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Q166HardRectifiers & Converters

A step-down delta-star transformer, with per-phase turns ration of 5, is fed from a 3-phase, 1100 V, 50 Hz source. The secondary of this transformer through a 3-pulse type rectifier feeds a R load of 10 Ω. Find the maximum value of the load current (phase).

Answer & step-by-step solution
Correct answer: A. √2 × 22 A
Quick explanation

The primary is delta-connected, so its phase voltage is 1100 V. With a per-phase turns ratio of 5, the star-secondary phase voltage is 1100/5 = 220 V rms. The three-pulse rectifier's maximum load voltage is the phase-voltage peak, so I_max = √2×220/10 = √2×22 A.

V₁,phase = V₁,line = 1100 V (delta) V₂,phase = V₁,phase/5 = 220 V rms V₂,peak = √2 × 220 V Imax = V₂,peak/R = (√2×220)/10 = √2 × 22 A

Concept & Reasoning

Because the primary is delta, V1,phase = V1,line = 1100 V. The per-phase turns ratio N1/N2=5 gives V2,phase=220 V rms. A three-pulse midpoint rectifier connected to the star secondary uses the phase-to-neutral secondary voltages, whose peak is √2V2,phase. For a 10 Ω resistive load, the maximum instantaneous load current is therefore √2×22 A. This calculation is a good example of why line/phase voltage must be resolved before applying transformer ratio and rectifier relations.

How to Solve It in the Exam

Convert line/phase voltage before applying turns ratio.

Important Exam Point

Always write the delta/star line-to-phase relationships; this avoids √3 errors.

Quick Trick

Convert line/phase voltage before applying turns ratio.

Common Mistake

Using line voltage directly on the star-connected secondary without checking the phase relation.

Solution diagram – A step-down delta-star transformer, with per-phase turns ration of 5, is fed…
Exam tipAlways write the delta/star line-to-phase relationships; this avoids √3 errors.
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