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(B) Let the capacitance of each capacitor be $C = 10\,\mu F$ and battery voltage be $V = 200\, V$.Case $1$: Switch $S$ is open.The circuit consists of

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$4500$

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(B) Let the capacitance of each capacitor be $C = 10\,\mu F$ and battery voltage be $V = 200\, V$.Case $1$: Switch $S$ is open.The circuit consists of three capacitors in parallel connected in series with one capacitor. The equivalent capacitance is $C_{eq1} = \frac{(3C) 	imes C}{3C + C} = \frac{3C^2}{4C} = \frac{3}{4}C = \frac{3}{4} 	imes 10 = 7.5\,\mu F$.The initial charge drawn from the battery is $q_i = C_{eq1} V = 7.5 	imes 200 = 1500\,\mu C$.Case $2$: Switch $S$ is closed.The capacitor in parallel with the switch gets short-circuited. The circuit now consists of three capacitors in parallel. The equivalent capacitance is $C_{eq2} = 3C = 3 	imes 10 = 30\,\mu F$.The final charge drawn from the battery is $q_f = C_{eq2} V = 30 	imes 200 = 6000\,\mu C$.The charge that flows through $AB$ is the change in charge drawn from the battery: $\Delta q = q_f - q_i = 6000 - 1500 = 4500\,\mu C$.

Four capacitors of capacitance $10\,\mu F$ and a battery of $200\, V$ are arranged as shown. How much charge will flow through $AB$ after the switch $S$ is closed? (in $\mu C$)

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