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Question

An uncharged capacitor of 50μF is connected to a battery of 10V. What amount of energy supplied by the battery is lost as heat while charging the capacitor ?

A
5mJ
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B
2.5mJ
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C
1.25mJ
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D
10mJ
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Solution

The correct option is B 2.5mJ
Given:
C=50μF; 10V

Work done by battery is given by

W=QV=CV2

And, energy stored in capacitor will be

E=12CV2

E=12×50×106×102=2500μJ

Heat dissipated (H) in the connecting wires =12CV2

H=12×50×106×102=2500μJ=2.5mJ
Key concept-Now, work done by battery = Energy stored in the capacitor + Heat dissipated in the connecting wires.

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