Question

A full wave p-n diode rectifier uses a load resistance of $1300$ $\Omega$. No filter is used. If the internal resistance of each diode is $9$ $\Omega$, then the efficiency of this full wave rectifier is

• A. $80.64\%$
• B. $40.6\%$
• C. $13.9\%$
• D. $100\%$

Answer 1

The correct option is A $80.64\%$

The efficiency $\eta$ of full wave rectifier is given by:
$\eta =\frac{0.812R_L}{r_f+R_L}$

$\eta =\frac{0.812}{1+\frac{r_f}{R_L}}$

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

$\eta =\left(\frac{0.812}{1+\frac{r_f}{R_L}}\right)\times 100$

$\eta =\left(\frac{0.812}{1+\frac{9}{1300}}\right)\times 100$

$\eta =\left(\frac{0.812}{1+0.0069}\right)\times 100$

$\eta =\left(\frac{0.812}{1.0069}\right)\times 100$

$\eta =\left(0.8064\right)\times 100=80.64\%$