Question

Q23
21:45 m a
The applied AC power to a half-wave rectifier is 200 W. The DC power
output obtained is 50 W. The rectification efficiency is​

Answer 1

Given:

The applied AC power to a half-wave rectifier is 200 W. The DC power output obtained is 50 W.

To find:

The rectification efficiency is​

Solution:

From given, we have,

The applied AC power to a half-wave rectifier is 200 W.

⇒ Pac = 200 W

The DC power output obtained is 50 W.

⇒ Pdc = 50 W

The rectification efficiency is​ given by,

η = Pdc/Pac

= 50/200 × 100

= 25%

∴ η = 25%, the rectification efficiency.