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Question

A 10 μF capacitor charged to 20 V is connected across a 15 V battery through a switch such that positive terminal of battery is connected to negative charged plate of capacitor. When switch is closed, work done by battery of 15 V in long time will be
  1. 2250 μJ
  2. 3000 μJ
  3. 30000 μJ
  4. 5250 μJ


Solution

The correct option is D 5250 μJ

Intial charge Q on capacitor is Q=CV=10×20=200 μC

On closing switch final charge on capacitor is q=Cε=10×15 μC=150 μC
So charge flown through battery is Δq=q+Q=350 μC
So work of battery is W=εΔq
W=15×350 μJ=5250 μJ

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