Question

A three-phase slip ring induction motor develops full load torque at a slip of 0.04 when the slip rings are short circuited. If the rotor resistance is increased 3 times by inserting external resistance in the rotor circuit, then the slip will be:
1. 0.04
2. 0.01
3. 0.12
4. 0.06

Answer 1

Correct Answer - Option 3 : 0.12

Concept:

Rotor Resistance Control:

For induction motor, the torque in rotor resistance control methods is given as

\({\rm{{\rm T}}} = \frac{{{K_t}s}}{{{R_2}}}\)

where

s = slip

R₂ = Rotor Resistance

R₂¹ = Rotor resistance become three times = 3R₂

For full load torque \(s\; \propto \;{R_2}\)

Calculation:

Given

s₁ = 0.04

\(\begin{array}{l} {s_2} = \frac{{{R_2}{^1} }}{{{R_2}}} \times {s_1}\\ = \frac{3R_2}{R_2} \times 0.04 \end{array}\)

= 0.12