Question

A 230 V, 50 Hz, 4-pole, single phase induction motor is rotating in the clockwise (forward) direction at a speed of 1420 rpm then the backward slip and effective rotor resistance in backward branch of equivalent circuit respectively will be_____ (Assuming rotor resistance at standstill to be $7.5\Omega$).

• A. 0.025, 3.85 $\Omega$
• B. 1.947, 3.85 $\Omega$
• C. 0.053, 7.2 $\Omega$
• D. 1.947, 1.93 $\Omega$

Answer 1

The correct option is D 1.947, 1.93 $\Omega$

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

$\text{Forward slip,}$ $s_f=\frac{N_s-N_r}{N_s}=\frac{1500-1420}{1500}=0.053$

$\text{Backward slip,}$ $s_b=(2-s)=(2-0.053)=1.947$

The effective rotor resistance in backward branch
$=\frac{R_2}{2(2-s)}=\frac{7.5}{2(2-0.053)}$

$=1.926\Omega \simeq 1.93\Omega$