[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?

[Stonebridge]

(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

[Zishi]

(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)?

[Stonebridge]

(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.

[Zishi]

(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)