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Impedance of a parallel RC circuit? 
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#1
Nov1110, 03:42 AM

P: 37

Hi there,
I have a circuit in which consists of a resistance (rs) in series with parallel combination of capacitor and resistance(rc). I want to calculate value of rs. So i did this by calculating total impedance Ztotal which is V/I and then substracting impedance of parallel rc circuit. Ztotal=rs+Zcrc where Zcrc is the impedance of the parallel rc. I know the impedance for the parallel rc circuit but i want to get rid off the jth term which is present in the reactance term of the capacitor. How it can be done?? Xc=1/jwc then here how to eliminate the jth term because i will subtract this impedance of parallel rc circuit from the total impedance to calculate rs. thanks 


#2
Nov1110, 03:55 AM

P: 4,513

For a fraction with a complex value in the denominator, multiply both the numerator and denominator by the complex conjugate of the denominator. This will result in a real value in the denominator.



#3
Nov1110, 03:58 AM

P: 37

thanks but is it ok if i square the equation and then finally square root to get the Xc back?
Xc^2=1/(jwc)^2=1/w^2*c^2 Xc=1/wc ???? is it right??? 


#4
Nov1110, 12:03 PM

Mentor
P: 40,730

Impedance of a parallel RC circuit?



#5
Nov1210, 03:01 AM

P: 37

thanks but actually i am calculating this impedance of a battery circuit. The input voltage is not a pure dc due to some ripples in the signal so how i should treat this signal as sc or dc??? if dc then can i get rid off the complex jth term??
thanks 


#6
Nov1210, 04:12 AM

P: 4,513

Maybe you should describe the circuit you are interested in, in more detail. A battery itself, or a battery associate Rs and Cs will not have an AC voltage component after it's settled down, so it's difficult to know what you are talking about. Is there a switch or another source of emf?



#7
Nov1210, 04:27 AM

P: 37

thanks..... Actually i have an equivalent circuit of a car battery (during charging phase) which consists of a resistance(r1) in series with parallel combination of capacitor and resistance(r2). The input is voltage and current from the rectifier but its not a pure dc as it has ripples. So should i consider the circuit as ac or dc?
the reason is i am struggling with the impedance component of the parallel rc circuit. If it is dc then what will be the reactance formula. I know the reactance term will be infinite because of zero frequency but i calculated the frequency from the voltage signal and its varying. So if we use Xc=1/jwc then should this jth term to be considered or not? My final aim is to calculate r1 and my idea of doing it is calculating the impedance. So my logic is Ztotal=r1+Zxcr2 i.e. total impedance Ztotal= V/I r1=series resistance and Zxcr2=Impedance of parallel r2 and c r2=r2; Xc=i/jwc so what is the right way of calculating r1 if all other terms are given??? thanks 


#8
Nov1210, 05:06 AM

P: 4,513

If you have ripple you don't have zero frequency.
First of all you want to model your unregulated voltage source as V_{dc} + V_{ac}cos(2 pi f t) together with a series resistance. The series resistance could be negligable. I don't know. The voltage source has both AC and DC parts so the AC ripple may be a concern. It's difficult to tell at this point. It's not that hard to deal with R and C in parallel. w = 2 pi f, where f = frequency. In general, for Z1 in parallel with Z2: Z = (Z1 Z2)/(Z1+Z2). Just like resistors in parallel. Z1=R Z2=1/jwC Z = (R/jwC)/[R+(1/jwC)] To simplify the bottom, multiply the top and bottom terms by jwC. Z = R/(jwRC+1) = R/(1+jwRC) Now we do the complex conjugate thing to get a real denominator by multiplying top and bottom by 1jwRC. Z = R(1jwRC)/[(1+jwRC)(1jwRC)] Z = R(1jwRC)/[1+(wRC)^{2}] 


#9
Nov1210, 05:28 AM

P: 37

Dear phrak thanks a lot for an awesome reply...but what do you mean by:
"First of all you want to model your unregulated voltage source as Vdc + Vaccos(2 pi f t) together with a series resistance? " do you mean that if the incoming voltage is 14v then 14+14cos(2pift)?? if yes then pi=3.14 and what will be t?? the current time during the simulation or something else??? yes you are right that the series resistance would be negligible but my concern is to find the series resistance(r1) only. the values that i have to my model circuit is the continuous value of incoming voltage and current from the rectifier. So if i divide incoming voltage by incoming current then is it the total impedance of the circuit??? 


#10
Nov1210, 05:35 AM

P: 4,513

[QUOTE=jak9;2981417]Dear phrak thanks a lot for an awesome reply...but what do you mean by:
"First of all you want to model your unregulated voltage source as Vdc + Vaccos(2 pi f t) together with a series resistance? " do you mean that if the incoming voltage is 14v then 14+14cos(2pift)?? OK, so your DC value is 14 volts in average. But you have a bit of ripple. Measure the ripple. Do you have an oscilliscope? Let's say it ripples from top to bottom 1.2 volts. Then your voltage source is 14 + 2 cos( 2 pi f t). Tell me. what is your ripple frequency and what is the value of C? 


#11
Nov1210, 05:55 AM

P: 37

ok to make it more clear my circuit is dealing with battery charging and as the voltage is not constant because of ripples. So the frequency i calculated is by f=1/t and its around 100hz or sometimes more or less by 10. The capacitor is a double layer capacitor and the value i got is 207360F which is unacceptable and the new value of the capacitor now is 0.6678F.
so in short C=0.6678, f=100hz and by the from where this 2 comes 14+2cos(pift)?? 


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