*** MY inserted comments ***


"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"3) #1 with 50% chopper, [ILavg =0.5 ma... PL = 0.5 mW]"

At 1/2-scale, the *** AVERAGE *** current is 0.0005 amp.
P = 0.0005 squared x 1000 = 0.00025 watt.

***
Power is calculated from *RMS* values of voltage and/or current. Do you
disagree with this last statement? You previously quoted the RMS character
of the sine wave, so I know you understand the concept, yet you seem to be
denying that the pulsed wave need not follow the same laws. Or is it that
that you disagree that the 50-50 pulse has the .707 RMS value.


As the meter reads the same, with the chopper or resistor, the series
drop from a series resistor in place of the chopper is the same.

***
It is not. In the chopper case, the series drop is NEVER 0.5 volt.
It is either zero or one volt - changing with time. This mucks about with
your reasoning. Load resistance and *RMS* load current are used to
calculate the power, no? We do this for the sine wave, why not the pulse.

The load resistance inside the meter is the same. Therefore, the power in
both cases is identical.

**** Last sentence is incorrect.



Lossy resistance or lossless resistance produce the same effect in the
load.
Best regards, Richard Harrison, KB5WZI

********************
I am afraid this is incorrect. Yes, The AVERAGE current is, as you say,
the same, but the Effective or RMS current is as I said = 0.000707. Do you
wish to address your interpretation of the pulse RMS formula and the
apparent agreement with my power analysis copied again here?

----------------------------------------------------------------------------
--
When the switch is closed, Power in the load is PL= 1.0 mW. When open,
PL= 0.0. Therefore...the average Power in the load must be half or 0.5 mW.

----------------------------------------------------------------------------
--

If you stay with your chopper PL=0.25mw claim, then what do you say
about my argument, which uses the average power as a measure rather than the
average current? Do you claim that power does not average?
Why is the average current the correct parameter to arrive at the
correct answer and the average power not? And what do you say about the
commonly accepted square pulse formula P=Imax * SQUARE-ROOT(Duty-cycle)?

Imax - 1ma. ROOT(.5) = .707

I am quite puzzeled by your conclusion. Consult with any respected expert
of your choice.
73 & good DX (if desired) (:-) , Steve
--
Steve N, K,9;d, c. i My email has no u's.