How and WHY does the RMS of current gve us the equivalent DC?
I've been though this answer here:-root mean square for dc but it doesn't answer my question.
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It's an arguably limited equivalence, but a convenient one. The power dissipation in a linear resistor R due to an alternating current $I\cdot\sin(\omega t)$ is the same as the power dissipation in the same resistor due to a direct current with the corresponding r.m.s. value, which is $I/\sqrt 2$.
To see why, just consider the instantaneous power dissipation rate in the alternating case, that is $I^2\cdot R\cdot \sin^2(\omega t)$ . With the math equivalence $\sin^2(\omega t) = (1 - \cos(2 \omega t))/2$ , of which the mean value is of course just $1/2$, the mean power dissipation rate is $(I^2/2).R$, i.e. the same as what would be produced by a direct current $I/\sqrt 2$. I hope that helps.
terry-s
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Thanks! So it's basically just defined that way, right? – math and physics forever Aug 31 '22 at 11:19
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1@fhhh : Well if attention is confined to questions of power dissipation then it's a physical equivalence and not just a definition. But for any other, non-power, purposes, if the rms value is taken to be equivalent to the a.c. value then yes, I guess you could call it conventional and a matter of definition. – terry-s Aug 31 '22 at 11:34
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Ok.THanks a lot! – math and physics forever Aug 31 '22 at 11:36
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Do you mind if I accept this answer in a day? I't answers my doubt, but I'd like to wait for a bit – math and physics forever Aug 31 '22 at 11:37
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That's fine and you're welcome. HTH! – terry-s Aug 31 '22 at 11:37
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1@fhhh, Re, "So it's...just defined that way..?" No. "defined" means that somebody just decided what some word should mean, and everybody else agreed. Terry S. showed you the mathematical reason why it has to be that way. – Solomon Slow Aug 31 '22 at 14:06