I read in sci.electronics.design that Wayne <
[email protected]>
wrote (in said:
The resistor is in series with the capacitor and the signal gen is
across one lead of the capacitor and one lead of the resistor. In other
word, if they were both resistors it would be a voltage divider. The
resistor or the capacitor can be ground. Forgive me for asking a dumb
question but why are V_c^2 + V_r^2 = V_s^2 to the power of 2?
Because the two voltages have a phase difference of 90 degrees (or, in
practice, very nearly; the capacitor won't be perfectly loss-free).
If this
was a voltage divider then you would have V?=Vs(R/R+Rc)?
In method 2 could you give me an example with the following figures?
Vs=10v p-p
Vr=4v p-p
Phase = 45deg (with respect Vs)
It's helpful to use a 'phasor diagram' to see what is going on. The
current I is in phase with the voltage across the resistor, and the
voltage across the capacitor is at 90 degrees to the current. So our
diagram is (*use Courier font*):
/|
/ |
/ |
/ |
Vs = 10 / |
/ | Vc =?
/ |
/ |
/ |
/ 45 |
/__________| ----------> Current I
Vr = 4 V
Now, I hope you can see that this is not, in theory, possible. The angle
at the apex of the triangle must be 45 degrees as well and the triangle
must be isosceles. So Vc must be 4 V as well, and then by Pythagoras
(which is the same as the squared values in Tim's equation), Vs =
sqrt(4^2 + 4^2) = sqrt(32) = 5.66 V.
One practical explanation, if your figures are correct, is that the
capacitor is very lossy, thus having a lot of resistance of its own.
This add to the horizontal side of the triangle (without altering the 4
V across the resistor). Another 3.07 V brings the figure right. The Vc
must also be 7.02 V, of course, because of the 45 degrees. We then get
by Pythagoras, Vs = sqrt(7.07^2 + 7.07^2) = sqrt(100) = 10.
It's a very poor capacitor if this is true. There are other
possibilities. What frequency are you working at and what are the
resistor and capacitor values? Have you actually measured 10 V at the
signal generator, or is that just what the output control says? If the
signal generator source impedance isn't very much smaller than the
impedance of your RC circuit, the Vs value will be way off, indeed it
may well be 5.66 V!