Discussion
Star-delta starting of an induction motor reduces the starting current to
Answer: C
Correct answer: one third of its direct-on-line value
In star each winding sees \( V_L/\sqrt{3} \), so the phase current falls by \( \sqrt{3} \); and the line current in star equals the phase current, whereas in delta it is \( \sqrt{3} \) times the phase current. The two factors combine to give
\[ \frac{I_{star}}{I_{delta}} = \frac{1}{3} \]The torque, proportional to the square of the voltage, falls to one third as well. The one-ninth figure is a common trap: it is the ratio of the phase currents multiplied out incorrectly.
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