Question

A single phase, 230 V, 50 Hz, 4 pole, capacitor start induction motor has the following standstill impedance.
Main winding: $Z_m=(8.0+j5.0)\Omega$ and
auxiliary winding: $Z_a=(9.0+j6.0)\Omega$.
The value of the starting capacitor required to produce ${90}^{\circ }$ phase difference between the current in the main and auxiliary windings will be _____ $\mu F.$

• A. 156.03

Answer 1

The correct option is A 156.03
$\text{Main winding impedance:}$
$Z_m=(8.0+j5.0)\Omega$
$=9.434\angle {32}^{\circ }\Omega$

$\text{Auxiliary winding impedance:}$
$Z_a=(9.0+j6.0)\Omega$
$=10.82\angle {33.69}^{\circ }\Omega$




Consider the above circuit:
$\overrightarrow{I_m}$ lags behind voltage $\overrightarrow{V}$ by angle ${\theta }_m$

$\Rightarrow {\theta }_m={32}^{\circ }$

Total impedance of auxiliary winding circuit,
$\overrightarrow{Z_{aw}}=\overrightarrow{Z_a}+\frac{1}{j\omega C}$
$=9-j(\frac{1}{\omega C}-6)=Z_{aw}\angle -{\theta }_A$

$\text{Where,}$ ${\theta }_A=ta{n}^{-1}\frac{(\frac{1}{\omega C}-6)}{9}$

from the circuit diagram, we can draw the phase as below:

$\text{Where,}$ ${\theta }_A+{\theta }_m={90}^{\circ }$

$ta{n}^{-1}\left(\frac{\frac{1}{\omega C}-6}{9}\right)+{32}^{\circ }={90}^{\circ }$

or, $ta{n}^{-1}\left(\frac{\frac{1}{\omega C}-6}{9}\right)={58}^{\circ }$

or, $\frac{1}{\omega C}-6=14.4$

$\frac{1}{\omega C}=20.4$

$C=\frac{1}{2\pi \times 50\times 20.4}=156.03\mu F$