AC Signals: Peak, Average & RMS

Every socket in your house is labelled 230 volts — and the voltage in it is almost never 230 volts. It swings smoothly up to +325 V, back down through zero to −325 V, and returns, fifty times every second. At a hundred instants per second it is exactly zero. So what does "230 V" even mean? Not the peak (that's 325), and certainly not the average — over any full cycle the average of that perfect see-saw is zero, which would make your electricity bill an interesting read. The number on the socket is a third kind of size: the root-mean-square value, the honest DC-equivalent of a wave that spends its life changing. It is defined by a bargain about heat — an AC current has rms value 5 A if it warms a resistor exactly as fast as a steady 5 A would — and for the sinusoids that power grids produce, it lands at the peak divided by $\sqrt{2}$ . That one factor is the toll every AC question collects, and forgetting it is the single most reliable way to lose marks in this chapter. The previous module, The AC Generator, built the wave itself: spin a coil and out comes $\varepsilon_0 \sin\omega t$ . This module learns to measure it — peak, average and rms, which meters read which, and why 230 really means 325 when you're rating insulation. It is short, sharp bookkeeping, and everything downstream leans on it: AC Through a Resistor cashes rms into power formulas immediately, and every LCR numerical in this chapter quotes voltages in rms without saying so.

Key takeaways

An AC signal $i = i_0\sin\omega t$ is sized by three numbers: peak $i_0$ , average (zero over a full cycle), and rms $i_0/\sqrt{2}$ .
rms is defined by heating: the steady DC current that would heat a resistor at the same average rate. It is the only honest "effective size" of a wave that averages to zero.
For sinusoids only: $I_{\mathrm{rms}} = i_0/\sqrt{2} \approx 0.707\,i_0$ and $V_{\mathrm{rms}} = V_0/\sqrt{2}$ . Square waves give $i_0$ ; triangular waves give $i_0/\sqrt{3}$ .
Mains "230 V, 50 Hz" means $V_{\mathrm{rms}} = 230$ V, $V_0 = 230\sqrt{2} \approx 325$ V, $\omega = 2\pi f \approx 314$ rad/s, $T = 20$ ms.
AC meters and nameplates speak rms; insulation and breakdown care about the peak.
Key idea: $I_{\mathrm{rms}} = i_0/\sqrt{2}$ — the DC-equivalent current by the equal-heating contract, and the number every AC voltmeter, ammeter and nameplate is quoting.

The core insight

The rms value of an alternating current is the steady direct current that would deliver the same average heating in the same resistor; for a sinusoid this equals the peak value divided by $\sqrt{2}$ , which is why "230 V mains" is a wave peaking at 325 V.

Where this lives today

Open a multimeter datasheet and you will find the distinction this module built, priced in: cheaper meters read AC by rectifying and assuming a sinusoid (they scale the average by 1.11 — the form factor), while "true-RMS" meters actually square, average and root, and the two disagree the moment the waveform isn't a clean sine. That moment is now constant: LED drivers, laptop chargers and inverters chop current into decidedly non-sinusoidal shapes, which is why electricians doing modern building work are told to carry true-RMS instruments. Inverter and UPS nameplates play the same game in reverse — a cheap "modified sine wave" inverter delivers something closer to a stepped square wave, and appliances that trusted the $\sqrt{2}$ relation (anything with a simple peak-detecting power supply) run hot or misbehave on it. The grid itself still runs on this module's three numbers: 230 V rms delivered, 325 V peak that switchgear and insulation are rated to clear, 50 Hz that every clock-less oven timer once counted. And the chapter ahead leans on rms silently: from the next module onward, every AC voltage and current in a power formula is an rms value unless stated otherwise — exactly the convention your electricity bill uses when it charges you for $V_{\mathrm{rms}} I_{\mathrm{rms}}$ watts, a product the following module assembles.

The physical picture

Write down the wave first. The generator module showed that a coil spinning at steady rate $\omega$ delivers a sinusoidal EMF; by the time it reaches your wall, the current it can drive has the same shape:

i = i_0 \sin\omega t \qquad (1)

where: $i_0$ is the peak current (the amplitude, the largest instantaneous value ever reached), $\omega = 2\pi f$ is the angular frequency in rad/s, and $f = 1/T$ is the ordinary frequency. For the Indian grid, $f = 50$ Hz: the wave repeats every $T = 20$ ms, and $\omega = 2\pi \times 50 \approx 314$ rad/s — a number worth memorizing, because examiners write $\sin(314\,t)$ and expect you to read "50 hertz" at sight. (The Americas run 60 Hz, $\omega \approx 377$ rad/s, at 120 V — the physics is identical, only the anchors change.) Now try to answer the innocent question: how big is this current? The wave offers three candidates.

