A 30 kW, 4-pole, 400 V, 50 Hz, wound rotor induction motor with Y-connected windings drives a constant torque load. With shorted sliprings, the machine runs at 1476 rpm. When an external resistance of 0.27 Ω per phase is connected in series in the rotor circuit, the speed drops to 1404 rpm. Neglecting rotational losses, the actual per phase rotor winding resistance is _______ Ω(Round off to two decimal places)

The correct option is 0.09.

Let us understand why this option is correct by deriving the solution step-by-step.

First, we determine the synchronous speed ($N_s$) of the induction motor using the formula:
$N_s=\frac{120f}{P}$
where:
- Frequency, $f=50\text{Hz}$
- Number of poles, $P=4$
Substituting these values, we get:
$N_s=\frac{120\times 50}{4}=1500\text{rpm}$

Now, we analyze the two operating conditions of the induction motor driving a constant torque load.

Condition 1: Shorted sliprings (no external resistance)
Let the actual per-phase rotor winding resistance be $r_2$.
The rotor speed under this condition is $N_1=1476\text{rpm}$.
The slip $s_1$ is calculated as:
$s_1=\frac{N_s-N_1}{N_s}=\frac{1500-1476}{1500}=\frac{24}{1500}=0.016$

Condition 2: With external rotor resistance
When an external resistance of $R_{\text{ext}}=0.27\Omega$ per phase is connected in series with the rotor circuit, the total rotor resistance per phase becomes $r_2+0.27$.
The rotor speed drops to $N_2=1404\text{rpm}$.
The new slip $s_2$ is calculated as:
$s_2=\frac{N_s-N_2}{N_s}=\frac{1500-1404}{1500}=\frac{96}{1500}=0.064$

For a three-phase induction motor, the electromagnetic torque $T$ is given by:
$T\propto \frac{s{E}_2^2{R}_r}{R_r^2+{\left(s{X}_2\right)}^2}$
At low values of slip (which is typically the case under normal running conditions), the term ${\left(s{X}_2\right)}^2$ in the denominator is extremely small compared to $R_r^2$ and can be neglected. Under this assumption, the torque equation simplifies to:
$T\propto \frac{s}{R_r}$
Since the motor drives a constant torque load, the torque remains constant ($T_1=T_2$). Therefore, we have:
$\frac{s_1}{r_2}=\frac{s_2}{r_2+R_{\text{ext}}}$

Substituting the values of $s_1$, $s_2$, and $R_{\text{ext}}$:
$\frac{0.016}{r_2}=\frac{0.064}{r_2+0.27}$

Simplifying the ratio by dividing both sides of the numerator by 0.016:
$\frac{1}{r_2}=\frac{4}{r_2+0.27}$

Cross-multiplying to solve for $r_2$:
$r_2+0.27=4r_2$
$3r_2=0.27$
$r_2=0.09\Omega$

Thus, the actual per-phase rotor winding resistance is indeed 0.09 Ω.