• Jul 27th 2009, 11:39 PM
tommoturbo
hello there struggling to get this right in my head

i have an equation

Fr=1/2Pi sqrtLC and need to make Lc the subject

it is written as Fr equals 1 over 2pi times sqroot of LC (ie both 2pi and sqrtLc both divided by 1)

i got Lc2=1/2pi Fr not convinced im right though

any help appreciated

ian
• Jul 27th 2009, 11:46 PM
pomp
Quote:

Originally Posted by tommoturbo
hello there struggling to get this right in my head

i have an equation

Fr=1/2Pi sqrtLC and need to make Lc the subject

it is written as Fr equals 1 over 2pi times sqroot of LC (ie both 2pi and sqrtLc both divided by 1)

i got Lc2=1/2pi Fr not convinced im right though

any help appreciated

ian

I'm sorry but this is really poorly presented, try using parenthesis to make your expressions clearer. (better yet learn Latex)

Is your original equation: $Fr = \frac{1}{2 \pi \sqrt{LC}}$ ?
• Jul 27th 2009, 11:55 PM
tommoturbo
Yes thats exactlly the equation (sorry im no good at that sort of stuff)
• Jul 27th 2009, 11:56 PM
bananaxxx
$Fr = \frac{1}{2 \pi \sqrt{LC}}$

Multiply both sides by $\sqrt{LC}$ to give
$Fr\sqrt{LC} = \frac{1}{2 \pi }$

Divide both sides by $Fr$ to give
$\sqrt{LC} = \frac{1}{2 \pi Fr}$

Now square both sides to give
$LC = \left(\frac{1}{2 \pi Fr}\right)^2$
so
$LC= \frac{1^2}{\left(2 \pi Fr\right)^2}$
so
$LC= \frac{1}{4 \pi^2 \left(Fr\right)^2}$

Is that what you were after?

Si
• Jul 28th 2009, 12:01 AM
tommoturbo
wow thanks for the quick response, thats exactly what i am after, i was almost getting it but forgetting to multiply 2pi to get 4pi.

Much appreciated

thanks Ian
• Jul 28th 2009, 12:11 AM
bananaxxx
No problem.

Just remember the golden rule:
$(ab)^2=a^2b^2$

But $(a+b)^2\neq a^2+b^2$!!!!

In fact $(a+b)^2= a^2+2ab+b^2$

Si