A one microfarad capacitor of a TV is subjected to 4000 V potential difference. The energy stored in capacitor is :
8J
16 J
4×10−3J
2×10−3J
A
2×10−3J
B
8J
C
16 J
D
4×10−3J
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Solution
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Given, capacitor C=10−6F and potential difference V=4000V The energy stored in capacitor =12CV2=12×10−6×(4000)2=8J
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