A capacitor $C_1$ is charged up to a voltage $V\, = 60\,V$ by connecting it to battery $B$ through switch $( 1)$, Now $C_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C_2\, = 3.0\,\mu F$ and $C_3\,= 6.0\,\mu F$ through a switch $(2)$ as shown in the figure. The sum of final charges on $C_2$ and $C_3$ is......$\mu C$
JEE MAIN 2018, Diffcult
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$Q_{1}=60\, \mu C$
Equivalent capcitance of $C_{2}$ and $C_{3}$
$C^{\prime}=\frac{C_{2} C_{3}}{C_{2}+C_{3}}=\frac{3 \times 6}{3 \times 6}\, \mu F=2\, \mu F$
Common potential diference to $C_{1}$ and $C^{\prime}$ combined.
$v^{\prime}=\frac{C_{1} v_{1}+0}{C_{1}+C^{\prime}}=\frac{60}{1+2}=3$ volts
Charge of $C_{2}, C_{3}$ system $=C^{\prime} v=2 \times 20=40\, \mu C$
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