Reduction in supply voltage by 10% will change the torque of an induction motor by

Concept:

Torque of induction motor is

\(\rm T=\frac{3}{\omega_s}.\frac{V^2}{\left(\frac{r_2}{s}\right)^2+x^2_2}×\frac{r_2}{s}\)

i.e, Torque is directly proportional to square of the voltage

∴ T ∝ V²

Calculation:

Case 1:

Torque = T₁

Voltage = V₁

Case 2:

Reduction in supply voltage by 10%

Torque = T₂

Voltage = V₂ = 0.9 V₁

\(\frac{T_2}{T_1}=\frac{V_2^2}{V_1^2}\)

\(\frac{T_2}{T_1}=\frac{{(0.9V_1)}^2}{{V_1}^2} = 0.81\)

T₂ = 0.81 T₁

% Change in T =  \(\frac{T_1-0.81T_1}{T_1}\times 100 \%= 19\%\)

Therefore, the correct answer is an option (2)