Find the average load voltage of the rectifier circuit, if the de
Find the average load voltage of the rectifier circuit, if the device is switched to ON at α, angular angle.
A. <span class="math-tex">\(\frac{{{V}_{rms}}}{\pi \sqrt{2}}\cos \alpha\)</span>
B. <span class="math-tex">\(\frac{{{V}_{rms}}}{\pi \sqrt{2}}(1+\cos \alpha )\)</span>
C. <span class="math-tex">\(\frac{{{V}_{rms}}}{\pi }\left( 1+\cos \alpha \right)\)</span>
D. <span class="math-tex">\(\frac{{{V}_{rms}}}{2\pi }\left( 1+\cos \alpha \right)\)</span>
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Right Answer is: B
SOLUTION
The given circuit represents a half wave-controlled rectifier with a purely resistive load.
Circuit diagram:
Waveforms:
From the above waveform, the average output voltage is given by:
\({{V}_{0}}=\frac{1}{2\pi }\mathop{\int }_{0}^{2\pi }{{V}_{m}}\sin \left( \omega t \right)d\left( \omega t \right)\)
\({{V}_{0}}=\frac{{{V}_{m}}}{2\pi }\left[ -\cos \left( \omega t \right) \right]_{\alpha }^{\pi }\)
\({{V}_{0}}=\frac{{{V}_{m}}}{2\pi }\left( 1+\cos \left( \alpha \right) \right)\)
\({{V}_{0}}=\frac{\sqrt{2}{{V}_{rms}}}{2\pi }\left( 1+\cos \left( \alpha \right) \right)\)
\({{V}_{avg}}=\frac{{{V}_{rms}}}{\pi \sqrt{2}}(1+\cos \alpha )\)