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## Re: [TowerTalk] tuners and power rating

 To: towertalk@contesting.com Re: [TowerTalk] tuners and power rating Steve Hunt Wed, 01 Dec 2010 12:43:51 +0000 mailto:towertalk@contesting.com>
 ```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. > > _______________________________________________ _______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mailman/listinfo/towertalk ```
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