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Homework Help: RC circuit analysis using phasors!

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data

    I can't post pictures as I do normally, so you'll have to deal with the cumbersome attachment method! It's just a simple RC circuit. Resistor on top, capacitor on right side with the sinusoidally varying voltage source on the left.

    [tex]V_s(t)\,=\,25\,cos\left(2000\,\pi\,t\,-\,30^{\circ}\right)[/tex]

    Obtain an expression for [itex]v_c(t)[/itex], the voltage across the capacitor.

    NOTE: Picture of circuit is attached.



    2. Relevant equations

    [tex]A\,cos\left(\omega\,t\,+\,\phi_0\right)\,\,\longrightarrow\,\,A\,e^{j\,\phi_0}[/tex]

    [tex]j\,=\,e^{j\,\frac{\pi}{2}}[/tex]

    Various equations and conversions from the text, Fundamentals of Applied Electromagnetics (Ulaby, 5 ed.) PROBLEM 1.21



    3. The attempt at a solution

    I obtained the voltage equation for the RC circuit as:

    [tex]R\,i(t)\,+\,\frac{1}{C}\,\int\,i(t)\,dt\,=\,v_s\,(t)[/tex]

    Now, I convert to phasor form:

    [tex]R\,\tilde{I}\,+\,\frac{1}{C}\,\left(\frac{1}{j\,\omega}\,\tilde{I}\right)\,=\,V_s[/tex]

    The book gives a conversion to get the phasor expression for the current:

    [tex]\tilde{I}\,=\,V_0\,e^{j\,\left(\phi_0\right)}\,\frac{j\,\omega\,C}{j\,\omega\,C\,R\,+\,1}[/tex]

    I apply this formula, given for the RC circuit by the text, and I get this:

    [tex]\tilde{I}\,=\,25\,e^{-j\,30^\circ}\,\frac{j\,\left(2000\,\pi\right)\,\left(200\,X\,10^{-12}\right)}{j\,\left(2000\,\pi\right)\,\left(200\,X\,10^{-12}\right)\,left(1\,X\,10^6\right)\,+\,1}[/tex]

    [tex]\tilde{I}\,=\,25\,e^{-j\,30^\circ}\,\frac{j\,4\,X\,10^{-7}\,\pi}{j\,0.4\,+\,1}[/tex]

    Then, after doing some manipulation of the complex number:

    [tex]\tilde{I}\,=\,0.0000229\,+\,j\,0.0000180\,=\,2.29\,X\,10^{-5}\,+\,j\,1.8\,X\,10^{-5}[/tex]

    I know that this is probably not correct, but I went ahead and used it in the subsequent equation (also given in the text) to find the capacitor voltage phasor:

    [tex]\tilde{V_c}\,=\,\frac{\tilde{I}}{j\,\omega\,C}[/tex]

    After some plug & chug…

    [tex]\tilde{V_c}\,=\,14.3\,-\,j\,18.2[/tex]

    This is NOT the answer given in the text! I know I didn't convert back from phasor form, but still... The correct answer is:

    [tex]V_c(t)\,=\,15.57\,cos\left(2000\,\pi\,t\,-\,81.5^{\circ}\right)[/tex]

    Where did I go wrong? Can anyone get me started on the right track here?
     

    Attached Files:

    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 8, 2008 #2
    Any ideas? This problem has been bothering me for a week or so :)
     
  4. Mar 8, 2008 #3
    What you want to do is express everything in impedance (Z).

    Since Vs is in cosine form ... phasor of it would be 25{-30 or 25cos30 - jsin30 (if { means phasor)

    Then you want Z of c and r in the circuit.. given by the equations..
    Zr = R and Zc= -1/(jwc)

    Since impedances are expression with units of resistance and you have a series circuit just get Zeq by summing the Zl and Zr. Then to change to phasors... Zeq = R + -1/(jwc) = some form with real and imaginary => change that to phasor...

    then to get I(t) which is the same for C and R, you have I(t)=V/Zeq which gives another phasor expression..

    finally, to get Vc(t) you just use I(t)Zc
     
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