The total energy stored in the three capacitors shown in the figure,…

The correct answer is Option (3) → 48

The total energy stored in a network of capacitors equals the energy of its equivalent capacitor, \( U=\frac{1}{2}CV^{2} \). Capacitances in series combine by adding their reciprocals, while capacitances in parallel add.

From the figure, the two \( 6\,\mu\mathrm{F} \) capacitors are in series, and this pair is in parallel with the \( 3\,\mu\mathrm{F} \) capacitor across the 4 V cell.

The series pair gives:

    \( \begin{aligned} C_s &= \frac{C_1C_2}{C_1+C_2} \\ &= \frac{6\times6}{6+6}\,\mu\mathrm{F} \\ &= 3\,\mu\mathrm{F} \end{aligned} \)

Adding the \( 3\,\mu\mathrm{F} \) capacitor in parallel:

    \( \begin{aligned} C_{\mathrm{eq}} &= C_s+C_3 \\ &= 3\,\mu\mathrm{F}+3\,\mu\mathrm{F} \\ &= 6\,\mu\mathrm{F} \end{aligned} \)

The total energy stored is:

    \( \begin{aligned} U &= \frac{1}{2}C_{\mathrm{eq}}V^{2} \\ &= \frac{1}{2}\times6\,\mu\mathrm{F}\times(4\,\mathrm{V})^{2} \\ &= 48\,\mu\mathrm{J} \end{aligned} \)