Question 8.A.2: The forward-field impedance of a one-quarter-hp four-pole 11......
The forward-field impedance of a one-quarter-hp four-pole 110-V 60-Hz single-phase induction motor for a slip of 0.05 is given by
Z_{f}=12.4+j16.98\ \OmegaAssume that
X_{m}=53.5~\OmegaFind the values of the rotor resistance and reactance.
Step-by-Step
We have
Z_{f}=\frac{j0.5X_{m}[0.5(R_{2}^{\prime}/s)+j0.5X_{2}^{\prime}]}{0.5(R_{2}^{\prime}/s)+j0.5(X_{2}^{\prime}+X_{m})}Thus
12.4+j16.98=\frac{j26.75[(R_{2}^{\prime}/s)+j X_{2}^{\prime}]}{(R_{2}^{\prime}/2)+j(X_{2}^{\prime}+53.5)}As a result, cross multiplying and separating real and imaginary parts, we obtain
\begin{array}{c}{{0.3652\frac{R_{2}^{\prime}}{s}-0.4636X_{s}^{\prime}=24.8}}\\ {{0.4636\frac{R_{2}^{\prime}}{s}-0.3652X_{s}^{\prime}=33.96}}\end{array}Solving the two equations, we get
X_{2}^{\prime}=2.605\;\Omega \\ {\frac{R_{2}^{\prime}}{s}}=71.208Thus
R_{2}^{\prime}=3.5604~\OmegaRelated Answered Questions
Question: 8.A.3
Verified Answer:
We need the backward-field impedance
Z_{b}=...
Question: 8.A.1
Verified Answer:
The equivalent circuit of the motor yields
...
Question: 8.2
Verified Answer:
In Example 8.1, we obtained
Z_{i}=26.841\an...
Question: 8.1
Verified Answer:
Z_{f}=\frac{j\,36.7(30+j1.2)}{30+j37.9}=22....