Question

A 20 kW, 6-pole, 50-Hz, 3-phase slip-ring induction motor runs at 960 revolutions per minute on full load with a rotor current per phase of 40 A. Allowing 300 W for the copper loss in the short circuiting gear and 1100 W for mechanical losses, the resistance per phase of the three-phase rotor winding will be _____$\Omega$. (upto three decimal places)

• A. 0.121

Answer 1

The correct option is A 0.121
$\text{Given Pole, P = 6}$

$f=50Hz$

$\text{Synchronous speed,}{N}_s=\frac{120f}{P}=\frac{120\times 50}{6}=1000rpm$

$\text{slip, s}=\frac{1000-960}{1000}=0.04$

$\text{Mechanical power developed,}$
$P_{mech}=20000+1100$
$=21100W$

$\text{Rotor copper loss =}\frac{s}{1-s}\times \text{mechanical power developed}$
$=\frac{0.04}{(1-0.04)}\times 21100\text{...(i)}$
$\text{Also rotor copper loss}$ $=3I_2^2\times R_2+300$
$=3\times (40{)}^2\times R_2+300$ $\text{...(ii)}$

$\text{from equation (i) and (ii) we get}$

$4800\times R_2+300=\frac{0.04}{(1-0.04)}\times 21100$

$4800R_2=579.17$

$R_2=0.121\Omega$