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Question

Consider the situation shown in figure, the switch S open for a long time and then closed. Then work done by the battery will be:

A
CE22
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B
CE24
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C
2CE2
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D
CE28
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Solution

The correct option is A CE22
When the switch is open, the capacitors are in series. So equivalent capacitance will be

Ceq=C×CC+C=C2

Charge flown by battery is

q=CeqV

q=C2E

Thus, the same charge will be appearing on both capacitors.


After the switch S is closed, the potential difference across second capacitors becomes zero instantly. Therefore, it will have zero charge or no charge.


Potential of point A and B is the same, since they lie on the same conducting wire.

VA=VB

ΔV=0

Charge on first capacitor is,

q=CVq=CE

Therefore, charge supplied (q'') by battery will be,

q′′=qq

q′′=CECE2

q′′=CE2

Work done by battery is,

W=(q′′)×E

W=CE2×E=CE22

Hence, option (a) is correct.

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