i)(a) The peak value of load current $I_{m}$ can be calculated as

$I_{m}=\frac{V_{m}}{R_{s}+R_{f}+R_{L}}=\frac{30}{0+30+500}$=56.6 mA ................(i)

b) The dc value of the load current is

$I_{dc}=\frac{V_{m}}{\pi\times R_{s}+R_{f}+R_{L}}=\frac{30}{\pi\times 530}$=18 mA ................(ii)

c) $I_{m}=\frac{I_{m}}{2}=\frac{56.6}{2}$=283 mA ............(iii)

ii) The dc output voltage

$V_{dc}=I_{dc}\times R_{L}=18\times 10^{-3}\times 500$=9V ...............(iv)

iii)$V_{rms}=\frac{V_{m}}{\sqrt 2}=\frac{30}{\sqrt 2}$=21.21 V ...........(v)

iv)Efficiency of …