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JEE Advanced 2021 · Paper 1, Physics question 8

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Question Stem for Question Nos. 7 and 8

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μq_1\ \muC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 μq_2\ \muC.

Circuit: a 1 V cell on the left and a 2 V cell on the right, both with the positive terminal at the top and the negative terminal on a common bottom wire. A two-way switch S at the top left connects the top wire either to P (the positive terminal of the 1 V cell) or to Q (a wire straight down to the bottom wire, bypassing the cell). From S the top wire runs through a 1 Ω resistor to a node, then through a 2 Ω resistor to the positive terminal of the 2 V cell. A 1 μF capacitor connects the node to the bottom wire.

The magnitude of q2q_2 is ____.

Answer and worked solution

Official answer: 0.6

Steady state with S at Q. The 1 Ω resistor is now connected straight to the bottom wire, at 0 V. At the node:
0−V1+2−V2=0  ⟹  V=23 V.\frac{0-V}{1}+\frac{2-V}{2}=0\implies V=\frac23\text{ V}.

Charge.
q2=CV=23 μC≈0.67 μC.q_2=CV=\frac23\ \mu\text{C}\approx0.67\ \mu\text{C}.

Topics:Capacitors and capacitanceKirchhoff's laws
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