# 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?

• If I remember correctly, there should be a proof that states how RMS is the closest approximation of the actual value of a signal in the time duration of interest. Using that, we could probably prove that $P=I_{rms}V_{rms}=\frac{1}{T2-T1} \int_{T1}^{T2}V(t)I(t)dt$. Unfortunately, it seems I lost the book which had the proof of that. :( – AndrejaKo Jan 28 '14 at 20:10
• RMS current times RMS voltage does not equal mean power. It equals (mean) apparent power. If you've got non-resistive loads, this can make a difference. – Someone Somewhere May 29 '16 at 19:25

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 very simple proof (in the discrete sampling case in the question) is by substitution of E/R for I in the RMS equation

$$x_{\mathrm{rms}}=\sqrt{\dfrac1n(x_1^2+x_2^2+x+\cdots+x_n^2)}.$$

and very simple algebra.

And yes, this is true because it is specified that we have a purely resistive load so there is no phase angle issue and no harmonic present in I that is not also present in E.

EDIT

definition of RMS for discrete points (from Wikipedia): $$x_{\mathrm{rms}} = \sqrt{ \frac{1}{n} \left( x_1^2 + x_2^2 + \cdots + x_n^2 \right) }$$

so $$V_{RMS} = \sqrt{ \frac{1}{n} \left( V_1^2 + V_2^2 + \cdots + V_n^2 \right) }$$

and $$I_{RMS} = \sqrt{ \frac{1}{n} \left( I_1^2 + I_2^2 + \cdots + I_n^2 \right) }$$

and by Ohm’s Law $$I_i = V_i/R$$ substitution:

$$I_{RMS} = \sqrt{ \frac{1}{n} \left( (V_1/R)^2 + (V_2/R)^2 + \cdots + (V_n/R)^2 \right) }$$

then:

$$I_{RMS} = \sqrt{ \frac{1}{n} \left( V_1^2/R^2 + V_2^2/R^2 + \cdots + V_n^2/R^2 \right) }$$

Pulling out the 1/R^2

$$I_{RMS} = \frac{1}{R}\sqrt{ \frac{1}{n} \left( V_1^2 + V_2^2 + \cdots + V_n^2 \right) }$$

so:

$$V_{RMS} * I_{RMS}$$ is:

$$1/R( \frac{1}{n} \left( V_1^2 + V_2^2 + \cdots + V_n^2 \right))$$

distributing the 1/R:

$$( \frac{1}{n} \left( V_1^2/R + V_2^2/R + \cdots + V_n^2/R \right))$$

Using Ohm’s Law substitution again:

$$( \frac{1}{n} \left( V_1I_1 + V_2I_2 + \cdots + V_nI_n \right))$$

which is:

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

• If the algebra is simple, can you show us? You can use LaTeX markup to typeset the math. – Phil Frost Jan 28 '14 at 20:57
• Thanks for the encouragement. I hadn't used LaTex since 1983. – George White Jan 28 '14 at 21:36

The key is that for a resistive load, the voltage and current are in phase.

If the voltage and current are both $\sin(t)$, then then their product is given by the equality $\sin^2(t) = 1/2 + 1/2 \sin(2t)$. The power is a sine wave of twice the frequency, which oscillates about $1/2$. This is its average over time (the "mean" of the "square"). The root of the mean square is $\sqrt{1/2} = 1/\sqrt{2} = \sqrt{2}/2 \approx 0.707$. That's where we get that magic number.

The root mean square voltage or current are the DC equivalent voltage and current that will produce the same power dissipation over time. If the average power dissipation is $1/2$ W, then such a power dissipation can be steadily produced by $\sqrt{2}/2$ VDC multiplied by $\sqrt{2}/2$ A DC.

If current and voltage are out of phase 90 degrees (pure reactive load), then we can think of one as being $\cos(t)$ and the other being $\sin(t)$. The applicable equality is then $\sin(t)\cos(t) = 1/2 \sin(2t)$. The power waveform is no longer "biased" to oscillate around $1/2$; its average is zero: power flows into and out of the load on alternate half cycles, as the power waveform swings positive and negative.

So to answer the question, the RMS voltage and current are defined based on the mean power: each one is derived from the square root of the mean power. Multiplying two values together that are obtained from the square root of the mean power, recovers mean power.

• I think Stephen Colling's answer is the best. It does not rely on the details of the waveform and covers the continuos case. Also, "The root mean square voltage or current are the DC equivalent voltage and current that will produce the same power dissipation over time" seems to answer the question by assuming the answer and then going in a circle to get the answer. – George White Jan 29 '14 at 1:03

Lets simplify more this issue without math. Take this simple circuit that is produce a square waveform with a period of 10 sec. The voltage is like this and current is Then the power waveform will be When switch is open no power is delivered to the resistor so the total energy is 10 watts X 5 seconds= 50 Joules, and it is the same as we apply 5 watts in 10 seconds and this is the average power. The average voltage is 5 volts and the average current is 0.5 ampere. Doing simple calculation, the average power results 2.5Watt or 25 Joules which is not true.

So let’s make this trick WITH THIS ORDER:

1. First square the voltage (and current)

2. Second take the average of the square

3. Then take the square root of the average

The square of the voltage waveform will be And the average is 50V^2 (not 50^2 volt). From this point forget about the waveform. Only values. Square root of the above value is 7,071…volt RMS. Doing the same to the current will found 0,7071..A RMS And the average power will be 7,071V x 0,7071A= 5 Watt

If you are try to do the same with RMS power the result will be a meanigless 7,071Watt.

So the only equivalent heating power is the average power and the only way to calculate is to use the rms values of voltage and current

• Can't we calculate average power dissipated in a resistor as the average of the instantaneous power? Where is the mathematical proof that the OP requested? – Joe Hass Jan 28 '14 at 22:40
• For some complex waveforms off course we have to integrate them using time intervals close to zero for exact average values. I avoide to use any math at all, that's why I use square wave which is very easy to see the meaning of average. RMS is also an average value. – GR Tech Jan 28 '14 at 22:54
• It seems to me that you show that the actual average power is 5 watts and that RMS V * RMS I = 5 watts demonstrating, for this case, that the OP is correct. You also show that, in this case, average V * average I = 2.5 watts. – George White Jan 28 '14 at 23:17
• OK I understand. Language problem again. What I was try to say is that the calculation Vavg x Iavg is not correct. Thanks for discourage me! – GR Tech Jan 28 '14 at 23:21
• If "RMS is also an average value" then why isn't the RMS value of the power line voltage equal to 0.0V just like the average value? – Joe Hass Jan 29 '14 at 0:31