Question 6.5: The following parameters are available for a 60-Hz four-pole...

Z_{f} =\frac{j36.7\left(30+j1.2\right) }{30+j37.9} =22.796\angle 40.654°
=17.294+j14.851\Omega

The result above is a direct application of Eq. (6.32). Similarly, using Eq. (6.33), we get

Z_{f} =\frac{j\left(X_{m} /2\right)\left[\left(R^{′}_{2}/2s\right)+j\left(X^{′}_{2}/2\right) \right] }{\left(R^{′}_{2}/2s\right)+j\left[\left(X_{m}+ X^{′}_{2}\right)/2 \right] }              (6.32)

Z_{b}=\frac{j\left(X_{m}/2 \right){\left[R^{′}_{2}/2\left(2-s\right) \right] +j\left(X^{′}_{2}/2\right) } }{\left[R^{′}_{2}/2\left(2-s\right)+j\left[\left(X_{m} +X^{′}_{2}\right)/2 \right] \right] }                          (6.33)

Z_{b}=\frac{j36.7\left[\left(1.5/1.95\right)+j1.2 \right] }{\left(1.5/1.95\right) +j37.9} =1.38 \angle 58.502°\Omega
=0.721+j1.766\Omega

We observe here that |Z_{f}| is much larger than |Z_{b}| at this slip, in contrast to the situation at starting (s = 1), for which Z_{f} = Z_{b}.

The input impedance Z_{i} is obtained as

Z_{i}= Z_{1}+Z_{f}+Z_{b}=19.515+j18.428
=26.841\angle 43.36° \Omega