Total power periodic voltage develops in a resistor
Seperate all the harmonics into individual sinewaves. Calculate power in the resistor (i.e. V^2/R) for each. Add all the powers up. Job done.

Alternatively take all the individual RMS voltages from each sinewave and add them up like this Total RMS = \$\sqrt{A^2 + B^2 + C^2 + etc}\$.

Then divide that by 2 ohms and you get the same value as I first mentioned.

If you add 2V DC to the signal still treat like an example of superimposition - it does not affect any harmonics or the power dissipated by the harmonics.


  Is effective value for a zero harmonic 2 or a square root of 2?


It's 2V not sqrt(2)