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In full wave rectifier load resistance is $2000 Ω$ and diode resistance in forward bias is $40 Ω$ , then efficiency percentage of rectifier is

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56.54%

79.6%

64.7%

40.2%

answer is B.

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Detailed Solution

$η = \frac{0.812 \times R_{L}}{R_{L} + r_{f}} \times 100 = \frac{0.812 \times 2000}{2040} \times 100 = 79.6 %$

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