Hi
I have calculated the following network using the cosine law. ___SIGGEN (Vs)_ ¦ ¦ ¦--/\/\/\/\-----¦ ¦----¦ ¦ Vs Vc ¦
Hi
I have calculated the following network using the cosine law. ___SIGGEN (Vs)_ ¦ ¦ ¦--/\/\/\/\-----¦ ¦----¦ ¦ Vs Vc ¦
-- GND Vs^2=Vr^2+Vc^2 - 2VrVc Cos(theta1)
Homework ?
Wayne wrote:
Theta1 would be the angle between Vr and Vc and this would be 90o. Vr and Vc form a right angle. So there is no cosine term. The voltage divider rule still applies with z=r-jx so that Vr=Vs*r/(r-jx) and Vc=Vs*(-jx)/(r-jx) then use 1/(r-jx)=(r+jx)/(r^2+x^2) and Vs=Vr+Vc to get the "format" you want.
Use a Smith Chart. Piece of cake unless your teacher wants you to show an absolute result with working.
-- "What is now proved was once only imagin'd." - William Blake, 1793.
Wayne,
I'm not sure what this is supposed to represent. I think you meant to use Vr under the resistor, not Vs. So it looks like you have a signal generator, Vs, feeding a series combination of a resistor and a capacitor.
For what you are doing I am assuming the real part is along the x-axis and the imaginary part is along the y-axis. That is the usual orientation for the complex plane in doing electronics where Z = R + jX (X is positive for inductive impedance and negative for capacitive impedance)
This gives an impedance with R = |Z|cos(theta) and X = |Z|sin(theta)
So you can write |Z|cos(theta)+j|Z|sin(theta) as the complex impedance.
|Z| = sqrt(R^2 + X^2) it also follows that tan(theta) = X/R
Since Vr = RVs/(R^2 + X^2)^(1/2) and you know Vr/Vs = .25 you can figure out what X is.
R/(R^2 + X^2)^(1/2) = .25, R = .25(R^2 + X^2)^(1/2), Squaring both sides you get R^2 = (1/16) (R^2 + X^2) Solving for X^2 gives X^2 = 15R^2. So X = (R)(sqrt 15)
This gives a theta of 75deg not 80deg.
On the other hand, if your angle is correct, it looks like you should have measured about .17v not .25v.
tim
Homework is $20 a question, $125/hour. Paypal or bank check ...
-- Nicholas O. Lindan, Cleveland, Ohio Consulting Engineer: Electronics; Informatics; Photonics. Remove spaces etc. to reply: n o lindan at net com dot com
---------------- You don't. It's not worth the bother and teaches you nothing.
theta1 is the angle of which voltage? With respect to what? I am guessing that it is the phase of Vc with respect to the voltage Vr (check sign of angle )
Look at the circuit- apply KVL in phasor form. ---done.
-- Don Kelly dhky@peeshaw.ca
-- "Fred Bloggs" wrote in message news:41719BBF.3000001@nospam.com...
This is a psudo capasitace. It is an electrochemical cell and I am simulating it as a RC network..
Some heavy home work!!!!!!!!!!!
Cheers
Prolly; and he was asleep in simple Geometry...
In that case I stand by my original posting but suggest that a) your measurements are not correct b) your model is not sufficient
Glenn
Wayne wrote:
Tim
Your write, sorry, mis-place an 's'. The circuit should read:
Hi
I have calculated the following network using the cosine law. ___SIGGEN (Vs)_ ¦ ¦ ¦--/\/\/\/\-----¦ ¦----¦ ¦ Vr Vc ¦
-- GND Vs^2=Vr^2+Vc^2 - 2VrVc Cos(theta1)
"Theta1 would be the angle between Vr and Vc and this would be 90o"
ONLY 90o at resonance frequency remember....
-- "Fred Bloggs" wrote in message news:41719BBF.3000001@nospam.com...
have you taken the capacitance / resistance of the measurment tool into account ? at low C or high R they are important too....
In that case I stand by my original posting but suggest that a) your measurements are not correct b) your model is not sufficient
Glenn
Wayne wrote:
Yes, by subtraction.
How do you know that the "capacitor" doesn't have some loss component? I suspect that this is the case.
Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
If that is the case then he uses Z=Vs/I which is again a measured phasor for I- he will not be able to get away with just magnitude measurement. Then he uses Zunknown=Z-R and there you have it. Zunknown= Ru-jXu.
Don I don't. I am going to try and work out what my electrochemical cell looks like in terms of a RC network.
Wa
If you are modeling an electrochemical cell, i can guarantee losses...
He has a magnitude and a phase angle listed. I suspect that the phase angle is that of the voltage across the "capacitor" with respect to that across the resistor (i.e. with respect to the current.) I don't recall him any impedance values- just the voltages. theta1=80deg
I asked Wayne for clarification as to what is what he has adrawn two Vs values and doesn't specify which voltage is associated with theta1. so the problem is: a) 1 @ 80 =0.25+jVc assuming a pure capacitor - this doesn't work as has been pointed out. b) 1 @ ?? =0.25 +|Vc| @80 does have a valid solution for |Vc| and for ??
Take your choice.
-- Don Kelly dhky@peeshaw.ca
Right- well he does know R- so he measures magnitude and phase of Vr wrt Vs and then I=Vr/R.
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