A full wave p – n diode rectifier uses a load resistor of 1500Ω. No filter is used. The forward bias resistance of the diode is 10Ω. The efficiency of the rectifier is:

80.6%

Here, $r_i =10Ω, R_L =1500Ω$ 

Efficiency of the full wave rectifier is

$η=\frac{P_{dc}}{P_{ac}}=\frac{(2I_m/π)^2R_L}{(I_m/\sqrt{2})^2(r_f+R_L)}=\frac{0.812R_L}{r_f+R_L}=\frac{0.812×1500}{10+1500}=0.806=80.6\%$