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(C) When the switch $S$ is open,the capacitors are in series with the $9 \ V$ battery. The equivalent capacitance is $C_{eq} = \frac{3 \mu F 	imes 6

Vedclass · Accepted Answer

$27 \mu C$

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(C) When the switch $S$ is open,the capacitors are in series with the $9 \ V$ battery. The equivalent capacitance is $C_{eq} = \frac{3 \mu F 	imes 6 \mu F}{3 \mu F + 6 \mu F} = 2 \mu F$. The charge on each capacitor is $q = C_{eq}V = 2 \mu F 	imes 9 \ V = 18 \mu C$. The potential at $X$ is $V_X = 9 \ V - \frac{18 \mu C}{3 \mu F} = 3 \ V$. The potential at $Y$ is $V_Y = 0 \ V$ (assuming the negative terminal is at $0 \ V$).When the switch $S$ is closed,the circuit becomes two parallel branches. The left branch has a $3 \mu F$ capacitor in series with a $3 \ \Omega$ resistor,and the right branch has a $6 \mu F$ capacitor in series with a $6 \ \Omega$ resistor. The potential at $X$ becomes $9 \ V$ because it is connected to the positive terminal through the switch. The potential at $Y$ is $0 \ V$. The charge on the $3 \mu F$ capacitor is $q_1 = 3 \mu F 	imes 9 \ V = 27 \mu C$. The charge on the $6 \mu F$ capacitor is $q_2 = 6 \mu F 	imes 9 \ V = 54 \mu C$. The total charge at node $X$ before closing was $0$ (net charge on the plates connected to $X$). After closing,the charge on the plates connected to $X$ is $-(27 \mu C + 54 \mu C) = -81 \mu C$. The charge that flows from $Y$ to $X$ is the change in charge on the plates connected to $X$,which is $27 \mu C$.

$A$ circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed,the total amount of charge that flows from $Y$ to $X$ is

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