Re: simple reactance calculations

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Barton B. Anderson" <bartb@xxxxxxxxxxxxxxxx>

The equation you showed was Xc = 1/(2*pi*f*C). If you know you want 5 
ohms of reactance, then you to solve for C since that is the unknown.

C = 1/(2*pi*f*Xc)

With Xc of 5 ohms at 60Hz, C = 1/(2*pi*60*5) = 530.5 uF

Does Xc increase when C is decreased? Yes.

With Xc of 10 ohms at 60Hz, C = 1/(2*pi*60*10) = 265.3 uF

Take care,
Bart


Tesla list wrote:

> Original poster: "Langer Giv'r" <transworldsnowboarding19@xxxxxxxxxxx>
>
> Hello, I am having some trouble finding the capacitive reactance for 
> a capacitive current limiter.  Im sure the problem is a small error 
> that i am frequently overseeing but still I am stumped.  I want to 
> get about 5 ohms reactance from a capacitor, but i do not know what 
> capacitance to use
>
> with 0.95uF, and using the reactance formula, it says i get 2792 
> ohms of reactance, which is obviously wrong.  1 / ((2)(pi)(f)(Capacitance))
>
> Does anyone have a solution for this?  and also, is it true that 
> capacitive reactance increases as capacitance is decreased?
>
> thanks
>
> _