A 400 pF capacitor is charged with a 100 V battery. After disconnecting battery this capacitor is connected with another 400 pF capacitor. Then find out energy loss.
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a
1 x 10-6 J
b
2 x 10-6 J
c
3 x 10-6 J
d
4 x 10-6 J
answer is A.
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Detailed Solution
Energy loss = C1C22C1+C2V1-V22 =400 x 10-124100-02J=1 x 10-6 J