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This question belongs to Electrical Engineering Semiconductor Diode

Semiconductor Diode

?

A half-wave rectifier has an input voltage of 240 V r.m.s. If the step-down transformer has a turns ratio of 8:1, what is the peak load voltage? Ignore diode drop.

A) 27.5 V B) 86.5 V C) 30 V D) 42.5 V

Answer & Solution

Correct Answer: Option D

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Question posted by Examveda

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Lavito Achumi 3 years ago

Secondary Voltage=240/8 = 30V
Peak Load Voltage= √2 x 30 = 42.5V

Om Jariwala 5 years ago

The

.... filter circuit

results in the best voltage regulation

1. choke input

2. capacitor input

3. resistance input

4. none of the above

Harihara Parida 6 years ago

Emf/ turn are constant in a transferred.
Secondary vtg = 240/8= 30
In half wave rectifier pic= ✓2 x Vm
✓2*30=42.5.

Engr Farman 6 years ago

V1/V2 =N1/N2
Vrms = 0.707Vo

Engr Farman 6 years ago

Vrms= 0.707 Vo
Vo=240/8 = 30

Jayanth Sparrow 8 years ago

I want clarify in this

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