Finding a Unknown Inductor

[ljcox]

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

This is correct.

 

[mvs sarma]

Cant be ljcox ,
because against an equation F and L have to be inverse square
Fr =1 /(2*pi*Sqrt(L*C))
Sqrt(L*C) = 1 /(2*pi*Fr)
L*C = 1/ (2*pi*Fr)^2
L = 1 / {((2*pi*Fr)^2)*C}
L= 1 / {4*3.141*3.141 *Fr*Fr*C)

 

[ljcox]

Yes, you're right. I should have looked at it more carefully.

It should be L = 1/{C * (2pi x fr)^2}

 

[ericgibbs]

Yes, you're right. I should have looked at it more carefully.

It should be L = 1/{C * (2pi x fr)^2}

hi Len,
We now agree
Knowing that Fr= 1/(2Π*√LC)

I suggested to the OP that he moved the '2Π' to the left side of the equation, so that it became 2Π*F and substituted ω for 2ΠF.

This would give ω = 1/(√LC)

Then its a simple matter of squaring both sides of the equation to, ω^2 = 1/LC....

Transposing gives, L = 1/(ω^2 * C) ..... (1)

E

 

[meowth08]

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.

If you are not sure of your answer, all you need to do is substitute any number to your variables in the original equation say 2 for L, 3 for C. From here, you can get fr. Now, from your derived equation, substitute the value of fr that you got, and 3 for C. If you get L equal to 2, your derivation might be right.

Note: These are not always true all the time. You have to use different values for your variables to verify your equation.