Question

In the circuit shown in the figure, a $12\text{volt}$ battery has been connected across the points $\text{X}$ and $\text{Y}$. The work done by battery will be:

• A.
  $180\mu \text{J}$
• B.
  $384\mu \text{J}$
• C.
  $424\mu \text{J}$
• D.
  $212\mu \text{J}$

Answer 1

The correct option is B

$384\mu \text{J}$
First of all let us simplify the given circuit and replace the whole circuit by a single equivalent capacitance $C_{eq}$ connected across $\text{X}$ & $\text{Y}$.

Using point potential techique,


Therefore we have ${}^{\prime }{3}^{\prime }$ points named $1,2$ & $3$ with different potentials.

Redrawing the circuit gives,



For series connection for $1-2$ and $2-3$ , equivalent capacitance,

$C_{13}=\frac{2\times 1}{2+1}=\frac{2}{3}\mu \text{F}$


Now, the final equivalent capacitance,

$C_{XY}=\frac{2}{3}+2=\frac{8}{3}\mu \text{F}$

Therefore, we have final equivalent circuit,


Total charge supplied by the battery,

$q=(C_{XY}\times V)=\frac{8}{3}\times 12=32\mu \text{C}$

Hence, the workdone by battery will be,

$W_{battery}=qV$

$\Rightarrow W_{battery}=\left(32\mu \text{C}\right)\times \left(12\text{V}\right)$

$\therefore W_{battery}=384\mu \text{J}$

Hence, option (b) is correct answer.

Why this question?: Whenever workdone by battery is asked, it should immediately strike that we require total charge $\left(q\right)$ supplied by battery in the circuit. In order to find $q$ supplied, we need to simplify circuit and find $C_{eq}$ first.