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Multiplication of \$V_{rms}\$ and \$I_{rms}\$ gives average power. What things we get if we multiply \$V_{avg}\$ and \$I_{avg}\$??

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    \$\begingroup\$ Quite often zero... \$\endgroup\$
    – Andy aka
    Sep 24, 2019 at 10:05
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    \$\begingroup\$ @kishore: In case Andy's comment is not clear, the average of a sinewave is zero. \$\endgroup\$
    – Transistor
    Sep 24, 2019 at 11:48

2 Answers 2

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Multiplication of Vrms and Irms gives average power into a resistive load only. Into other loads which result in out of phase waveforms, capacitive, inductive, rectifier for instance, the result is not average power.

If you multiply Iavg by Vavg, the caveats are similar. For well behaved waveforms like DC, the result is average power. It's easy however to construct other waveforms into other loads for which it's not average power.

In all cases, you get average power from averaging the product of the instantaneous voltage and the instantaneous current. In some well behaved cases, you can take short-cuts.

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Multiplying \$I_{RMS}\$ by \$V_{RMS}\$ gives apparent power. This is the hypotenuse of the power triangle, where the other two sides are the real (average) power (in watts) and the complex power (in VARs). If, and only if, the load is purely resistive then the complex power is zero and the apparent power will equal the real power.

In general, multiplying \$I_{AVG}\$ and \$V_{AVG}\$ gives you nothing meaningful.

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  • \$\begingroup\$ "In general, multiplying IAVG and VAVG gives you nothing meaningful." - except in DC circuits, where rms current can be deceptive. \$\endgroup\$ Sep 24, 2019 at 20:59
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    \$\begingroup\$ @BruceAbbott Which is why I said "in general". Only in specific cases, such as dc circuits, do you get a meaningful result. But in dc circuits we wouldn't usually even talk about average current and average voltage...there is just the current and the voltage. \$\endgroup\$ Sep 24, 2019 at 22:22

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