In the circuit shown in figure (A), the capacitors and batteries are ideal and circuit is at steady state. The capacitors C1, C2 and C3 are disconnected from the circuit shown in figure (A) with their charges intact and reconnected as in figure (B) with polarities as indicated. Initially charges on capacitor C1, C2 & C3 before connecting to the circuit A was zero on each

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On closing switches S1 and S2 the potential difference (Vd – Ve) between the points d and e becomes 2.0 V

The charge on the plate C3 connected to e after closing the switches is 6μC

In figure A, the work done by 2 V battery is 50μJ

The work done by the 10 V battery in charging the capacitors is 90μJ

answer is A.

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Detailed Solution

KVL, +2−(x−y)2+y3=0and 10−x−y3=0After Solving, x=9μC and y=3μCCharge passing through 10V battery is 9μC∴Work done by this battery=qV=90μJCharge passing through 2V battery is 6μC∴Work done by this battery=qV=12μJCharge conservation,+3−6−9=−q1−q2+q3All the three capacitors are in parallel connection V is same ,q1C1=q2C2=−q3C3⇒q11=q22=−q33After solving, ⇒q1=+2μC, q2=+4μC, q3=−6μCVd-Ve=42=2V

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