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Electronic Devices and Circuits - Engineering

Q1:

What is the necessary a.c. input power from the transformer secondary used in a half wave rectifier to deliver 500 W of d.c. power to the load?

A 1232 W

B 848 W

C 616 W

D 308 W

ANS:A - 1232 W

Given,

Dc load power = 500 W,
We need to find ac input power from the transformer.
As we know, the formula for the efficiency of half wave rectifier is as below,
efficiency = dc load power/ac input power.
We know that the efficiency of half-wave rectifier =40.6% or 0.406.
Therefore,

0.406=500/ac input power.
Ac input power = 500/0.406.
Ac input power = 1231.52.
Therefore, the answer is 1232.