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.
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
The Attempt at a Solution
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...
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 ??