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Physics II course sequence · Interactive lesson

Sinusoidal AC and rms values

Every AC number you will ever quote — mains voltage, a charger's output, the rating stamped on a fuse — is an rms number. This lesson builds that value from scratch: average the square, take the root, and check what shape of waveform you were averaging before reaching for √2.

UPII-11.05 · Lesson 5 of 8 in this unit. Position is not a completion score.
Starting model: Sinusoidal AC and rms valuesTop panel: v(t) = V₀ sin(ωt + φ) over a fixed 40 ms window, peak 325 V at 50 Hz, with a faint line at the peak and dashed lines at ±V_rms = ±229.8 V. Bottom panel: the same wave squared and scaled, v²/V₀², never negative, swinging at twice the frequency about its dashed mean of exactly ½ — whose square root is the 0.707 fixing the rms lines above.v(t) = V₀ sin(ωt + φ)peak V₀ = 325 Vrms = 229.8 Vf = 50 Hzφ = 45°v² / V₀² peaks at 1, mean ½40 ms window, left to right
Start with a picture. Change the model in the lab below.
1

Build the model

A sine averages to zero, and still boils the kettle.

The key idea

The rms value of an alternating voltage or current is the square root of its mean square over a whole cycle: the steady DC value that would dissipate the same average power in the same resistor.

Example. A 230 V rms supply peaks at 230√2 = 325 V, and across a 53 Ω heater it delivers 230²/53 = 998 W on average, exactly as a 230 V battery would, while the instantaneous power swings between 0 and 2.0 kW.

The complete physical model and assumptions

An alternating source is fixed by three numbers and nothing else: a peak value V₀, an angular frequency ω = 2πf, and a phase constant φ, so that v(t) = V₀ sin(ωt + φ). The sine is not an arbitrary choice — a coil turning steadily in a uniform field produces exactly this shape, and because circuit laws are linear a sinusoid in gives a sinusoid out at the same frequency. The trouble is that this source averages to zero over a full cycle, so the plain mean is useless as a rating: a DC meter across the mains reads nothing while the kettle boils. The fix is to average something that cannot cancel. Square the waveform, average the square over a whole number of cycles, then take the root: Vᵣₘₛ = √⟨v²⟩. The definition is built so the resistor results survive untouched, Pavg = Vᵣₘₛ²/R, which is what makes Vᵣₘₛ the DC value with the same heating effect. Doing that integral for a sine gives ⟨v²⟩ = V₀²/2 and hence Vᵣₘₛ = V₀/√2. The limits sit inside the derivation, not outside it: the √2 belongs to the sine's shape and not to the word AC, and the averaging window must hold whole cycles. And because squaring erases the sign, rms keeps no record of phase — which is exactly why it cannot, on its own, tell you the power delivered to anything but a resistor.

1Three numbers, and what each one does

Write the source as v(t) = V₀ sin(ωt + φ). The peak value V₀ sets the height in volts, the angular frequency ω sets how fast the phase advances in rad s⁻¹, and the phase constant φ says where in the cycle the clock was started. Only ω appears in the circuit equations, but instruments report the frequency f in hertz, so convert: ω = 2πf and T = 1/f. European mains at 50.0 Hz gives T = 20.0 ms and ω = 314 rad s⁻¹. Phase is read the same way in either currency. Two sinusoids of the same frequency keep a fixed phase difference, and 30° of it corresponds to (30/360) × 20.0 ms = 1.67 ms of lead or lag at 50 Hz. A phase difference between waves of different frequencies is not constant, so the idea only earns its keep at a single frequency.

2The cycle average is zero, so it cannot be the rating

Integrate a full period of the source: ∫₀ᵀ V₀ sin(ωt + φ) dt = 0, because the negative half exactly undoes the positive half. That is not a technicality. A moving-coil meter, which responds to the mean, reads zero across a live socket, and no average of v itself can appear in a rating that must predict heating. The rectified mean ⟨|v|⟩ = 2V₀/π = 0.637 V₀ is at least non-zero, but it is still the wrong quantity: power in a resistor goes as v², not as |v|, so an average taken before squaring throws away the very nonlinearity that decides the heating. Whatever average we choose has to be taken after the square.

