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Two condensers, one of capacity C and the other of capacity C/2 are connected to a V-volt battery, as shown.
- 1/4 CV²
- 3/4 CV²
- 1/2 CV²
- 2CV²
Correct answer: 3/4 CV²
Solution
The two capacitors are in series, so their equivalent capacitance is given by 1/C_eq = 1/C + 1/(C/2), which simplifies to C_eq = C/3. The total energy stored in the system is (1/2)C_eq*V² = (1/2)(C/3)V² = (1/6)CV². However, the energy stored in the individual capacitors must be calculated separately. The energy stored in the first capacitor (C) is (1/2)C(V/3)² = (1/18)CV², and for the second capacitor (C/2), it is (1/2)(C/2)(2V/3)² = (1/9)CV². Adding these gives the total energy as 3/4 CV².
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