Fwd: RE: [TowerTalk] Walt Maxwell responds to Steve Best

[i4jmy@iol.it]

When two REAL sources are paralleled, the impedance of the “virtual" 
generator obtained is the parallel of the two individual impedances at 
point were paralleled. 
The developed voltage across the virtual generator is actually that 
value producing a power equal to P1+P2 if the load itself (conjugate) 
matches the source, but goes up to the double (2*V) when the load is an 
infinite number or to zero V when load is a short.
(The load impedance is here intended as the resulting impedance at the 
point were generators are paralleled) 
It's also possible to calculate separately the delivered power (and 
developed voltage) of each generator across the load, but in this case 
the second generator must be present in the circuit for the 
calculations and considered as another load, with negative impedance, 
in parallel with the first one.


73,
Mauri I4JMY

> ---------- Initial message -----------
> 
> From    : owner-towertalk@contesting.com
> To      : "Jim Reid" <kh7m@hsa-kauai.net>, 
<w8ji@contesting.com>,       <towertalk@contesting.com>, "Walter 
Maxwell" <w2du@journey.com>
> Cc      : 
> Date    : Thu, 11 May 2000 23:26:55 -0400
> Subject : RE: [TowerTalk] Walt Maxwell responds to Steve Best
> 
> 
>
> 
> W2DU:
> 
> Steve said: "Two "in-phase" powers (P1 and P2) never add to become
> the algebraic sum of P1 + P2. "
> 
> The above statement is totally untrue. When the
> corresponding voltages and currents of the forward and
> reflected powers are in phase the two powers add
> algebraically. In the paragraph below Steve talks about
> powers where the voltage and currents are out of phase.
> This is irrelevant, because when the antenna tuner is
> correctly adjusted to deliver maximum output power the
> voltage and current phases are exactly equal at zero degrees.
> 
> Steve Response:
> 
> How many times do we need to do this calculation?  Take two voltages 
of zero
> phase.  Calculate the power for each into any impedance.  Now add the 
two
> voltages together in 
y would add in a transmission line.  
Now
> calculate the total power.  Is it the sum of P1 + P2?  NO!!  Let's do 
it
> again.  100 volts is delivered to a 50 ohm load.  P = 200 watts.  50 
volts
> is delivered to a 50 ohm load.  P = 50 watts.  Add both voltages.  
150 volts
> is delivered to a 50 ohm load.  P = 450 watts.  Power is a function of
> voltage squared.  It should be obvious that (V1 + V2)^2 is greater 
than V1^2
> + V2^2.






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