A capacitor having capacity of 2.0μF is charged to 200V and then the plates of the capacitors are connected to a resistance wire. The heat produced in joule will be :
2×10−2
4×10−2
4×104
4×1010
A
4×10−2
B
2×10−2
C
4×104
D
4×1010
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Solution
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Heat produce in a wire is equal to energy stored in capacitor. H=12CV2 =12×(2×10−6)×(200)2 =10−6×200×200 =4×10−2J.
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