Question 9.1: A 2.5-kW 120-V 60-Hz capacitor-start motor has the following......
A 2.5-kW 120-V 60-Hz capacitor-start motor has the following impedances for the main and auxiliary windings (at starting):
Z_{main} = 4.5 + j3.7 ~Ω \quad \text{main winding}\\Z_{aux} = 9.5 + j 3.5 ~Ω \quad \text{auxiliary winding}
Find the value of starting capacitance that will place the main and auxiliary winding currents in quadrature at starting.
The currents \hat{I}_{main} and \hat{I}_{aux} are shown in Fig. 9.4a and b. The impedance angle of the main winding is
\mathcal{\phi}_{main}=\tan ^{-1} \left( \frac{3.7}{4.5} \right) = 39.6°
To produce currents in time quadrature with the main winding, the impedance angle of the auxiliary winding circuit (including the starting capacitor) must be
\mathcal{\phi}= 39.6° – 90.0° = -50.4°
The combined impedance of the auxiliary winding and starting capacitor is equal to
Z_{total}= Z_{aux}+ j X_c = 9.5 + j (3.5 + X_c) Ω
where X_c=- \frac{1}{ωC} is the reactance of the capacitor and ω=2π60≈377 ~rad/sec. Thus
\tan ^{-1} \left( \frac{3.5 +X_c}{9.5} \right) = -50.4° \\ \frac{3.5 + X_c}{9.5} = \tan (-50.4°) = – 1.21
and hence
X_c = -1.21 × 9.5 – 3.5=-15.0 ~Ω
The capacitance C is then
C=\frac{-1}{ωX_c}= \frac{-1}{377 ×(-15.0)} = 177 ~μF
