# Is it true that the average of the product of current and voltage is always real power?

I’ve had this idea that I recently realized is an assumption. It seemed to me that the average of the instantaneous power (I*V) is always the real power dissipated by the load. However, if you have an AC voltage centered around ground with a resistive load, you’ll get a current waveform with the same shape and phase as the voltage. If you average the product of the current and voltage in this situation, you’d get zero. So here’s my question:

If you have an AC voltage waveform centered around gnd over some complex load that causes distortion/deformation in the current signal so that the current is not purely sinusoidal, how would you measure the power?

What if the AC voltage waveform was slightly offset?

• Your initial description how AC power over resistive load averages to zero is wrong. For positive AC cycles, instantaneous voltages and currents are positive, so their product, the power is also positive. For negative AC cycles, instantaneous voltages and currents are negative, so their product, the power, will again be positive. The average power is positive. – Justme Feb 22 '20 at 8:15
• The instantaneous power is I*V. To get the average power, you need to integrate. – chrylis -cautiouslyoptimistic- Feb 22 '20 at 18:56
• @chrylis-onstrike- that gets you energy (joules). You must average to get power or integrate then divide by the time frame which you integrated over. – Andy aka Feb 23 '20 at 9:43
• @Andyaka Correct, of course, but the division is the simple bit. – chrylis -cautiouslyoptimistic- Feb 23 '20 at 14:21
• Which is why the OP wrote “average” in his question and not “integrate”. – Andy aka Feb 23 '20 at 14:42

If you average the product of the current and voltage in this situation, you’d get zero.

Think again...

I think you need to look at the top left diagram (resistive load).

If you have an AC voltage waveform centered around gnd over some complex load that causes distortion/deformation in the current signal so that the current is not purely sinusoidal, how would you measure the power?

Quite simply instantaneous v multiplied by instantaneous i then averaged will always give you real power consumed (i.e. what you would be billed on by your utility company).

What if the AC voltage waveform was slightly offset?

Exactly the same answer: Average(v x i) is power.

However, if you have an AC voltage centered around ground with a resistive load, you’ll get a current waveform with the same shape and phase as the voltage. If you average the product of the current and voltage in this situation, you’d get zero.

Er, no.

Consider a positive half cycle. Positive voltage, positive current, power product is positive.

Then comes a negative half cycle. Negative voltage, negative current, power product is positive.

Average of two positive powers will still be positive.

For a purely resistive load, the power dissipated by the resistance is

$$P_R = I_R^2 \cdot R = \frac{V_R^2}{R}$$

Note that for a sinusoidal voltage, both the positive and negative voltages are squared, and therefore the resistor dissipates "positive power" whether the voltage across it is positive or negative. Because the voltage is squared, the average power can be calculated using the sine wave's positive half cycle or its negative half cycle; both yield the same result. Given

$$V_R = A\;\mathrm{sin}(\omega t)$$

the average power dissipated by the resistor is

\begin{align*} P_{AVG} &= \frac{1}{T} \int_{0}^{T}P_R\;\mathrm{dt}\\ &= \frac{1}{T} \int_{0}^{T} \frac{V_R^2}{R}\;\mathrm{dt}\\ &= \frac{1}{R\,T} \int_{0}^{T} \left \{ A\,\mathrm{sin}(\omega\,t) \right \}^2 \; \mathrm{dt} \; \bigg \rvert_{T=\pi/\omega}\\ &= \frac{A^2\,\omega}{R\,\pi} \int_{0}^{\pi/\omega} \mathrm{sin}^2(\omega\,t)\,\mathrm{dt}\\ &= \frac{A^2}{2\,R}\\ \end{align*}

What if the AC voltage waveform was slightly offset?

I assume you mean, "What if the sine wave is centered about a non-zero DC voltage?"

$$V_R = A\;\mathrm{sin}(\omega\,t) + V_{DC}$$

MATLAB to the rescue!

%% Housekeeping
clear
clc

%% Symbols
syms A w t R V_DC

%% Equations
V_R = A * sin(w*t) + V_DC;
I_R = V_R / R;
P_R = V_R * I_R;
T = pi/w;   % 1/2 cycle
P_AVG = 1/T * int(P_R, t, 0, T);

%% Solutions
simplifyFraction(P_R)      % Result 1
simplifyFraction(P_AVG)    % Result 2


Result 1 $$P_R(t) = \frac{ \left ( A\;\mathrm{sin}(\omega\,t) + V_{DC} \right )^2}{R}$$

Result 2 $$P_{AVG} = \frac{\pi\,A^2 + 8\,A\,V_{DC} + 2\,\pi\,V_{DC}^2}{2\,R\,\pi}$$

If you have an AC voltage waveform centered around gnd over some complex load that causes distortion/deformation in the current signal so that the current is not purely sinusoidal, how would you measure the power?

If the load has both resistive and reactive components, then you must take into account the phase angle between the voltage and current when calculating or measuring the power. In this situation there are three possible power calculations: real power (Watts), reactive power (Volt-Amps-Reactive, VAR), and apparent power (Volt-Amps, VA). Meters exist that can measure a signal's true power, reactive power, and apparent power. For more information, try an Internet search using the keywords "power triangle" or "power factor".

Power measurement for arbitrary waveforms is usually performed with a power meter (true average power), or a wattmeter (active power), or a True RMS multimeter (RMS responding and RMS indicating) and some math (<- usually yields approximate results, at best), etc.

Yes it's true.

if you have an AC voltage centered around ground with a resistive load, you’ll get a current waveform with the same shape and phase as the voltage. If you average the product of the current and voltage in this situation, you’d get zero.

Multiply two negative quantities together and you get a positive number.

If you have an AC voltage waveform centered around gnd over some complex load that causes distortion/deformation in the current signal so that the current is not purely sinusoidal, how would you measure the power?

Multiply voltage and current, and take the average of the result.

What if the AC voltage waveform was slightly offset?

It still works the same.

Power is real even when the load is non resistive. Because power is calculated from impedance, which is the real part of RLC circuits.

• Incorrect or at best ambiguous. Real Power is zero when there is no resistance. – Andy aka Feb 23 '20 at 9:45

First off, it's E not V. In a nonreactive resistive load power is E^2 *R for all instantaneous values. Or RMS equivalent where you don't have to integrate the values you multiply the peak (not peak to peak) voltage by .707 and that yields RMS or DC equivalent power. The positive and negative values are absoluted by the square function. In a highly reactive load and one that distorts the waveform you have to do vector analysis to determine the resistive component of the load. It is impossible to use voltage and current to determine real power without considering resistance ( which can also be done using E and I phase angles). You can have a tank circuit which has huge currents and voltages and virtually no power.

• "First off, it's E not V." - no, it's U! :) E is the symbol for electric field, so V is used for voltage eg. hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html#c1 – Bruce Abbott Feb 23 '20 at 19:44
• "...multiply the peak voltage by .707..." no, not in general. This only applies for certain (sinusoidal) waveforms. IEEE Std 280 specifies the use of $V$, $E$ and possibly $U$ for voltage. – Elliot Alderson Feb 23 '20 at 22:06