A sinusoidal source has peak voltage 12 V and frequency 50 Hz. (a) If connected directly across a 100 Ω resistor, find r.m.s. voltage and average power. (b) It is instead connected to an ideal full-wave bridge and 1000 μF smoothing capacitor supplying approximately constant current 0.02 A. Estimate ripple using ΔV ≈ IΔt/C, with recharging every half-cycle. (c) Explain the effect of doubling capacitance.

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Vrms 8.485 V; direct resistor power 0.72 W; ripple 0.2 V; doubling capacitance approximately halves ripple. For a sinusoid Vrms = Vpeak/√2 = 12/√2 = 8.485 V. This is the steady d.c. voltage that would produce the same mean heating in the stated resistor. Average resistive power is Vrms²/R = 12²/(2 × 100) = 0.72 W. Peak voltage squared divided by resistance would instead give the instantaneous maximum power. Full-wave rectification produces two recharging peaks per cycle, so Δt ≈ 1/(2f) = 0.01 s. With C = 0.001000 F, ripple ≈ 0.02 × 0.01/0.001000 = 0.2 V. The capacitor supplies charge to the load between peaks, so its voltage falls as charge is removed. Doubling capacitance halves the approximate fall for the same charge loss. This small-ripple model neglects diode drops, finite charging intervals and source resistance, which matter in a real circuit.

From sinusoidal supply to smoothed output 1

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