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JEE Advanced 2021 Paper 1

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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}

\mu \{C\}

. Then

{S}

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

{q}_{2} \mu {C}

Section: Physics

Question : 7 of 57

Marks: +1, -0

The magnitude of

q_{1}

Solution:

When switch is connected to position

P

From KVL,

V_{A}-1 \cdot i_{1}-1+2-2 i_{1}=V_{A} \Rightarrow 3 i_{1}=1

\therefore \;\; i_{1}=\frac{1}{3}\ \mathrm{A}

Again

V_{A}-1 \cdot i_{1}-1=V_{B}

or,

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

Potential drop across capacitor

\Delta V=\frac{4}{3}\ \mathrm{V}

\therefore \;\;

Charge on capacitor,

q_{1}=C \Delta V=1 \times \frac{4}{3}\ \mu\mathrm{C}

q_{1}=1.33\ \mu\mathrm{C}

When switch is connected to position

Q

From KVL,

V_{A}-1 \cdot i_{2}+2-2 i_{2}=V_{A} \Rightarrow 3 i_{2}=2

\therefore \;\; i_{2}=\frac{2}{3}\ \mathrm{A}

Again,

V_{A}-i_{2} \times 1=V_{B}

V_{A}-V_{B}=i_{1} \times 1=\frac{2}{3}\ \mathrm{V}

Potential difference across capacitor

\Delta V=\frac{2}{3}\ \mathrm{V}

\therefore \;\;

Charge on capacitor,

q_{2}=C \Delta V=1 \times \frac{2}{3}=0.67\ \mu\mathrm{C}

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