A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is ?
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2 Answers
283we can replace a charged capacitor by uncharged capacitor with a batter
So there will be two uncharged capacitor of Ceq= 3C
and two battery placed in opposite polarity so εnet/ss]= 2V -V = V
so Energy = 1/2 * 3C*V2
1161After connection, total charge will be 5CV, So final distribution of charge on capacitors will be: 5CV/3 and 10CV/3 and potential drop across any capacitor will ve 5V/3. Now you can find energy easily which will be: (25/6) CV2.
- Sahil Jain Sir can you please check the ans againUpvote·0· Reply ·2014-06-16 02:43:54
- Akash Anand Sorry..my mistake. This would be the answer when positive plates of both capacitor would be connected.
For this case:
After connection, total charge will be 3CV, So final distribution of charge on capacitors will be: CV and 2CV and potential drop across any capacitor will ve V. Now you can find energy easily which will be: (3/2) CV^2.
