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In stator resistance starter, if applied voltage across motor terminals is reduced by 50%, then torque is reduced to ________ of the full voltage value.
In an induction motor, the developed starting torque is directly proportional to the square of the applied stator voltage. When the applied voltage is reduced to 50% (or 0.5 times) of its full value, the starting torque decreases by the square of this factor.
In an induction motor, the developed starting torque is directly proportional to the square of the applied stator voltage. When the applied voltage is reduced to 50% (or 0.5 times) of its full value, the starting torque decreases by the square of this factor.
Ts∝V2 — Starting torque is proportional to the square of the applied voltage.
Tnew=(VfullVnew)2×Tfull=(0.5)2×Tfull=0.25×Tfull — Calculation of torque reduction.
The starting torque Ts of an induction motor is given by the relation Ts∝V2, where V is the applied phase voltage. If the voltage is reduced to k times its original value, the new torque Tnew becomes k2 times the original full-voltage torque Tfull. Since k=0.5, the resulting torque is (0.5)2=0.25 or 25% of the original value.
Stator resistance starters reduce the voltage applied to the motor terminals during starting to limit the inrush current.
A major drawback of this method is the severe reduction in starting torque due to the square-law relationship with voltage.
Because of this, stator resistance starting is generally used only for small, light-load applications.
Simple and inexpensive construction
Provides smooth acceleration
High power loss in the starting resistors
Drastic reduction in starting torque
Small induction motors
Applications where high starting torque is not required
If the voltage is halved (50%), the torque becomes (1/2)2=1/4=0.25, which is 25%.
Option B (12.5%) would be the result if torque were proportional to the cube of voltage, which is incorrect.
Option C (50%) and Option D (75%) are incorrect as they do not follow the T∝V2 relationship.
A is correct — The starting torque is proportional to the square of the voltage, so reducing voltage to 50% reduces the torque to (0.5)2=0.25 or 25%.
Always remember that for an induction motor, starting torque follows a square law with voltage (T∝V2), whereas starting current is directly proportional to voltage (I∝V).
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