In the circuit shown here C_1 = 6,mu F,C_2 = | vedclass

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(C) $1$. When switch $S_1$ is closed,capacitor $C_1$ gets charged by the battery $B = 20\,V$. The charge on $C_1$ is $Q_1 = C_1 	imes V = 6\,\mu F

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) $1$. When switch $S_1$ is closed,capacitor $C_1$ gets charged by the battery $B = 20\,V$. The charge on $C_1$ is $Q_1 = C_1 	imes V = 6\,\mu F 	imes 20\,V = 120\,\mu C$.$2$. When $S_1$ is opened,the charge $Q_1 = 120\,\mu C$ remains stored on $C_1$. Capacitor $C_2$ is initially uncharged $(Q_2 = 0)$.$3$. When switch $S_2$ is closed,the charge $Q_1$ is shared between $C_1$ and $C_2$ until they reach a common potential $V'$.$4$. The common potential $V'$ is given by $V' = \frac{Q_{total}}{C_{total}} = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{120\,\mu C + 0}{6\,\mu F + 3\,\mu F} = \frac{120}{9}\,V = \frac{40}{3}\,V$.$5$. The final charge on $C_2$ is $Q_2' = C_2 	imes V' = 3\,\mu F 	imes \frac{40}{3}\,V = 40\,\mu C$.

In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ 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 $C_2$ in $\mu C$?

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