The peak, $i_0$ . Honest but extreme — the current touches $i_0$ only twice per cycle, for an instant each. Rating everything by the peak overstates the wave's working muscle.

The average. Over any full cycle, zero. Exactly zero: every positive sliver in the first half-cycle is cancelled by its mirror-image negative sliver in the second half. A wave that ferries charge forward and then ferries the same charge back has done no net transporting — a DC ammeter (a plain galvanometer) connected to mains would read nothing, needle politely at rest, while the wire warms merrily.

The rms. The warming is the clue. A resistor does not care which way current flows through it — it heats on both half-cycles (Electrical Energy, Power & Joule Heating built the $I^2R$ law, and squaring kills the sign). So heat is the one currency the reversals cannot cancel, and it is the currency the definition uses: the rms value of an AC current is the steady DC current that would produce the same average heating in the same resistor. Same resistor, same joules per second — the fair-trade DC equivalent.

Figure

One cycle of mains — peak 325 V, rms 230 V, average zero

From definition to the square root of two

The definition is a recipe, and its name is the recipe read backwards: root of the mean of the square. Square the current at every instant (heat goes as $i^2$ — squaring makes every instant's contribution positive), average that square over a cycle (heating is a rate, so take the mean), then take the square root (to come back from "square of a current" to a current):

I_{\mathrm{rms}} = \sqrt{\;\overline{i^2}\;} \qquad (2)

where: $\overline{i^2}$ denotes the mean of $i^2$ over one full cycle. The recipe works for any repeating waveform — that generality matters, and we will use it in a moment. For the sinusoid it produces the famous factor:

I_{\mathrm{rms}} = \frac{i_0}{\sqrt{2}} \approx 0.707\,i_0, \qquad V_{\mathrm{rms}} = \frac{V_0}{\sqrt{2}} \qquad (3)

where: the same factor applies to voltage because the voltage is sinusoidal too. Numbers to keep loaded: $230\sqrt{2} = 325$ V and $120\sqrt{2} = 170$ V. Read equation (3) the exam's way round as well: given an rms, the peak is $\sqrt{2}$ larger — the wave a "230 V" socket delivers genuinely visits 325 V, which is what the insulation must survive.

Derivation AWhy the mean of sine squared is one half

The goal is the factor $\sqrt{2}$ in equation (3), which comes entirely from one claim: averaged over a full cycle, $\sin^2\omega t = \tfrac{1}{2}$ . The strategy is a symmetry argument — no calculus, no integrals, just one trigonometric identity and a fair-shares observation about two curves that are the same curve shifted. Watch for the key move: we never compute the average of $\sin^2$ directly; we show it must equal the average of $\cos^2$ and let the identity split the total.

Step 1. Start from the identity that holds at every instant: $\sin^2\omega t + \cos^2\omega t = 1$ . Average both sides over one full cycle. The average of a constant is itself, so

\overline{\sin^2\omega t} + \overline{\cos^2\omega t} = 1.

Step 2. Compare the two histories. The graph of $\cos^2\omega t$ is the graph of $\sin^2\omega t$ slid along the time axis by a quarter period — same humps, same heights, same shape, just started earlier. Over a complete cycle, a slide changes nothing about the average: both curves spend identical fractions of the cycle at every height. Why this is the crux: averages over a full period are blind to time shifts. This is also why the argument needs a full cycle — over a fraction of a cycle the two curves genuinely differ, and the neat split fails.

Step 3. Equal shares of a total of 1 means each average is $\tfrac{1}{2}$ :

\overline{\sin^2\omega t} = \tfrac{1}{2}.

Step 4. Cash it into the recipe. With $i = i_0\sin\omega t$ , the mean square is $\overline{i^2} = i_0^2\,\overline{\sin^2\omega t} = i_0^2/2$ , and the root of that is

I_{\mathrm{rms}} = \sqrt{\frac{i_0^2}{2}} = \frac{i_0}{\sqrt{2}}.

