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A capacitor is charged in four equal voltage steps. It is first charged to V0/4 and held for a time T >> RC, then the voltage is raised to V0/2 (without discharging) and held for T >> RC, then to 3*V0/4, and finally to V0. If the total energy dissipated across the resistance throughout this process is (1/N)*C*V0², find N.
- 1
- 2
- 4
- 8
Correct answer: 8
Solution
When a capacitor already at voltage V_i is charged to V_f, the energy dissipated in the resistor is (1/2)*C*(V_f - V_i)². With four equal steps each of size V0/4, the total dissipation is 4*(1/2)*C*(V0/4)² = C*V0²/8, so N = 8.
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