Bottlehead Forum
General Category => Technical topics => Topic started by: Deke609 on November 09, 2018, 01:09:02 PM
-
I've dipped my toe into the math and am immediately confused. I'm struggling to understand:
(a) why the DC voltage equivalent of an AC current isn't |avg voltage of 1/2 AC sine wave| -- or stated differently, the sum of the absolute values of the average voltage of both +ve and -ve, all divided by 2 --- which is (0.637*Vpk); and,
(b) why the work done by an AC current isn't directly proportional to its avg voltage value
But VAC avg isn't used to calculate the "effective value" of an AC current source (which I take to mean its DC equivalent). Instead the effective value = VAC rms = 0.707 * VAC pk.
I'm clearly missing something - but when i search for answers online, all I come across are explanations to the effect that the average value of a sine wave is 0 b/c its +ve and -ve cancel each other out -- which is obviously true, but I would have thought that currents of the same absolute magnitude do the same work, regardless of polarity (whether "pushing" or "pulling"). So the "cancels out" explanation doesn't answer my problem.
What do I need to read up on to figure this out? Once upon a time I was comfortable with the integration/differentiation of simple x/y functions-- if this requires something more complicated, I may have to put figuring this out on hold (perhaps permanently :D).
Many thanks,
Derek
-
"Twlnkle, twinkle, little star,
power equals I-squared R"
This handy mnemonic reminds us that power is proportional to current squared. (Power is also voltage squared over resistance.) The rms voltage is the square root of the average voltage-squared, hence Root-Mean-Squared.
-
I bet the best text's on this are not electronics as much as ones dealing with power generation and distribution...John
-
Thanks PJ. So, if RMS is based on Ohm's law, couldn't average power also be calculated using either P = Isquare * R, or P = I * V. In the case of P = I * V, wouldn't average power equal the average of the sum of instants between t-zero and t-one of Iavg * Vavg, so simply Iavg * Vavg? That seems a simpler calculation. Is using RMS simply a convention? And if so, does its use serve a particular purpose?
Edit: And just to clarify, I mean the average of the absolute values of I and V over a period t - so that the +ve and -ve don't cancel each other out.
-
I bet the best text's on this are not electronics as much as ones dealing with power generation and distribution...John
Hi John -you may well be right. I've been reading the U.S. Navy's texts on electricity and electronics from the 1950s and found them pretty good at explaining things in everyday world terms -- until I came to Average Voltage and Voltage RMS. After two pages of explaining V avg, the text effectively stated, "But this value is only used for taking voltage readings of AC with voltmeters, and is otherwise useless ... So let's look at V rms instead". Which left me thinking, "Huh? Why is V avg useless?".
Maybe I'll be able to find a similar text for training power line and generator technicians from the 1950s.
-
Consider the difference between a sine wave and a square wave when attempting to apply Ohm's Law, then you'll start to see why we need the adjustments.
-
Thanks PB. I will ponder that.
-
Thanks PJ. So, if RMS is based on Ohm's law, couldn't average power also be calculated using either P = Isquare * R, or P = I * V. In the case of P = I * V, wouldn't average power equal the average of the sum of instants between t-zero and t-one of Iavg * Vavg, so simply Iavg * Vavg? That seems a simpler calculation. Is using RMS simply a convention? And if so, does its use serve a particular purpose?
Edit: And just to clarify, I mean the average of the absolute values of I and V over a period t - so that the +ve and -ve don't cancel each other out.
You are correct that power is the average of I*V, but that is NOT the same as average I times average V.
Very simple example: Imagine a 1-ohm resistor, with 2 volts applied for one second, and zero volts for the next second. In the first second, current is 2 amps so power is 4 watts; it's zero in the second second so the average power over those two seconds must be 2 watts. But over those two seconds, the average voltage is one volt, and current is one amp, so Vavg*Iavg is 1 watt.
-
Thanks PJ. I see my error. Pavg over time period t = mean of all instances of I*V over time period t. That being the case, it probably makes most sense to calculate average power based on either Isquared or Vsquared - this way there is only one variable (or variable function) and one constant (assuming R is constant).
cheers and thanks,
Derek