i)Limit to line secondary voltage

$V_{rms}=\frac{2v_{m}}{\sqrt 2}$=50 V

Hence $V_{m}=\frac{50\sqrt 2}{2}$=35.35 V

No load dc voltage

$V_{dc}=\frac{2V_{m}}{\pi}=0.636 V_{m}=22.50 v$ ......................(i)

ii) DC load current

$I_{dc}=\frac{0.636 V_{m}}{R_{5}+R_{f}+R_{L}}$ =4.49 mA ................(ii)

iii) DC load voltage

$V_{dc}=\frac{2V_{m}}{\pi}-I_{dc}(R_{f}+R_{5})$

2250-$4.49\times 10^{-3}\times 10=22.45 V$ ................(iii)

iv)DC power delivered to load

$P_{dc}=I_{dc}V_{dc}$

$4.49\times 10^{-3}\times 22.50$=0.101 W ...........................(iv)

v) AC power

$P_{ac}=I^{2}_{rms}(R_{s}+R_{f}+R_{L})$

=$(\frac{I_{m}}{\sqrt 2})^{2}(R_{s}+R_{f}+R_{L})=\frac{V_{m}^{2}}{2\times (R_{s}+R_{f}+R_{L})}$

=$\frac{35.35^{2}}{2\times 5010}$=0.125 W ..........................(v)

vi)Regulation from no load to full load

=$\frac{R_{f}+R_{s}}{R_{L}}\times 100 %$

=$\frac{10}{5000}\times 100 %$=0.2% ..............................................(vi)

vii) AC Ripple voltage across load

$V_{r}=0.482 \times V_{dc}$

=$0.482\times 22.45$

=10.82 V ................................................(vii)

viii) Power conversion efficiency

$eta=\frac{P_{dc}}{P_{ac}}=\frac{0.101}{0.125}$=88% ..........................(viii)

ix) PIV across each diode $2 V_{m}=2\times 35.3$=70.7 V ................(ix)

x) Transformer secondary rating

$\frac{P_{dc}}{TUF}=\frac{0.101}{0.693}$=00.145 VA ...............(x)

xi) Ripple frequency =$2\times main frequency$

$2\times 50$=100 Hz.............................(xi)