Question

A 3 phase, 4 pole, 400 V, 10 kW slip ring induction motor has rotor resistance = 0.16 Ω per phase and stator resistor of 0.27 Ω per phase. The voltage across the slip rings at standstill is 141 V per phase. The motor develops a torque of 62.4 Nm at a slip of 0.08 and the rotor current is 6 A. What is the rotor current if a voltage is injected to the rotor so that the motor runs of slip s = 0.02 and develops same torque?
1. 6 A
2. 5 A
3. 4 A
4. 3 A

Answer 1

Correct Answer - Option 4 : 3 A

Given that, I₂ = 6 A

S₂ = 0.08

\(S{'_2} = 0.02\)

In an induction motor, the torque is related as

\(T \propto \frac{{3I_2^2{R_2}}}{s}\)

So, \(\frac{{I_2^2}}{{{s_2}}} = \frac{{{{\left( {I_2^1} \right)}^2}}}{{s_2^2}}\)

\( \Rightarrow \frac{{{6^2}}}{{0.08}} = \frac{{{{\left( {I_2^1} \right)}^2}}}{{0.02}}\)

\( \Rightarrow I_2^1 = 3A\)