1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

Finding a Unknown Inductor

Discussion in 'Homework Help' started by Trav, Dec 12, 2011.

  1. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    fr = 1/(2π√LC)
    Does anybody know how to rearrange this formula to find L? having some trouble doing it.
     
  2. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,187
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    A Clue would be.

    Multiply both sides of the Equation by 2Π

    Then square both sides of the Equ.

    Can you show these steps and the following steps in transposing the formula.??
     
  3. Jon Wilder

    Jon Wilder Active Member

    Joined:
    Oct 22, 2010
    Messages:
    859
    Likes:
    82
    Location:
    Fresno, CA
    Delete
     
    Last edited: Dec 12, 2011
  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. ljcox

    ljcox Well-Known Member

    Joined:
    Dec 25, 2003
    Messages:
    3,205
    Likes:
    28
    Location:
    Melbourne Australia

    One wonders what they are teaching in schools & universities these days.

    A girl I was helping with maths recently needed her calculator to do 6 - 8 + 1
     
  6. mvs sarma

    mvs sarma Well-Known Member

    Joined:
    Oct 29, 2006
    Messages:
    3,512
    Likes:
    76
    Location:
    Hyderabad, India.
    i wait for a day when these new generation would take computer help to chew their own food, LoL
     
  7. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    I've tried alot of methods and im pretty sure one of them is right what i basically need is clarification before i put it onto paper.
     
  8. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,187
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    Post what you have so far.
     
  9. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

    Joined:
    Nov 17, 2003
    Messages:
    39,228
    Likes:
    641
    Location:
    Derbyshire, UK
    It's VERY basic school algebra.
     
  10. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    Fr = 1/ 2pi sqrtLC
    2pi x fr = 1/sqrtLC
    (2pi x fr)^2 = 1/LC
    (2pi x fr)^2/C = L

    or

    1/Fr = 2pi sqrtLC
    1/Fr(2pi) = sqrtLC
    1/ (fr(2pi))^2 = LC
    1/ C(fr(2pi))^2 = L
     
    Last edited: Dec 13, 2011
  11. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,187
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    hi,
    Fr = 1/ 2pi sqrtLC
    2pi x fr = 1/sqrtLC
    (2pi x fr)^2 = 1/LC
    how did you get from here (2pi x fr)^2 = 1/LC to here (2pi x fr)^2/C = L .??????????
     
  12. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    Take the 1 over so it becomes x making it irrelevent then taking the C over to make it a division.
     
  13. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,187
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    You may know that ω = 2Πf

    So what if you started with ω^2 = 1/(LC)
     
  14. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    Would it not be the same as anything x1 is the same figure.
     
  15. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,187
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    So what is your final equation.???
     
  16. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    L = (2pi x fr)^2 / C
     
  17. mvs sarma

    mvs sarma Well-Known Member

    Joined:
    Oct 29, 2006
    Messages:
    3,512
    Likes:
    76
    Location:
    Hyderabad, India.
    (2pi * fr)^2 = 1/LC
    Perhaps from here

    L*C = 1 / 4*(pi ^2) *( F^2)*C

    and changing values of pi^2 , i simplify it to
    L= 1 /39.51*C*F^2

    for frequency L and C both are inverse square proportion only.
     
    Last edited: Dec 13, 2011
  18. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

    Joined:
    Jan 4, 2007
    Messages:
    21,187
    Likes:
    644
    Location:
    Ex Yorks' Hants UK
    hi,
    If you think thats correct, try to transpose it back into the original formula f= 1/(2Π√LC))
     
  19. Trav

    Trav New Member

    Joined:
    Dec 8, 2011
    Messages:
    9
    Likes:
    0
    Would my formula be 1/ 4(pi^2)(F^2)xC ?
     
  20. Jon Wilder

    Jon Wilder Active Member

    Joined:
    Oct 22, 2010
    Messages:
    859
    Likes:
    82
    Location:
    Fresno, CA
    It's really simple...let's walk through it.

    The original equation...

    F = 1 / (2pi sqrt(LC))

    If we were to put this equation into words, it would say -

    Frequency is the reciprocal of 2pi times the square root of the product of L and C

    So the first thing we need to do to rearrange is reciprocate the frequency itself -

    L = 1 / F

    Now we need to divide the frequency by 2pi -

    L = (1/F) / 6.28

    Once we've done that, we now need to square it to reverse the square root function -

    L = ((1/F) / 6.28) ^ 2

    Then the last thing we do is divide this by the value of C -

    L = ((1/F) / 2pi) ^ 2 / C

    Once you've solved for L, you can then perform the F = 1 / (2pi sqrt(LC)) to check your work.

    Make sense?
     
    Last edited: Dec 13, 2011
  21. crutschow

    crutschow Well-Known Member Most Helpful Member

    Joined:
    Mar 14, 2008
    Messages:
    10,592
    Likes:
    477
    Location:
    L.A., USA Zulu -8
    Eric asked "how did you get from here (2pi x fr)^2 = 1/LC to here (2pi x fr)^2/C = L .??????????
    The reply was:
    I believe the algebraic rule you are trying to use is "performing the same function on both sides of the equation does not change the equation".

    Thus you first can take the reciprocal (1 over) of both sides of the equation giving 1 / ((2pi x fr)^2) = LC.

    The you can can divide both sides by C to isolate L giving L = 1 / (C x ((2pi x fr)^2))
     

Share This Page