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For half wave rectifier secondary voltage V=30 sin$\omega t$,load resistance 500$\Omega$ used. The diode forward resistance 30$\Omega$neglecting transformer secondary resistance Find $I_{m} I_{dc}$
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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)

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