A half-wave rectifier has a load RL = 900 Ω…

A half-wave rectifier has a load $R_L = 900\ \Omega$. First take an ideal diode. Find the efficiency, which is the direct power in the load divided by the alternating power that the transformer gives. Then take a diode that has a constant forward resistance $r_f = 100\ \Omega$ when it conducts and is otherwise ideal. The current is $i = I_m \sin \omega t$ in the positive half cycle. Find the efficiency now. The direct power is $I_{dc}^2 R_L$ with $I_{dc} = I_m/\pi$. The power from the transformer is $I_{rms}^2 (R_L + r_f)$ with $I_{rms} = I_m/2$.