Four capacitors of capacitance 10, mu F and a | vedclass

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(A) When the switch $S$ is open,the three capacitors on the left are in parallel,giving $C_p = 10 + 10 + 10 = 30\, \mu F$. This combination is in seri

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$4.5 	imes 10^{-3}\, C$

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(A) When the switch $S$ is open,the three capacitors on the left are in parallel,giving $C_p = 10 + 10 + 10 = 30\, \mu F$. This combination is in series with the fourth capacitor $(10\, \mu F)$.Equivalent capacitance $C_{eq} = \frac{30 	imes 10}{30 + 10} = \frac{300}{40} = 7.5\, \mu F$.Initial charge $q_i = C_{eq} V = 7.5\, \mu F 	imes 200\, V = 1500\, \mu C$.When the switch $S$ is closed,the fourth capacitor is short-circuited. Only the three parallel capacitors remain in the circuit.Equivalent capacitance $C_{eq}' = 10 + 10 + 10 = 30\, \mu F$.Final charge $q_f = C_{eq}' V = 30\, \mu F 	imes 200\, V = 6000\, \mu C$.Charge flowing through $AB$ is $\Delta q = q_f - q_i = 6000\, \mu C - 1500\, \mu C = 4500\, \mu C = 4.5 	imes 10^{-3}\, C$.

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?

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