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Alternating Current

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Senior 6 · Physics

Alternating Current

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Lesson 1 of 4

Sinusoidal Quantities, Phasors and RMS Values

25 min

What "230 V ~ 50 Hz" really means

Look at the rating plate on a kettle or a phone charger sold in Rwanda and you will find the symbol ~ next to the voltage. It means alternating. Rwanda's mains supply is 230 V at 50 Hz (REG designs its low-voltage network for 400 V/230 V, and the Rwanda Grid Code sets a nominal frequency of 50 Hz, normally held between 49.5 Hz and 50.5 Hz): the voltage between the live and neutral pins reverses direction 100 times a second, and it is not 230 V for most of that time. Its highest value is about 325 V. So what is the "230"? In Unit 3 you showed that a coil turning in a magnetic field gives $\varepsilon = \varepsilon_0 \sin\omega t$ . This lesson gives you the language for such quantities (peak, period, phase and phasor) and answers the question: what single number fairly describes a voltage that never stays still?

Describing a sinusoidal quantity

An AC generator, a signal generator in the laboratory and the national grid all produce voltages that vary sinusoidally with time:

v = V_0 \sin(\omega t + \phi)

Each symbol has a name and a meaning:

Symbol	Name	Meaning and unit
$v$	instantaneous value	the voltage at the instant $t$ , in V
$V_{0}$	peak value (amplitude)	the greatest value of $v$ , in V
$2 V_{0}$	peak-to-peak value	from a crest to a trough, in V
$T$	period	the time for one complete cycle, in s
$f = \frac{1}{T}$	frequency	cycles per second, in Hz
$\omega = 2\pi f$	angular frequency	in $\mathrm{rad\,s^{-1}}$
$\omega t + \phi$	phase	the angle that fixes where in its cycle the quantity is, in rad
$\phi$	phase constant	the phase at $t = 0$ , in rad

Figure 1.1: An alternating voltage. Its peak is $V_{0}$ , its peak-to-peak value is $2 V_{0}$ , and one cycle takes a period $T$ .

Definition

An alternating current (AC) is a current that reverses its direction periodically. A sinusoidal current has the form $i = I_0 \sin(\omega t + \phi)$ , where $I_{0}$ is the peak value and $\omega = 2\pi f$ is the angular frequency. Two sinusoids of the same frequency with phase constants $\phi_1$ and $\phi_2$ have a phase difference $\phi_1 - \phi_2$ ; the one with the larger phase leads. Example: $i = 2.0\sin(100\pi t)\,\mathrm{A}$ has $I_0 = 2.0\,\mathrm{A}$ and $f = 50\,\mathrm{Hz}$ .

Worked example 1: reading a sinusoid

The mains voltage can be written $v = 325\sin(100\pi t)\,\mathrm{V}$ , with $t$ in seconds. (a) State the peak value and find the frequency and period. (b) Find $v$ at $t = 2.5\,\mathrm{ms}$ . (c) Find the first two times after $t = 0$ at which $v = 162.5\,\mathrm{V}$ .

Solution

(a) $V_0 = 325\,\mathrm{V}$ . $\omega = 100\pi\,\mathrm{rad\,s^{-1}}$ , so $f = \frac{\omega}{2\pi} = 50\,\mathrm{Hz}$ and $T = \frac{1}{f} = 0.020\,\mathrm{s} = 20\,\mathrm{ms}$ .

(b) The phase is $100\pi \times 2.5 \times 10^{-3} = 0.25\pi\,\mathrm{rad}$ (45°). So $v = 325\sin(0.25\pi) = 325 \times 0.707 = 230\,\mathrm{V}$ (3 s.f.). Your calculator must be in radians for this step.

(c) $\sin(100\pi t) = \frac{162.5}{325} = 0.500$ , so $100\pi t = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ . Then $t = \frac{1}{600}\,\mathrm{s} = 1.67\,\mathrm{ms}$ and $t = \frac{5}{600}\,\mathrm{s} = 8.33\,\mathrm{ms}$ . There are two answers because the voltage passes 162.5 V once on the way up and once on the way down.

Phasors: a rotating picture of a sinusoid

Adding sine curves point by point is slow. Engineers use a neater picture. Draw an arrow of length $V_{0}$ from the origin and let it turn anticlockwise at the angular velocity $\omega$ . At time $t$ it makes the angle $\omega t$ with the horizontal reference line, and its vertical component is $V_0\sin\omega t$ : exactly the instantaneous value $v$ . The rotating arrow is called a phasor.

