RLC Calculations

Thread Starter

roadey_carl

Joined Jun 5, 2009
137
Hello,

If you go to this website http://www.sengpielaudio.com/calculator-XLC.htm

Type in a resistance and the frequency and it will give you the inductance and the capacitance.

I’m particular interested in finding out what the calculation/relationship between the resistance and the frequency to find out capacitance. Does anybody know the calculation?


thanks
 

edgetrigger

Joined Dec 19, 2010
133
It is very simple. It is not resistance that you feed in it is called as reactence. All the formulae are provided in the weblink that you have mentioned.

Now when you type a reactance and frequency, you can calculate L and C at that frequency.

XL is called as inductive reactence and Xc is called as capacitive reactence.
and the formulae[ XL = 2∏fL, XC = 1/2∏fC ] is given in that website.

At resonance the reactence will be same for both cacitence and inductance.
That is XL = XC = R

Substitute R in XL and enter the frequency f in the formula for inductive reactence [XL = 2∏fL] the value of L can be found out.

Similarly substitute the value of R in XC and the frequency value of C can be found out.
 
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edgetrigger

Joined Dec 19, 2010
133
Further if you have L and C you can find out R( which is either XL or XC) , also f can be foundout by substituting the value of L and C in the formula of f at resonance which is f = 1/(2*∏√(LC))


Another Case:

If you have L and f then, use value of L and f in the resonance frequency formula f = 1/(2*∏√(LC)) to findout C , then find out either XL or XC using relavent reactence formulae[ XL = 2∏fL, XC = 1/2∏fC ].

Same with C and f, find L using resonant frequency formula f = 1/(2*∏√(LC)) to findout L , then find out either XL or XC using relavent reactence formulae[ XL = 2∏fL, XC = 1/2∏fC ].
 
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Thread Starter

roadey_carl

Joined Jun 5, 2009
137
I think I get it... or at least some of it.... I'll have another go in the morning when my brain is fresh!

Thanks for your time!
 

Thread Starter

roadey_carl

Joined Jun 5, 2009
137
Further if you have L and C you can find out R( which is either XL or XC) , also f can be foundout by substituting the value of L and C in the formula of f at resonance which is f = 1/(2*∏√(LC))


Another Case:

If you have L and f then, use value of L and f in the resonance frequency formula f = 1/(2*∏√(LC)) to findout C , then find out either XL or XC using relavent reactence formulae[ XL = 2∏fL, XC = 1/2∏fC ].

Same with C and f, find L using resonant frequency formula f = 1/(2*∏√(LC)) to findout L , then find out either XL or XC using relavent reactence formulae[ XL = 2∏fL, XC = 1/2∏fC ].

Its taken me a while but I understand most of that apart from this bit :D

if L = 24401.636 & F= 50

C = 1/(2*∏√(24401.636*50) This should equal my capacitance?
I Can't seem to understand this section could you maybe show me a calculation so I can understand it?
 

edgetrigger

Joined Dec 19, 2010
133
Its taken me a while but I understand most of that apart from this bit :D

if L = 24401.636 & F= 50

C = 1/(2*∏√(24401.636*50) This should equal my capacitance?
I Can't seem to understand this section could you maybe show me a calculation so I can understand it?

if you have L and f, then use it to first find XL using formula XL = 2∏fL

this XL will be equal to XC, substitute XL in XC = 1/2∏fC and find C.

as you know XL = XC = Reactance R

in case of your example

XL = 7.666 M ohms is the Inductive ractance which is equal to XC = R

XC = 1/2∏fC

therefore C = 1/2∏x50x7.666M
C = .4152229 nano farad
R = XL = XC = 7.666 M
 

Thread Starter

roadey_carl

Joined Jun 5, 2009
137
Thats awesome edgetrigger! I've always struggled rearranging formulas but I understand this! Thanks for talking the time out to explain :D
 
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