RC circuit analysis using phasors

In summary: Vc(t) ...In summary, the problem involves finding the expression for v_c(t), the voltage across the capacitor in a simple RC circuit with a sinusoidally varying voltage source. To do this, we first convert the circuit to phasor form and express everything in impedance (Z). Then, using the impedance equations for the resistor and capacitor, we calculate the equivalent impedance (Zeq) for the circuit. From this, we can get the phasor expression for the current (I) and use it to calculate the phasor expression for the voltage across the capacitor (Vc). Finally, we convert back to time domain to get the final expression for v_c(t).
  • #1
VinnyCee
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0

Homework Statement



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.



Homework 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



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?
 

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  • #2
Any ideas? This problem has been bothering me for a week or so :)
 
  • #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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