# RLC circuit conversions and angles

Discussion in 'Homework Help' started by notoriusjt2, Jan 27, 2011.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0

I am trying to solve the questions from the link above. I need a clarification on how to find the impedence angles

Question #6.) This one I can solve OK
XL=942.48Ω
Z=1771.5Ω
R-parallel=2092Ω
X-parallel=3329.74Ω
therefore
L=1.326H

Zt=1771.5Ω
θ=tan^-1(R/XL)=32.14 degrees

Question #7.)
Xc=1178.93Ω
Z=1066.55Ω
R-series=455Ω
X-series=964.88Ω
therefore
C=3.29uF

Zt=1066.55Ω
θ=tan^-1(Xc/R)=64.75

the only reason I was able to find θ on question #7 is because I looked at the answer, then tried to figure out what method they used to get it. But why is...
Question #6.)θ=tan^-1(R/XL)
and
Question #7.)θ=tan^-1(Xc/R)

what is the correct method in finding the angles?

2. ### Papabravo Expert

Feb 24, 2006
10,338
1,850
In the z-plane we plot resistance on the horizontal axis and we plot reactance on the vertical (jω) axis. In all cases the angle, measured from the positive horizontal axis, is computed as the arctangent of the reactance over the resistance. The answer key is wrong or there is something else going on.

3. ### amilton542 Active Member

Nov 13, 2010
494
64
Last edited: Jan 27, 2011
4. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
so both of these angles should be calculated using the following?

θ=tan^-1(X/R)

5. ### amilton542 Active Member

Nov 13, 2010
494
64
the easiest method to calculate a vector angle for complex impedances is using polar and rectangular notation which also consists of trigonometry.have you not gone through the links i gave you its all in there