Question

Three capacitors having capacitance $20\mu \text{F},30\mu \text{F}$ and $40\mu \text{F}$ are connected in series with a $12\text{V}$ battery. The amount of work done by the battery in charging the combination will be:

• A. $1.5\times {10}^{-3}\text{J}$
• B. $2.25\times {10}^{-3}\text{J}$
• C. $1.33\times {10}^{-3}\text{J}$
• D. $3\times {10}^{-3}\text{J}$

Answer 1

The correct option is C $1.33\times {10}^{-3}\text{J}$


Here $q_1=q_2=q_3$ (capacitors in series), let it be equal to $q$.

The whole arrangement may be replaced by a single equivalent capacitance $C_{eq}$ with the same charge $q$.


$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{20}+\frac{1}{30}+\frac{1}{40}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{6+4+3}{120}=\frac{13}{120}$

$\therefore C_{eq}=\frac{120}{13}\mu \text{F}$

Hence, work done by battery,

$W=\left(C_{eq}V\right)V[\because q=C_{eq}\times V]$

$\Rightarrow W=\frac{120}{13}\times {10}^{-6}\times 12\times 12$

$\Rightarrow W=\frac{17280}{13}\times {10}^{-6}=1.33\times {10}^{-3}\text{J}$

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