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Q.

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

3 x 10-6 J

b

1 x 10-6 J

c

4 x 10-6 J

d

2 x 10-6 J

answer is A.

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Detailed Solution

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Energy loss = C1C22C1+C2V1-V22

                   =400 x 10-124100-02J=1 x 10-6 J

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