Question 2.10: In Fig. 2-13, vS = Vm sin ωt and the diode is ideal. Calcula......

Only one cycle of v_S need be considered. For the positive half-cycle, i_D > 0 and, by voltage division,
v_L = \frac{R_L}{R_L  +  R_S} (V_m \sin ωt) ≡ V_{Lm} \sin ωt

For the negative half-cycle, the diode is reverse-biased, i_D = 0, and v_L = 0. Hence,
V_{L0} = \frac{1}{2 \pi} \int_0^{2\pi} v_L  (ωt)  d(ωt) = \frac{1}{2 \pi} \int_0^{\pi} V_{Lm} \sin ωt  d(ωt) = \frac{V_{Lm}}{\pi}

Although the half-wave rectifier gives a dc output, current flows through R_L only half the time, and the average value of the output voltage is only 1/\pi = 0.318 times the peak value of the sinusoidal input voltage.    The output voltage can be improved by use of a full-wave rectifier (see Problems 2.28 and 2.50).
When rectifiers are used as dc power supplies, it is desirable that the average value of the output voltage remain nearly constant as the load varies.    The degree of constancy is measured as the voltage regulation,
\text{Reg} ≡ \frac{(\text{no-load}  V_{L0})  –  (\text{full-load}  V_{L0})}{\text{full-load}  V_{L0}}          (2.6)

which is usually expressed as a percentage.   Note that 0 percent regulation implies a constant output voltage.