Question

A 4-pole, 3-phase, 400 V, 50 Hz induction motor has following parameters for its circuit model (referred to stator side on equivalent - star basis)
$r_1=1.2\Omega$ $x_1=1.16\Omega$
$r_2^{\prime }=0.4\Omega$ $x_2^{\prime }=1.16\Omega$
$X_m=35\Omega$
for a speed of 1440 rpm, the input current to the motor is _____A.

• A. 21.09

Answer 1

The correct option is A 21.09
Given,
$N_r=1440rpm$

$N_s=\frac{120\times f}{P}=\frac{120\times 50}{4}=1500rpm$

$\text{Slip, s =}$ $\frac{1500-1440}{1500}=0.04$

$\text{Rotor resistance at s = 0.04,}$

$r_2^{\prime }=\frac{0.4}{0.04}=10\Omega$

$\text{Consider the circuit diagram shown below,}$

$Z_1=\frac{j35(10+j1.16)}{10+j(35+1.16)}$

$=\frac{352.45\angle {96.6}^{\circ }}{37.51\angle {74.5}^{\circ }}=9.4\angle {22.1}^{\circ }\Omega$

$=(8.703+j3.53)\Omega$

$\text{Total impedance of circuit,}$
$\overrightarrow{Z_1}=(8.703+j3.53)+(1.2+j1.16)$
$=9.90+j4.69=10.955\angle {25.35}^{\circ }\Omega$

$\text{Stator current,}$ $I_1=\frac{231}{10.955}=21.09A$