A 3-phase, 415 V, 4-pole, 50 Hz induction motor draws 5 times the rated current at rated voltage at starting. It is required to bring down the starting current from the supply to 2 times of the rated current using a 3- phase autotransformer. If the magnetizing impedance of the induction motor and no load current of the autotransformer is neglected, then the transformation ratio of the autotransformer is given by ______. (round off to two decimal places)

The correct answer is 0.63.

Concept:
When an induction motor is started using a 3-phase autotransformer, the voltage applied to the motor terminals is reduced to a fraction x (where x is the transformation ratio of the autotransformer) of the rated supply voltage.
Since the motor current is directly proportional to the applied voltage:
The starting current drawn by the motor from the autotransformer secondary is:
$I_{st,motor}=x{I}_{sc}$
where $I_{sc}$ is the short-circuit (starting) current of the motor at full rated voltage.

The starting current drawn from the supply (primary side of the autotransformer), neglecting no-load and magnetizing currents, is given by:
$I_{st}=x{I}_{st,motor}=x^2{I}_{sc}$

Calculation:
Let the rated (full-load) current of the induction motor be $I_{fl}$.
Given that the starting current at rated voltage (direct-on-line starting current, $I_{sc}$) is 5 times the rated current:
$I_{sc}=5I_{fl}$

We want to limit the starting current from the supply to 2 times the rated current:
$I_{st}=2I_{fl}$

Substituting these values into the starting current relation:
$2I_{fl}=x^2\left(5I_{fl}\right)$

Solving for $x^2$:
$x^2=\frac{2I_{fl}}{5I_{fl}}=\frac{2}{5}=0.4$

Taking the square root to find the transformation ratio x:
$x=\sqrt{0.4}\approx 0.6324$

Rounding off to two decimal places, we get:
$x=0.63$