RMS Values

The size of a current that averages to zero.

8 minUpdated 22 October 2026rms currentrms voltage

Lesson at a glanceShow the key points

Mean over a cycle (or long time) is zero.
Mean square i₀²/2; i_rms = i₀/√2 (39.3).
i_rms gives the same heating as a steady current; AC ratings are rms (230 V ↔ 325 V peak).

An alternating current averages to zero, yet it lights bulbs perfectly well. The right way to describe its size is the root-mean-square value.

The average is zero

For i = i₀ sin(ωt + φ), the average over a time t is

\bar i = \frac1t\int_0^t i_0\sin(\omega t + \phi)\,dt = -\frac{i_0}{\omega t}\left[\cos(\omega t + \phi) - \cos\phi\right]

Over a whole period this is exactly zero, and over a long time it tends to zero (the bracket stays between −2 and 2 while ωt grows).

Root mean square

But i² is never negative. Its average over a period:

\overline{i^2} = \frac1T\int_0^T i_0^2\sin^2(\omega t + \phi)\,dt = \frac{i_0^2}{2T}\int_0^T\left[1 - \cos 2(\omega t + \phi)\right]dt = \frac{i_0^2}{2}

rms current(39.3)

i_{rms} = \sqrt{\overline{i^2}} = \frac{i_0}{\sqrt2}

also called the virtual current; likewise V_rms = V₀/√2

Interactive

Mean, mean-square and rms

Over a whole cycle a symmetric alternating current averages to zero, so the mean tells us nothing about its heating. Square it first: i² is never negative, and its average (dashed) is the mean square. The square root of that is the rms value: a steady current of that size would heat a resistor just as much. For a sine wave it is i₀/√2.

mean0.000× i₀

mean square0.500× i₀²

rms (= i₀/√2)0.707× i₀

Waveform

Why rms is the useful value

A current i = i₀ sin(ωt + φ) through a resistor R produces heat i²R dt in each instant; over a period,

U = \int_0^T i_0^2R\sin^2(\omega t + \phi)\,dt = i_{rms}^2RT

So a steady current equal to i_rms heats a resistor exactly as much as the alternating current does. That’s why AC voltages and currents are quoted as rms values: “230 V AC” means V_rms = 230 V, and the peak is 230√2 ≈ 325 V. (For 220 V supplies the peak is 311 V.)

Worked example

Peak to rms

An alternating current has a peak of 4.0 A and a frequency of 50 Hz. Find its rms value and the time it takes to rise from zero to its peak.

Worked example

A resistor on the mains

A 46 Ω resistor is connected to a 230 V, 60 Hz supply. Find (a) the rms current, (b) the peak current, (c) how long the current takes to fall from its peak to its rms value.

Worked example

rms of a ramp

A current grows steadily as i = i₀(t/τ) from t = 0 to τ. Find its rms value over that interval.

What to remember

1Mean of a sinusoidal current over a cycle (or long time) is zero.
2Mean square i₀²/2; rms = i₀/√2 (39.3), the heating-equivalent steady current.
3AC ratings are rms values: 230 V rms ↔ 325 V peak.

Check yourself

A 230 V rms supply's voltage swings between…

Which steady current produces the same heat in a resistor as i = 10 sin ωt A?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadRoot mean squareWikipedia

PreviousAC & the Generator Next Resistors, Capacitors & Inductors on AC

RMS Values

The size of a current that averages to zero.

8 minUpdated 22 October 2026rms currentrms voltage

Lesson at a glanceShow the key points

Mean over a cycle (or long time) is zero.
Mean square i₀²/2; i_rms = i₀/√2 (39.3).
i_rms gives the same heating as a steady current; AC ratings are rms (230 V ↔ 325 V peak).

An alternating current averages to zero, yet it lights bulbs perfectly well. The right way to describe its size is the root-mean-square value.

The average is zero

For i = i₀ sin(ωt + φ), the average over a time t is

\bar i = \frac1t\int_0^t i_0\sin(\omega t + \phi)\,dt = -\frac{i_0}{\omega t}\left[\cos(\omega t + \phi) - \cos\phi\right]

Over a whole period this is exactly zero, and over a long time it tends to zero (the bracket stays between −2 and 2 while ωt grows).

Root mean square

But i² is never negative. Its average over a period:

\overline{i^2} = \frac1T\int_0^T i_0^2\sin^2(\omega t + \phi)\,dt = \frac{i_0^2}{2T}\int_0^T\left[1 - \cos 2(\omega t + \phi)\right]dt = \frac{i_0^2}{2}

rms current(39.3)

i_{rms} = \sqrt{\overline{i^2}} = \frac{i_0}{\sqrt2}

also called the virtual current; likewise V_rms = V₀/√2

Interactive

Mean, mean-square and rms

mean0.000× i₀

mean square0.500× i₀²

rms (= i₀/√2)0.707× i₀

Waveform

Why rms is the useful value

A current i = i₀ sin(ωt + φ) through a resistor R produces heat i²R dt in each instant; over a period,

U = \int_0^T i_0^2R\sin^2(\omega t + \phi)\,dt = i_{rms}^2RT

Worked example

Peak to rms

An alternating current has a peak of 4.0 A and a frequency of 50 Hz. Find its rms value and the time it takes to rise from zero to its peak.

Worked example

A resistor on the mains

A 46 Ω resistor is connected to a 230 V, 60 Hz supply. Find (a) the rms current, (b) the peak current, (c) how long the current takes to fall from its peak to its rms value.

Worked example

rms of a ramp

A current grows steadily as i = i₀(t/τ) from t = 0 to τ. Find its rms value over that interval.

What to remember

1Mean of a sinusoidal current over a cycle (or long time) is zero.
2Mean square i₀²/2; rms = i₀/√2 (39.3), the heating-equivalent steady current.
3AC ratings are rms values: 230 V rms ↔ 325 V peak.

Check yourself

A 230 V rms supply's voltage swings between…

Which steady current produces the same heat in a resistor as i = 10 sin ωt A?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadRoot mean squareWikipedia

PreviousAC & the Generator Next Resistors, Capacitors & Inductors on AC

The average is zero#

Root mean square#

Why rms is the useful value#

Check yourself

Keep exploring

The average is zero#

Root mean square#

Why rms is the useful value#

Check yourself

Keep exploring

The average is zero

Root mean square

Why rms is the useful value

The average is zero

Root mean square

Why rms is the useful value