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Old March 5th 04, 09:07 PM
Richard Harrison
 
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Steve Nosko wrotew:
"3) #1 with 50% chopper, [ILavg =0.5 ma...
PL = 0.5 mW]"

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

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

Lossy resistance or lossless resistance produce the same effect in the
load.

Best regards, Richard Harrison, KB5WZI

 
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