Question

Find the workdone by the battery when switch ${\text{S}}_n$ is closed as shown in figure.

• A. $\frac{18000}{3}\mu \text{J}$
• B. $\frac{1800}{3}\mu \text{J}$
• C. $\frac{1600}{3}\mu \text{J}$
• D. $\frac{16000}{3}\mu \text{J}$

Answer 1

The correct option is D $\frac{16000}{3}\mu \text{J}$

When the switch is open, capacitors $C_1$ and $C_2$ are in parallel and their equivalent capacitance is in series with $C_3$. So, the effective capacitance of the system is,

$\frac{1}{C_{eqv}}=\frac{1}{(10+10)}+\frac{1}{10}=\frac{3}{20}$

$\Rightarrow C_{eqv}=\frac{20}{3}\mu \text{F}$

So, the charge supplied by the battery,

$Q_i=C_{eqv}V=\frac{20}{3}\left(20\right)=\frac{400}{3}\mu \text{C}$

When switch ${\text{S}}_n$ is closed, capacitor $C_3$ becomes short circuited and the effective capacitance of the system becomes.

$C_{eqn}=C_1+C_2=20\mu \text{F}$

Now, charge supplied by the battery,

$Q_f=Ceqv\times V=20\left(20\right)=400\mu \text{C}$

Thus, workdone by the battery,

$W=V(Q_f-Q_i)$

$=20(400-\frac{400}{3})$

$=\frac{16000}{3}\mu \text{J}$

Hence, option (d) is the correct answer.