Question

A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input.

(a) 368.98mW

(b) 275.2mW

(c) 145.76mW

(d) 456.78mW

The question was posed to me during a job interview.

My question comes from Full-wave Rectifier in division Application of Diodes of Electronic Devices & Circuits

Answer 1

Right option is (b) 275.2mW

To explain I would say: The AC power input PIN=IRMS^2(RF+Rr).

IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA

So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.