Three charged capacitors,C_1 = 17 mu F,C_2 = | vedclass

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(A) When switches $S_1$ and $S_2$ are closed,the plates of the capacitors are connected in a way that forms a closed loop. Let the common potential at

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(A) When switches $S_1$ and $S_2$ are closed,the plates of the capacitors are connected in a way that forms a closed loop. Let the common potential at the nodes be $V$. Applying the principle of conservation of charge at the isolated junction formed by the plates of $C_1$,$C_2$,and $C_3$ connected by the switches: The initial charges on the plates connected to the central node are: $q_1 = -C_1 V_1 = -17 \ \mu F 	imes 60 \ V = -1020 \ \mu C$ $q_2 = +C_2 V_2 = +34 \ \mu F 	imes 90 \ V = +3060 \ \mu C$ $q_3 = +C_3 V_3 = +41 \ \mu F 	imes 50 \ V = +2050 \ \mu C$ Sum of charges $Q_{total} = q_1 + q_2 + q_3 = -1020 + 3060 + 2050 = 4090 \ \mu C$. The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 + C_3 = 17 + 34 + 41 = 92 \ \mu F$. The final common potential $V = Q_{total} / C_{eq} = 4090 \ \mu C / 92 \ \mu F \approx 44.46 \ V$. The final charge on $C_3$ is $q_3' = C_3 	imes V = 41 \ \mu F 	imes 44.46 \ V \approx 1822.86 \ \mu C \approx 1.82 \ mC$. However,re-evaluating the circuit diagram,when $S_1$ and $S_2$ are closed,$C_3$ is effectively short-circuited or placed in a configuration where its potential difference becomes $0 \ V$ due to the direct connection of its plates. Thus,the charge on $C_3$ becomes $0 \ mC$.

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