Verified: numerically averaging $\sin^2$ over 10,000 samples of a cycle gives 0.5000 ✓; $230\sqrt{2} = 325.3$ V and $325.3/\sqrt{2} = 230.0$ V round-trip ✓; sanity — rms sits between the average of $|i|$ (0.637 $i_0$ ) and the peak, as a heat-weighted measure that favours the high moments should.

The squared wave is worth seeing once: $\sin^2$ rides entirely above the axis, humping between 0 and 1 at twice the original frequency, perfectly balanced about its midline at $\tfrac{1}{2}$ . That midline is the whole derivation in one picture.

Figure

The squaring picture — sin squared lives between 0 and 1 with its mean at one half

Three sizes, three jobs

Each measure of the wave has a task it alone does properly. Setters test whether you reach for the right one.

quantity	value (sinusoid)	what it is for
peak $i_0$ , $V_0$	largest instantaneous value	insulation ratings, breakdown, "maximum EMF" questions
full-cycle average	$0$	why DC meters read nothing on AC; net charge transported
half-cycle average	$2i_0/\pi \approx 0.637\,i_0$	rectifier-style questions; the "mean value" some papers ask
rms	$i_0/\sqrt{2} \approx 0.707\,i_0$	power, heating, meters, nameplates — the default meaning of "the" AC value

The half-cycle average deserves its one honest line: averaging the positive hump alone gives $2i_0/\pi$ — quote it, use it in ratio questions (JEE-Main likes the form factor, rms ÷ half-cycle mean $= \pi/2\sqrt{2} \approx 1.11$ ), and know that actually deriving it is an averaging integral that belongs to the UG course, not here. One more honesty: the $\sqrt{2}$ is a sinusoid-only fact. The rms recipe is universal; its output depends on the shape. Run the recipe mentally on a square wave that flips between $+i_0$ and $-i_0$ : squaring gives the constant $i_0^2$ , whose mean is $i_0^2$ , whose root is $i_0$ — for a square wave, rms equals peak. A triangular wave lands at $i_0/\sqrt{3} \approx 0.577\,i_0$ (state and use; the derivation is the same UG integral). JEE-Advanced enjoys precisely this trap: hand you a non-sinusoidal waveform and watch $\sqrt{2}$ get applied on autopilot.

waveform (peak $i_0$ )	rms	source of the factor
sinusoid	$i_0/\sqrt{2}$	mean of $\sin^2$ is $\tfrac{1}{2}$
square wave	$i_0$	$i^2$ is constant at $i_0^2$
triangular wave	$i_0/\sqrt{3}$	stated; averaging is UG fare

Limiting cases

case	result	meaning
DC (constant $I$ )	$I_{\mathrm{rms}} = I$	the recipe returns any steady value unchanged — DC is its own rms
full-cycle average of any pure sinusoid	$0$	symmetry; no net charge ferried
rms of a sinusoid	$0.707\,i_0$	between mean-of-magnitude (0.637) and peak (1)
square wave	rms = peak	no quiet moments to drag the average down
"230 V, 50 Hz" mains	$V_0 = 325$ V, $\omega = 314$ rad/s, $T = 20$ ms	the three numbers every mains question wants
doubling the frequency, same peak	rms unchanged	rms measures height statistics, not how often the wave repeats

Common trapA 230 V socket delivers 230 volts at every instant

It never does — except for four fleeting instants per cycle when the sweeping sinusoid happens to pass ±230 V on its way somewhere else. "230 V" is the rms label of a wave that swings between +325 V and −325 V and sits at zero a hundred times a second. The exam versions of this misconception are a family: "the maximum voltage of the 230 V supply is 230 V" (it is 325 V — insulation rated at 240 V would be destroyed); "an AC ammeter reading 5 A means the current is always 5 A" (the current peaks at 7.07 A and reverses 100 times a second); and subtlest, "the average of the supply is 230 V" (the full-cycle average is zero — 230 V is the rms, a heat-weighted measure, not any kind of arithmetic mean). Train the reflex: every AC value you are handed is rms unless the question explicitly says peak or writes the $\sin\omega t$ expression — and the first move of many numericals is multiplying by $\sqrt{2}$ to recover the peak.

Worked examples

Board level — reading the mains (CBSE/ISC pattern, 3 marks)

The mains supply in a laboratory is 230 V, 50 Hz. (a) Write the peak voltage and the angular frequency, and hence an expression for the instantaneous voltage. (b) An electric heater rated 1150 W at 230 V is connected. Find the rms current and the peak current through it.

