Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Getting from non-imaginary to imaginary

  1. Apr 21, 2015 #1
    I was reading the derivation of capacitor reactance and I understand it up to the point where it is converted to polar coordinates. How do you get from
    [tex]X=\frac{sin(wt)}{wCcos(wt)}[/tex]
    to
    [tex]X=\frac{1}{jwC}[/tex]

    This implies that
    [tex]\frac{sin(wt)}{cos(wt)}=-j[/tex]
    And I'm confused how that is derived.

    Thanks

    Edit: reactance is X not Z
     
  2. jcsd
  3. Apr 21, 2015 #2
    Could you give some more background? Perhaps a reference? I've never seen reactance derived like this.
     
  4. Apr 21, 2015 #3
    Assuming a sinosoidal voltage across the capacitor, the current will be the derivative of that waveform times C [itex](I=C\frac{dV}{dt})[/itex]. So to find reactance, you divide V by I:

    [itex]V=sin(wt)[/itex]
    [itex]I=wCcos(wt)[/itex]
    [itex]\frac{V}{I}=X=\frac{sin(wt)}{wCcos(wt)}[/itex]

    But X is also defined as
    [itex]X=\frac{1}{jwC}[/itex]

    I found it here
     
  5. Apr 21, 2015 #4
    If you define the reactance as the ratio of the instantaneous values then is a time dependent quantity. This is not the usual definition of reactantce, which is a constant for a given component. The two definitions are not equivalent.
     
  6. Apr 21, 2015 #5
    I think the root of my confusion comes down to not knowing why we use i. We couldn't use any letter (A+By) to represent the 2nd dimension because at some point the characteristics of i become significant (squaring it gives -1).

    But how is the i derived in 1/jwc
     
  7. Apr 21, 2015 #6
    It is not "derived". It comes from using complex representation for voltage and current.
    You don't need it, it is just easier to do the math this way. If you start with real quantities (U and I) then the reactance will be real as well (just X=1/wC).

    In complex representation
    [itex]u=U_0 e^{i \omega t} [/itex]
    [itex]i=I_0 e^{i( \omega t+\phi)} [/itex]
    The relationship between u and i is
    [itex] \frac{du}{dt}=\frac{1}{C}i [/itex]
    Plugging in and simplifying the [itex]e^{i \omega t}[/itex] you get
    [itex]U_0 i \omega =\frac{I_0}{C} e^{i \phi} [/itex]
    This holds if [itex]\phi=90^o [/itex] and [itex]U_o=\frac{I_0}{i \omega C}[/itex] or Uo=Io*X

    Now try to do the same in sin and cos representation and see what you get.
     
  8. Apr 21, 2015 #7
    Wait, wouldn't euler's forumula get you there from my original definition anyways?

    According to euler:
    [itex]cos(x)=.5(e^{jx}+e^{-jx})[/itex]
    [itex]sin(x)=.5(e^{j(90-x)}+e^{-j(90-x)})[/itex]

    And if you divide these, it comes out to 1/j, right?
     
  9. Apr 21, 2015 #8
    I don't see how. See yourself if it does.
    For me looks like a function of x, not a constant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Getting from non-imaginary to imaginary
  1. Imaginary time (Replies: 22)

Loading...