![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](https://www.gkseries.com/blog/wp-content/uploads/2023/08/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.jpg)
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.
![](https://www.gkseries.com/blog/wp-content/uploads/2023/08/image-34.png)
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