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Homework Help: Archived RC circuit question

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Given an RC circuit which has a Capacitor[itex] (C=6\times 10^{-6} F) [/itex] and a resistor [itex](R=5 \Omega) [/itex]conected in series to an a.c. voltage source of the form [itex]v=V_{0} e^{j\omega t} [/itex]with a [itex]V_{0}[/itex]=1 Volt. Frequency f=10kHz
    a)What is the phase of the current with respect to the applied voltage?
    b) What is the magnitude of the current?

    2. Relevant equations


    [itex]Z_{C}[/itex]=[itex]\frac{1}{j\omega C}[/itex]

    Z=x+jy where j=[itex]\sqrt{-1}[/itex]

    Z=ze[itex]^{j \phi}[/itex] with z=[itex]\sqrt{x^{2}+y^{2}}[/itex] and [itex] \phi =tan^{-1}[/itex][itex] \frac{y}{x} [/itex]

    [itex] \omega = 2\pi \times f[/itex]

    3. The attempt at a solution

    [itex] \frac{1}{\omega C} = 2.65[/itex]

    Z=R+ [itex] \frac{1}{j\omega C}[/itex]


    Which can be written as [itex]Z=ze^{j\phi}[/itex]


    [itex]\phi =tan^{-1}\frac{-2.65}{5}=-27.9 degrees[/itex]

    Since [itex]i=\frac{V_{0}}{z}e^{j(\omega t-\phi)}[/itex] then I can say the current leads the voltage by 27.9 degrees.

    And I can also say the magnitude of the current i is equal to [itex]\frac{V_{0}}{z}=\frac{1}{5.66}[/itex]

    How did I go?? Am I on track anywhere at all or have I made a bit of a mess of it?
    Last edited by a moderator: Feb 6, 2016
  2. jcsd
  3. Feb 6, 2016 #2


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    Staff: Mentor

    The calculations look fine. The results are good.
  4. Feb 7, 2016 #3
    Everything look allright.
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