Question

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

$\text{Main winding}$ ${\overrightarrow{Z}}_{lm}=3+j2.7\Omega$;
$\text{Auxiliary winding}$ ${\overrightarrow{Z}}_{la}=7+j3\Omega$

The value of capacitor in $\mu$F to be connected in series with auxiliary winding to achieve a phase difference of $\alpha ={67}^{\circ }$ between the currents of two winding at start is

• A. 147.75
• B. 508.145
• C. 295.5
• D. 259.07

Answer 1

The correct option is B 508.145

$\text{Phase angle of main winding current}$

$\angle I_m=-\angle Z_m=-\angle ta{n}^{-1}\frac{2.7}{3}=-{42}^{\circ }$

Phase angle of auxiliary winding current with capacitor in series,
$\angle I_a=-\angle ta{n}^{-1}\frac{(3-\frac{1}{\omega C})}{7}$

$\text{Now phase difference,}$ $\alpha ={67}^{\circ }$ (Given)

${67}^{\circ }=\angle I_a-\angle I_m$

${67}^{\circ }=-ta{n}^{-1}\frac{(3-\frac{1}{\omega C})}{7}+{42}^{\circ }$

$ta{n}^{-1}\left(\frac{3-\frac{1}{\omega C}}{7}\right)=-{25}^{\circ }$

$\frac{3-\frac{1}{\omega C}}{7}=-0.466$

$3-\frac{1}{\omega C}=-3.2641$

$\frac{1}{\omega C}=3+3.2641$

$\frac{1}{\omega C}=6.2641$

$C=\frac{1}{2\pi \times 50\times 6.2641}=508.145\mu F$