Question 2.SP.14: The circuit of Fig. 2-22(a) is to be used as a dc power supp......

Let $T$ denote the period of $v_S$.    For $R_L = 10  Ω$,
$v_L = \begin{cases} \frac{R_L}{R_L  +  R_S}v_s = \frac{10}{10  +  10}10 = 5  \text{V} \quad\quad 0 \leq t \lt T/2 \\  0  (\text{diode blocks}) \quad\quad\quad   \quad\quad\quad T/2 \leq t \lt T \end{cases}$

and so              $V_{L0} = \frac{5(T/2)  +  0 (T/2)}{T} = 2.5  \text{V}$

For $R_L = 1  kΩ$,

and so              $V_{L0} = \frac{9.9(T/2)  +  0}{T} = 4.95  \text{V}$

Then, by (2.6) and using $R_L = 10  Ω$ as full load, we have
$\text{Reg} ≡ \frac{(\text{no-load} V_{L0})  –  (\text{full-load} V_{L0})}{\text{full-load} V_{L0}}$          (2.6)
$\text{Reg} = \frac{4.95  –  2.5}{2.5} (100 \%) = 98 \%$

This large value of regulation is prohibitive for most applications.    Either another circuit or a filter network would be necessary to make this power supply useful.