Two capacitors of capacitances of \( 6\,\mu\mathrm{F} \) and \( 12\,\mu\mathrm{F} \) are connected in series with a 12 V battery. The total energy stored in the combination is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 288\,\mu\mathrm{J} \)
The two capacitors in series act as one capacitor of equivalent capacitance \( C_{\mathrm{eq}} \), and the energy stored in it is \( U = \tfrac{1}{2} C V^{2} \).
\( \displaystyle \frac{1}{C_{\mathrm{eq}}} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12} \Rightarrow C_{\mathrm{eq}} = 4\,\mu\mathrm{F} \)
\( \displaystyle U = \frac{1}{2} C_{\mathrm{eq}} V^{2} = \frac{1}{2} \times 4\times10^{-6} \times 12^{2} \)
\( \displaystyle U = 2\times10^{-6} \times 144 = 288\times10^{-6}\,\mathrm{J} = 288\,\mu\mathrm{J} \)