RMS Values and AC Power

An rms value is simply the steady DC value that would do the same heating as an AC wave. The name tells you how to compute it, read backwards: square the wave, take the mean, then take the root (root mean square). It's what lets a voltage that keeps flipping direction be called "120 V", and it's what you need to find real power. For a sine wave, $V_{\text{rms}} = V_0/\sqrt2$: take $V_0$ (the peak) and scale it down by $\sqrt2$ (because the wave spends most of its time below the peak). The power formula $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$ is plain volts times amps with one extra piece, $\cos\phi$ (how in step the current is with the voltage). The square roots and cosines make it look worse than it is; it's really just an average plus a correction for timing.

What "120 volts" means. A 1500 W space heater plugged into a 120 V outlet sees a voltage that swings between $+170$ V and $-170$ V sixty times a second, passing through zero 120 times a second. So what does "120 V" refer to? It's the steady DC voltage that would heat the heater at the same average rate. Finding it is an average value problem.

RMS, derived. In a resistor, $p(t) = v^2/R = \frac{V_0^2}{R}\cos^2\omega t$. Average it over one period using $\cos^2\theta = \frac12(1 + \cos2\theta)$: the $\cos2\omega t$ part averages to zero, leaving

So the effective voltage is the square root of the mean of $v^2$. That's exactly the recipe in the name root mean square: $V_{\text{rms}} = V_0/\sqrt2 \approx 0.707V_0$. A 120 V outlet peaks at $120\sqrt2 = 170$ V, a 230 V outlet at 325 V. Currents work the same way, and AC meters read rms. The $\sqrt2$ came from averaging $\cos^2$, so for other waveforms you have to redo the integral: a square wave of height $V_0$ has $V_{\text{rms}} = V_0$, a triangle wave $V_0/\sqrt3$.

When current and voltage are out of step. In the last lesson we saw that inductors and capacitors shift the current's timing. So what does that do to power? Let $v = V_0\cos\omega t$ and $i = I_0\cos(\omega t - \phi)$. The product-to-sum identity gives

The second term swings at twice the line frequency and averages to zero. It's energy flowing into an inductor's or capacitor's field and back out, twice per cycle. The first term is the steady part:

• A resistor has $\phi = 0$: all of its power is real.
• A pure inductor or capacitor has $\phi = \pm90^\circ$ and $\cos\phi = 0$. It borrows energy for a quarter cycle and returns it the next, for zero average power.
• In general $\cos\phi = R/|Z|$, from the angle of $Z = R + jX$ in the phasors lesson, and since $V_{\text{rms}} = I_{\text{rms}}|Z|$, the power is $I_{\text{rms}}^2|Z|\cdot\frac{R}{|Z|} = I_{\text{rms}}^2R$. Only resistance dissipates.

Why utilities care about power factor. If the back-and-forth part averages to zero, why worry about it? Because the wires still have to carry it. The product $S = V_{\text{rms}}I_{\text{rms}}$, the apparent power in volt-amperes, sets the current, and current is what heats cables and transformers ($I^2R$ losses) and what they must be sized for. A motor with power factor 0.7 draws 1.4 times the current of a heater with the same real power. The cure is power factor correction: a capacitor in parallel with the motor supplies the motor's back-and-forth current locally, so the line carries only the useful in-phase part.

The phasor shortcut. With rms phasors, $P_{\text{avg}} = \operatorname{Re}\left[\tilde V\tilde I^{*}\right]$, where $^*$ is the complex conjugate. The conjugate subtracts the current's angle from the voltage's, leaving $\cos\phi$.

A common mistake. It's tempting to think power is always volts times amps, $V_{\text{rms}}I_{\text{rms}}$, because that's how it worked for DC. It isn't when there's a phase angle: that product is the apparent power, and the real power carries the extra $\cos\phi$. Also keep peak and rms values apart. Household ratings are rms, while the amplitude in $V_0\cos\omega t$ is a peak. And check whether your calculator expects degrees or radians before taking that cosine.

Power engineering runs on these formulas. Utilities bill large customers for a poor power factor and install capacitor banks at substations to correct it. Data centers size their power supplies in volt-amperes rather than watts, and an amplifier's rated output is an rms power. Wireless chargers and RFID readers use conjugate matching to push the most power across the gap. The 2ω pulsing of single-phase power is why large motors run on three-phase supplies, whose three pulsing powers add up to a constant.