A series combination of two capacitors \( C_1 = 3\,\mu\mathrm{F} \) and \( C_2 = 6\,\mu\mathrm{F} \) is connected in parallel with a third capacitor, \( C_3 = 4\,\mu\mathrm{F} \). This system is connected to a 10 V battery. The total energy stored in the system is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3\times10^{-4}\,\mathrm{J} \)
First find the equivalent capacitance: capacitors in series add as reciprocals, and capacitors in parallel add directly. The energy stored is then \( U = \tfrac{1}{2}CV^{2} \).
\( \displaystyle C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{3\times6}{3+6} = 2\,\mu\mathrm{F} \)
\( \displaystyle C_{\mathrm{eq}} = C_{12} + C_3 = 2 + 4 = 6\,\mu\mathrm{F} \)
\( \displaystyle U = \tfrac{1}{2}\,C_{\mathrm{eq}}V^{2} = \tfrac{1}{2}\times6\times10^{-6}\times(10)^{2} = 3\times10^{-4}\,\mathrm{J} \)