Electronic – Mathematical proof that RMS voltage times RMS current gives mean power


I know this is true because I read it in a reputable source. I also understand intuitively that power is proportional to the square of voltage or current for a resistive load, and that the "S" in RMS is for "square". I am seeking a hard mathematical proof.

Let \$I_i\$ denote the current at instant \$i\$, and likewise \$V_i\$ denotes the voltage at that instant. If we can measure voltage and current at all the instants, and there are \$n\$ instants, then mean apparent power is:

$$ P = \frac{1}{n} \sum_{i=i}^n I_i V_i $$

What is an elegant mathematical proof that

$$ P = I_{RMS} V_{RMS} $$

achieves the same result for resistive loads?

Best Answer

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.