Question

For $C_1=C_2=C_3=4\mu \text{F}$, what is the work done by the battery till the capacitors are fully charged?

• A. $12\mu \text{J}$
• B. $24\mu \text{J}$
• C. $48\mu \text{J}$
• D. $6\mu \text{J}$

Answer 1

The correct option is B $24\mu \text{J}$

Equivalent capacitance of the given circuit will be

$C_{eq}=\frac{C_1\times C_2}{C_1+C_2}+C_3$

$\Rightarrow C_{eq}=\frac{4\times 4}{4+4}+4=2+4=6\mu \text{F}$

Charge supplied to capacitors is given by
$Q=CV$

$\therefore Q=C_{eq}V=6\times {10}^{-6}\times 2=12\times {10}^{-6}\text{C}$

Work done by battery is

$W=Q\Delta V=12\times {10}^{-6}\times 2=24\mu \text{J}$

Hence, option $\left(b\right)$ is correct.

Key concept- Work done by a battery in a circuit is its voltage times charge flown through it.