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