QJEEAdv21Phbca1: JEE Advanced 2021 Physics - Class 12 JEEAdv21Phbca1 Solution: In the circuit shown below, the switch S is connected to…

Question

Verified Solution • 2025–26 Session moderate

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

\mu

C. Then S is switched to position Q. After a long time, the charge on the capacitor is q₂

\mu

C.

The magnitude of q₁ is ________________.

Accepted Answer

Switch connected to position P

{V_A} - 1\,.\,{i_1} - 1 + 2 - 2{i_1} = {V_A}

3{i_1} = 1

{i_1} = {1 \over 3}A

Now,

{V_A} - 1\,.\,{i_1} - 1 = {V_B}

{V_A} - {V_B} = 1 + {i_1} = {4 \over 3}V

Potential drop across capacitor,

\Delta V = {4 \over 3}V

\therefore

Charge on capacitor,

{q_1} = C\Delta V = 1 \times {4 \over 3}\mu C

{q_1} = 1.33\mu C

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