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JEE Advanced 2021 · Paper 1, Physics question 7

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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 q1q_1 is ____.

Answer and worked solution

Official answer: 1.3

Steady state with S at P. In the steady state no current flows into the capacitor. Let the node above the capacitor be at potential VV. Both cells have their positive terminal on the top wire, so Kirchhoff's current law at the node gives
1−V1+2−V2=0  ⟹  V=43 V.\frac{1-V}{1}+\frac{2-V}{2}=0\implies V=\frac43\text{ V}.

Charge.
q1=CV=1 μF×43 V≈1.33 μC.q_1=CV=1\,\mu\text{F}\times\frac43\text{ V}\approx1.33\ \mu\text{C}.

Topics:Capacitors and capacitanceKirchhoff's laws
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