A sinusoidal source has peak voltage 28 V and frequency 90 Hz. (a) If connected directly across a 260 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 0.036 A. Estimate ripple using ΔV ≈ IΔt/C, with recharging every half-cycle. (c) Explain the effect of doubling capacitance.

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Vrms 19.8 V; direct resistor power 1.508 W; ripple 0.2 V; doubling capacitance approximately halves ripple. For a sinusoid Vrms = Vpeak/√2 = 28/√2 = 19.8 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 = 28²/(2 × 260) = 1.508 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) = 0.005556 s. With C = 0.001000 F, ripple ≈ 0.036 × 0.005556/0.001000 = 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 9

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