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 percent. 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

Question

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 percent. 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.

Accepted Answer

The equivalent circuit of the motor yields

[latex]Z_{f} =\frac{\left\{\left[1.78/2\left(0.5\right) \right]+j0.64 \right\}\left(j12.5\right) }{\left[1.78/2\left(0.05\right) \right]+j\left(0.64+12.5\right) } =5.6818+j8.3057[/latex]

[latex]Z_{b} =\frac{\left\{\left[1.78/2\left(1.95\right) \right]+j0.64 \right\}\left(j12.5\right) }{\left[1.78/2\left(1.95\right) \right]+j\left(0.64+12.5\right) } =0.4125+j0.6232[/latex]

As a result of the problem specifications

[latex]P_{i} =490 W[/latex] [latex]\cos \phi =0.57[/latex] [latex]V=110[/latex]

[latex]I_{i} =\frac{P_{i}}{V \cos \phi}=\frac{590}{110\left(0.57\right) } = 7.815\angle -55.2488°[/latex]

Thus the input impedance is

[latex]Z_{i}=\frac{V}{I_{i}} =\frac{110}{7.815} \angle 55.2498°=8.023+j11.5651[/latex]

The stator impedance is obtained as

[latex]Z_{1}=Z_{i}-\left(Z_{f}+Z_{b}\right) =1.9287+j2.636\Omega[/latex]

Question 6.7: A single-phase induction motor takes an input power of 490 W...

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