Equivalent Impedance for RC Circuit

Discussion in 'Homework Help' started by hibell, May 8, 2012.

  1. hibell

    Thread Starter New Member

    May 8, 2012
    Maybe I am reading this wrong, but for some reason, I cannot think of how to do this problem:

    A capacitor C and a 5-ohm resistor are in parallel with each other at an angular frequency ω of 10^3 rad/s. (a) Find the value of C if Re(Z_in) = 1 Ω. (b) At the same frequency, the above parallel combination is replaced by an equivalent circuit consisting of a series combination of R_1 and C_1. Find the values of R_1 and C_1.

    The answer to (a) is given as C = 400 μF and answer to (b) is given as R_1 = 1 Ω and C_1 = 500 μF.

    Any help is appreciated.
  2. WBahn


    Mar 31, 2012
    Q1) Given a resistor, R, and capacitor, C, what is the equivalent impedance, Z, if they are placed in parallel? For this problem, I would recommend working in rectangular coordinates, at least for your answers.

    Q2) What is the real part of the of the expression for Z found in #Q1?

    Q3) What value of C is needed, together with R=5Ω and ω=1krad/s, to make this equal to 1Ω?

    Q4) Given a different resistor, Rs, and capacitor, Cs, what is the equivalent impedance of the series combination of these two?

    Q5) What do the values of Rs and Cs need to be in order to make the answer to #Q4 equal the answer to #Q1 (using the value of C found in #Q3)?
    hibell likes this.
  3. hibell

    Thread Starter New Member

    May 8, 2012
    1. Z_{in} = \frac{\frac{R^2}{j\omega C} + \frac{R}{\omega ^2 C^2}}{R^2 + \frac{1}{\omega ^2 C^2}}

    2. Re[Z_{in}] = \frac{\frac{R}{\omega ^2 C^2}}{R^2 + \frac{1}{\omega ^2 C^2}}

    3. Solving for C when Re[Z_{in}] = 1 yields C = 400 μF.

    Will post the series solution shortly. I see the error in my ways. Thanks WBahn for the much progress made due to your very helpful hand. My error was that I was making a silly algebraic mistake! =}
    Last edited: May 9, 2012
  4. hibell

    Thread Starter New Member

    May 8, 2012
    4. Z_{eq} = R_{s} + \frac{1}{j\omega C_{s}

    5. Substituting C = 400 μF into Z_{in} found in (1) yields Z_{in} = 1 + j2.

    Solving for R_{s} and C_{s} shown below:

    R_{s} = Re[Z_{in}] = 1

    \frac{1}{\omega C_{s}} = Im[Z_{in}] = 2 ---> C_{s} = 500 \mu F

    Thank you again for the clear steps WBahn! :D