In a half wave rectifier, the output is taken across a $90 Ω$ load resistor. If the resistance of diode in forward biased condition is $10 Ω$ , the efficiency of rectification of ac power into dc power is

Question

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Answer 1

The efficiency $η$ of half wave rectifier is given by, $η = \frac{4}{π ^{2}} \frac{R _{L}}{r _{f} + R _{L}}$

where, $R_{L}$ is the resistance of load resistor and $r_{f}$ is the resistance of diode in forward biased condition.

Hence, the percent efficiency is, $η = (\frac{0 . 4 0 5 6 ( 9 0 )}{1 0 + 9 0}) \times 100$

$η = (\frac{3 6 . 5 4}{1 0 0}) \times 100 = 36.54 %$

Answer 2

The efficiency

η

of half wave rectifier is given by,

η = \frac{4}{π ^{2}} \frac{R _{L}}{r _{f} + R _{L}}

where,

R_{L}

is the resistance of load resistor and

r_{f}

is the resistance of diode in forward biased condition.

Hence, the percent efficiency is,

η = (\frac{0 . 4 0 5 6 ( 9 0 )}{1 0 + 9 0}) \times 100

η = (\frac{3 6 . 5 4}{1 0 0}) \times 100 = 36.54 %