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7-7. A 208-V, two-pole, 60-Hz, Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components are R = = 0.200 N R2 = 0.120 N Хм 3D 15.0 N X = 0.410 N X2 = 0.410 2 %3D Pmech = 250 W P, P. = 180 W misc core For a slip of 0.05, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG

7-7. A 208-V, two-pole, 60-Hz, Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components are R = = 0.200 N R2 = 0.120 N Хм 3D 15.0 N X = 0.410 N X2 = 0.410 2 %3D Pmech = 250 W P, P. = 180 W misc core For a slip of 0.05, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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7-7. A 208-V, two-pole, 60-Hz, Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components are R = 0.200 N R2 = 0.120 N Хм 15.0 Q %3D X = 0.410 2 X2 = 0.410 2 P, mech = 250 W = 0 P misc P. = 180 W core For a slip of 0.05, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG
Transcribed Image Text:7-7. A 208-V, two-pole, 60-Hz, Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components are R = 0.200 N R2 = 0.120 N Хм 15.0 Q %3D X = 0.410 2 X2 = 0.410 2 P, mech = 250 W = 0 P misc P. = 180 W core For a slip of 0.05, find (a) The line current IL (b) The stator copper losses PSCL (c) The air-gap power PAG
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