3Average the square: where 0.707 comes from

Square the source: v² = V₀² sin²(ωt + φ). Use the identity sin²θ = ½(1 − cos 2θ) to get v² = (V₀²/2)[1 − cos(2ωt + 2φ)]. The cosine term is itself a sinusoid, of twice the frequency, so over a whole number of periods it averages to zero and only the constant survives: ⟨v²⟩ = V₀²/2. Take the root and Vᵣₘₛ = V₀/√2 ≈ 0.707 V₀. The same result follows without integration: sin²θ + cos²θ = 1 everywhere, and over a full cycle the two averages must be equal, so each is ½. Notice what the algebra also delivered — the square of a sinusoid oscillates at 2ω about its mean, and that doubled frequency is about to matter.

4Heating equivalence, and the 100 Hz ripple underneath it

For a resistor, p(t) = v²/R = (V₀²/2R)[1 − cos(2ωt + 2φ)]. Averaging gives Pavg = V₀²/2R = Vᵣₘₛ²/R, and with Ohm's law Pavg = Iᵣₘₛ² R = Vᵣₘₛ Iᵣₘₛ. That is the whole point of the definition: every DC power formula survives if rms values are inserted, so a 230 V rms supply into a 53 Ω heater delivers 998 W exactly as a 230 V battery would. What does not survive is steadiness. The instantaneous power swings from 0 to 2Pavg at 2f — 100 Hz on a 50 Hz supply — which is why filament lamps and cheap LEDs flicker at 100 Hz, why a single-phase motor's torque pulses, and why a full-wave rectifier's smoothing capacitor is sized against 100 Hz ripple rather than 50 Hz.

5The √2 belongs to the shape, not to the alternation

Repeat the recipe on other repeating waveforms and the factor changes. A symmetric square wave spends all its time at ±V₀, so ⟨v²⟩ = V₀² and Vᵣₘₛ = V₀: crest factor 1. A triangular wave gives V₀/√3 = 0.577 V₀. A half-wave rectified sine keeps only half the cycle, halving the mean square to V₀²/4, so Vᵣₘₛ = V₀/2 — while a full-wave rectified sine, which keeps every squared value, is back at V₀/√2. This is why meters disagree. An average-responding meter measures ⟨|v|⟩ and multiplies by the sine form factor 0.707/0.637 = 1.11; fed a square wave it reads 11% high, and fed a triangle 3.8% low. A true-rms meter squares, averages, and roots, and is right for all of them.

6What rms keeps and what it discards

Squaring destroys the sign, and with it the phase. Two 5.0 A rms currents of the same frequency can combine into anything from 0 to 10.0 A rms depending on their relative phase, so rms values never simply add. What does add is the mean square, and only when the parts are uncorrelated — a DC offset with an AC ripple, or harmonics of different frequency — giving Vᵣₘₛ² = VDC² + V₁rms² + V₂rms² + … . The definition also assumes a window holding a whole number of cycles: average over three-quarters of a cycle and the answer depends on where the window opened. For a signal whose amplitude drifts, rms is only meaningful over a window long compared with T and short compared with the drift.

2

Key equations

Read the units, assumptions and limits before substituting.

Sinusoidal source

v(t) = V₀ sin(ωt + φ)

Three numbers fix the entire waveform: its height, its rate, and where in the cycle t = 0 falls.

V₀ peak voltage in V, ω in rad s⁻¹, φ in rad, t in s

Angular frequency, period and frequency

ω = 2πf = 2π/T

Turns the cycles per second an instrument reports into the ω an equation needs.

f in Hz, T in s, ω in rad s⁻¹; 50.0 Hz gives T = 20.0 ms and ω = 314 rad s⁻¹

Root mean square, by definition

Vᵣₘₛ = √[(1/T)∫₀ᵀ v(t)² dt]

Works on any repeating waveform, including every one with no √2 in sight.

T is one period, or any whole number of them; Vᵣₘₛ carries the unit of v

Sinusoid: where the √2 comes in

Vᵣₘₛ = V₀/√2 ≈ 0.707 V₀

Read backwards it warns you: a 230 V rms socket puts 325 V across the insulation.

