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Re: Ok.. where am I going wrong
Sam,
Tesla List wrote:
> Original Poster: Sam Laur <slaur-at-sekunda.pp.utu.fi>
>
> > Original Poster: "Coiler" <mycroft-at-access1-dot-net>
> >
> > reason limiting works anyway. I had just hoped to take advantage
> > of the fact that this smoothing resistor would have voltage drop
> > across it proportinal to the current passing through it. I am just
>
> It has been a _long_ time since I've last thought about AC analysis
in-depth,
> so excuse me if I'm wrong... but with a mostly reactive load, isn't it
> very easy to have the current in a circuit out-of-phase with the voltage
> in the same circuit? Would this resistor then have a voltage drop
proportional
> to the out-of-phase circuit current? So there isn't necessarily that much
> "heating power" in the circuit, after all.
In series, inductive reactance (XL) is at 90 deg., capacitive reactance
(XC) is at
-90 deg., and resistance is at 0 deg. It is necessary to find the sum of these
oppositions to current which we call Impedance (Z) for the total opposition
and
therefore passing current. The current through the resistor is 0 deg. The
current
through the ballast is 90 deg. Because we can have capacitive and inductive
reactance's, it is necessary to calculate the total "net" reactance and add by
phasors to the total resistance to arrive at the total opposition for the
entire
series circuit known as the total impedance. i.e., Z = SQRT (R^2 + X^2).