Electronic – Low DC power consumption: RMS or mean

Ohm's law $$ 1: V(t) = I(t)R $$

Instantaneous power dissipation is product of voltage and current $$ 2: P(t) = V(t)I(t)\\ $$

Substitute 1 into 2 to get instantaneous power through a resistor in terms of voltage or current: $$ 3: P(t) = I^2(t)R = \frac{V^2(t)}{R}\\ $$

Average power is definitionally the integral of instantaneous power over a period, divided by that period. Substitute 3 into that to get average power in terms of voltage and current. $$ 4: P_{avg}=\frac{\int_0^T{P(t)dt}}{T}=\frac{R\int_0^T{I^2(t)dt}}{T}=\frac{\int_0^T{V^2(t)dt}}{RT}\\ $$

Definition of RMS current $$ 5: I_{RMS}=\sqrt{\frac{\int_0^T{I^2(t)dt}}{T}}\\ $$ Square both sides $$ 6: I_{RMS}^2 =\frac{\int_0^T{I^2(t)dt}}{T}\\ $$ Multiply by R to find equation 4 for average power $$ 7: I_{RMS}^2R =\frac{R\int_0^T{I^2(t)dt}}{T}=P_{avg}\\ $$ Definition of RMS voltage $$ 8: V_{RMS}=\sqrt{\frac{\int_0^T{V^2(t)dt}}{T}}\\ $$ Square both sides $$ 9: V_{RMS}^2=\frac{\int_0^T{V^2(t)dt}}{T}\\ $$ Divide by R to find equation 4 for average power $$ 10: \frac{V_{RMS}^2}{R}=\frac{\int_0^T{V^2(t)dt}}{RT}=P_{avg}\\ $$ Multiply expressions 7 and 10 for average power $$ 11: P_{avg}^2=V_{RMS}^2I_{RMS}^2\\ $$ Square root of both sides $$ 12: P_{avg} = V_{RMS}I_{RMS}\\ $$ Q.E.D.

The rms voltage is given by

\$\sqrt{\frac{1}{t_1-t_0}\int_{t_0}^{t_1}v^2(t) \mathrm dt}\$

The mean voltage is given by

\$\frac{1}{t_1-t_0}\int_{t_0}^{t_1}v(t) \mathrm dt\$

As you can see, these are not the same, except in special cases. A pure positive dc voltage is one such special case.

However, if the dc component of voltage is much bigger than any ac components, the rms and mean will be very close to each other. This could apply to the case of "noise with a huge offset".