Question

A 20 kW, 6-pole, 50 Hz, 3-phase slip ring induction motor runs at 950 rpm on full load with a rotor current per phase of 38 A. Assuming 2 kW mechanical losses, the rotor resistance per phase (in ohms) is ________

• A. 0.267

Answer 1

The correct option is A 0.267
$\text{Gross mechanical power developed = 20 + 2 = 22 kW}$

$\text{Rotor copper loss =}$ $\frac{s}{1-s}\times 22\times {10}^3$

$s=\frac{1000-950}{1000}=\frac{50}{1000}=0.05$

$\text{Rotor copper loss =}$ $\frac{0.05}{1-0.05}\times 22\times {10}^3$

$=1157.9W$

$3I_2^2{R}_2=1157.9$

$3(38{)}^2\times R_2=1157.9$

$R_2=0.267\Omega$