Question

A 3-phase, 4 pole, 50 Hz induction motor has rotor copper losses equal to 300 W. The torque developed by this motor at 4% slip will be _____ N-m.

• A. 47.75

Answer 1

The correct option is A 47.75
$\text{We know,}{P}_{cu}=s{P}_g\text{.... (copper loss in terms of air gap power)}$

$\Rightarrow P_g=\frac{P_{cu}}{s}=\frac{300}{0.04}=7500W$

$\text{Now,}$ $P_m=P_g(1-s)=7500(1-0.04)=7200W$

$\text{Here synchronous speed,}$ $N_s=\frac{120f}{P}=\frac{120\times 50}{4}=1500rpm$

$\text{Torque developed =}\frac{P_g}{{\omega }_s}$$=\frac{7500}{2\pi \times \frac{1500}{60}}=47.75N-m$

$\text{Alternatively, torque can be calculated as}$

$T=\frac{P_m}{{\omega }_s(1-s)}=\frac{7200}{2\pi \times \frac{1500}{60}(1-0.04)}$
$=47.75N-m$