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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 CE22When 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′=CV⇒q′=CE Therefore, charge supplied (q'') by battery will be, q′′=q′−q ⇒q′′=CE−CE2 ⇒q′′=CE2 Work done by battery is, W=(q′′)×E ∴W=CE2×E=CE22 Hence, option (a) is correct.

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