Question

A 6 pole, 50 Hz induction motor has an equivalent rotor resistance of 0.01 $\Omega$/phase. If its stalling speed is 900 rpm, the resistance that must be inserted in rotor windings per phase to obtain maximum torque at starting is

• A. 0.1 $\Omega$/phase
• B. 0.09 $\Omega$/phase
• C. 0.33 $\Omega$/phase
• D. 0.03 $\Omega$/phase

Answer 1

The correct option is B 0.09 $\Omega$/phase

$\text{Synchronous speed,}{N}_s=\frac{120\times 50}{6}=1000\text{rpm}$

$\text{Stalling speed = 900 rpm}$

$\text{Slip at stalling speed,}s=\frac{1000-900}{1000}=0.1$

$\text{Slip at maximum torque;}$
$S_{mT}=\frac{R_2}{X_2}=\frac{0.01}{X_2}$

$\because$ $0.1=\frac{0.01}{X_2}$

$X_2=0.1\Omega$

To obtain maximum torque at starting,
Let rotor resistance $=R_2^{\prime }$
at starting s = 1

$S_{mT}=\frac{R_2^{\prime }}{X_2}$

$\Rightarrow 1=\frac{R_2^{\prime }}{0.1}$

$\Rightarrow R_2^{\prime }=0.1\Omega$/phase

The external resistance to be added

$R_{ext}=0.1-0.01$

$=0.09\Omega /\text{phase}$