Answer:
[tex]\theta = 45^{\circ}[/tex]
Explanation:
There are a couples of ways of solving this problem.
Let's start with the easiest one which is using the following rule:
The angle of the impedance is equal to the power angle and also both of them are equal to the angle between the voltage and current. Therefore:
[tex]\theta=50^{\circ}-5^{\circ}=45^{\circ}[/tex]
This problem can be solved too using Phasors:
A voltage expressed in a cosinusoidal form and with a positive sign can be directly converted into a phasor. This is done by indicating the voltage and the phase shift angle:
[tex]Acos(\omega t +\theta)=A\angle \theta[/tex]
Hence:
[tex]v(t)=20cos(100t+50^{\circ} )=V=20\angle 50^{\circ}[/tex]
[tex]i(t)=15cos(100t-5^{\circ} )=I=15\angle -5^{\circ}[/tex]
Electric power is defined as:
[tex]P=VI=(20\angle 50^{\circ} )*(15\angle -5^{\circ} )=300\angle45^{\circ}[/tex]
So, the power angle is:
[tex]\theta =45^{\circ}[/tex]
