Question

The rotor current at starting of a three phase, 460 V, 1710 rpm, 60 Hz, four pole, squirrel-cage. induction motor is six times the rotor -current at full load. The starting torque =______ % of full load torque.

• A. 180

Answer 1

The correct option is A 180
$\text{Synchronous speed,}$
$N_s=\frac{120\times f}{P}=\frac{120\times 60}{4}=1800rpm$

$\text{Full load slip,}$
$s_{fl}=\frac{N_s-N_r}{N_s}=\frac{1800-1710}{1800}=0.05$
$T=\frac{I_2^2{R}_2}{{\omega }_{syn}}\propto \frac{I_2^2{R}_2}{s}$

$\frac{T_{st}}{T_{fl}}={\left[\frac{I_{2\left(st\right)}}{I_{2\left(fl\right)}}\right]}^2\times s_{fl}$

$T_{st}=\left[\frac{6I_{fl}}{I_{fl}}\right]\times 0.05\times T_{fl}=180\%of{T}_{fl}$