A 3-phase star connected slip ring induction motor has the following parameters referred to the stator:

$R_s=3\mathrm{\Omega },X_s=2\mathrm{\Omega },X_r^{\prime }=2\mathrm{\Omega },R_r^{\prime }=2.5\mathrm{\Omega }$

The per phase stator to rotor effective turns ratio is 3:1. The rotor winding is also star connected. The magnetizing reactance and core loss of the motor can be neglected. To have maximum torque at starting, the value of the extra resistance in ohms (referred to the rotor side) to be connected in series with each phase of the rotor winding is ____________ (rounded off to 2 decimal places).

The correct answer is 0.26 (with the accepted official range being 0.26 to 0.29).

1. Parameter Identification
From the given problem, the parameters of the star-connected induction motor referred to the stator are:
Stator resistance: $R_s=3\Omega$
Stator reactance: $X_s=2\Omega$
Rotor reactance referred to stator: $X_r^{\prime }=2\Omega$
Rotor resistance referred to stator: $R_r^{\prime }=2.5\Omega$
Stator-to-rotor effective turns ratio: $\frac{N_s}{N_r}=3$

2. Condition for Maximum Starting Torque
To determine the total rotor resistance per phase (referred to the stator side) that yields maximum torque at starting, we find the value of $R_{r,total}^{\prime }$ that maximizes the starting torque equation. Mathematically, this condition is given by:
$R_{r,total}^{\prime }=\sqrt{R_s^2+{(X_s+X_r^{\prime })}^2}$

3. Calculating the Total Required Rotor Resistance
Substituting the given parameter values into the formula:
$R_{r,total}^{\prime }=\sqrt{3^2+{(2+2)}^2}$
$R_{r,total}^{\prime }=\sqrt{9+16}=\sqrt{25}=5\Omega$

4. Extra Resistance Referred to Stator
The total rotor resistance consists of the intrinsic rotor resistance and the added external series resistance:
$R_{r,total}^{\prime }=R_r^{\prime }+R_{ext}^{\prime }$
$5=2.5+R_{ext}^{\prime }$
$R_{ext}^{\prime }=5-2.5=2.5\Omega$

5. Referring the Extra Resistance to the Rotor Side
To find the actual external resistance $R_{ext}$ to be added to the rotor circuit, we refer the stator-referred value back to the rotor side using the turns ratio:
$R_{ext}=\frac{R_{ext}^{\prime }}{{(N_s/N_r)}^2}$
$R_{ext}=\frac{2.5}{3^2}=\frac{2.5}{9}\approx 0.28\Omega$
Rounding to two decimal places, this evaluates to approximately 0.28 Ω, which lies within the acceptable range of 0.26 to 0.29 (with 0.26 designated as the correct key).