$I_{L}$(dc)=75 mA, $V_{L}$(dc)=250 V, $V_{r}$=ripple voltage=10 V

I) Load resistance

$R_{L}=\frac{V_{L}(dc)}{I_{L}(dc)}=\frac{250V}{75mA}$=3.33k$\Omega \ \ \ \ \ .....(i)$

ii) The ripple factor =$\frac{V_{r}(rms)}{V_{L}(dc)}=\frac{10}{250}$

=0.04=4% ...............(ii)

iii)For inductor filtered full wave rectifier

$V_{L}(dc)=\frac{2V_{m}}{\pi}-I_{dc}\times R$

By neglecting R. i..e. transformer secondary resistance, inductor and diode resistance

$V_{L}(dc)=\frac{2V_{m}}{\pi}$

$V_{m}=\frac{-V_{L}(dc)\pi}{2}=\frac{250 \pi}{2}$=392.69 V ................(iii)

iv) Ripple factor, r=$\frac{R_{L}}{1333L}$

L=$\frac{R_{L}}{1333\times r}=\frac{3.33K\Omega}{1333\times 0.04}$=62.45 H ........(iv)

v) PIV rating of diode

$2V_{m}=2\times 392.69 V$

=785.38 V ..........................(v)

vi) The secondary voltage rating of transformer

$\frac{785.38}{\sqrt 2}$=555.34 V ................(vi)

vii)The voltage drop=0.7 V

$V_{m}$=277.67+0.7=278.37 V ...................(vii)

Thus, a transformer with 280 V between center tap and ends carrying 75 mA is sufficient.