Question

An alternating voltage of 360 V, 50 Hz is applied to a full-wave rectifier. The internal resistance of each diode is 100 W. If RL = 5 kW, then what is the peak value of output current?

(a) 0.9 A

(b) 0.07 A

(c) 0.097 A

(d) 1.097 A

The question was posed to me during an interview.

My question is from Semiconductor Electronics topic in division Semiconductor Electronics : Materials, Devices and Simple Circuits of Physics – Class 12

Answer 1

The correct option is (c) 0.097 A

The explanation: The required equation is as follows:

Ipeak=Irms × √2=\(\frac {V_{rms} \times \sqrt {2}}{R_L+2r_p}\)

Ipeak=\(\frac {360 \times \sqrt {2}}{5000 + 200}\)

Ipeak=\(\frac {360 \times 1.414}{5200}\)

Ipeak=0.097 A