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Equivalent Impedance problem

  1. Jan 26, 2007 #1
    1. The problem statement, all variables and given/known data
    First of all, sorry the poor quality of the drawing... lol made on paint!
    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 lea 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

    3. The attempt at a solution

    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 !!!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

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  3. Jan 27, 2007 #2


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    Can't see your picture. but i'll try answering it anyway...for the first bit it appears that you just have trouble simplifying complex numbers, so here is one way to do it.
    [tex]\displaymath{\frac{1}{a+j b}=\frac{1}{a+j b}\times \frac{a-j b}{a- j b}=\frac{a}{a^2+b^2}-j \frac{b}{a^2+b^2}}[/tex]
    now you have separated real and imaginary part. Alternatively, you can turn a+j b in to phasor form ie. [tex]r e^{j\theta}[/tex] where r is absolute value of a +jb, while the angle is the Arg(a+jb) then 1/(a+j b) is simply
    [tex](1/r) e^{-j \theta}[/tex]

    finding range of freq... ie. [tex]\omega[/tex] in your [tex]1/j\omega C[/tex] etc. clearly different [tex]\omega[/tex] values give you different total reactance in the circuit, shall led to an equation to be solved in [tex]\omega[/tex] perhaps. the assumption is probably there to help simplify the expression you may end up.
  4. Jan 27, 2007 #3


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    firstly, it is always easier to work with the symbols [tex]L_1, L_2, R_c, R_1, R_2, C[/tex] than their numerical values, simplify the expression first THEN sub in the numerical values.

    so, for part a) you simply has
    [tex]Z_{AB} = R_1+R_2+j\omega (L_1+L_2) + Z_{eq}[/tex] where
    [tex]Z_{eq}[/tex] is the parallel combination of [tex]C[/tex] and [tex]R_c[/tex], which is simply
    [tex]\displaymath{\frac{R_c}{j\omega C R_c +1}}[/tex]
    if you don't wanna simplify this any further that's it... otherwise, use the method suggested before to break [tex]Z_{eq}[/tex] into real and imaginary part and then factor out [tex]j\omega[/tex]..etc. And THEN sub it all the numerical values for the final answer

    For Part b) you want to assume [tex]R_c\rightarrow \infty[/tex] so [tex]1/R_c \rightarrow 0[/tex] and so now [tex]Z_{eq}= 1/j\omega C[/tex]

    you then use
    [tex]Z_{AB} = R_1+R_2+j\omega (L_1+L_2) + Z_{eq}[/tex] again with new [tex]Z_{eq}[/tex] to proceed. do u actually want [tex]X_{AB} >10 R_{AB}[/tex] or [tex]X_{AB} < 0[/tex] (for capacitive) or do you actually mean if [tex]Z_{AB} = R_{AB} + j X_{AB}[/tex], you are seeking solutions when: [tex]|X_{AB}| > 10 R_{AB}[/tex] AND [tex]X_{AB} < 0[/tex], I guess the second condition can be easily satisfied given your values of [tex]L_1, L_2, C[/tex].

    in other words you are finding the freq in which the parasitics effects can be ignored.

    check your numbers again, for you may have made a mistake
  5. Jan 29, 2007 #4


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    the frequencies you are working out is the angular frequency (rad/s) of the input signal that drives the circuit, so you have found the [tex]\omega[/tex] of [tex]A_0 \sin(\omega t+\phi)[/tex], now if [tex]\omega <0[/tex], then you really just have a phase shift, for you can factor the -ve sign out and then introduce a new phase that comes with a +ve angular freq.

    what is more crucial however may lie in the wording of the question itself... now something seems funny is that you want capacitive like but as you can see reactance X_ab depends on L1, L2 as well as C. So as freq becomes large X_C becomes small while X_L part becomes big. Now if you just look at X_ab,
    as a function of w, you will see that X_ab flips it sign when w reaches about ~0.2x10^7 , so suddenly it is no longer capacitive (imagery part is +ve), it is inductive. Note values of w in the range worked out seems to yield a +ve X_ab...(you should double check that.. as I don't have maple so can't check you numebrs) which is not quite right if you want the thing to be capacitive.

    I believe you have to solve w under two conditions: |X_ab| > 10 R_ab and X_ab <0, there may be a small window where both can be true.. I can't tell by inspection, must work the numbers out.

    The equation you have solved was simply X_ab > 10 R_ab without the absolute sign, so the answer you've got would naturally be true only if X_ab is +ve... for R_ab is +ve!

    my advice (after all that crap)... is try solving -X_ab > 10R_ab for w and see whether there is a solution of w, if so, you are done... if not, perhaps best if you post the exact wording of the questions again.
  6. Jan 29, 2007 #5
    so yep...
    i have a solution for -Xab
    Maples returns this :
    RealRange(-infinity,Open(-2236117.978)), RealRange(Open(0.),Open(2236017.978))

    so [-infinity,-2236117.978[
    and ]0,2236017.978[

    Do i consider the negatives values ??
  7. Jan 29, 2007 #6


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    ah, that's make more sense... you have a finite +ve range that tells you that if freq goes too high, the device will fail to behave like a capacitor because of parasitics. As to whether to ignore negative values: I have to admit I am not 100% sure on this, but it may depends highly on how you define your phasors, the convention is course [tex]\omega[/tex] goes in counter clockwise on the complex plane. So I would say, you just take the +ve range, and ignore the -ve range which is just from the maths and no physical "meaning"...well, perhaps they do have meaning...[tex]\omega[/tex] goes the other way, but that will inevitably changes the definition of all phasors that you may define. for instance, L would now lags C by 180 and not the other way round. :frown:
  8. Jan 30, 2007 #7
    I would do it this way.

    1. Write the equation for Zab in terms of the angular frequency w. This equation should contain only one unknown, i.e., w.

    2. Separate the real and imaginary part of the above equation, i.e., Zab = Rab + jXab.

    3. Apply the constraint Xab > 10*Rab given in the question to find the range of values for w.

    I believe there is no need for any assumptions like Zeq = 1/jwC at this point.
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