In the circuit shown, the energy stored in the capacitor is n   μJ . The value of n is _____.

The capacitor works as an open circuit in the given configuration.

The potential difference between the two plates of the capacitor will be the difference across the resistors having resistances $3 \mathrm{\Omega }$ and $4 \mathrm{\Omega }$.

Hence, the potential difference $\left(∆V\right)$ across the capacitor can be calculated as follows:

$\begin{array}{rcl}\Delta V& =& \left(\frac{4}{6}\times 12-\frac{3}{12}\times 12\right) \mathrm{V}\\ & =& 5 \mathrm{V}\end{array}$

The formula to calculate the energy stored in the capacitor is given by

$U=\frac{1}{2}C(∆V{)}^2 ...\left(1\right)$

Substitute the values of the known parameters into equation (1) to calculate the required energy stored.

$\begin{array}{rcl}U& =& \frac{1}{2}\times 6 \mathrm{\mu F}\times {\left(5 \mathrm{V}\right)}^2\\ & =& 75 \mathrm{\mu J}\end{array}$