Question 8.A.1: A single-phase induction motor takes an input power of 490 W......
A single-phase induction motor takes an input power of 490 W at a power factor of 0.57 lagging from a 110-V supply when running at a slip of 5%. Assume that the rotor resistance and reactance are 1.78 Ω and 1.28 Ω, respectively, and that the magnetizing reactance is 25 Ω. Find the resistance and reactance of the stator.
Step-by-Step
The equivalent circuit of the motor yields
Z_{f}=\frac{\{\{1.78/2(0.05)\}+j0.64\}(j12.5)}{[1.78/2(0.05)]+j(0.64+12.5)}=5.6818+j8.3057 \\ Z_{b}=\frac{\{\{1.78/2(1.95)\}+j0.64\}(j12.5)}{[1.78/2(1.95)]+j(0.64+12.5)}=0.4125+j0.6232As a result of the problem specifications:
P_{i}=490\,\mathrm{W}\qquad\cos\phi=0.57\qquad V=110 \\ l_{i}={\frac{P_{i}}{V\cos\phi}}={\frac{590}{110(0.57)}}=7.815\angle{\mathrm{-}}55.2488{\mathrm{~deg}}Thus the input impedance is
Z_{i}={\frac{V}{I_{i}}}={\frac{110}{7.815}}\,\angle55.2498\ \mathrm{deg}=8.023+j11.5651The stator impedance is obtained as
Z_{1}=Z_{i}-(Z_{f}+Z_{b})=1.9287+j2.636\,\OmegaRelated Answered Questions
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