In an induction motor, if the supply voltage is halved, the starting torque becomes 

Concept:

The torque equation of a three-phase induction motor is given by,

\(T = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{sE_2^2{R_2}}}{{\left( {R_2^2 + {s^2}X_2^2} \right)}}} \right)\)

At starting slip is equal to one.

\(T_{st} = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{E_2^2{R_2}}}{{R_2^2 + X_2^2}}} \right)\)

By the above expression, we can say that starting torque of an induction motor is directly proportional to the square of applied voltage, when all other factors are constant.

∴ T_st ∝ E²

Where,

R₂ & X₂ are rotor side resistance & reactance respectively

N_s is synchronous speed is rps

Application:

If the supply voltage is halved, then the starting torque becomes one fourth of its previous value.