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Re: AC electricity



At 02:20 PM 1/17/01 -0400, Tim O'Donnell wrote:
I know that AC delivers energy - that is obvious.
But I still have trouble understanding how this is possible
since half the time the potential is positve and the other
have it is negative. It seems they should cancel out.

Voltage is not energy flow.

(I strongly recommend using the word "voltage" rather than "potential".
There exist distributions of voltage that strongly violate the definition
of what a potential is.)

I know and can do the rms (root mean squared - although
shouldn't it be squared mean root) for figuring out various
quantities, but I still don't think I have a basic
understanding on "how" it really works.

So far so good. RMS voltage is not the right answer in general. For
example, the RMS voltage across a capacitor tells you nothing about the
power dissipated in the capacitor.

By the way, RMS means root of the mean of the square.
This complies with the usual conventions for precedence of operators.

Anyway, the right answer is that the power (energy per unit time) is given
by voltage times current:
P = V I

Do the dimensional analysis. This formula checks out.

For a resistor, current is proportional to voltage, so P = V I is
consistent with the special-case formula alluded to above: instantaneous
power proportional to voltage squared.

For a capacitor, if the voltage looks like a sine wave the current looks
like a cosine wave. The power (P = V I) is a double-frequency sine wave
and averages to zero.

Bottom line: As a general rule, P = V I will keep you out of trouble for
ordinary circuits. This includes cases where Kirchhoff's laws are an
adequate approximation, which excludes circuits that involve _radiated_ power.