Eav

exaltus

Joined Jan 24, 2005
1
Why is the average voltage Eav = Emax* 2 /Pie,

My understanding is 2*Emax = Peak to Peak voltage

therefore the E average Eav= 2Emax / 2 pie R,

R,radius should also equal to E av,

Therefore E av = 2 Emax / 2 pie Eav

Therefore (Eav= sq. rt of Emax /pie)

Brandon

Joined Dec 14, 2004
306
Originally posted by exaltus@Jan 24 2005, 04:39 PM
Why is the average voltage Eav = Emax* 2 /Pie,
My understanding is 2*Emax = Peak to Peak voltage

therefore the E average Eav= 2Emax / 2 pie R,

R,radius should also equal to E av,

Therefore E av = 2 Emax / 2 pie Eav

Therefore (Eav= sq. rt of Emax /pie)
[post=4778]Quoted post[/post]​
Its better to just integrate when you want to find the average voltage.
i.e. (1/T) int(F(x),dx,0,T) For say sin we have 2 average we can take. The 1/2 cycle or full cycle. Full cycle we know will be 0 since its sin.

1/2 cycles: 1/pi int(sin(x),dx,0,pi)=1/pi [-cos(x)]0,pi = 2/pi

This happens to be Emax (1) * 2 /pi for the sinusoidal.

When you get away from sinusoidal signals, you can't apply this relation anymore. Better to just stick with the derivation of average than trying to remember difference versions of the same thing.