Question

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 dc power is

• A. $40.6\%$
• B. $81.2\%$
• C. $73.08\%$
• D. $36.54\%$

Answer 1

Answer: (D)

$36.54\%$

The efficiency $\eta$ of half wave rectifier is given by, $\eta =\frac{4}{{\pi }^2}\frac{R_L}{r_f+R_L}$

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, $\eta =\left(\frac{0.4056\left(90\right)}{10+90}\right)\times 100$
$\eta =\left(\frac{36.54}{100}\right)\times 100=36.54\%$