Root Mean Square of a Sinusoid - With Derivation

[@Axotron]

If you want to be even more complete in the mathematical derivation steps, you could use the definition of RMS based on average power at 1:29 and show the equation for instantaneous power in a resistive load: P = U^2/2 = (Vp sin(theta))^2/R = Vp^2 sin(theta)^2 / R. The average power is then [integral from 0 to 2*pi of P]/2*pi. You already show the primitive function of sin(x)^2 at 4:05 (labelled "nasty") and could use that directly by plugging in 2*pi and 0 to solve the integral: [integral from 0 to 2 pi of sin^x] = 1/2 * (2*pi - sin(4*pi)/2) - 1/2 * (2*0 - sin(2*0)/2) = pi. (This can alternatively even be shown graphically by plotting sin^2(x) = 1/2 - cos(2x)/2, which clearly has an average of 1/2, so has an integral of pi from 0 to 2*pi.)
With the integral figured out, one can proceed to see what DC voltage is required for the same average power:
P_avg = [integral from 0 to 2*pi of P]/2*pi = Vp^2*[integral from 0 to 2*pi of sin^2(x)]/2*pi/R = Vp^2 * pi /2*pi/R =Vp^2/2/R
For a DC voltage V_rms, the power is V_rms^2/R.
Setting these expressions equal yields V_rms^2 = Vp^2/2 and taking the square root gives the desired answer V_rms = Vp/sqrt(2)
Sorry for nitpicking... Your animations and explanations are nice.

[@oldhackee3915]

Thanks