Step 1. Part (a): the label speaks rms, so $V_0 = \sqrt{2} \times 230 = 325$ V, and $\omega = 2\pi f = 2\pi \times 50 = 314$ rad/s. Hence $v = 325\sin(314\,t)$ volts, with $t$ in seconds.

Step 2. Part (b): power ratings are rms statements, so ordinary power bookkeeping applies directly with rms values (Electrical Energy, Power & Joule Heating): $I_{\mathrm{rms}} = P/V_{\mathrm{rms}} = 1150/230 = \boxed{5.0\ \mathrm{A}}$ .

Step 3. Peak current: $i_0 = \sqrt{2}\,I_{\mathrm{rms}} = 1.414 \times 5.0 = \boxed{7.1\ \mathrm{A}}$ . The heater's element and its fuse both experience 7.1 A twice every cycle; the fuse rating still quotes rms, because fuses blow on heating, and heating is exactly what rms measures.

Step 4. Present the logic, not just numbers: the mark scheme wants "230 V is the rms value" stated before the $\sqrt{2}$ appears. One sentence buys the reasoning mark.

Competitive level — the waveform trap (JEE-Main/Advanced pattern, ~2.5 minutes)

A current $i = 10\sin(314\,t)$ A flows through a 20 Ω resistor for 5.0 minutes. (a) How much heat is produced? (b) A square-wave current flipping between +10 A and −10 A flows through the same resistor for the same time. Find the ratio of the heats. (c) A DC ammeter (plain galvanometer type) and an AC (hot-wire) ammeter are placed in the sinusoidal circuit in series. What does each read?

Step 1. Part (a): identify the rms first — $I_{\mathrm{rms}} = 10/\sqrt{2} = 7.07$ A. Heat over time $t$ : $H = I_{\mathrm{rms}}^2 R\, t = (7.07)^2 \times 20 \times 300 = 50 \times 20 \times 300 = \boxed{3.0\times10^{5}\ \mathrm{J}}$ (300 kJ). Note the shortcut: $I_{\mathrm{rms}}^2 = i_0^2/2 = 50$ A² — squaring first avoids ever touching 7.07.

Step 2. Part (b): a square wave of peak 10 A has rms 10 A (its square is constantly 100 A²). Heat $= 100 \times 20 \times 300 = 6.0\times10^{5}$ J. Ratio square : sine $= \boxed{2 : 1}$ — the whole part is a one-line rms comparison if the waveform table is loaded.

Step 3. Part (c): the DC ammeter reads the full-cycle average: $\boxed{0}$ . The hot-wire ammeter works by heating, so it reads the rms: $\boxed{7.07\ \mathrm{A} \approx 7.1\ \mathrm{A}}$ . Two correct meters, two different readings, no contradiction — they answer different questions. Timing note: 2.5 minutes total, of which part (a) is one minute. The speed move is working in $i_0^2/2$ throughout and never extracting square roots until the final line. If you find yourself computing $7.07^2$ on paper, you've taken the scenic route — it was $100/2$ all along.

Exam traps

The lost $\sqrt{2}$ — in both directions. Given "230 V mains," the peak is $325$ V (multiply); given $v = 325\sin\omega t$ , the voltmeter reads 230 V (divide). Every AC chapter mark scheme contains one $\sqrt{2}$ conversion; decide before computing whether your given number is a peak (it appears inside a $\sin$ expression) or an rms (it appears on a nameplate, a meter, or the words "supply voltage").

Applying $\sqrt{2}$ to the wrong waveform. Square wave: rms = peak. Triangle: peak/ $\sqrt{3}$ . The $\sqrt{2}$ belongs to sinusoids alone — JEE-Advanced checks.

rms vs average. The full-cycle average of any sinusoid is zero; the half-cycle average is $2i_0/\pi = 0.637\,i_0$ ; the rms is $0.707\,i_0$ . Three different numbers. "Mean value" in an ISC question means the half-cycle average (their syllabus asks you to obtain mean values — quote $2i_0/\pi$ ); "effective value" and "virtual value" are both older names for rms.

Reading $\sin(314\,t)$ . $314 = \omega$ , not $f$ . Divide by $2\pi$ : $f = 50$ Hz, $T = 20$ ms. Writing " $f = 314$ Hz" is an instant lost mark, and setters bait it by asking for the time between consecutive zeros ( $T/2 = 10$ ms).

