written 6.2 years ago by |
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 half wave rectifier is
$\eta=\frac{4}{\pi^{2}}\frac{R_{L}}{R_{s}+R_{r}+R_{L}}$
=$\frac{4}{\pi^{2}}\times \frac{500}{530}$=38.23% .....................................(vi)
v) Percentage regulation
% regulation=$\frac{R_{s}+R_{f}}{R_{1}}\times 100=\frac{30}{500}\times 100=6%$ .....................(vii)