EP Basics: Power calculation Alternating current power: RMS value or mean value?

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Should one actually use the RMS value (also RMS – Root Mean Square; Ger.: quadratischer Mittelwert) or the average value to specify the alternating current power of signals, systems, or devices?

This depends on how you define effective power. After all, you don't want to calculate the RMS value of the alternating current curve, because the result of this calculation would have no physical significance.

In contrast, you use the RMS values of voltages and/or currents to calculate the average power, i.e., the mean power value, which in turn yields meaningful results.

Let's look at a practical example: What is the power output when a sinusoidal alternating voltage of 1 V_RMS is applied to a resistance of 1 Ω?

P = U²/R = 1²/1 = 1 W (1)

This is well known, and thus there are no discrepancies here.

Calculate effective power

But how does the calculation of the effective power compare? In Image 1, a sinusoidal alternating voltage with an RMS value of 1 V is shown. The peak-to-peak value (from peak to peak) is therefore U_p-p = 1 V_RMS 2 √2 = 2.828 V. The voltage thus swings from –1.414 to +1.414 V.

The curve in Image 2 represents the power dissipated at a resistance of 1 Ω for this sinusoidal voltage of 1 V_RMS according to P = U²/R.

The instantaneous power curve has an offset of 1 W and swings from 0 to 2 W. The RMS value of this power curve is 1.225 W.

One way to calculate this value is indicated in Equation 2:

RMS = (Offset² + (AC_RMS)²)^1/2 (2)

RMS = (1 W² + (2 W / 2√2)²)^1/2 = (1 + 0,5)^1/2 = 1.225 W

This can be verified with a numerical series in MATLAB or Excel.

Calculate average power

In contrast, the average value of the power curve in Image 2 is 1 W, which can be derived simply visually: The curve oscillates symmetrically around the 1-W line. By calculating the numerical average of the data points forming this curve, you obtain the same result. The average power corresponds to the power obtained when using the effective voltage.

Thus, the power dissipated at a resistance of 1 Ω for a sinusoidal voltage of 1 V_RMS is not 1.225 W, but 1 W. Consequently, the average power – that is, the mean value – provides the correct value, and thus the average power is of physical relevance.

The effective power (according to the definition used here) has no meaningful significance, thus no physical or electrical relevance, but merely represents a quantity that can be calculated for exercise purposes.

It is also a trivial exercise to conduct the same analysis with a sinusoidal current of 1 A_RMS through a 1-Ω resistor. The result is the same.

Power dissipation: Average value and RMS value in ICs

Power supplies for integrated circuits typically provide a direct voltage, which is why effective power in supplying ICs is not an issue. With direct current, the average and effective values are identical to the DC value.

Conclusion: According to the definition explained here, you must be careful about which power you use for time-varying powers such as voltages and currents (e.g., noise, RF signals, oscillators): Use the effective voltage and effective current to calculate the average power, as this provides meaningful power specifications. (kr)