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#11
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Richard Harrison wrote:
Roy Lewallen, W7EL wrote: "The average power is therefore relatively small, much smaller than the product of RMS volts times RMS amps." RMS is short for root-mean-square. RMS is synonymous with the "effective value" of a sinusoidal waveform. Therefore, the average power for the time period of one complete cycle or any number of complete cycles is the product of the effective volts times the effective amperes. No, I'm sorry, that isn't true. The average power isn't the product of the product of the RMS voltage times the RMS current, except in the single circumstance of their being in phase. See page 19 of "Alternating Current Fundamentals" for derivations of the proof. I don't have this book, but I know that in the past you've quoted from books without having fully understood the context of the quote. I'm sure that's the case here. Any electrician or technician should know that for sinusoidal waveforms, Pavg = Vrms * Irms * cos(theta) where theta is the phase angle between V and I. And hopefully you can see with a few moments and a calculator that if theta = 90 degrees, Pavg = zero regardless of V and I. Average power is exactly the product of rms volts times rms amps in usual circumstances. Perhaps your "usual circumstances" are that the load is purely resistive. But that's not "usual circumstances" for a host of applications. Roy Lewallen, W7EL |
#12
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How about P = Square(V) / R watts
or P = Square(I) * R watts for no phase angles. |
#13
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Reg Edwards wrote:
How about P = Square(V) / R watts or P = Square(I) * R watts for no phase angles. Those are fine by me. Just for the record, though: V is the voltage across R, I is the current through R. This is important when there are other components, hence possibly other values of V and I, in the circuit. P can be average and V and I RMS; or P, V, and I can all be functions of time -- the formulas are ok in either case. And finally, just for the fuss-budgets, we're assuming R isn't varying with time. Picky as this might seem, it's awfully important to make perfectly clear just what a formula applies to and what it doesn't. Otherwise it's sure to be misapplied in situations where it isn't valid. (Clarification reduces the chance of misapplication from certain to only highly likely.) Roy Lewallen, W7EL |
#14
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Roy Lewallen, W7EL wrote:
"V is the voltage across R, I is the current through R." Yes. The voltage drop across a resistor is always in-phase with the current through the resistor as there is no energy storage in a resistor. Best regards, Richard Harrison, KB5WZI |
#15
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On Wed, 15 Feb 2006 17:13:45 -0600 in rec.radio.amateur.antenna,
(Richard Harrison) wrote, RMS is short for root-mean-square. RMS is synonymous with the "effective value" of a sinusoidal waveform. Therefore, the average power for the time period of one complete cycle or any number of complete cycles is the product of the effective volts times the effective amperes. That is true for the Special Case where your load is a pure resistance. In general, no. |
#16
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Reg Edwards wrote:
How about P = Square(V) / R watts or P = Square(I) * R watts for no phase angles. Actually, the "R" implies an impedance of R+j0, i.e. a phase angle of zero. -- 73, Cecil http://www.qsl.net/w5dxp |
#17
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Owen Duffy wrote:
"Leaving aside your new confusing term "effective value", if you multiply Vrms by Irms in an AC circuit you get Apparent Power (units are volt amps or VA." Exactly, except the term "effective value is as old as a-c power calculations. The value of the a-c volt was chosen to produce the same effect, lamps as bright, heat as warm, as d-c does. My electronic dictionary says: "rms amplitude - Root-mean-square amplitude, also called effective amplitude. The value assigned to an alternating current or voltage that results in the same power dissipation in a given resistance as dc current or voltage of the same numerical value.' Best regards, Richard Harrison, KB5WZI |
#18
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#19
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Owen Duffy wrote:
"Work this example through with your textbook." Don`t need the textbook. I`ve been working these for over 60 years. Pythagoras gave us the solution in ancient times when electricity was produced by rubbing an amber rod with an animal pelt. The impedance is close enough to 120 ohms. I=E/Z E= 1 ampere Power=Isquared x R Power = 85 watts Best regards, Richard Harrison, KB5WZI |
#20
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