How do I calculate Phase Angles?

[shintashi]

So I got to the part of my math/engineering textbook on vectors and scalars, and I've got to calculate phase angles and voltage. I see this equation, and I've tried doing it different ways but I don't get correct answers.

tanθ=X_L−X_C / X_R
Angle θ represents the phase angle between the current and the voltage.

I thought I was supposed to square each value first,
then subtract C from L,
then divide by R
then get the square root of the result,
then push the inverse button + Tangent on the calculator to change the result into an angle in degrees.

I didn't do well with vector spaces in school and have trouble with them now.

Also, am I supposed to do any RMS stuff like root 2 over 2, since its AC, and implied? Or am I overthinking it?

 

[cnh1995]

X_L and X_c are 180° out of phase. You can subtract one from the other directly, depending on which one is greater. You don't need rms values for impedance.

 

[jim hardy]

It helps to draw the picture first.
That makes it into an elementary trigonometry problem. Which you solve one step at a time...
1. Find what is impedance . You have to add in Rectangular coordinates (unless angle is same).
2. Divide voltage by that impedance. You have to divide in polar co-ordinates .

You'll get good at rectangular-polar conversion. Pythagoras Rules!

Doing these beginner exercises with a slide rule instills the "one step at a time" thinking method.
If you're using a calculator, force yourself to think in steps by writing down each step and its result , until the process becomes automatic..

 

[shintashi]

thank you both! I was definitely overthinking it. The numbers aren't squared when doing the fraction portion to calculate phase angle.
it's just the bigger number minus the smaller number in C & L, and then divide by R. Once i have that value,
1. type in the (probably decimal) value into the calculator
2. push INVerse on the calculator
3. push TANgent. Boom, Phase angle.

Then for calculating the voltage, its Pythagoras: a^2 + b^2 = c^2,
1. with a = the result from C - L, then squared
2. b = Resistance, then squared
3. add these two together, which is the c^2 value
4. square root that value and I get my voltage.

I checked this process and i got the same results as the answers in the back of the book.