Hi All,

Referring to Question 9 of the worksheet titled "Series and parallel AC circuits" (http://www.allaboutcircuits.com/worksheets/ac_s_p.html),
I get the phase angle at -42.08 by doing the following steps:

1.) Calculated Xc = 0 - j1354.51Ω
2.) Calculated the phase difference at -42.08° using θ = arctan Xc/R

The worksheet answer is -47.9°, which seems like it's a complimentary
angle to -42.08°

Could anybody please help me understand why the phase difference is
not -42.08° as calculated in step 2 above?

Sincerely,
tjterblanche

 

because you are basically measuring across the capacitor, you will be referring to the reactive component voltage, which for the angle you would then use arcsine(Xc/Z).

oops, that's wrong, you would use arcsine(R/Z)

 

It seems like you're calculating the complementary angle to 42.08° when you use θ=arcsine R/Z?

IMHO there are two ways to draw the phasor triangle for any circuit with a single reactive component.

1.) You start by drawing the reactive component's phasor first (from the origin); You then draw the resistive phasor with the end of the reactive phasor as its origin;

2.) You draw the resistive phasor first; You then draw the reactive phasor with the end of the resistive phasor as its origin.

Both these phasor diagrams will have the exact same hypotenuse (magnitude and angle), but their right angles are at opposite sides of the hypotenuse.

Which phasor diagram do you use and why, because it determines the angle that θ points to?

 

I wanted to include this as an example:

The VERY FIRST phasor diagram on the Series Resistor-Capacitor Circuits page (http://www.allaboutcircuits.com/vol_2/chpt_4/3.html) shows the phase angle between \(E_{R}\) and \(E_{T}\) to be -79.3°.

The way I understand it, θ=arcsine R/Z would calculate the phase angle between \(E_{C}\) and \(E_{T}\)

 

I think your reading too much into it. There is only one reactive component, and by convention, the right angle is to the left.