A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60° and output AC power of 2127 Watts, find the rectification efficiency

Correct Answer: 28 %
Vo = (Vm/2π) x (1+cosα) = 77.64 V Pdc = Vo2xR = 602.8 W Rectification efficiency = Pdc/Pac = 28.32 %.