A 10,mu F capacitor and a 20,mu F capacitor a | vedclass

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(B) Initially,when connected in series,the charge $Q$ on each capacitor is the same.Equivalent capacitance $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac

Vedclass · Accepted Answer

$\frac{800}{9}\,V$

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(B) Initially,when connected in series,the charge $Q$ on each capacitor is the same.Equivalent capacitance $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{10 	imes 20}{10 + 20} = \frac{200}{30} = \frac{20}{3}\,\mu F$.The total charge $Q = C_{eq} V = \frac{20}{3} 	imes 200 = \frac{4000}{3}\,\mu C$.The potential difference across $C_1$ is $V_1 = \frac{Q}{C_1} = \frac{4000/3}{10} = \frac{400}{3}\,V$.The potential difference across $C_2$ is $V_2 = \frac{Q}{C_2} = \frac{4000/3}{20} = \frac{200}{3}\,V$.When reconnected in parallel with positive plates together,the total charge $Q_{total} = Q_1 + Q_2 = \frac{4000}{3} + \frac{4000}{3} = \frac{8000}{3}\,\mu C$ (since both capacitors were charged with the same magnitude of charge $Q$ in series).The common potential $V = \frac{Q_{total}}{C_1 + C_2} = \frac{8000/3}{10 + 20} = \frac{8000}{3 	imes 30} = \frac{800}{9}\,V$.

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