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Q.

A capacitor of 5 μF is charged to a potential of 100 V. Now, this charged capacitor is connected to a battery of 100 V with the positive terminal of the battery connected to the negative plate of the capacitor. For the given situation, mark the correct statement(s). 

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a

Work done on the battery is 0.1 J.

b

Heat dissipated in the circuit is 0.1 J. 

c

The charge flowing through the 100 V battery is 1000 μC.

d

The charge flowing through the 100 V battery is 500 μC. 

answer is B, C.

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Detailed Solution

Initial condition just after the connection of battery:

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Condition after a long time: 

It means battery supplies 1000 μC charge from its positive terminal and an equal an opposite charge enters from its negative terminal, i.e., charge flow through battery is 1000 μC. Work done by the battery is

Wbattery =100V×103μC=0.1J

From energy conservation law, Ui+Wbattery =Uf+ΔH

Ui=Uf so ΔH=0.1J

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A capacitor of 5 μF is charged to a potential of 100 V. Now, this charged capacitor is connected to a battery of 100 V with the positive terminal of the battery connected to the negative plate of the capacitor. For the given situation, mark the correct statement(s).