Given:

Pole (P) = 8

Frequency (f) = 50 Hz

Rotor resistance (r_{2}) = 0.08 Ω

**Rotor speed at maximum torque = 650 RPM**

Let's consider the slip at maximum torque as s_{m,}

Slip at maximum torque is given by

**\(s_m = \dfrac {r_2}{x_2}\)**

Consider additional rotor resistance to be added = r_{2}'

Rotor reactance = x_{2}

Synchronous speed N_{s} = 120f/P = 120 × 50 / 8 = 750 RPM

As the speed of the motor at maximum toque condition is 650 RPM

Slip at maximum torque is given by

**\(s_m = \dfrac {N_s - N_r}{N_s} =\dfrac {750-650}{750}= 0.1333\)**

At maximum torque

\(s_m =\dfrac {r_2}{x_2} ⇒ 0.1333 = \dfrac {0.08}{x_2}\)

**x**_{2} = 0.6 Ω

To get the maximum torque at starting the slip at maximum torque should be equal to slip at starting (s=1).

⇒ \(1 =\dfrac {{r_2+r_2'}}{ x_2}\)

x_{2} = r_{2} + r_{2}'

0.6 = 0.08 + r_{2}'

**r**_{2}' = 0.52 Ω

**∴ The value of additional rotor resistance to be added (r2') at the time of starting is 0.52 Ω.**