ( equal to the voltage provided by the battery) so the voltages across A and B have to increase...but that would mean that the current through A and B increases.....?!?
     
  6. Apr 9, 2016 #5

    Pr=Pbat=dEth/dt=(dq/dt)*Vr=I*Vr
    I got his from my textbook. We haven't learned it in class yet, so I am confused about it and the textbook does not help..
     
  7. Apr 9, 2016 #6

    cnh1995

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    What happens to the total resistance of the circuit when the switch is closed?
     
  8. Apr 9, 2016 #7

    cnh1995

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    Right.
     
  9. Apr 9, 2016 #8
    But how is that possible??? A and B are right after the positive terminal of the battery so no matter the situation they should receive I total...?
     
  10. Apr 9, 2016 #9

    cnh1995

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  11. Apr 10, 2016 #10

    phyzguy

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    Of course A and B receive all of Itotal, but Itotal is different when the switch is closed than when it is open.
     
  12. Apr 10, 2016 #11
    l think I see. Itotal=Vtotal*Req. Switch open: Req =3R (the bulbs are identical and in series). Switch closed: Req is smaller than 3R (C is now in parallel with A and B). So I is smaller when the switch is closed. Correct?
     

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  13. Apr 10, 2016 #12

    cnh1995

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    Yes.
    No. C is eliminated from the circuit. Look up the term "short circuit".
    Are you sure? Equivalent resistance decreased after closing the switch.
     
  14. Apr 10, 2016 #13

    cnh1995

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    I=V/R and not V*R.
     
  15. Apr 10, 2016 #14
    True! So the current increases.
     
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