Capacitor question


    Rep:
    Question:


    I'm facing a problem in part (b). Half of the charge is equal to 0.06 C - so ½QV(½ times 0.06 times 24 = 0.72 J) should give the correct value of energy dissipated in the bulb. But the mark scheme gives 1.08 J as the answer. Where am I going wrong?

    Rep:
    (Original post by Zishi)
    Question:


    I'm facing a problem in part (b). Half of the charge is equal to 0.06 C - so ½QV(½ times 0.06 times 24 = 0.72 J) should give the correct value of energy dissipated in the bulb. But the mark scheme gives 1.08 J as the answer. Where am I going wrong?
    How to do this...
    Initial energy on capacitor is 0.5CV2 where V is 24V
    After half the charge leaves, the voltage will be 12V (V=Q/C)
    New energy in capacitor is 0.5CV2 where V=12V
    Subtract the 2nd from the 1st for the energy in the bulb

    Rep:
    (Original post by Stonebridge)
    How to do this...
    Initial energy on capacitor is 0.5CV2 where V is 24V
    After half the charge leaves, the voltage will be 12V (V=Q/C)
    New energy in capacitor is 0.5CV2 where V=12V
    Subtract the 2nd from the 1st for the energy in the bulb
    Alright. So the voltage drops to half because the charge drops to half (according to Q=CV)?

    Rep:
    (Original post by Zishi)
    Alright. So the voltage drops to half because the charge drops to half (according to Q=CV)?
    Absolutely. The pd on a capacitor is directly proportional the the amount of charge on it.

    Rep:
    (Original post by Stonebridge)
    Absolutely. The pd on a capacitor is directly proportional the the amount of charge on it.
    Thank you so much. (PRSOM)

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: April 19, 2012

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