Physics/Lessons/Magnetic Fields/Alternating Currents
Physics · Magnetic Fields · A2 Level

Alternating Currents

An alternator's output constantly changes direction and magnitude, so "the current" or "the voltage" of an AC supply isn't a single fixed number the way it is for a battery. This lesson defines the peak and root-mean-square (rms) values that let us describe an AC supply meaningfully, and uses them to work out the power it delivers.

7 min read

Peak and instantaneous values

An alternating voltage varies sinusoidally with time, $V=V_0\sin(\omega t)$, where $V_0$ is the peak voltage — the maximum value reached each cycle — and $V$ is the instantaneous voltage at time $t$. If the supply drives current through a fixed resistance $R$, the current follows the same sinusoidal shape and is in phase with the voltage: $I=I_0\sin(\omega t)$, with peak current $I_0=V_0/R$.

A graph of voltage against time showing one full period of a sine wave, oscillating between plus V-naught and minus V-naught. Dashed horizontal lines mark the peak value V-naught, the trough minus V-naught, and the rms value V-naught over root two. A double headed arrow on the left shows the peak to peak range of 2 V-naught, and the x axis is marked at quarter, half, three quarter and full period.
An alternating voltage $V=V_0\sin(\omega t)$ oscillates between $+V_0$ and $-V_0$. The rms value, $V_0/\sqrt2$, sits between zero and the peak — it's the equivalent steady voltage that would deliver the same average power.

Why we need an rms value

The average of a symmetric sine wave over a full cycle is zero — it spends as much time negative as positive, so a simple mean tells you nothing useful about how much power it delivers. Power dissipated in a resistor depends on $I^2R$, though, and $I^2$ is always positive whichever way the current is flowing. So instead of averaging $I$ directly, we average $I^2$, then take a square root to get back to a current-like quantity — the root-mean-square, or rms, value.

Derivationrms current, I_rms = I₀ / √2
1
Instantaneous power in a resistor: $P=I^2R=I_0^2R\sin^2(\omega t)$.
Squaring $I=I_0\sin(\omega t)$.
2
Average $\sin^2(\omega t)$ over a full cycle is $\tfrac12$.
$\sin^2$ oscillates symmetrically between 0 and 1, spending equal "weight" above and below $\tfrac12$ — a standard result from the identity $\sin^2\theta=\tfrac12(1-\cos2\theta)$, since $\cos2\theta$ averages to zero.
3
Average power: $\overline{P} = \tfrac12 I_0^2 R$.
4
Define $I_{\text{rms}}$ as the steady (DC) current that would give the same average power: $I_{\text{rms}}^2R=\tfrac12 I_0^2R$.
5
Cancel $R$ and take the square root: $\displaystyle I_{\text{rms}} = \frac{I_0}{\sqrt2}$.
▪ derived

The same argument applies equally to voltage, giving $V_{\text{rms}}=\dfrac{V_0}{\sqrt2}$. These are the values you'd actually measure with a standard AC voltmeter or ammeter, and the values quoted for mains supplies.

UK mains supply

UK mains electricity is quoted as 230 V — this is $V_{\text{rms}}$, not the peak voltage. The actual peak voltage is $V_0=V_{\text{rms}}\times\sqrt2\approx325\text{ V}$, and the supply alternates at a frequency of 50 Hz.

Power in AC circuits

Because $I_{\text{rms}}$ and $V_{\text{rms}}$ are defined precisely so that they reproduce the same average power as an equivalent DC supply, the familiar DC power equations carry straight over, provided rms values are used throughout:

$\overline{P} = I_{\text{rms}}V_{\text{rms}} = I_{\text{rms}}^2R = \dfrac{V_{\text{rms}}^2}{R}$

Using peak values instead would overstate the power by a factor of 2 — a common error to watch for. Always check whether a quoted current or voltage is a peak or an rms value before substituting into a power equation.

Key takeaways

  • An alternating supply is described by $V=V_0\sin(\omega t)$ and $I=I_0\sin(\omega t)$, with $V_0$ and $I_0$ the peak values.
  • Because a sine wave averages to zero over a cycle, power delivery is instead described using root-mean-square values: $I_{\text{rms}}=\dfrac{I_0}{\sqrt2}$ and $V_{\text{rms}}=\dfrac{V_0}{\sqrt2}$.
  • rms values are defined as the equivalent steady (DC) values that would deliver the same average power — and are what AC meters actually read.
  • Average power: $\overline{P}=I_{\text{rms}}V_{\text{rms}}=I_{\text{rms}}^2R=V_{\text{rms}}^2/R$ — using peak values here overstates the power by a factor of 2.
  • UK mains is 230 V rms at 50 Hz, corresponding to a peak voltage of about 325 V.
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