The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Capacitor question

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?
(edited 12 years ago)
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
Reply 2
Avatar for SalBhatti
SalBhatti
OP
13
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)?
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.
Reply 4
Avatar for SalBhatti
SalBhatti
OP
13
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)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you