Question

For a 3-phase induction motor, what fraction/multiple of supply voltage is required for a direct-on-line starting method such that starting current is limited to 5 times the full-load current and motor develops 1.5 times full-load torque at starting time?
1. 1.632
2. 1.226
3. 0.816
4. 0.456

Answer 1

Correct Answer - Option 2 : 1.226

The full load torque is directly proportional to the square of the voltage.

T_fl ∝ V²

Let the reduced terminal voltage at starting is xV.

Now the starting torque is

T_st ∝ (xV)²

\( \Rightarrow \frac{{{T_{fl}}}}{{{T_{st}}}} = \frac{1}{{{x^2}}}\)

\( \Rightarrow x = \sqrt {\frac{{{T_{st}}}}{{{T_{fl}}}}} \)

Given that, the motor develops 1.5 times full-load torque at the starting time.

T_st = 1.5 T_fl

\(x = \sqrt {1.5} = 1.226\)