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 AThe 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

An alternating voltage of 360 V, 50 Hz is applied to a full-wave recti

1.	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 AThe 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» 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

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