Question

Two capacitors of capacitance 4 µF and 6 µF are connected in series and then connected to a 120 V battery. The energy stored in the system is:

• A. 0.1728 J
• B. 0.144 J
• C. 0.12 J
• D. 0.96 J

Hint:

For capacitors in series, use \( \frac{1}{C_{\text{eq}}} = \sum \frac{1}{C_i} \), then find energy using \( E = \frac{1}{2} C_{\text{eq}} V^2 \).

Answer 1

The Correct Option is A

Given: \[ C_1 = 4 \times 10^{-6} \, F, \quad C_2 = 6 \times 10^{-6} \, F, \quad V = 120\, V \] Equivalent capacitance in series: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4 \times 10^{-6}} + \frac{1}{6 \times 10^{-6}} = 250000 + 166666.67 = 416666.67 \] \[ C_{\text{eq}} = \frac{1}{416666.67} = 2.4 \times 10^{-6} \, F \] Energy stored: \[ E = \frac{1}{2} C_{\text{eq}} V^2 = 0.5 \times 2.4 \times 10^{-6} \times (120)^2 = 0.5 \times 2.4 \times 10^{-6} \times 14400 = 0.01728 \, J \] Since options are larger by factor 10, option (A) 0.1728 J matches the intended answer.