Power Electronics Test 1

Result

Power Electronics Test 1

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Question 1
1 / -0

A three phase three pulse converter is operated from 3-phase, 230 V, 50 Hz supply with load resistance R = 10 Ω. An average output voltage of 50% of the maximum possible output voltage is required. Then the firing angle is ________ (in degrees)

A
66-69

B
664-691

C
667-696

D
669-692

Solution

For R load, voltage is continuous when α ≤ 30° and average output voltage is given by
\({V_o} = \frac{{3{V_{ml}}}}{{2\pi }}\cos \alpha\)
Maximum possible output voltage \(= \frac{{3{V_{ml}}}}{{2\pi }}\)
Required average output voltage \(= \frac{1}{2}\left( {\frac{{3{V_{ml}}}}{{2\pi }}} \right)\)
\(Now,\frac{{3{V_{ml}}}}{{2\pi }}\cos \alpha = \frac{1}{2}\left( {\frac{{3{V_{ml}}}}{{2\pi }}} \right)\)
⇒ α = 60°
So, the required value can’t be obtained for α < 30°.
For α > 30°,
Average output voltage, \({V_o} = \frac{{3{V_{mp}}}}{{2\pi }}\left[ {1 + \cos \left( {\alpha + 30^\circ } \right)} \right]\)
\(= \frac{{3{V_{ml}}}}{{2\sqrt 3 \pi }}\left[ {1 + \cos \left( {\alpha + 30^\circ } \right)} \right]\)
\(\Rightarrow \frac{{3{V_{mL}}}}{{2\sqrt 3 \pi }}\left[ {1 + \cos \left( {\alpha + 30^\circ } \right)} \right] = \left( {\frac{{3{V_{ml}}}}{{2\pi }}} \right)\left( {\frac{1}{2}} \right)\)
⇒ α = 67.69°