AC Fundamentals (Peak, RMS, Average)

PHYAlternating Current — Basics

AC quantities oscillate, so we describe them by a peak (amplitude) value, an RMS value (the DC-equivalent for heating), and an average value over a cycle. Students read $I_0$ and $\omega$ off an equation like $i = I_0\sin\omega t$ , convert between peak and RMS, and extract frequency/period.

Getting $\sqrt 2$ vs 2 and $\omega$ vs $f$ right is the core skill examiners check.

Instantaneous sinusoidal current

i = I_0 \sin(\omega t)

I_0

= peak current (A),

\omega

= angular frequency (rad/s),

t

= time (s).

RMS values

I_{rms} = \dfrac{I_0}{\sqrt{2}} \approx 0.707\,I_0, \qquad V_{rms} = \dfrac{V_0}{\sqrt{2}}

valid for pure sinusoids; RMS is the steady DC value that gives the same heating in a resistor.

Frequency and time period

\omega = 2\pi f, \qquad T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}

f

= frequency (Hz),

T

= period (s);

\omega

is the angular frequency in rad/s.

Average over a half cycle

I_{avg} = \dfrac{2 I_0}{\pi} \approx 0.637\,I_0

average of

\sin

over half a cycle; over a FULL cycle the average of a symmetric AC is zero.

From $i = 10\sin(314t)$ : peak $I_0 = 10\,\text{A}$ and $\omega = 314\,\text{rad/s}$ , so $f = \omega/2\pi \approx 50\,\text{Hz}$ and $T = 1/f = 0.02\,\text{s}$ .
RMS is what AC meters read and what mains ratings (e.g. 220 V) refer to; the peak is $V_0 = \sqrt2\,V_{rms} \approx 311\,\text{V}$ for 220 V mains.
The full-cycle average of a sinusoidal AC is zero (equal positive and negative halves), which is why the RMS (root-mean-square) value is used for heating.
Half-cycle average $= 2I_0/\pi$ , used for rectified output and for average-value numericals.
The factor $\sqrt2$ (peak/RMS) and $\pi/2$ (peak/half-cycle-average) are specific to SINUSOIDS — they change for other waveforms.
Angular frequency $\omega$ (rad/s) and frequency $f$ (Hz) differ by a factor $2\pi$ ; the argument of $\sin$ is always $\omega t$ , not $ft$ .
Form factor $= I_{rms}/I_{avg} = \pi/(2\sqrt2) \approx 1.11$ for a sinusoid (occasionally asked in MCQs).

Where the marks go

Reading 314 as the frequency — it is $\omega$ ; the frequency is $314/2\pi \approx 50\,\text{Hz}$ .
Using the half-cycle average $2I_0/\pi$ when the RMS $I_0/\sqrt2$ is wanted, or quoting 'zero' (the full-cycle average) when a half-cycle average was asked.
Dividing peak by 2 instead of $\sqrt2$ for RMS.
Forgetting $T = 1/f$ and instead writing $T = f$ .

How the board asks it

Numericalreading $I_0$ and $\omega$ off $i = I_0\sin\omega t$
An alternating current is given by $i = 10\sin(314\,t)$ A. Find (i) the peak value of the current, (ii) the frequency, and (iii) the time period of the AC.
Numericalpeak/RMS conversion via the $\sqrt2$ factor
The RMS value of the voltage of domestic AC mains is $220\,\text{V}$ . Calculate the peak value of this voltage.
Define / staterms value as the DC-equivalent for heating
Define the root-mean-square (RMS) value of an alternating current, and state its relation to the peak value for a sinusoidal AC.
Give reasonsthe full-cycle average of a sinusoidal AC is zero
Give reasons why the average value of a sinusoidal alternating current over one complete cycle is zero, yet such a current can still produce heat in a resistor.
Derive / provehalf-cycle average $= 2I_0/\pi$
Obtain an expression for the average (mean) value of an alternating current $i = I_0\sin\omega t$ over the positive half cycle, and hence show that it equals $2I_0/\pi$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.