JEE Main 2025 Jan 29 Shift 2, Physics Q14: Capacitors And Capacitance

JEE Main2025Jan 29, Shift 2Physics

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.

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
D
$q_{1} = 20 μ$ C, $q_{2} = 10 μ$ C

Solution

Initially $C_{1}$ carries charge $Q_{0} = C_{1} V_{0} = 6 \times 5 = 30 μ$ C; $C_{2}$ is empty.

After the switch closes, the two capacitors are connected in parallel and share a common potential $V$ . Charge is conserved on the isolated network: $C_{1} V + C_{2} V = Q_{0}$ , i.e. $(6 + 12) V = 30 \Rightarrow V = \frac{5}{3}$ V.

$q_{1} = C_{1} V = 6 \cdot \frac{5}{3} = 10 μ$ C, $q_{2} = C_{2} V = 12 \cdot \frac{5}{3} = 20 μ$ C.

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JEE Main 2025 Jan 29 Shift 2, Physics Q14: Capacitors And Capacitance

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