Question

A slip ring induction motor drives a constant torque load. If the supply voltage reduces to 1/2 times its previous stator voltage, then slip and rotor current gets modified by factors of, (Assume voltage control method)

• A. 2 and $\frac{1}{\sqrt{2}}$ respectively
• B. 4 and $\frac{1}{2}$ respectively
• C. $\sqrt{2}$ and 2 respectively
• D. 4 and 2 respectively

Answer 1

The correct option is D 4 and 2 respectively

$\text{We know that,}$

$\text{Torque,}T=\frac{3}{{\omega }_{syn}}\times \frac{V^2}{R_2^{\prime }}s\text{(for low slip)}$

$\text{Now, T = constant,}$ $T\propto V^{2}s$

$\text{(or)}$ $V_2^2{s}_2=V_1^2{s}_1$

$\text{(or)}$ $s_2={\left(\frac{V_1}{V_2}\right)}^2{s}_1$

$\text{(or)}$ $s_2=4s_1$

$\text{hence slip increases 4 times,}$

$\text{Also,}$ $T=\frac{3I_2^{\prime 2}}{{\omega }_{syn}}\times \frac{R_2^{\prime }}{s}=$ $\text{constant.}$

$\text{(or)}$ $I_2^{\prime 2}\propto s$

$\frac{I_2^{\prime }}{I_1^{\prime }}=\sqrt{\frac{s_2}{s_1}}=\sqrt{\frac{4}{1}}=2$

$\text{Hence, current increases by 2 times.}$