In the circuit shown here, C1=6 μ F , C2=3 μ F and battery B=20 V. The switch S1 is first closed. It is then opened and afterwards S2 is closed. What is the charge finally on C2 ?

Question

In the circuit shown here, C1=6 μ F , C2=3 μ F and battery B=20 V. The switch S1 is first closed. It is then opened and afterwards S2 is closed. What is the charge finally on C2 ? (A) 120 μ C (B) 80 μ C (C) 40 μ C (D) 20 μ C

Tardigrade Admin · Accepted Answer

Common potential, V=(6 × 20+3 × 0/(6+3)) =(120/9) volt So, charge on 3 μ F capacitor, Q2=3 × 10-6 × (120/9)=40 μ C

Q. In the circuit shown here, $C_{1} = 6 μ F, C_{2} = 3 μ F$ and battery $B = 20 V$ . The switch $S_{1}$ is first closed. It is then opened and afterwards $S_{2}$ is closed. What is the charge finally on $C_{2}$ ?

$120 μ C$

$80 μ C$

$40 μ C$

$20 μ C$

Common potential, $V = \frac{6 \times 20 + 3 \times 0}{( 6 + 3 )}$
$= \frac{120}{9}$ volt
So, charge on $3 μ F$ capacitor,
$Q_{2} = 3 \times 1 0^{- 6} \times \frac{120}{9} = 40 μ C$

Q. In the circuit shown here, C1​=6μF,C2​=3μF and battery B=20V. The switch S1​ is first closed. It is then opened and afterwards S2​ is closed. What is the charge finally on C2​ ?

120μC

80μC

40μC

20μC