Jim,
I disagree!
If we restrict ourselves to talking about the series-form complex load
impedance, the reactive component will *not* affect the current required
through the load for a particular power dissipation. For example, to
dissipate 150W in a load of 150+j0 you need a current of 1A; you also
need the same 1A if you want to dissipate 150W in a load of 150+j1000.
Once you start talking about parallel components, things are different.
Your example of a 150 Ohm resistor in parallel with a 150 Ohm inductor
represents a load of 75+j75 Ohms - the resistive part is no longer 150 Ohms.
73,
Steve G3TXQ
On 01/12/2010 04:14, jimlux wrote:
>
> The current could be higher or lower. Say you've got a 150 ohm resistor
> in parallel with a 150 ohm inductor. The current that flows through the
> inductor also flows through the feedline, and contributes to the loss.
>
> Or, if you've got the resistor and inductor in series. Now the current
> flow will be even less. (by a factor of 70%).
>
> that's the difference between apparent power (Irms * Vrms) and active
> and reactive power.
>
>
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