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A full-wave p–n diode rectifier uses a load resistor of $1500 Ω$ . No filter is used. The forward bias resistance of the diode is $10 Ω$ . The efficiency of the rectifier is

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$40.6 %$

$81.2 %$

$80.66 %$

$40.2 %$

answer is C.

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Detailed Solution

Efficiency $η = \frac{0.812 \times R_{L}}{R_{L} + R_{f}} \times 100$

$= \frac{0.812 \times 1500}{1500 + 10} \times 100$

$= 80.66 %$

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