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Homework Help: Complex Numbers and AC Electronics.

  1. Nov 24, 2007 #1
    Okay. I'm just looking for someone to check my answer to the following question!

    I'm not sure this is the correct forum for such a question, if not, feel free to have it moved :)


    An a.c. supply of amplitude [itex]20V[/itex] and a frequency of [itex]5 kHz[/itex] is connected across a Resistor [itex]R = 4.7 k\Omega[/itex] and a capacitor [itex]C = 10nF = 10 \times 10^{-9}F[/itex] connected in series. If the output resistance of the voltage source [itex]R_s[/itex] is assumed to initially be [itex]0\Omega[/itex], find:

    a) The Impedance of the circuit in both [itex]R + jX[/itex] and [itex]Z\angle\theta[/itex] form.

    b) The magnitude of the current flowing in the circuit.

    c) The phase difference between the applied voltage and the current.

    d) The voltage across the resistor and the voltage across the capacitor.

    e) Show that [itex]V_{s} + V_{c}[/itex] is equal to the voltage applied to the circuit.

    f) It is suspected that [itex]R_{s}[/itex], the output resistance of the voltage course, is not actually [itex]0\Omega[/itex]. When the circuit is investigated experimentally, it is fond that the phase difference between the applied voltage and the current is actually [itex]31\degree[/itex]. From this observation, calculate R_{s}.
    By the way this is an Engineering problem, so we're using [itex]j[/itex] for the complex numbers jargon :).

    My answer to a)

    [itex]Z_{total} = Z_{C} + Z_{R}[/itex]

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

    [itex]C = 10\times 10^{-9}[/itex], [itex]\omega = 2\pi f = 2\pi(5000) = 31415.9[/itex]

    [itex]Z_{C} = \frac{-j}{31415.9 \times 10 \times 10^{-9}}[/itex]

    [itex]Z_{C} = -j(\frac{1}{13.14 \times 10^{-4}})\Omega[/itex]

    [itex]Z_{C} = -j3184.7\Omega[/itex]

    [itex]Z_{R} = 4700\Omega[/itex]

    [itex]Z_{total} = 4700 -j3184.7[/itex] Which is my answer in R + jX format.

    To get it in [itex]Z\angle\theta[/itex] format I construct an Argand diagram, and deduce from it that:

    Using Pythagoras:

    [itex]Z^{2} = 4700^{2} + 3184^{2}[/itex]

    [itex]Z = \sqrt{4700^{2} + 3184^{2}}[/itex]

    [itex]Z = 5677.35\Omega[/itex]

    Using Trigonometry:

    [itex]\theta = tan^{-1}\frac{3184.7}{4700}[/itex]

    [itex]\theta = 34.12\degree[/itex]

    But I see from the Argand Diagram that this is a negative degree.

    [itex]Z = 5677.35\angle-34.12\degree[/itex]

    Let's see who can spot all of Mush's mistakes on that one then :)?
  2. jcsd
  3. Dec 1, 2007 #2


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    Science Advisor
    Homework Helper

    This is the correct forum for your problem.
    A little hint: If you had a specific question with your homework, you would get more replies.
    Without checking your math, for what you've shown, conceptually you are on the right track.
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