Answer :

0.52

Given:

Pole (P) = 8

Frequency (f) = 50 Hz

Rotor resistance (r₂) = 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

$s_m={\displaystyle \frac{r_2}{x_2}}$

Consider additional rotor resistance to be added = r_2'

Rotor reactance = x₂

Synchronous speed Ns = 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

$s_m={\displaystyle \frac{N_s-N_r}{N_s}}={\displaystyle \frac{750-650}{750}}=0.1333$

At maximum torque

$s_m={\displaystyle \frac{r_2}{x_2}}\Rightarrow 0.1333={\displaystyle \frac{0.08}{x_2}}$

x₂ = 0.6 Ω

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

⇒$1={\displaystyle \frac{r_2+r_2^{\prime }}{x_2}}$

x₂ = r₂ + r₂'

0.6 = 0.08 + r₂'

r₂' = 0.52 Ω

∴ The value of additional rotor resistance to be added (r_2') at the time of starting is 0.52 Ω.