please help me transpose this formula!! f =1/(2*pi*sq. root overL*C) To make L the subject.?

Answer 1

f = 1/(2*pi)[(L*C)^(1/2)]

2*pi*f = 1/sqrt(L/C)

Let w = 2*pi*f (angular velocity)

w = 1/sqrt(LC)

wsqrt(LC) = 1

w^2 = 1/LC

LC = 1/w^2

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

Answer 2

1) multipply both sides by sqr root (L.c)
sqr root(l.c)f=1/2.pi

2) divide both sides by f
sqr root(l.c)=1/2.pi.f

3) sqr both sides (to get rid of root)
l.c=(1/2.pi.f) ^2

4) divide both sides by c

l=((1/2.pi.f)^2)/c

hope this helps.

Answer 3

f = 1 / (2*pi*sqrt(L*C))
f*2*pi*sqrt(L*C) = 1
sqrt(L*C) = 1 / (f*2*pi)
L*C = (1/f*2*pi)^2
L = ((1/f*2*pi)^2)/C