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.
Choose the applied voltage as a reference for phase angles.
Phase angle of the main winding current
\angle \bar{I}_{m}=-\angle \bar{Z}_{1 m}=-\angle(3+j 2.7)-42^{\circ}The phase angle of the auxiliary winding current with capacitor in series
\angle \bar{I}_{a}=-\angle[(7+j 3)-j / \omega C]Now \alpha=\angle \bar{I}_{a}-\angle \bar{I}_{m}
90^{\circ}=-\tan ^{-1}\left(\frac{3-\frac{1}{\omega C}}{7}\right)-\left(-42^{\circ}\right)
Or \tan ^{-1}\left(\frac{3-\frac{1}{\omega C}}{7}\right)=-48^{\circ}
Or \frac{3-\frac{1}{\omega C}}{7}=-1.11
For \omega=2 \pi \times 50 rad / s, this yields
C=295.5 \mu F