The starting current of a three-phase induction motor is equal to three times the full-load current. If the full-load slip is 2%, then the starting torque as a percentage of full-load torque is

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This question was previously asked in

BPSC AE Paper 6 (Electrical) 25 Mar 2022 Official Paper

Answer 1

Concept:

The ratio between the starting torque to full load torque is given by:

\(\frac{{{T}_{st}}}{{{T}_{fl}}}={{s}_{fl}}\times {{\left( \frac{{{I}_{st}}}{{{I}_{fl}}} \right)}^{2}}\)

Where Tst is the starting torque

Tfl is the full load torque

sfl is the slip at full load

Ist is the starting current

Ifl is the rated current

Calculation:

The starting current of a 3-ϕ induction motor is 3 times the rated current. i.e.

Ist = 3 Ifl

Full load slip (sfl) = 2 %

\(\frac{{{T}_{st}}}{{{T}_{fl}}}={0.02}\times {{\left( \frac{{3}}{{1}} \right)}^{2}}\)

= 18 %