In the circuit shown below, initially the switch S1 is open, the capacitor C1 has a charge of 6 coulomb, and the capacitor C2 has 0 coulomb

Q. In the circuit shown below, initially the switch S1 is open, the capacitor C1 has a charge of 6 coulomb, and the capacitor C2 has 0 coulomb. After S1 is closed, the charge on C2 in steady state is coulomb.

Sol:

When switch open,

Q1 = 6 C, Q2 = 0 C, C1 = 1 F, C2 = 2 F

Q1 = C1V1

⇒ V1 = 6 V

Q2 = C2V2

⇒ V2 = 0 V

Equivalent capacitance,

Ceq = C1 + C2 = 3 F

By law of conservation of charge

Ceq Veq = V1 C1 + C2 V2

⇒ (3) (Veq) = (1) (6) + (2) (0)

⇒ Veq = 2 V

After switch closed, Q2 = C2 Veq = (2) (2) = 4 C