The capacitor C1 in the circuit shown has a voltage of 20

The capacitor C₁ in the circuit shown has a voltage of 20 V across it before the switch S is closed at time t = 0. C₁ = 2 µF and C₂ = 3 µF. The voltage across C₂ after S is closed is—

Initial charge on C₁ = 2 × 10^{– 6} × 20 = 40 × 10^{– 6} C Since charge remain the same so,
C = C₁ + C₂ = 5 µF

V=	Q	+ 10 =	40 × 10^–6	= 8 V
	C		5 × 10^–6

Alternate method
q₀ = C₁ V₀
q₀ = q₁ + q₂

V=	C	V₀ =	2	× 20 = 8 V
	C₁ + C₂		5