In a half wave rectifier- the output is taken across a 90- Omega load resistor- If the resistance of diode in forward biased condition is 10- Omega- the efficiency of rectification of ac power into power is81-2-36-54-73-08-40-6-

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Toppr Admin · Accepted Answer

The efficiency-xA0-x3B7-xA0-of half wave rectifier is given by-xA0-x3B7-4-x3C0-2RLrf-RLwhere- RL-xA0-is the resistance of load resistor and rf-xA0-is the-xA0-resistance of diode-xA0-in-xA0-forward biased condition-

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
$81.2 %$
$36.54 %$
$73.08 %$
$40.6 %$

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.

In a half wave rectifier, the output is taken across a 90Ω load resistor. If the resistance of diode in forward biased condition is 10Ω, the efficiency of rectification of ac power into dc power is81.2%36.54%73.08%40.6%

The efficiency η of half wave rectifier is given by, η=4π2RLrf+RLwhere, RL is the resistance of load resistor and rf is the resistance of diode in forward biased condition.

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
$81.2 %$
$36.54 %$
$73.08 %$
$40.6 %$

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.