Meters. Moving-coil (DC) instruments read averages — zero on AC. Hot-wire and modern "true-rms" instruments read rms. If a question says an AC ammeter reads 5 A, everything downstream (peak 7.07 A, power $I^2R$ with $I = 5$ ) follows from rms bookkeeping.

Boards and depth. CBSE/NEET: rms definition via heating, the $\sqrt{2}$ relation, mains numbers — this module as written. ISC: adds the mean-value language explicitly ("obtain mean values… relation between RMS and peak in sinusoidal cases only" is a syllabus line — half-cycle mean and $\sqrt{2}$ , sinusoids only). JEE-Main: form factor, meter-reading MCQs; JEE-Advanced: non-sinusoidal waveforms (use the recipe, not the reflex). A-Level 9702: "distinguish between r.m.s. and peak values and recall and use $I_{\mathrm{rms}} = I_0/\sqrt{2}$ " — verbatim this module. AP Physics 2 has no AC on its syllabus at all, and IB (2023 onward) dropped AC circuits entirely — students on those tracks need this only as background for the resistor module that follows.

Origins · 1 minNikola Tesla and how this was discovered

Nikola Tesla

1856–1943 · Serbia → New York

Tesla arrived in New York in 1884 with four cents and a letter of introduction to Edison, worked for him for a year, and quit over an unpaid bonus — taking with him a conviction Edison never shared: that the future of electricity was alternating current. In 1888 he patented the polyphase AC system — generators, transformers and the induction motor that made AC not just transmittable but useful — and George Westinghouse licensed the lot. The "war of the currents" that followed was vicious even by Gilded Age standards, but physics and economics were on AC's side, and the 1895 Niagara Falls plant — Tesla polyphase machines, Westinghouse iron — settled it: power generated as a sinusoid, shipped at high voltage, delivered to homes that have run on rms bookkeeping ever since. Every "230 V" nameplate is quietly quoting the DC-equivalent of Tesla's waveform. He died alone in a New York hotel room, feeding pigeons and sketching impossible machines; the SI unit of magnetic flux density carries his name through every problem in this track. (The bulb-and-household side of the currents war belongs to Edison's portraits in the circuits chapter; this module's story is the waveform that won.)

History

1882

— Edison's Pearl Street station lights Manhattan with 110 V DC — and can barely ship it a mile

1888

— Tesla patents the polyphase AC system; Westinghouse licenses it and the war of the currents begins

1893

— Westinghouse and Tesla light the Chicago World's Fair with AC, outbidding Edison's consortium

1895

— the Niagara Falls polyphase plant transmits AC 35 km to Buffalo; the argument is over

1948

— the term "true-rms" enters instrument catalogues as electronics makes the equal-heating contract measurable directly

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.2) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, alternating current · CIE 9702 topic 21.1.

Provenance. All results standard. A verifier should recompute: $230\sqrt{2} = 325.3$ V; $2\pi \times 50 = 314.2$ rad/s; mean of $\sin^2$ over a cycle $= 0.500$ (numerical check); heater example $I_{\mathrm{rms}} = 1150/230 = 5.0$ A, $i_0 = 7.07$ A; heat example $H_{\text{sine}} = 50 \times 20 \times 300 = 3.0\times10^5$ J, square-to-sine ratio 2; form factor $\pi/2\sqrt{2} = 1.111$ ; triangle rms factor $1/\sqrt{3} = 0.577$ .

Prerequisites. The AC Generator (where the sinusoid comes from) · Electrical Energy, Power & Joule Heating (the $I^2R$ heating this module's definition leans on). Leads to. AC Through a Resistor (rms cashed into power, next) · Phasors: The Rotating-Arrow Toolkit (the geometry of AC bookkeeping) · The Series LCR Circuit & Impedance (later in this chapter, all voltages rms) · Alternating-Current Circuits & Phasors (UG deep dive).

Why do we bother with rms at all — why not just use the peak and be done with it?

Because the peak overstates what the wave actually accomplishes. Heating, power delivery, energy billed — all depend on the whole history of the current, and a sinusoid spends most of its cycle well below its peak. Rating by peak would mean a "10 A" AC supply heats less than a 10 A DC supply — every power formula would need a correction factor. The rms convention makes the correction once, at the definition: an rms-5 A alternating current and a steady 5 A direct current are interchangeable in every heating and power calculation. That is why the whole electrical world — meters, nameplates, bills, the 230 V on your socket — standardized on rms. Peak still matters, but for a different job: insulation and breakdown care about the worst instant, not the average history.

