Hi can i found the R.M.S of the the function Code ( (Unknown Language)): 10sin(wt)+20 by divide it by the root of 2 ? Thank you

If the peak voltage is 10v then the rms voltage is 10/sqrt(2). You're right. The function still has the same frequency and phase angle.

The r.m.s. of any function of DC and harmonics is given by the formula X r.m.s = sqrt(x0^2 + x1^2/2 + x2^2/2 + ... ) where : x0 is the DC value. x1,x2,.. are the amplitude of the harmonics. so the answer to ur question is X r.m.s = sqrt( 20^2 + 10^2/2) = 21.213 this can be driven from the definition of the r.m.s.

I see that sawabyplus understood the function to have a dc value. I assumed that you had put the second parenthesis in the wrong place so that we have 10sin(wt+20) volts or amps, whatever -- you didn't specify. If sawabyplus is right in assuming V = 20 + 10sin(wt), then we need the value of omega and t to solve for the instantaneous rms voltage. You did not provide this information so we cannot compute the rms voltage as it varies sinusoidally. However, if my hunch is right, then we can write V = 7.07sin(wt + 20)

you are right PRS, it might have been a mistake. but what's meant by the instantaneous rms value ?, isn't the root mean square a constant that describes the function better than the simple average in average power calculations? what would be the idea behind mean if instantaneous? and the most important, when to use that instantaneous expression? and what is physics behind it?

things are mixed up a little bit with me if my function was Code ( (Unknown Language)): 10sin(wt+30)+20 R.M.S = 21.213 and σ =30 degree ?

Sawabyplus, instantaneous value of the voltage was meant, not instantaneous rms value. You're right the latter makes no sense.

zaidquis, dc voltage is constant, not changing. I may be wrong but I think that only sinewaves are expressed as rms voltages or currents.