Three charged capacitors, $C_1$ = $17\ μF$, $C_2$ = $34\ μF$, $C_3$ = $41\ μF$and two open switches, $S_1$ and $S_2$ are assembled into a network with initial voltages and polarities, as shown. Final status of the network is attained when the two switches, $S_1$ and $S_2$ are closed. In the figure, the final charge on capacitor $C_3$ in $mC$, is closet to:
Diffcult
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Circuit is short when both switch $\mathrm{S}_{1} $ and $ \mathrm{S}_{2}$ is closed
So $\mathrm{V}_{\mathrm{C}_{3}}=0$
$\mathrm{q}=0$
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