How do half-wave and full-wave rectifiers compare in operation and in the average output voltage for an ideal transformer with peak input Vm?

• A. Half-wave conducts only one half-cycle; V_out average = Vm/π. Full-wave bridge conducts both halves; average = 2Vm/π.
• B. Half-wave conducts both halves; V_out average = Vm/π. Full-wave bridge conducts one half-cycle; average = Vm/2.
• C. Half-wave conducts only one half-cycle; average = Vm. Full-wave bridge conducts both halves; average = Vm.
• D. Half-wave conducts one quarter-cycle; average = Vm/2. Full-wave bridge conducts all time; average = 0.

Answer: (A)

The main idea is how much of the AC waveform contributes to the DC output when you rectify it with an ideal transformer. The secondary voltage is v = Vm sin(ωt). In a half-wave rectifier, current only flows during the positive half-cycle, so the output is Vm sin(ωt) for 0 to π and zero for π to 2π. The average value over a full cycle is the integral of that output divided by the period: (1/2π) ∫0^π Vm sinθ dθ = Vm/π. In a full-wave bridge, both halves of the waveform are used, with the negative half flipped to positive, so the output is |Vm sin(ωt)| over 0 to 2π. The average becomes (1/2π) ∫0^{2π} |sinθ| Vm dθ = (Vm/2π) × 4 = 2Vm/π. So the half-wave average is Vm/π, and the full-wave bridge average is 2Vm/π. With an ideal transformer and no smoothing capacitor, these are the true DC averages of the pulsating outputs.