Figure 1.2: As the phasor turns, the height of its tip traces out the sine wave. The marked point is at $\omega t = 40^\circ$ .

When several quantities have the same frequency, their phasors all turn together, so the angles between them never change. We therefore draw them at one instant, usually $t = 0$ , and the diagram stays true for all time. A quantity that leads by $\phi$ is drawn $\phi$ anticlockwise from the reference; one that lags is drawn clockwise. Because the vertical component of a vector sum is the sum of the vertical components, adding sinusoids of the same frequency is the same as adding their phasors as vectors. This is the key idea of Lessons 2 and 3.

Definition

A phasor is a vector whose length equals the peak (or rms) value of a sinusoidal quantity and whose angle from a reference direction equals the quantity's phase. Phasors rotate anticlockwise at $\omega$ ; the sinusoid is the projection of the phasor on the vertical axis. Example: $v = 12\sin(\omega t + \frac{\pi}{3})$ is a phasor of length 12 V drawn $60^\circ$ anticlockwise from the reference.

Worked example 2: adding two voltages with phasors

Two components in series have voltages $v_1 = 3.0\sin\omega t$ and $v_2 = 4.0\sin(\omega t + \frac{\pi}{2})$ , in volts. Find the total voltage $v = v_{1} + v_{2}$ in the form $V_0\sin(\omega t + \phi)$ .

Figure 1.3: Adding the phasors of $v_{1}$ and $v_{2}$ nose to tail.

Solution

$v_{2}$ leads $v_{1}$ by $90^\circ$ , so its phasor is at right angles to that of $v_{1}$ . By Pythagoras, $V_0 = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{V}$ . The phase angle is $\phi = \tan^{-1}\frac{4.0}{3.0} = 53.1^\circ = 0.927\,\mathrm{rad}$ . So

v = 5.0\sin(\omega t + 0.927)\,\mathrm{V}

Check at $t = 0$ : $v_1 + v_2 = 0 + 4.0 = 4.0\,\mathrm{V}$ , and $5.0\sin(0.927) = 5.0 \times 0.800 = 4.0\,\mathrm{V}$ . Note that the peak of the sum (5.0 V) is less than $3.0 + 4.0 = 7.0\,\mathrm{V}$ , because the two peaks happen at different times.

Mean and root mean square values

Over a whole cycle a sinusoidal current flows forwards as much as backwards, so its mean value is zero. A moving-coil meter reading the mean would show nothing, yet the current plainly heats a kettle. For heating, the direction does not matter: the power in a resistor is $P = i^{2} R$ , and $i^{2}$ is never negative.

Deriving the rms value

For $i = I_0\sin\omega t$ in a resistance $R$ , the instantaneous power is $p = I_0^2R\sin^2\omega t$ . Using $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ , the mean of $\sin^2\omega t$ over one period $T = \frac{2\pi}{\omega}$ is

\langle \sin^2\omega t \rangle = \frac{1}{T}\int_0^T \frac{1 - \cos 2\omega t}{2}\,dt = \frac{1}{T}\left[\frac{t}{2} - \frac{\sin 2\omega t}{4\omega}\right]_0^T = \frac{1}{2}

since $\sin 2\omega T = \sin 4\pi = 0$ . So the mean power is $\langle p \rangle = \frac{1}{2}I_0^2R$ . A steady direct current $I$ gives the same heating if $I^2R = \frac{1}{2}I_0^2R$ , that is

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

Figure 1.4: The mean of $i^{2}$ is $\frac{1}{2}I_0^2$ , so the "root of the mean of the square" is $\frac{I_0}{\sqrt{2}}$ .

The name says how to compute it for ANY waveform: square the current, take the mean over a cycle, then take the square root, $I_{\text{rms}} = \sqrt{\langle i^2 \rangle}$ . The factor $\frac{1}{\sqrt{2}}$ holds only for a sinusoid.

The mean over half a cycle

Rectifier circuits use the mean of one half-cycle:

\langle i \rangle_{\text{half}} = \frac{2}{T}\int_0^{T/2} I_0\sin\omega t\,dt = \frac{2I_0}{\omega T}\Big[-\cos\omega t\Big]_0^{T/2} = \frac{2I_0}{2\pi}\times 2 = \frac{2I_0}{\pi} \approx 0.637\,I_0

AC voltmeters and ammeters are scaled to read rms values, and every AC rating ("230 V", "13 A fuse") is an rms value unless it says otherwise.

