The 2019-2023 General License question pool requires you to compute RMS voltage from peak voltage in an AC sine wave voltage:

G5B09: What is the RMS voltage of a sine wave with a value of 17 volts peak?

A. 8.5 volts

B. 12 volts

C. 24 volts

D. 34 volts

The Root Mean Square (RMS) voltage is relevant to AC voltage signals. With the assumption that the AC signal is a nicely behaving sine wave oscillation, the computation of RMS voltage from a given peak voltage value becomes trivial. But first, let’s define this RMS thing.

A voltage applied to a load or a resistance will dissipate power. For instance, plug in a lamp to a wall socket and the 120 VAC applied across the light bulb will dissipate power, and that power dissipation will manifest as light and heat from the bulb’s filament. The amount of power (P) in watts dissipated may be computed with the voltage (E) and resistance (R) as

P = E² / R

If the light bulb has direct current (DC) flowing through it with the EMF (voltage) pushing along in only a single direction, it is simple to use the constant voltage value to make this calculation. But what about the AC case in which the voltage is pushing first in one direction and then reversing to apply EMF in the opposite direction, many times each second, and varying in voltage over each cycle's time? What voltage value is to be used in this more complex AC case? The peak voltage of the sine wave form? The complete peak-to-peak voltage difference in the AC signal? The overall average voltage… wait…. that’s zero, isn’t it, averaging positive and negative values? Hmmm…

The solution comes from the concept of power dissipation. In the AC case the voltage used in computation should be a type of average value that results in the same power dissipation as a DC voltage of the same value. This average value voltage is the RMS voltage for the AC signal. So, for instance, an AC signal with an RMS voltage of 12 volts will dissipate the same amount of power as would a 12 volt DC signal. Keep in mind that the AC signal’s peak voltage will exceed the RMS value (> 12 volts in this example) and fall below the RMS value in other portions of the signal waveform.

Computing RMS voltage for a non-sinusoidal, complex waveform is a difficult prospect. Fortunately, we’re concerned only with that nicely oscillating sine wave form that smoothly transitions from positive voltage to negative voltage in a regular cycle. For sine wave voltage signals there is a single and consistent relationship between the peak voltage [E(Peak)] of the sine wave and the RMS voltage: The RMS voltage [E(RMS)] is:

E(RMS) = 0.707 x E(Peak)

So, the calculation in our question becomes very simple since the peak voltage is provided as 17 volts.

E(RMS) = 0.707 x E(Peak) E(RMS) = 0.707 x 17 volts E(RMS) = 12 volts

Note, you should also use RMS voltage for calculation of power dissipation in AC circuits, as in the first equation above. That is, square E(RMS) and divide by R. Also, see question G5B12 for a related application of E(RMS) in a power calcuation.

The answer to General Class question G5B09, “What is the RMS voltage of a sine wave with a value of 17 volts peak?” is “B. 12 volts.”

-- Stu WØSTU

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