Question

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

$\text{Main winding,}$ $\overline{Z_m}=(3+j2.7)\Omega$

$\text{Auxiliary winding,}$ $\overline{Z_A}=(7+j3)\Omega$

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 current of the two windings at start will be ____$\mu F$.

• A. 295.5

Answer 1

The correct option is A 295.5
Choose the applied voltage as a reference for phase angle,

$\text{Phase angle of main winding current}$
$=-{42}^{\circ }(\angle I_m=-\angle \overline{Z_m})$

The phase angle of the auxiliary winding current with capacitor in series,
$\angle \overline{I_a}=-\angle \left[\right(7+j3)-\frac{j}{\omega C}]$

$\text{Now,}$ $\alpha =\overline{\angle I_a}-\angle \overline{I_m}$

${90}^{\circ }=-ta{n}^{-1}\left[\frac{3-\frac{1}{\omega C}}{7}\right]-\left({42}^{\circ }\right)$

$-ta{n}^{-1}\left[\frac{3-\frac{1}{\omega C}}{7}\right]={48}^{\circ }$

$\frac{3-\frac{1}{\omega C}}{7}=-1.11$

$\frac{1}{\omega C}=3+7.77=10.77$

$C=\frac{1}{2\pi \times 50\times 10.77}=295.5\mu F$