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In the circuit shown here $C_1=6 \mu F, C_2=3 \mu F$ and battery $B=20 \mathrm{~V}$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$
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The correct answer is:
$40 \mu \mathrm{C}$
Common potential $V=\frac{6 \times 20+3 \times 0}{(6+3)}=\frac{120}{9} \mathrm{Voft}$ So, charge on $3 \mu F$ capacitor
$Q_2=3 \times 10^{-6} \times \frac{120}{a}=40 \mu C$
$Q_2=3 \times 10^{-6} \times \frac{120}{a}=40 \mu C$
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