Two capacitors of capacitances of 6 μF and 12 μF are connected in series with a 12 V battery. The total energy stored in the combination is

288 μJ

The correct answer is Option (4) → 288 μJ

$\text{Given: } C_1 = 6~\mu\text{F},~ C_2 = 12~\mu\text{F},~ V = 12~\text{V}$

$\text{Capacitors in series: } \frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$

$\frac{1}{C_\text{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4}$

$C_\text{eq} = 4~\mu\text{F}$

$\text{Energy stored: } U = \frac{1}{2} C_\text{eq} V^2$

$U = \frac{1}{2} \cdot 4 \times 10^{-6} \cdot (12)^2$

$U = 2 \times 10^{-6} \cdot 144$

$U = 2.88 \times 10^{-4}~\text{J}$

$\text{Answer: } U = 2.88 \times 10^{-4}~\text{J}$