Question 9.39: The standstill impedances per phase of the inner and outer c...

Let S be the slip at which the two cages develop equal torques.

During running condition,

Now, \frac{I_{2i} }{I_{2o} } = \frac{Z_{2o} }{ Z_{2i}}

or \left\lgroup\frac{I_{2i} }{I_{2o} }\right\rgroup ^{2} = \left\lgroup\frac{Z_{2o} }{ Z_{2i}}\right\rgroup ^{2} = \frac{\left(2/ S\right)^{2} + \left(0.5\right)^{2} }{\left(0.5/ S\right)^{2} + \left(2\right)^{2}}

Power developed in the two cages

and T_{2i} = T_{2o}

or \frac{T_{2i}}{T_{2o}} = 1

or \frac{\left[\left(2/ S\right)^{2} + \left(0.5\right)^{2} \right] \left(0.5 / S\right) }{\left[\left(0.5/ S\right)^{2} + \left(2\right)^{2} \right] \left(2 / S\right)} = 1    or  \frac{\left\lgroup\frac{4}{S^{2}} + 0.25 \right\rgroup \times 0.5}{\left\lgroup\frac{0.25}{S^{2}}+ 4 \right\rgroup \times 2} = 1

or \frac{4}{S^{2}} + 0.25 = \left\lgroup\frac{0.25}{S^{2}}+ 4 \right\rgroup 4    or  \frac{4}{S^{2}} + 0.25 = \frac{1}{S^{2}}+ 16

or \frac{3}{S^{2}} = 15.75    or  S = \sqrt{\frac{3}{15.75}} = 0.4364 = \mathbf{43.64\%}