phasor to complex

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)
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Reply to
Wayne
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Homework ?

Wayne wrote:

Reply to
Jan-Erik Söderholm

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.

Reply to
Fred Bloggs

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.
Reply to
Paul Burridge

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

Reply to
tim gorman

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

Reply to
Nicholas O. Lindan

---------------- 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
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Reply to
Don Kelly
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"Fred Bloggs"  wrote in message
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Reply to
Don Kelly

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

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

Cheers

Reply to
Wayne

Prolly; and he was asleep in simple Geometry...

Reply to
Robert Baer

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:

Reply to
Glenn

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)
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Reply to
Wayne

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

ONLY 90o at resonance frequency remember....

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"Fred Bloggs"  wrote in message
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Reply to
peterken

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:

Reply to
peterken

Yes, by subtraction.

Reply to
Wayne

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.

Reply to
Fred Bloggs

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

Reply to
Wayne

If you are modeling an electrochemical cell, i can guarantee losses...

Reply to
Robert Baer

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
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Reply to
Don Kelly

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

Reply to
Fred Bloggs

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