The main and auxiliary winding impedances of a 50 Hz, capacitor-start single-phase induction motor are:

Main winding $\bar{Z}_{l m}=3+j 2.7$

Auxiliary winding $\bar{Z}_{l a}=7+j 3$

Determine the value of the capacitor to be connected in series with the auxiliary winding to achieve a phase difference of $\alpha=90^{\circ}$ between the currents of the two windings at start.

Question

The main and auxiliary winding impedances of a 50 Hz, capacitor-start single-phase induction motor are:

Main winding [latex] \bar{Z}_{l m}=3+j 2.7[/latex]

Auxiliary winding [latex] \bar{Z}_{l a}=7+j 3[/latex]

Determine the value of the capacitor to be connected in series with the auxiliary winding to achieve a phase difference of [latex] \alpha=90^{\circ}[/latex] between the currents of the two windings at start.

Accepted Answer

Choose the applied voltage as a reference for phase angles.

Phase angle of the main winding current

[latex] \angle \bar{I}_{m}=-\angle \bar{Z}_{1 m}=-\angle(3+j 2.7)-42^{\circ}[/latex]

The phase angle of the auxiliary winding current with capacitor in series

[latex] \angle \bar{I}_{a}=-\angle[(7+j 3)-j / \omega C][/latex]

Now [latex] \alpha=\angle \bar{I}_{a}-\angle \bar{I}_{m}[/latex]

[latex] 90^{\circ}=- an ^{-1}\left(\frac{3-\frac{1}{\omega C}}{7} ight)-\left(-42^{\circ} ight)[/latex]

Or [latex] an ^{-1}\left(\frac{3-\frac{1}{\omega C}}{7} ight)=-48^{\circ}[/latex]

Or [latex] \frac{3-\frac{1}{\omega C}}{7}=-1.11[/latex]

For [latex] \omega=2 \pi imes 50 rad / s[/latex], this yields

[latex] C=295.5 \mu F[/latex]

Question 10.4: The main and auxiliary winding impedances of a 50 Hz, capaci...

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