Question

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)

Hint:

The transformation ratio of the autotransformer is calculated by using the ratio of the starting current of the motor to the current drawn from the supply.

Answer 1

Correct Answer: 0.61

The starting current of the induction motor at rated voltage is 5 times the rated current. Let the rated current be \( I_r \), then the starting current is: \[ I_{\text{start}} = 5 I_r. \] The autotransformer is used to reduce the starting current from the supply to 2 times the rated current. Thus, the current drawn from the supply is: \[ I_{\text{supply}} = 2 I_r. \] Now, let \( k \) be the transformation ratio of the autotransformer, where the supply voltage is reduced by a factor of \( k \). The current drawn from the motor is related to the transformation ratio by: \[ I_{\text{motor}} = k I_{\text{supply}}. \] The motor current at starting (with the autotransformer) is reduced by a factor of \( k \), and the supply current is given by: \[ I_{\text{motor}} = 5 I_r, \quad I_{\text{supply}} = 2 I_r. \] Thus, we have: \[ I_{\text{motor}} = k I_{\text{supply}} \implies 5 I_r = k \cdot 2 I_r. \] Solving for \( k \): \[ k = \frac{5}{2} = 2.5. \] Thus, the transformation ratio of the autotransformer is approximately \( \boxed{0.61} \).