Finding a Unknown Inductor

[Trav]

fr = 1/(2π√LC)
Does anybody know how to rearrange this formula to find L? having some trouble doing it.

 

[ericgibbs]

fr = 1/(2π√LC)
Does anybody know how to rearrange this formula to find L? having some trouble doing it.

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.??

 

[Jon Wilder]

Delete

 

[ljcox]

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

 

[mvs sarma]

i wait for a day when these new generation would take computer help to chew their own food, LoL

 

[Trav]

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.

 

[ericgibbs]

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.

Post what you have so far.

 

[Nigel Goodwin]

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.

It's VERY basic school algebra.

 

[Trav]

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

 

[ericgibbs]

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

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 .??????????

 

[Trav]

Take the 1 over so it becomes x making it irrelevent then taking the C over to make it a division.

 

[ericgibbs]

Take the 1 over so it becomes x making it irrelevent then taking the C over to make it a division.

You may know that ω = 2Πf

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

 

[Trav]

Would it not be the same as anything x1 is the same figure.

 

[ericgibbs]

Would it not be the same as anything x1 is the same figure.

So what is your final equation.???

 

[Trav]

L = (2pi x fr)^2 / C

 

[mvs sarma]

(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.

 

[ericgibbs]

L = (2pi x fr)^2 / C

hi,
If you think thats correct, try to transpose it back into the original formula f= 1/(2Π√LC))

 

[Trav]

Would my formula be 1/ 4(pi^2)(F^2)xC ?

 

[Jon Wilder]

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?

 

[crutschow]

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:

Take the 1 over so it becomes x making it irrelevent then taking the C over to make it a division.

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))