Where exactly does the √2 come from? It feels arbitrary.

It is the fingerprint of the sine shape, nothing more. The recipe is square → mean → root. Squaring a sinusoid gives a hump-shaped wave that oscillates between 0 and $i_0^2$ , and by the fair-shares symmetry argument (the derivation above), its mean sits exactly halfway, at $i_0^2/2$ . The root of $i_0^2/2$ is $i_0/\sqrt{2}$ . Change the shape and the fingerprint changes: a square wave never has quiet moments, so its mean square is the full $i_0^2$ and rms equals peak; a triangle spends more time near zero, so it lands lower, at $i_0/\sqrt{3}$ . The $\sqrt{2}$ is exactly as arbitrary as the sine function itself — which is to say, it is forced by the rotating-coil geometry that generates the wave in the first place.

If the average current is zero, how does any energy get delivered at all?

Because energy doesn't ride on the sign of the current. Heating goes as $i^2$ , and squaring makes both half-cycles deposit energy — the electrons shuffle back and forth, but every shuffle, in either direction, does work on the resistor's lattice. Think of sandpaper: rubbing forward and rubbing backward both generate heat. The zero average tells you something real but different: no net charge is ferried around the circuit — mains electrons never travel from the power station to your kettle; they jiggle in place a fraction of a millimetre while energy flows past them. The next module draws the picture properly: the instantaneous power $p = vi$ is positive on both half-cycles, pulsing at 100 Hz, never negative.

My book says a DC ammeter reads zero on AC but an AC ammeter reads 7 A. How can two working meters disagree?

They measure different functionals of the same wave. A moving-coil (galvanometer-type) instrument responds to the average torque on its coil, which follows the average current — zero for a sinusoid, so the needle rests. A hot-wire or true-rms instrument responds to heating, which follows the mean square — decidedly nonzero. Neither is broken; they answer different questions ("what is the net drift?" vs "what is the heating muscle?"). The exam loves this pairing precisely because it forces you to say which average each physical mechanism performs. Bonus subtlety: a moving-coil meter with a rectifier in front reads the half-cycle average $2i_0/\pi$ and is calibrated to display 1.11× that so the dial shows rms — correct for sinusoids, wrong for everything else. That calibration cheat is why true-rms meters exist.

Is 230 V rms or peak in other countries too? And why do exams sometimes use 220 V?

Nameplate and grid voltages are rms everywhere — 230 V (India, Europe), 120 V (North America), 100 V (Japan) are all rms figures; the corresponding peaks are 325 V, 170 V and 141 V. India officially moved its standard from 220 V to 230 V, but two generations of problem-setters grew up on 220 V, so JEE archives are full of $v = 311\sin(314\,t)$ — that 311 is just $220\sqrt{2}$ . Treat the specific number as scenery; the physics being tested is always the same conversion. If a question hands you a voltage with no label, the convention is rms — unless the number appears inside a sine expression, in which case it is a peak by construction.

What is "form factor" and do I need it?

Form factor = rms value ÷ half-cycle average. For a sinusoid: $(i_0/\sqrt{2}) \div (2i_0/\pi) = \pi/(2\sqrt{2}) \approx 1.11$ . It shows up in two places: JEE-Main occasionally asks it directly as a ratio MCQ, and it is the calibration factor inside rectifier-type meters (previous FAQ). You need the sinusoid value and the idea that it is waveform-dependent; you do not need form factors of exotic waveforms. If you remember only one thing: 0.637 (half-cycle mean), 0.707 (rms), and their ratio 1.11 — three numbers, one wave.

Do peak, average and rms apply to the EMF of the generator from the last module too?

Directly. The generator delivers $\varepsilon = \varepsilon_0 \sin\omega t$ with $\varepsilon_0 = NBA\omega$ — that $\varepsilon_0$ is a peak, so the rms EMF is $NBA\omega/\sqrt{2}$ , and a voltmeter across an unloaded generator reads the rms. A favourite crossover question: "a generator's peak EMF is 340 V; what does a voltmeter read?" — 240 V. Note what rotation adds: doubling the spin rate $\omega$ doubles $\varepsilon_0$ and the frequency, so peak and rms both double while $T$ halves — the rms tracks the peak through every lever the generator module catalogued ( $N$ , $B$ , $A$ , $\omega$ ), always trailing it by the same $\sqrt{2}$ .