Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

Capacitor question

Announcements Posted on
Rate your uni — help us build a league table based on real student views 19-08-2015
  1. Offline

    ReputationRep:
    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?
  2. Offline

    (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
  3. Offline

    ReputationRep:
    (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)?
  4. Offline

    (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.
  5. Offline

    ReputationRep:
    (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)

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: April 19, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

New on TSR

Rate your uni

Help build a new league table

Poll
How do you read?
Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.