# Impedance equation

• Oct 9th 2012, 10:03 AM
Krislton
Impedance equation
Hi guys,

i need some help please with the following equation, I need to make c the subject of z=sqrt(r^2 + (xl - xc)^2).

My first step was remove the sqrt to become z^2=r^2 + (xl - xc)^2
Then expand the brackets so z^2=r^2 + (xl - Xc)(Xl -Xc)
then Multiply out the brackets so z^2 = r^2 + xl^2 - xc^2 - 2xlxc

then I'm a bit stuck where to go from there. I could move the r^2 to the other side but I'm still none the wiser on getting c as the subject.

i was also given xl= 2(pi)fl and xc = 1/ 2(pi)FC

z=104ohms r=100ohms f=50hz, l= 0.1 henry

any enlightenment would be greatly appreciated!

thanks

kris
• Oct 9th 2012, 02:55 PM
skeeter
Re: Impedance equation
$\displaystyle z^2 = r^2 + (x_L - x_C)^2$

$\displaystyle z^2 - r^2 = (x_L - x_C)^2$

$\displaystyle \sqrt{z^2 - r^2} = |x_L - x_C|$

if $\displaystyle x_L > x_C$ ...

$\displaystyle \sqrt{z^2 - r^2} = x_L - x_C$

$\displaystyle x_C = x_L - \sqrt{z^2-r^2}$

if $\displaystyle x_L < x_C$ ...

$\displaystyle \sqrt{z^2 - r^2} = -(x_L - x_C)$

$\displaystyle x_C = x_L + \sqrt{z^2-r^2}$

finally ...

$\displaystyle x_C = \frac{1}{2\pi f C}$

$\displaystyle C = \frac{1}{2\pi f x_C}$
• Oct 10th 2012, 12:28 AM
Krislton
Re: Impedance equation
Hi skeeter

Thanks for the input but how do I transpose so C is the subject? Both of the above still with have xc or a c on the both sides. Sorry for not understanding and I'm sure you correct it's just that I still don to now how to use the formula by your solution :)

Thanks,

Kris.
• Oct 10th 2012, 08:31 AM
Krislton
Re: Impedance equation
Ok I get how c equals that now but the question I've been given is calculated the value of c, but how is the value of c supposed t o be calculated if I'm only given z=104 ohms, R= 100 ohms, f = 50 hz, l = 0.1 Henry. It's just throwing me as I can input the values but I still don't know xc.

thanks
• Oct 10th 2012, 08:34 AM
Krislton
Re: Impedance equation
Double post sorry