Definition

The root mean square (rms) value of an alternating current is the value of the steady direct current that would dissipate heat at the same average rate in the same resistor. For a sinusoid, $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ and $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$ . Example: the 230 V mains has a peak of $230\sqrt{2} = 325\,\mathrm{V}$ .

Worked example 3: a 2.0 kW kettle

A kettle is rated 2.0 kW, 230 V. Find (a) the rms current, (b) the peak current, (c) its resistance and (d) the greatest instantaneous power it takes.

Solution

(a) The rating is an average power and an rms voltage: $I_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{2000}{230} = 8.70\,\mathrm{A}$ .

(b) $I_0 = \sqrt{2}\,I_{\text{rms}} = 1.414 \times 8.70 = 12.3\,\mathrm{A}$ .

(c) $R = \frac{V_{\text{rms}}^2}{P} = \frac{230^2}{2000} = 26.5\,\mathrm{\Omega}$ . (Dividing 230 V by the rounded 8.70 A gives 26.4 Ω; always work from unrounded values.)

(d) The power peaks when $v$ and $i$ peak together: $p_{\max} = V_0I_0 = 325 \times 12.3 = 4.0\,\mathrm{kW}$ . The power swings between 0 and 4.0 kW, 100 times a second, and averages 2.0 kW: the mean power is half the peak power.

Worked example 4: a waveform that is not a sine

Figure 1.5: A rectangular alternating current.

Find (a) the mean value and (b) the rms value of the current in Figure 1.5.

Solution

One cycle lasts $2.0 + 6.0 = 8.0\,\mathrm{ms}$ .

(a) $\langle i \rangle = \frac{(3.0)(2.0) + (-1.0)(6.0)}{8.0} = \frac{6.0 - 6.0}{8.0} = 0$ . The charge sent forwards equals the charge sent back.

(b) $\langle i^2 \rangle = \frac{(3.0)^2(2.0) + (1.0)^2(6.0)}{8.0} = \frac{18 + 6.0}{8.0} = 3.0\,\mathrm{A^2}$ , so $I_{\text{rms}} = \sqrt{3.0} = 1.7\,\mathrm{A}$ .

Using $\frac{I_0}{\sqrt{2}} = \frac{3.0}{1.414} = 2.1\,\mathrm{A}$ would be wrong: that shortcut is for sinusoids only.

Experiment 1.1: reading an AC signal on an oscilloscope

Aim: to measure the peak value, rms value and frequency of an alternating voltage, and to test $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$ .

Apparatus: a signal generator (or the low-voltage AC outlet of a laboratory power pack), a cathode-ray or digital oscilloscope, a digital multimeter set to AC volts, connecting leads.

Method:

Connect the signal generator to the Y-input of the oscilloscope and, in parallel, to the multimeter.
Switch the time base on. Adjust the Y-gain (volts per division) and time base (time per division) until one to three complete cycles fill most of the screen.
Measure the peak-to-peak height in divisions, halve it and multiply by the Y-gain to get $V_{0}$ .
Measure the length of one cycle in divisions and multiply by the time base to get $T$ ; then $f = \frac{1}{T}$ .
Compare $\frac{V_0}{\sqrt{2}}$ with the multimeter reading. Repeat for three different amplitudes.

Figure 1.6: A trace from the experiment.

Reading Figure 1.6: peak-to-peak height = 5.6 div, so $V_0 = \frac{5.6}{2} \times 1.0 = 2.8\,\mathrm{V}$ and $V_{\text{rms}} = \frac{2.8}{\sqrt{2}} = 2.0\,\mathrm{V}$ . One cycle spans 5.0 div, so $T = 5.0 \times 2\,\mathrm{ms} = 10\,\mathrm{ms}$ and $f = 100\,\mathrm{Hz}$ . Measuring peak to peak and halving is more precise than measuring from the centre line, because the trace's thickness and any offset cancel.

Warning

Safety. Never connect an oscilloscope, or any student apparatus, to the mains. Use only a signal generator or a laboratory supply of 12 V or less. The mains peak of 325 V can kill, and an oscilloscope's earthed input lead can short a live circuit.

Warning

Common mistakes. (1) Calculator in degrees when the phase is $100\pi t$ radians. (2) Taking the peak-to-peak height as $V_{0}$ : halve it first. (3) Using $\frac{I_0}{\sqrt{2}}$ for square, triangular or rectified waves; go back to $\sqrt{\langle i^2 \rangle}$ . (4) Saying the mean of a sinusoidal current is $0.637 I_{0}$ : that is the mean of a HALF-cycle; over a full cycle it is zero.

