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 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

Answer: (A)

Given: $C_1=1.0\mu F;C_2=3.0\mu F;C_3=6.0\mu F$

Capacitance $C_2$ and $C_3$ are in series, their equivalent capacitance is
$\frac{1}{C}=\frac{1}{C_2}+\frac{1}{C_3}$
$\Rightarrow \frac{1}{C}=\frac{C_2+C_3}{C_2{C}_3}$
$\Rightarrow C=\frac{C_2{C}_3}{C_2+C_3}=\frac{3.0\mu F\times 6.0\mu F}{3.0\mu F+6.0\mu F}=2\mu F$
Potential on $C_1$ when battery is connected to switch (1) is
$C_1\times 60V_{\text{out }}(1)$
Potential when battery is connected to switch (2) is
$\left(C_1+C\right)V^{\prime }(2)$
Equating Eqs. (1) and (2), we get
$\left(C_1+C\right)V^{\prime }=60C_1$
$\Rightarrow V^{\prime }=\frac{60C_1}{C_1+C}$
$\Rightarrow V^{\prime }=\frac{60\times 1\mu F}{1\mu F+2\mu F}=\frac{60\times 1\mu F}{3\mu F}=20V$
Final charge on $C_2$ and $C_3$ is $Q=C{V}^{\prime }=2\mu F\times 20=40\mu C$
Therefore, $Q=40\mu C$