Equivalent impedance and reactance of a capacitor

Discussion in 'Homework Help' started by xMrdibbsx, Jan 27, 2007.

  1. xMrdibbsx

    Thread Starter New Member

    Jan 27, 2007
    Problem 4.67 from Rizzoni ( Pricniples and applications of electrical engineering 5th eiditon)
    First of all, sorry the poor quality of the drawing... lol made on paint!
    LINK : [​IMG]
    Anyways here is the problem and all given data :
    The capacitor model we have used to far has been treated as an ideal circuit element. a more accurate model for a capacitor is shown in figure 4.67 ( drawing attached ). The ideal capacitor C has a large 'leakage' resistance, RC, in parrallel with it. RC models the leakage current trhough the capacitor. R1 and R2 represent the lead wire resistance, and L1 and L2 represent the lead wire inductances.

    a) If C= 1µF, RC = 100MΩ, R1=R2=1µΩ and L1=L2=0.1µH, find the equivalent impedance seen at the terminals A and B as a function of frequency ω.

    b) Find the range of frequencies for which Zab is capacitative, i.e. : Xab > 10Rab.
    Where Xab is the imaginary part of the equivalent impedance, and Rab the real part.

    Hint for b) Assume that Rc is much greather than 1/ωC so that you can replace Rc by an infinite resistance.

    2. Relevant equations
    Impedance formulas :
    Resistance = R
    Inductance = jωL
    Capacitor = 1 / jωC
    where j is the complex variable ( a + bj )

    Impedance in parallel add like this : 1 / ((1/Z1)+(1/Z2)+(1/Zn..))
    Impedancce in series add like this : Z1 + Z2 + Z3

    For a)
    For the equivalent impedance of the C and RC, i got this :
    1 / ((1/100MΩ)+(1/(1/jω1µF))) = 1x10^6 / (1+ jω)
    Equivalent resistance for all the circuit =
    1x10^6 / (1+ jω) + 2(jwL) + 2(R) =
    1x10^6 / (1+ jω) + 2(jw0.1µH) + 2(1µΩ)

    Now i just don't know how to simplify all this to get a nice answer.. I'm also not sure of what i did...

    For b)
    I have no clue.. I also don't understand the clue, or understand how it can help me solve this problem. Also since i have trouble simplifynig my previous equation i have a lot of trouble solving to find the range of frequencies.. can anyone help ??

    Thanks !!!
  2. beenthere

    Retired Moderator

    Apr 20, 2004

    I think if you work out the capacitive reactance for a couple of frequencies, you'll find that the Xc varies in a linear fashion with the change in frequency. Likewise for Xl. If you plot Xc and Xl, I think you'll see where the range of frequencies satisfies the question.
  3. xMrdibbsx

    Thread Starter New Member

    Jan 27, 2007
    The problem is i'm not even sure if my answer is correct...
    With the answer i have, when i try solving Xab > 10*Rab .. i get a complex answer for the range of frequencies... so i assume that's a wrong answer...

    What i did for b) is that i considred R = infinite. When i get the equivalent impedance of Rc and C, i get = (1)/((1/infinite)+(1/jωC))
    which equals to 1/jωC.
    Then i add up all the other impedances and get this answer :
    (1/jω10*10^-6)+(2*10^-7)j + 2*10^-6
    Now if i consider the complex part being 10 times bigger than the real part, (thats what they say to do in b) to find the range of frequencies...) I get a complex answer..
    So i guess i might be wrong calculating the overall equivalent impedance...
    Is it possible for you guys to review my calculations, maybe i made a mistake somewhere
  4. xMrdibbsx

    Thread Starter New Member

    Jan 27, 2007
    Allright, so i solved everything with maple.. but i don't know how to interpret the answers i have.

    Here for b) [​IMG]

    so that would say the frequencies for which its capacitative are : ] -0.2236017978*10^7,0 [ U ] 0.2236017978*10^7,infinity]

    Can i consider negative frequencies ??
    And i've also read that when the frequencie goes up, a capacitor loses its capatitative reactange ? So do i consider ] 0.2236017978*10^7,infinity]?
  5. chuckey

    Well-Known Member

    Jun 4, 2007
    .22 * 10^7 = 2.2 Mhz is in the shortwave band, so it depends on the actual use of the capacitor. It its used in a audio amplifier (20 - 20KHz) it can be disregarded. If it was used in a wideband amplifier for TV distribution (100 - 1000MHz), it would not work at all and would have to be shunted by a much smaller capacitor which does not have the internal inductance.
    Frank (I'll have a think about the maths now!)
  6. chuckey

    Well-Known Member

    Jun 4, 2007
    If you start of at a VERY low frequency, (like DC?) the circuits impedance is 100M + .2 X 10^-6 ohms~= 100Mohms, as the frequency rises the reactance of the capacitor falls and the whole thing acts as a cap. As the frequency gets higher, the reactance of the inductor rises, this actually reduces the total circuit impedance. Until at some frequency, XL=XC then the only impedance is the .2 micro ohms. Higher then this the circuit acts as an inductor. This makes the statement Xab > 10 X Rab, very difficults for me to work out because I would have thought that you would expect the impedence of the circuit to fall, so at a low frequency, its a cap when Xc << 100M ohm and it acts as a cap right up to wC = wL, when the circuit impedance falls to .2 micro Ohms.