[J & K CET 2004]
A) 100%
B) 25.20%
C) 40.2%
D) 81.2%
Correct Answer: D
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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