I've becomes a bit confused with the average power formulas. These formulas can be found on Wikipedia here and here. Let's suppose V(t) = 1V (DC) and we have a square wave for the current that switches from -1A to 1A. If I look at the first equation, I'd get that \$P_\mathrm{ave}=0\$W because the average value of a square wave is 0; however, if I look at the second equation, I'd find that \$P_\mathrm{ave}=1\$W because the RMS voltage is 1V and the RMS current is 1A.

I don't understand which equation is correct. They seem to be calculating different averages. If someone asks for the average power, which do they mean? What am I missing?

$$ P_\mathrm{ave} = \frac{1}{T_2 - T_1} \int_{T_1}^{T_2} V(t) I(t) \, \mathrm{d}t $$ $$ P_\mathrm{ave} = V_\mathrm{rms} I_\mathrm{rms} = \sqrt{\frac{1}{T_2 - T_1} \int_{T_1}^{T_2} V^2(t) \, \mathrm{d}t} \sqrt{\frac{1}{T_2 - T_1} \int_{T_1}^{T_2} I^2(t) \, \mathrm{d}t} $$


5 Answers 5


If someone asked for the average power dissipated in a device, what would that mean?

The average power is the time average of the instantaneous power. In the case you describe, the instantaneous power is a 1W peak square wave and, as you point out, the average over a period is zero.

But, consider the case of (in phase) sinusoidal voltage and current:

$$v(t) = V \cos \omega t $$

$$i(t) = I \cos \omega t $$

The instantaneous and average power are:

$$p(t) = v(t) \cdot i(t) = V_m \cos\omega t \cdot I_m \cos\omega t = \dfrac{V_m \cdot I_m}{2}(1 + \cos2\omega t) $$

$$p_{avg} = \dfrac{V_m \cdot I_m}{2}$$

(since the time average of sinusoid over a period is zero.)

In the above, we evaluated the time average of the instantaneous power. This will always give the correct result.

You link to the Wiki article on AC power which is analyzed in the phasor domain. Phasor analysis assumes sinusoidal excitation so it would be a mistake to apply the AC power results to your square wave example.

The product of the rms phasor voltage \$\vec V \$ and current \$\vec I \$ gives the complex power S:

$$S = \vec V \cdot \vec I = P + jQ$$

where P, the real part of S, is the average power.

The rms phasor voltage and current for the time domain voltage and current above are:

$$\vec V = \dfrac{V_m}{\sqrt{2}} $$

$$\vec I = \dfrac{I_m}{\sqrt{2}} $$

The complex power is then:

$$S = \dfrac{V_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} = \dfrac{V_m \cdot I_m}{2}$$

Since, in this case, S is purely real, the average power is:

$$P = \dfrac{V_m \cdot I_m}{2}$$

which agrees with the time domain calculation.

  • \$\begingroup\$ And just a reminder, gentle reader, that this result applies only to sinusoidal voltage and current. \$\endgroup\$
    – Joe Hass
    Oct 6, 2013 at 14:54
  • 1
    \$\begingroup\$ @JoeHass, phasor (AC) analysis presumes sinusoidal excitation. There is no phasor that represents, say, a square wave so, if one is working in the phasor domain, sinusoidal voltage and current are implicit. \$\endgroup\$ Oct 6, 2013 at 14:59
  • \$\begingroup\$ Yes, and since the original question involved a square wave I just wanted to make it clear that your solution could not be applied to the specific case described in the original question. Personally, since the OP was familiar with time series analysis I felt that jumping to phasor analysis might be confusing. \$\endgroup\$
    – Joe Hass
    Oct 6, 2013 at 15:11
  • \$\begingroup\$ @JoeHass, at your suggestion, I'll add a bit about the square wave. But, regarding the phasor analysis section, I included it precisely because the OP linked to the Wiki article on AC power. \$\endgroup\$ Oct 6, 2013 at 15:24

Multiplying the RMS voltage and current is not an average power calculation. The product of RMS current and voltage is the apparent power. Note also that RMS power and apparent power are not the same thing.

  • \$\begingroup\$ If someone asked for the average power dissipated in a device, what would that mean? So if there's a resistor, and it has some current and voltage through and across it, how would I calculate the average power? \$\endgroup\$
    – user968243
    Oct 6, 2013 at 13:30
  • \$\begingroup\$ The first formula you give above is correct. You find the instantaneous power as a function of time, integrate over the time interval of interest, and divide by the length of that interval. For a time-varying voltage with an average value of 0 volts, the average power of the resistor will be zero. That's why we use RMS power when talking about a.c. circuits. \$\endgroup\$
    – Joe Hass
    Oct 6, 2013 at 14:52
  • \$\begingroup\$ Joe, if the time average voltage across a resistor is zero, the average power delivered to the resistor need not be, and typically isn't, zero. For example, the time average of a sinusoidal voltage (over a period) is zero but the average power delivered to the resistor is not. This because the power is proportional to the square of the voltage and the time average of the square of the sinusoidal voltage is not zero. \$\endgroup\$ Oct 6, 2013 at 19:22
  • \$\begingroup\$ @AlfredCentauri You are right of course, when the voltage across a resistor is negative the current will also be negative (by the usual sign convention for passive elements), so the instantaneous power will also be positive. My apologies to all. \$\endgroup\$
    – Joe Hass
    Oct 6, 2013 at 20:03

For electrical calculations you'll almost always want to use the RMS power.

The confusion has to do with the difference between work and energy. Work = force X distance. If you drive 60 miles in one direction and then drive 60 miles in the opposite direction, mathematically you have done zero work, but we've used 120 miles worth of energy (gas).

Similarly, because the same number of electrons were moved the same distance (current) with the same force (voltage) in both directions (positive and negative), the net work is zero. That's not very helpful when you're interested in how much work we can get out of a machine, or how much heat we can get from a heater.

So, we go to RMS. It allows you to add the work done in the negative direction to the work done in the positive. It's mathematically the same as running your AC power through a rectifier and converting it to DC. You're squaring the values to make them all positive, averaging the values, and then taking the square root.

You could do the same by averaging the absolute values of voltage and current, but that's a nonlinear operation and doesn't allow us to use a nice equation.


I am actually struggling with the concept myself for calculating power efficiencies. Honestly, to calculate "Average Power" take instantaneous power \$ P(t) = V(t)*I(t) \$ and average it on the interval \$ P_\mathrm{ave} = \frac{1}{T_2 - T_1} \int_{T_1}^{T_2} V(t) I(t) \, \mathrm{d}t \$ like you did before. This is applicable to every case. This also means that the average power in your question is zero. The RMS value comes out wrong because of the nature of your current. I do not want to go into details but the way I see it, RMS power is misleading in most cases. Also RMS of voltage times RMS of current is the apparent power like someone mentioned before, but god alone knows what that means.

Also Prms = Pave when load is resistive. So a more general definition would be \$ Pave = Irms * Vrms * cos(\theta ) \$. So for resistive load \$ \theta \$ is zero Pave = Prms. Anyway i will really suggest you to use \$ P_\mathrm{ave} = \frac{1}{T_2 - T_1} \int_{T_1}^{T_2} V(t) I(t) \, \mathrm{d}t \$ which is true in every case (be it resistive inductive or of two random signals) and cannot go wrong.


I find it easier to think in term of energy.

In you example, when the current is positive, energy (power * time) is transferred from A to B. When the current is negative, energy is transferred from B to A.

If you are an observer between A and B, over a full cycle, no net energy is transferred, and thus the average power is zero (over a full cycle).


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