Whole cycles of a pure sine only; the same factor relates Iᵣₘₛ to the peak I₀

Heating equivalence in a resistor

Pavg = Iᵣₘₛ² R = Vᵣₘₛ²/R

This equivalence is what rms was defined to deliver, and it is exact only for a resistor.

Pavg in W, R in Ω, Iᵣₘₛ in A; the DC power formulas hold once rms values are used

Crest factor of a waveform

kc = V₀/Vᵣₘₛ

A property of the shape alone, so check it before converting any peak reading.

√2 ≈ 1.414 for a sine, 1 for a square wave, √3 ≈ 1.732 for a triangle, 2 for a half-wave rectified sine
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

325 V
100 to 350 V; steps of 25.
50 Hz
25 to 150 Hz; steps of 25.
45 °
−90 to 90 °; steps of 15.

Try this. Drag φ from 0 to 90° and the top curve slides from a sine to a cosine while the mean line below never moves: rms is the one number that throws phase away. Then raise V₀ and watch the dashed rms lines hold at 70.7% of the peak line, at every frequency.

Live results

RMS VOLTAGE Vᵣₘₛ
229.8 V
MEAN SQUARE ⟨v²⟩
52813 V²
PERIOD T
20.00 ms
v AT t = 0
229.8 V

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics II modelTop panel: v(t) = V₀ sin(ωt + φ) over a fixed 40 ms window, peak 325 V at 50 Hz, with a faint line at the peak and dashed lines at ±V_rms = ±229.8 V. Bottom panel: the same wave squared and scaled, v²/V₀², never negative, swinging at twice the frequency about its dashed mean of exactly ½ — whose square root is the 0.707 fixing the rms lines above.v(t) = V₀ sin(ωt + φ)peak V₀ = 325 Vrms = 229.8 Vf = 50 Hzφ = 45°v² / V₀² peaks at 1, mean ½40 ms window, left to right
Top panel: v(t) = V₀ sin(ωt + φ) over a fixed 40 ms window, peak 325 V at 50 Hz, with a faint line at the peak and dashed lines at ±Vᵣₘₛ = ±229.8 V. Bottom panel: the same wave squared and scaled, v²/V₀², never negative, swinging at twice the frequency about its dashed mean of exactly ½ — whose square root is the 0.707 fixing the rms lines above.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Peak voltage V₀: 325 V; Frequency f: 50 Hz; Phase constant φ: 45 °

Baseline 325; choose 100 to 350, in steps of 25.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The √2 is not a property of alternating current; it is a property of a full cycle of a sine. Students meet Vᵣₘₛ = V₀/√2 first and then apply 0.707 to square waves, rectifier outputs, pulse trains and spikes, where it can be wrong by tens of percent. Half-wave rectify a 170 V peak sine and the rms across the load is 85 V, not 120 V, because half of every cycle contributes nothing at all to the mean square. The second half of the same error is treating rms as an average of the waveform: the mean of |v| for a sine is 2V₀/π = 0.637 V₀, some 10% below 0.707 V₀. Root of the mean of the square — that order is the definition, and the three operations may not be swapped.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

An oscilloscope trace of a signal generator swings between +48.0 V and −48.0 V, and five complete cycles fill 8.00 ms of the sweep. Find the frequency, the angular frequency and the rms voltage, then write v(t) for a trace that is rising through zero at t = 0.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

A heating element is rated 1.50 kW when run from a 230 V rms, 50.0 Hz supply. Find its resistance and its peak current, then find the largest instantaneous power it ever reaches and the frequency at which that power pulsates.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

A transducer output v(t) = 3.00 + 4.00 sin(ωt) V drives a 10.0 Ω resistor. Find the true rms voltage and the average power, and say what goes wrong if the DC value and the rms of the AC part are added before squaring.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A diode in series with a resistor half-wave rectifies a sinusoidal supply of peak value 170 V, so the resistor sees v = 170 sin ωt V for half of every cycle and 0 V for the other half. What is the rms voltage across the resistor?
7

Review & continue

The rms value of an alternating voltage or current is the square root of its mean square over a whole cycle: the steady DC value that would dissipate the same average power in the same resistor.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Optional supporting lessons

These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.