A half-wave rectifier on a 50 Hz supply gives a peak output of 9.0 V. A 470 μF capacitor smooths the output across a 2.0 kΩ load. Estimate the minimum p.d. across the load, taking the capacitor to discharge for a whole period.
- Time between peaks = 1 ÷ 50 = 0.020 s (half-wave, one peak per cycle).
- RC = 2.0 × 103 × 470 × 10−6 = 0.94 s.
- V = V0e−t/RC = 9.0 × e−0.020/0.94 = 9.0 × e−0.0213.
- V = 9.0 × 0.979 = 8.8 V.
Answer: About 8.8 V (a ripple of about 0.2 V)