Power Electronic Problem

A full wave rectified circuit is given below.

If diode D2 is shorted out, what is the average voltage across RL?

A) 0 (i.e. short circuit)
B) 54 V
C) 108 V
D) 216 V

6 thoughts on “Power Electronic Problem

  1. I think the answer is B.

    The circuit above essentially acts like a half-wave rectifier because of the shorted diode. The average voltage would be Vavg=0.316*Vpk. With the two 10ohm resistors, the voltage is divided by 2. So peak voltage across load, Vpk = 240*√2 / 2 = 169.7. Vavg = 169.7*0.316=~54V.

    1. B.
      Average of sin function=(Vpeak/period of function) X (area under the sin function)

      Vpeak =sqrt2 x Vrms
      Vrms=240/2=120
      F(t) =sin(x)
      Limits: 0 to pi because When you short D1. The full wave rectifier will behave like a positive half wave rectifier. There will be 0 voltage from pi to 2pi.
      Hence:
      Average =(Sqrt 2 X 120)/(2pi) x {integral (sin(x),0,pi)}
      =54 volts.

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