A single-phase semiconverter is connected to 400 V- 50 Hz source- The converter is feeding a load of R- 50 - in series with large inductor which makes the load current ripple free- If the value of firing angle is 30- then current distortion factor will be -

For semiconverter, the current distortion factor CDF =2√(2) cos(α /2)/√(π(π α))=2√(2)cos(30 /2)/√(π( π π/6 )) =2.732/√(8.2246) =2.732/2.8678 =0.953 ...

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A single-phase semiconverter is connected to 400 V, 50 Hz source. The converter is feeding a load of $R = 50 Ω$ in series with large inductor which makes the load current ripple free. If the value of firing angle is 30°, then current distortion factor will be _____________.

0.953

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Solution

The correct option is A 0.953
For semiconverter, the current distortion factor

$C D F = \frac{2 \sqrt{2} c o s (α / 2)}{\sqrt{π (π - α)}} = \frac{2 \sqrt{2} c o s (30 / 2)}{\sqrt{π (π - \frac{π}{6})}}$
$= \frac{2.732}{\sqrt{8.2246}} = \frac{2.732}{2.8678} = 0.953$

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A single-phase semiconverter is connected to 400 V, 50 Hz source. The converter is feeding a load of R= 50 Ω in series with large inductor which makes the load current ripple free. If the value of firing angle is 30°, then current distortion factor will be _____________. 0.953

The correct option is A 0.953For semiconverter, the current distortion factor CDF=2√2cos(α/2)√π(π−α)=2√2cos(30/2)√π(π−π6) =2.732√8.2246=2.7322.8678=0.953

A single-phase semiconverter is connected to 400 V, 50 Hz source. The converter is feeding a load of $R = 50 Ω$ in series with large inductor which makes the load current ripple free. If the value of firing angle is 30°, then current distortion factor will be _____________.

0.953

The correct option is A 0.953
For semiconverter, the current distortion factor

$C D F = \frac{2 \sqrt{2} c o s (α / 2)}{\sqrt{π (π - α)}} = \frac{2 \sqrt{2} c o s (30 / 2)}{\sqrt{π (π - \frac{π}{6})}}$
$= \frac{2.732}{\sqrt{8.2246}} = \frac{2.732}{2.8678} = 0.953$