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Re: AC circuits and phase

On Fri, 25 Oct 1996, Chris Deacon wrote:

Here's a problem which I'm having trouble trying to resolve:

Consider a simple RC series circuit - a high pass filter. If I observe
the output signal across the resistor, and compare it to the input signal,
then the phase difference is given by tan(phi) = 1/wCR, which varies from
+pi/2 to 0 as frquency increases. I have no problem with that.

Now let's put an inductor in the circuit, between C and R. The text books
tell me that the phase difference is now tan(phi)=(XL-XC)/R. Obviously
the phase angle is zero at resonance, but this formula seems to tell me that
if the reactance of C is bigger than the reactance of L, then the sign of
phi is *negative*, apparently contradicting my previous paragraph.

Please will somebody help me resolve this apparent contradiction. I tried
deducing the phase myself from the ratio of the imaginary and real parts
of Z, and I got tan(phi)=-(XL-XC)/R, which would seem to satisfy my
observations, but I need to be convinced that this is not just wishful
thinking on my part!

Chris

+-----------------------------+-------------------------------------+
| Christopher Deacon | (709) 737-7631 |
| Dept of Physics and Physical| cdeacon@kelvin.physics.mun.ca !
| Oceanography | |
| Memorial University of Nfld | http://www.physics.mun.ca/~cdeacon |
+-----------------------------+-------------------------------------+


It would appear to me that you have it right.

tan(phi) = (XC - XL)/R

In this case i = Asin(wt + phi) if v = Bsin(wt). Sometimes there is
confusion about whether there is a (+) or and (-) sign on the "phi" in the
equation above.

W. Barlow Newbolt 540-463-8881 (telephone)
108 Parmly Hall 540-463-8884 (fax)
Washington and Lee University newbolt.w@fs.science.wlu.edu
Lexington, Virginia 24450 wnewbolt@liberty.uc.wlu.edu

"The best measure of a man's honesty isn't his income tax return. It's the
zero adjust on his bathroom scale"
Arthur C. Clarke