Connect with us

phasor to complex

Discussion in 'Electronic Design' started by Wayne, Oct 16, 2004.

Scroll to continue with content
  1. Wayne

    Wayne Guest

    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)
    theta1=80deg
    Vs=1v
    Vr=0.25v


    I am being dumb... but how do I get this into complex format e.g. a+jb.
    Please give an example using my data.

    Thanks


    Wayne
     
  2. You could have at least made an attempt to pretend it wasn't your homework.

    Gibbo
     
  3. Glenn Elmore

    Glenn Elmore Guest

    Looks to me like a pretty bad homework problem if that's what it is. It
    seems to be overspecified.

    A source of 1 volt which develops 0.25V across a resistor connected to
    any ideal capacitor can not yield an 80 degree phase shift. The current
    is common to the two and the angle between Vr and Vc has to be 90
    degrees, by definition. Unless I scribbled wrong, that means that |Vc| =
    ..968V and the angle = arctan (.25/.968)=14.5 degrees. The remaining
    angle is the complement of this, 75.5 degrees.

    I don't know where theta1 came from, but no angle in this situation is
    80 deg. If you force the phase shift across the capacitor to be 80
    degrees, you don't get 0.25V across the resistor. It takes a resistance
    significantly higher in value than the reactance of the capacitor to
    achieve this much phase shift. Such a resistor will have the bulk of the
    voltage drop and much more than .25V across it.

    I'll leave the details of that for a homework excercise.

    Glenn

    Gibbo wrote:
     
  4. Homework ?
     
  5. Use a Smith Chart. Piece of cake unless your teacher wants you to show
    an absolute result with working.
     
  6. Fred Bloggs

    Fred Bloggs Guest

    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.
     
  7. Homework is $20 a question, $125/hour. Paypal or bank check ...
     
  8. Don Kelly

    Don Kelly Guest

    ----------------
    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.
     
  9. Don Kelly

    Don Kelly Guest

    --
    How do you know that the "capacitor" doesn't have some loss component? I
    suspect that this is the case.

    Don Kelly

    remove the urine to answer
     
  10. tim gorman

    tim gorman Guest

    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
     
  11. Wayne

    Wayne Guest

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

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

    Cheers

    Wayne
     
  12. Robert Baer

    Robert Baer Guest

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

    Glenn Guest

    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


    is.It > seems to be overspecified.
    to > any ideal capacitor can not yield an 80 degree phase shift. The
     
  14. Wayne

    Wayne Guest

    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!!!!!!!!!!!
     
  15. peterken

    peterken Guest

    "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

    remove the urine to answer
     
  16. peterken

    peterken Guest

    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


    is.It > seems to be overspecified.
    to > any ideal capacitor can not yield an 80 degree phase shift. The
     
  17. What ?

    Could you clarify that statement. In the hope that in doing so you will realise
    how stupid it is and save someone else the bother of telling you.

    Gibbo
     
  18. Wayne

    Wayne Guest

    Yes, by subtraction.

    Wayne
     
  19. Fred Bloggs

    Fred Bloggs Guest

    Don Kelly wrote:


    How do you know that the "capacitor" doesn't have some loss component? I
    suspect that this is the case.

    Don Kelly

    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.
     
  20. Wayne

    Wayne Guest

    Don
    I don't. I am going to try and work out what my electrochemical
    cell looks like in terms of a RC network.


    Wayne
    Wa
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-