Question

Two capacitors $2\mu \text{F}$ and $6\mu \text{F}$ in series are connected in parallel to a third capacitor of $4\mu \text{F}$. This arrangement is then connected to a battery of emf $2\text{V}$. Find the total energy of system of capacitors?

• A. $22\mu \text{J}$
• B. $11\mu \text{J}$
• C. $5.5\mu \text{J}$
• D. $44\mu \text{J}$

Answer 1

The correct option is B $11\mu \text{J}$

As per the question capacitors $2\mu \text{F}$ and $6\mu \text{F}$ in series are connected in parallel to capacitor of $4\mu \text{F}$.

$\therefore$ Equivalent capacitance of this combination will be

$C_{\text{eq}}=\frac{2\times 6}{2+6}+4=1.5+4=5.5\mu \text{F}$

Emf of battery, $V=2\text{V}$

$\therefore$ energy stored in the combination of capacitors $U=\frac{1}{2}{C}_{\text{eq}}{V}^2$

$\Rightarrow U=\frac{1}{2}\times 5.5\times {10}^{-6}\times 2^2=11\times {10}^{-6}\text{J}$

$\therefore U=11\mu \text{J}$

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

Key Concept - Energy stored in a combination of capacitors is $\frac{1}{2}{C}_{\text{eq}}{V}^2$