JEE Main 2022 · 26 Jun · Shift 1 · Physics Q40

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PhysicsSingle correct+4 / -1Question 40 of 90

Two capacitors having capacitance $C_1$ and $C_2$ respectively are connected as shown in figure. Initially, capacitor $C_1$ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor $C_1$ is now connected to uncharged capacitor $C_2$ by closing the switch S. The amount of charge on the capacitor $C_2$ , after equilibrium, is :

Capacitors C1 and C2 in a single loop with an open switch S in the top wire.

\dfrac{C_1C_2}{(C_1+C_2)}V

\dfrac{(C_1+C_2)}{C_1C_2}V

(C_1+C_2)V

(C_1-C_2)V

Answer and worked solution

Official answer: A

Charge is shared until both have the same voltage: $V'=\dfrac{C_1V}{C_1+C_2}$ .

Charge on $C_2$ . $Q_2=C_2V'=\dfrac{C_1C_2}{C_1+C_2}V$ .

(A)

Topics:Series and parallel combinations Capacitors and capacitance

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JEE Main 2022 · 26 Jun · Shift 1, Physics question 40