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Power supply and power dissipated in a circuit

  1. Feb 28, 2012 #1
    Can Power supply be smaller than power dissipated?
    Because I saw this in one question
    attachment.php?attachmentid=44488&stc=1&d=1330433350.jpg
    attachment.php?attachmentid=44487&stc=1&d=1330433350.jpg
    There are 3 cells, the power supply of the two 4V cells cancel each other. (one is -4W and another one is 4W)
    So the total power supply should be 16W

    But in part c, the power dissipated is 4+8+2+6=20W

    Power supply < power dissipated?
     

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  2. jcsd
  3. Feb 28, 2012 #2

    Simon Bridge

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    No. Conservation of energy and all that.
    Not exactly - try the problem again, only this time remove the part with the 6ohm resistor.

    24W is supplied, but 4W is the rate work is being done against the current - leaving 20W of energy for the current to carry. 20W is dissipated. It balances.
    If you built this you'd notice that the "-4W" supply gets hot like all the others.

    You can think of this circuit as three PSUs with internal resistances driving a 6ohm load.
     
  4. Feb 28, 2012 #3

    Simon Bridge

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    Note: treating the bit without the 6Ohm resister as the supply, the Thévenin equivalent would have a voltage of 8V and resistance 2Ohms.

    You'll see this still gives 6W dissipated in the 6Ohm resister, and 2W in the Thévenin resister, while 8W is supplied. Again - everything balances.
     
  5. Feb 28, 2012 #4
    It seems the Thévenin equivalent should have 7.2V with 1.2 ohm resistance.
    attachment.php?attachmentid=44514&stc=1&d=1330494095.jpg
     

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  6. Feb 29, 2012 #5

    Simon Bridge

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  7. Feb 29, 2012 #6
    Using Thevenin equivalents isn't the easiest way here.

    What I did was.

    1. replace the 8v and 4v source by a 12 v source, and the 1ohm and 2ohm resistance that are in series with a 3 ohm resistance

    2. assume the potential at the bottom is 0, and the potential at the top right is U.

    3. work out the currents in the 3 branches: (V-12)/3, (V-4)/2, V/6 (all pointed downwards)

    4. the sum of the currents must be 0 (kirchhof)

    so U/3+U/2+U/6 - 12/3 - 4/2 = 0, so U=6 volt.

    then it's easy to work out the currents with the equations already given.

    The answer sheet is entirely correct, but it doesn't give the power in the top 4V source.
    This power is not equal to 4W, and it doesn't cancel the power of the other 4V source.
    I have no idea why you should think that.

    The same current is going through it as through the 8V source, so the power must be half the power of the 8V source.
     
  8. Feb 29, 2012 #7
    Thanks everyone, I understand it now.
     
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