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
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
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.
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.
Why voltage divided by 2?
I say it’s “A”, no current flows through resistor RL.