phasor to complex

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.

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.

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

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=80deg Vs=1v Vr=0.25v

This is a psudo capasitace. It is an electrochemical cell and I am simulating it as a RC network..

Some heavy home work!!!!!!!!!!!

"Theta1 would be the angle between Vr and Vc and this would be 90o"

ONLY 90o at resonance frequency remember....

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

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.

Right- well he does know R- so he measures magnitude and phase of Vr wrt Vs and then I=Vr/R.