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A 4 μF capacitor is charged to 400 volts and then its plates are joined through a resistance of 1kΩ. The heat produced in the resistance is
- 0.16 J
- 1.28 J
- 0.64 J
- 0.32 J
Correct answer: 0.64 J
Solution
The energy stored in the capacitor is given by the formula U = (1/2)CV². Substituting C = 4 μF = 4 × 10⁻⁶ F and V = 400 V, we get U = 0.5 × 4 × 10⁻⁶ × (400)² = 0.32 J. This energy is dissipated as heat in the resistor.
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