Two capacitors of capacitances 1 mu F and 2 | vedclass

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(B) Initial charges on the capacitors are:$q_1 = C_1 V_1 = 1 \ \mu F 	imes 20 \ V = 20 \ \mu C$$q_2 = C_2 V_2 = 2 \ \mu F 	imes 15 \ V = 30 \ \mu C$

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$\frac{10}{3} \ \mu C$

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(B) Initial charges on the capacitors are:$q_1 = C_1 V_1 = 1 \ \mu F 	imes 20 \ V = 20 \ \mu C$$q_2 = C_2 V_2 = 2 \ \mu F 	imes 15 \ V = 30 \ \mu C$When terminals $B$ and $C$ are connected and $A$ and $D$ are connected,the capacitors are connected in parallel with opposite polarities.The net charge in the circuit is $q_{net} = |q_2 - q_1| = |30 \ \mu C - 20 \ \mu C| = 10 \ \mu C$.The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 = 1 \ \mu F + 2 \ \mu F = 3 \ \mu F$.The common potential difference is $V_{com} = \frac{q_{net}}{C_{eq}} = \frac{10 \ \mu C}{3 \ \mu F} = \frac{10}{3} \ V$.The final charge on the $1 \ \mu F$ capacitor is $q_1' = C_1 V_{com} = 1 \ \mu F 	imes \frac{10}{3} \ V = \frac{10}{3} \ \mu C$.

Two capacitors of capacitances $1 \ \mu F$ and $2 \ \mu F$ are charged to potential differences $20 \ V$ and $15 \ V$ as shown in the figure. If terminals $B$ and $C$ are connected together and terminals $A$ and $D$ are connected together,then find the final charge on the $1 \ \mu F$ capacitor.

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