Steve Hunt wrote:
> 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.
OK.. I'll go for that, with the restriction.
>
> 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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