JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Capacitance

A capacitor,

C_{1} = 6 μ F

is charged to a potential difference of

V_{0} = 5 V

using a 5 V battery. The battery is removed and another capacitor,

C_{2} = 12 μ F

is inserted in place of the battery. When the switch ' S ' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges (

q_{1}

and

q_{2}

) on the capacitors

C_{1}

and

C_{2}

when equilibrium condition is reached.

Structure Circuit diagram — Circuit diagram

(A) $q_{1} = 15 μ C, q_{2} = 30 μ C$
(B) $q_{1} = 30 μ C, q_{2} = 15 μ C$
(C) $q_{1} = 10 μ C, q_{2} = 20 μ C$ ✓ Correct
(D) $q_{1} = 20 μ C, q_{2} = 10 μ C$

Answer: (C)

Solution

Before the switch is closed, capacitor

C_{1} = 6 μ F

carries the charge it picked up from the battery:

q_{1}^{'} = C_{1} V_{0} = 6 \times 5 = 30 μ C

C_{2} = 12 μ F

is completely uncharged.

Once the switch closes, the two capacitors are connected in parallel and charge redistributes between them until they reach a common potential

V

. Charge is conserved through this process:

q_{1} + q_{2} = q_{1}^{'} + q_{2}^{'} = 30 μ C

At the common potential,

q_{1} = C_{1} V = 6 V, q_{2} = C_{2} V = 12 V

Substituting into the conservation equation:

6 V + 12 V = 30 ⟹ 18 V = 30 ⟹ V = \frac{5}{3} V

q_{1} = 6 \times \frac{5}{3} = 10 μ C, q_{2} = 12 \times \frac{5}{3} = 20 μ C

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Capacitance

Solution

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