Question

A voltage source 𝑣(𝑡)=240 sin(314.16𝑡−20v°)𝑉 is connected to a load having an toa load having an impedance of Z. The resulting current through the load is 𝑖(𝑡)=15 sin(314.16𝑡+22.5°)A. Determine the circuit power factor and identify the elements that constitute the load, given that Z comprises of only two elements connected in series.

Answer 1

From the given values of current and voltage.

$V_{rms}=240/\sqrt2\newline I_{rms}=15/\sqrt2\newline Phase \space angle= 20+22.5=42.5$

This indicates the load is a capacitive load.

Magnitude of impedance$=V_{rms}/I_{rms}$

$Z=240/15=16\varOmega\newline R=Zcos\phi=16cos42.5=11.80\varOmega\newline X=Zsin\phi=10.81\varOmega\newline i.e Z=11.8 - j10. 81$

The capacitor value =$\frac{1}{X\omega}=\frac{1}{10.81\times314.16}=0.294\times10^{-3}F=0.294mF$