electricity: effective voltage?

[ShawnD1]

I'm having a problem with the concept of effective voltage. My problem will be best explained with pictures.

First we have the top part of our sine wave. V = sin(x)




Since the top is symmetrical, let's just use the first 90 degrees. To get our area, we'll integrate it from 0 to 90.
A = (-)cos(90) - (-)cos(0)
A = -0 + 1
A = 1




Since our effective voltage should be constant + or -, let's make it a box instead of a curve. To make this box, we'll just find the average area between 0 and 90 degrees.
A average = 1/2




Ok now that we have our box and its area, let's find the height of this box (usable voltage). Ok well the book says I square root this thing.... so it must be a square box! But how do I know it's a square......hmmm I guess I can't really say that I do know that.






Does anybody here know WHY this box is a square box? How do I know it's a square?

[nunyadam]

its not a square as to its shape. it is its mathmatical square.

[nunyadam]

http://www.ptti.com/html/help/Square_d.htm

[nunyadam]

http://www.ptti.com/html/help/square_r.htm

[paul9]

it's a LONG time since i was at college doing any of this but what i remember is that for the average, you would need to work out smaller squares in the spaces your large square did not encompass. after a short while, your squares will get too small to be practical. also you could try looking for 'RMS' or 'ROOT MEAN SQUARE' on google or a university website. i think it's best to think of the effective voltage as the time when it's non zero enough to drive your current through the circuit. in the uk i think our peak to peak ac supply is 330 volts and our supply is 240 volts rms. most multimeters are calibrated to show you rms voltage on ac, but an oscilloscope would show you the full peak voltages. btw i don't know why the effective voltage would only include positive values of voltage or only negative, unless you are using a half wave rectifier as opposed to full wave, eg a diode resistor and capacicitor( or just a diode) as opposed to a 4 diode bridge. edit: also you culd just as well use rectangles instead of squares to work out the square or average or area of this curve.