Question

A full-wave p-n diode rectifier uses a load resistor of $1500\Omega$. No filter is used. The forward bias resistance of the diode is $10\Omega$. the efficiency of the rectifier is

• A. $81.2\%$
• B. $40.6\%$
• C. $80.4\%$
• D. $40.2\%$

Answer 1

The correct option is C $80.4\%$

Given,

${\eta }_0=81.2$%

${\eta }_0=0.812$

$R_L=1500\Omega$

$R_f=10\Omega$

The efficiency of rectifier

$efficiency=\frac{output}{input}$

$\eta =\frac{{\eta }_0}{(1+\frac{R_f}{R_L})}\times 100$

$\eta =\frac{0.812}{(1+\frac{10}{1500})}\times 100$

$\eta =80.4$%

The correct option is C.