Semiconducting Devices Question 227
Question: The maximum efficiency of full wave rectifier is
[J & K CET 2004]
Options:
A) 100%
B) 25.20%
C) 40.2%
D) 81.2%
Show Answer
Answer:
Correct Answer: D
Solution:
For full wave rectifier $ \eta =\frac{81.2}{1+\frac{r _{f}}{R _{L}}} $
Therefore $ {n _{\max }}=81.2% $ (rf « RL)