Calculating AC Power

There is often confusion about which quantities to use and how to measure AC power.  I’m talking about power flowing from a source to a load, whether that’s line power, audio, or RF.  I’ll discuss non-sinusoidal waveforms here too.

First of all, here’s something you can always rely on:

Real power flowing to a load is the one-cycle average of the product of the instantaneous voltage across the load and the current into the load.

This holds regardless of waveform shape or whether the load is resistive or complex (with phase angle between voltage and current).  Measure the voltage across the load and the current into it with your oscilloscope and a current probe1, use the scope waveform multiplication function between the two traces to produce the instantaneous product, and use the scope’s one-cycle average calculation of that product waveform to get the real power delivered to the load.  Simple as that.  You can average over an integer number of cycles, too,  not just one.

“Wait a minute!!”, I hear you cry, “what about RMS?  I’ve always been told that it’s RMS power that’s important!”  Well, there’s actually no such thing as RMS power.  It’s a misnomer.  You can use RMS values of voltage and current under certain circumstances to calculate power, but that’s not “RMS power”.  You don’t take the RMS value of the voltage-current product, you take the average.

Waveforms figure
Purely reactive load
(V = red, I = blue, P = green)

At right is an example.   It shows voltage and current to a purely reactive (capacitive) load.  Voltage is red, current is blue (leading by 90°), and the instantaneous product of the two is that greenish color.  The RMS of the product sinusoid would be the peak value divided by \sqrt{2} or about .25.  This is obviously wrong, since no power is delivered to a reactive load.  Take the average of the instantaneous power waveform over one cycle, however, and you get the correct total power delivered to the load: zero.

So where does RMS come in?  With sinusoidal voltage and current signals, their RMS values can be a convenient way to calculate power.  Let’s do the math and see how.

The average of any function over a period of time is the integral of the function over the period divided by the length of time.  That’s just area over length, dimensionally.  For a function with period 2\pi, the average over one period \langle f \rangle is

\displaystyle \langle f \rangle =\frac{1}{2\pi} \int_{0}^{2\pi} f(t)dt

If you substitute f(t)=V_{pk}\sin(t) \cdot I_{pk}\sin(t-\theta) to calculate the average of the product of a voltage sinusoid with peak value V_{pk} and a current sinusoid with peak value I_{pk} having a phase shift of \theta relative to the voltage, you get

\displaystyle\begin{aligned} P&=\frac{1}{2\pi} \int_{0}^{2\pi} V_{pk}\sin(t) \cdot I_{pk}\sin(t-\theta)dt\\ &=\frac{1}{2\pi} V_{pk} I_{pk} \int_{0}^{2\pi}-\sin(t) [\cos(t) \sin(\theta)-\cos(\theta) \sin(t)]dt\\ &=\frac{1}{2\pi} V_{pk} I_{pk} \int_{0}^{2\pi}-\sin(t)\cos(t) \sin(\theta)dt+\int_{0}^{2\pi}\sin^2(t)\cos(\theta)dt \end{aligned}

Since the first integral equals zero (it’s the same as the \sin \cdot \cos product in the example above), we have

\displaystyle P&=\frac{1}{2\pi} V_{pk} I_{pk}\,\cos(\theta) \int_{0}^{2\pi}\sin^2(t)dt

Or, since the integral evaluates to \pi

\displaystyle P=\frac{1}{2}\, V_{pk} I_{pk}\,\cos\theta

Since the peak value of a sinusoid is equal to \sqrt{2} times the RMS, we can make the substitution and write

\displaystyle P= V_{RMS} I_{RMS}\,\cos\theta

Therefore, because of the particular properties of sinusoids, you can use RMS values of voltage and current (with phase angle) to simplify the expression for power with sinusoidal signals.  But keep in mind that this came from the average of the instantaneous voltage-current product over one cycle, and it also doesn’t hold for arbitrary waveforms.  In other words (where the i subscript indicates instantaneous values)

V_{RMS}\cdot I_{RMS}=\sqrt{\langle {V_i}^2 \rangle}\cdot \sqrt{\langle {I_i}^2 \rangle}\neq \langle V_i\cdot I_i \rangle

That is, in general the product of the RMS values of V and I does not equal the mean of their instantaneous product (the actual power), with or without a phase angle.

To add to the confusion, however, there’s another way you can use RMS values to calculate power.  If you know the current passing through a resistor and the resistor value, you can use the current RMS to calculate power by I^2R, regardless of the waveform shape.  In a sneaky way, this is actually the average of the instantaneous voltage-current product in disguise.  Here’s how.

The RMS value of an arbitrary current waveform with instantaneous value I_i can be written


(where \langle f \rangle indicates the mean of f over one period).  So {I_{RMS}}^2R can be written

\begin{aligned} P&=\left[\sqrt{\langle{I_i}^2\rangle}\right]^2R\\ &=\langle{I_i}^2\rangle R\\ &=\langle{I_i}\cdot\frac{V_i}{R}\rangle R\\ &= \langle{I_i}\cdot{V_i}\rangle \end{aligned}

where V_i is the instantaneous voltage across the resistor.  This is again the average (mean) of the instantaneous product of voltage across and current through the resistor.  A similar analysis holds for calculating resistor power as {V_{RMS}}^2/R also.

Power being the average of the instantaneous product of voltage and current makes intuitive sense too.  Power is only dissipated in the resistive part of a load.  The resistor representing the real part of the load dissipates the voltage across it times the current through it at any given instant.  But because of the thermal mass of the resistor the instantaneous power is averaged over time.2  There is no mechanism for producing the RMS of anything, that’s just a mathematical convenience.

To sum it up:

  • You can always calculate the real power delivered to a load as the average of the instantaneous product of the voltage across the load and the current through the load.
  • It’s safe to use RMS values in power calculations where sinusoidal voltage and current are used in V_{RMS}I_{RMS}\cos\theta, or with arbitrary voltage or current waveforms across/through a known resistance using {I_{RMS}}^2R or {V_{RMS}}^2/R.
  • Don’t use RMS values in V_{RMS}I_{RMS}\cos\theta where either waveform is non-sinusoidal, and don’t use the RMS value of the instantaneous voltage-current product.

1 Be sure to compensate for the skew difference between your voltage and current probes, or you will see a phase angle error between voltage and current.  Use a non-inductive resistor for a load driven with about the frequency you’re using, and adjust skew between the voltage and current scope channels to zero.

2 This points out the need to sometimes consider the peak instantaneous power, if the thermal mass of the load is not enough to average the instantaneous power.  This can happen at low frequencies or with physically small loads.