Question

A 40 kW induction motor with a star delta starter is supplied through a feeder connected on delta side of starter from 400 V, 50 Hz mains. With the feeder having only resistance drop and negligible reactance drop, the starting torque in motor is found to be the same with the star or delta connections of stator winding. On short circuit test on motor with delta winding following data are obtained (without feeder):
V = 200 V; I = 100 A; Power factor = 0.4.
The resistance of the feeder in ohm is (Assume turns ratio to be 1 for star delta starter)

• A. 0.67
• B. 2
• C. 4
• D. 1.154

Answer 1

The correct option is B 2

From short circuit per phase equivalent circuit parameters

$Z_{sc}=\frac{200}{100}=3.464\Omega$

$R_{sc}=Z_{sc}cos{\varphi }_{sc}=3.464\times 0.4=1.3856\Omega$

$X_{sc}=\sqrt{{3.464}^2-{1.3856}^2}=3.1748\Omega$

Let R be resistance of each feeder line.
Starting torque with star connection.

$T_{st}=\frac{3}{{\omega }_s}\frac{V_1^2}{(R+1.3856{)}^2+(3.175{)}^2}\times 1.3856$

Starting torque with delta connection winding parameter should be changed into equivalent star connection.

$T_{st}=\frac{3}{{\omega }_s}\frac{V_1^2}{{(R+\frac{1.3856}{3})}^2+{\left(\frac{3.175}{3}\right)}^2}\times \frac{1.3856}{3}$

For same starting torque,
$(R+1.386{)}^2+(3.175{)}^2=[{(R+\frac{1.3856}{3})}^2+{\left(\frac{3.175}{3}\right)}^2]\times 3$

$-2R^2+8.00=0$

$R^2=4$

$R=2\Omega$