RE: [SI-LIST] : Heating resistor: RMS or average current (was Re: Effect of low Zo for unterminated lines)

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From: Ingraham, Andrew ([email protected])
Date: Thu Oct 28 1999 - 04:40:42 PDT

>And, in a corollary fashion, using "average" amps or volts is also wrong.

That's correct.

>In the case of the power dissipation in a driver + a capacitive load,
>the average power delivered from the power supply to that whole
>thing is V*I(average), and most of the power is anticipated
>consumed in the driver.

If you mean (V*I) (average), then I agree with you.

If you mean V*(I(average)), then I disagree.

>While I agree with rms current and average power ideas, as stated below,
>what I was saying in that special case is, rather than getting into
>averaging the instantaneous power, you can get the answer
>of average power by the average current.

Wrong. Average current gives you the wrong answer.

Simple example: For an AC source with no DC component, such as a capacitor-
or transformer-coupled signal and a resistive load, the average current and
average voltage are both zero. But the power is not.

P(t) = I(t)(squared) * R (for constant R)
= I(t)^2 * R

Paverage = [integral of P(t)] / time
= R * [integral of I(t)^2] / time
= R * Irms^2
Irms is defined as sqrt{ [integral of I(t)^2] / time }.


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