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 $\mathrm{q}_{1} \mu$C. Then S is switched to position Q. After a long time, the charge on the capacitor is $\mathrm{q}_{2} \mu$C. The magnitude of $\mathrm{q}_{1}$ is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Current Electricity

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

q_{1} μ

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

q_{2} μ

C.

The magnitude of

q_{1}

is ____.

Answer: 1.33

Solution

When switch is at position P, in steady state no current flows through the capacitor.

Applying KVL in the left loop:

V_{A} - 1 \cdot i_{1} - 1 + 2 - 2 i_{1} = V_{A}

, giving

3 i_{1} = 1

, so

i_{1} = \frac{1}{3}

A.

Potential difference:

V_{A} - V_{B} = 1 \cdot i_{1} + 1 = \frac{1}{3} + 1 = \frac{4}{3}

V.

Charge on capacitor:

q_{1} = C Δ V = 1 \times \frac{4}{3} = 1.33

μC.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer

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