Final answer:
An auto-transformer with a turn's ratio of approximately 1.58 is used to start a 3-phase cage induction motor with twice the full-load current, resulting in a starting torque about 40% of the full-load torque.
Step-by-step explanation:
This starting method is more efficient and provides higher starting torque than using a stator resistance starter, which would yield only 25% starting torque under the same current limitation.
To start a 3-phase, cage induction motor with a maximum of twice the full-load current, we use an auto-transformer. Given that the motor takes a starting current 5 times its full-load current at normal voltage, the auto-transformer ratio required can be derived from the square root of the ratio of the starting current to the desired starting current. Therefore, the auto-transformer turn's ratio (n) necessary is the square root of (5/2), which equals approximately 1.58 or 158%. This means the auto-transformer should supply the motor with approximately 63.2% (100/158) of the line voltage during start-up.
The starting torque of a motor started with an auto-transformer is proportional to the square of the applied voltage ratio. Since the voltage is reduced to 63.2%, the starting torque would be approximately 40% (0.6322) of the full-load torque. Comparing this method to a stator resistance starter (which limits line current in the same manner), the starting torque using an auto-transformer is significantly higher since the resistance starter would reduce the starting torque to the square of the current ratio, which would be 25% for the same conditions. Therefore, an auto-transformer provides a more efficient start with higher starting torque.