A 240 V, 60 Hz squirrel-cage induction motor has a full-load slip of 0.02 and a full-load speed of 1764 rpm. The winding resistance of the rotor is 0.6 Ω and a winding reactance of 5 Ω at 60 Hz. Determine the rotor current at full-load. a. Ir =-3.53 A b. Ir= 7.89 A c. Ir= 237 A d. Ir= 1.18 A

Respuesta :

Answer:

full load current = 7.891151 A

so correct option is b. Ir= 7.89

Explanation:

given data

Energy E = V = 240 V

frequency =  60 Hz

full-load slip = 0.02

full-load speed = 1764 rpm

winding resistance = 0.6 Ω

winding reactance = 5 Ω

to find out

rotor current at full-load

solution

we will apply here full load current formula that is express as

full load current = [tex]\frac{S*E}{\sqrt{R^2+ (S*X)^2}}[/tex]    ...................1

here S is full-load slip and E is energy given and R is winding resistance and    X is winding reactance

put here value we get

full load current = [tex]\frac{0.02*240}{\sqrt{0.6^2+ (0.02*5)^2}}[/tex]  

full load current = 7.891151 A

so correct option is b. Ir= 7.89