# LCR circuit and frequency equation

1. Dec 9, 2011

### Howlin

1. The problem statement, all variables and given/known data
Im doing a lab experiment on LCR circuits and have gotten a graph of 1/capacatance against frequency^2. With my results I got the slope to be 0.89±0.01 and the intercept is 358708±43189

2. Relevant equations
the equation that was given was
frequency^2=(1/4*pi*L)[(1/C) - (R^2/2L)]
we then get
f^2 = 1/(4*pi*L*C) -R^2/(8*pi*L^2)

3. The attempt at a solution

since y =mx + c

the slope is
m=1/(4*pi*L) where m is 1/C
then c = -R^2/(8*pi*L^2)
we rearrrange the first equation to get L= 1/(4*pi*m) then when we know L we can get R
R^2 = -c*8*pi**l^2

When I put in my values for m for the first equation I get L = 2.85E-2 Henry
Then when to get R^2 I can as you cant the square of a negative number.

Would I get the positive of the R^2 value and then say the resistance is (plus/minus) 151 ohms?
Or can someone tell me what I am doing wrong
my lecturer does not know what is wrong so I kinda want to get it right