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

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
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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.