An 8 pole, 50 Hz, 3 phase, slip-ring induction motor has an effective rotor resistance of 0.08 Ω per phase. Its speed at maximum torque is 650 rpm. The additional resistance per phase that must be inserted in the rotor to achieve maximum torque at start is ______Ω . (Round off to 2 decimal places). Neglect magnetizing current and stator leakage impedance. Consider equivalent circuit parameters referred to stator.

The correct answer is 0.52.

Step 1: Calculate the synchronous speed ($N_s$) of the induction motor
The synchronous speed is given by the formula:
$N_s=\frac{120\times f}{P}$
Where:
f = supply frequency = 50 Hz
P = number of poles = 8
Substituting these values:
$N_s=\frac{120\times 50}{8}=750\text{rpm}$

Step 2: Find the slip at maximum torque ($s_m$)
The rotor speed at maximum torque is $N_r=650\text{rpm}$.
The slip at maximum torque is calculated as:
$s_m=\frac{N_s-N_r}{N_s}$
$s_m=\frac{750-650}{750}=\frac{100}{750}=0.1333$

Step 3: Calculate the rotor leakage reactance ($x_2$) per phase
Neglecting stator leakage impedance, the slip at maximum torque is given by:
$s_m=\frac{r_2}{x_2}$
Where $r_2=0.08\Omega$ is the initial rotor resistance per phase.
Solving for $x_2$:
$x_2=\frac{r_2}{s_m}=\frac{0.08}{0.1333}=0.6\Omega$

Step 4: Determine the additional rotor resistance required for maximum torque at starting
At start, the rotor speed $N_r=0$, so the starting slip is $s=1$.
To obtain maximum torque at start, the new slip at maximum torque must be equal to 1.
Let $r_2^{\text{'}}$ be the additional external resistance added per phase to the rotor circuit.
The condition for maximum torque at start becomes:
$s_{m,\text{new}}=\frac{r_2+r_2^{\text{'}}}{x_2}=1$
$r_2+r_2^{\text{'}}=x_2$
$0.08+r_2^{\text{'}}=0.6$
$r_2^{\text{'}}=0.6-0.08=0.52\Omega$

Thus, the additional resistance per phase that must be inserted in the rotor is 0.52 Ω.