power dissipation vs duty cycle.

[Kokoriantz]

If I apply a 50% duty cycle V volts on R, is the dissipation (V/2)²/R as the average voltage is half or V²/(2*R) as the power is averaged to half?

[bdunham7]

The latter.  Instantaneous power is I²R or V²/R.  You can average the power to get the...well, average power.  If you want to figure the power as a simple function of V and R, you need V_rms not V_avg.  For a 50% duty cycle with a top voltage of V and a bottom of 0, V_rms equals (sqrt(V²/2) (literally the root of the mean of the square).  Apply your power formula to that and R and it should all fall into place.