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)



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