RLC Calculations

Discussion in 'Math' started by roadey_carl, Dec 30, 2010.

  1. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
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    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
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
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    Are you refering to the second formula, the one of the Capacitiver Reactance maybe?
     
  3. edgetrigger

    Member

    Dec 19, 2010
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    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.
     
    Last edited: Dec 30, 2010
  4. edgetrigger

    Member

    Dec 19, 2010
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    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 ].
     
    Last edited: Dec 30, 2010
  5. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
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    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!
     
  6. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
    5

    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?
     
  7. Dave

    Retired Moderator

    Nov 17, 2003
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  8. edgetrigger

    Member

    Dec 19, 2010
    133
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    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
     
    roadey_carl likes this.
  9. roadey_carl

    Thread Starter Active Member

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

    Member

    Dec 19, 2010
    133
    19
    Happy that you figured it out. Great!
     
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