Question

A capacitor $C_1=1.0\mu F$ is charged up to a voltage $V=60V$ 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:

• A. $40\mu C$
• B. $36\mu C$
• C. $20\mu C$
• D. $54\mu C$

Answer 1

The correct option is A $40\mu C$

$Q_1=60\mu C$

Equivalent capacitance of $C_{2}and{C}_3$

$C^{\prime }=\frac{C_2{C}_3}{C_2+C_3}=\frac{3\times 6}{3+6}\mu F=2\mu F$

Common potential differfence to $C_{1}and{C}^{\prime }$ combined,

V' = $\frac{C_1{V}_1}{C_1+C^{\prime }}=\frac{60}{1+2}=3volts$

Charge of $C_2,C_3$ system = C' v = $2\times 20=40\mu C$