Tip

Remember the chain: peak × 0.707 = rms, and rms × 1.414 = peak. If the question mentions heating, power or a meter reading, it wants rms. If it mentions insulation, breakdown or the height of a trace, it wants the peak.

What is RMS value? (Khan Academy India). Watch how the heating effect of a current leads to the root-mean-square value, and why it is the peak divided by √2 only for a sine wave.

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Practice

A voltage is $v = 170\sin(120\pi t)\,\mathrm{V}$ . State its peak value and find its rms value, frequency and period.
Write an equation for the Rwandan mains voltage (230 V rms, 50 Hz), taking $v = 0$ and rising at $t = 0$ . Find $v$ at $t = 4.0\,\mathrm{ms}$ .
A 60 W lamp works at full brightness on the 230 V mains. Find the rms current, the peak current and the lamp's resistance.
Find the resultant of $v_1 = 6.0\sin\omega t$ and $v_2 = 8.0\sin(\omega t - \frac{\pi}{2})$ by drawing phasors. Does the resultant lead or lag $v_{1}$ ?
On an oscilloscope set to 5 V/div and 10 ms/div, a sine trace is 6.4 div from peak to peak and one cycle spans 2.0 div. Find the peak voltage, rms voltage and frequency.
A square-wave voltage switches between +12 V and −12 V. Find its rms value. Which heats a resistor more: this wave, or a sine wave of peak 12 V? By what factor?
Starting from $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ , show that the mean power in a resistor carrying $i = I_0\sin\omega t$ is $\frac{1}{2}I_0^2R$ .
Uwase says: "The mains is 230 V, so cable insulation rated for 250 V is always safe." Evaluate her reasoning.

Show answersHide answers

$V_0 = 170\,\mathrm{V}$ ; $V_{\text{rms}} = \frac{170}{\sqrt{2}} = 120\,\mathrm{V}$ ; $f = \frac{120\pi}{2\pi} = 60\,\mathrm{Hz}$ ; $T = 16.7\,\mathrm{ms}$ .
$v = 325\sin(100\pi t)\,\mathrm{V}$ . At 4.0 ms the phase is $0.40\pi$ rad (72°): $v = 325 \times 0.951 = 309\,\mathrm{V}$ .
$I_{\text{rms}} = \frac{60}{230} = 0.261\,\mathrm{A}$ ; $I_0 = 0.369\,\mathrm{A}$ ; $R = \frac{230}{0.261} = 882\,\mathrm{\Omega}$ .
$v_{2}$ lags by 90°, so its phasor points straight down. $V_0 = \sqrt{6.0^2 + 8.0^2} = 10\,\mathrm{V}$ , $\phi = -\tan^{-1}\frac{8.0}{6.0} = -53.1^\circ$ : $v = 10\sin(\omega t - 0.927)\,\mathrm{V}$ . It lags $v_{1}$ by 53.1°.
$V_0 = \frac{6.4}{2} \times 5 = 16\,\mathrm{V}$ ; $V_{\text{rms}} = 11\,\mathrm{V}$ (11.3 V); $T = 20\,\mathrm{ms}$ , $f = 50\,\mathrm{Hz}$ .
$v^2 = 144\,\mathrm{V^2}$ at every instant, so $V_{\text{rms}} = 12\,\mathrm{V}$ . The sine wave has $V_{\text{rms}} = 8.49\,\mathrm{V}$ . Power goes as $V_{\text{rms}}^2$ : $\frac{144}{72} = 2$ , so the square wave heats twice as fast.
$\langle p \rangle = I_0^2R\langle\sin^2\omega t\rangle = I_0^2R \cdot \frac{1}{T}\int_0^T \frac{1-\cos 2\omega t}{2}dt = \frac{1}{2}I_0^2R$ , because the cosine term integrates to zero over a whole number of cycles.
Unsafe reasoning. 230 V is the rms value; the voltage reaches $230\sqrt{2} = 325\,\mathrm{V}$ twice every cycle, and insulation must withstand the peak (plus surges). Cable for mains use should be rated well above 325 V; household cable is typically rated 300 V/500 V or more.

Key points

$v = V_0\sin(\omega t + \phi)$ , with $\omega = 2\pi f = \frac{2\pi}{T}$ .
A phasor is a rotating vector whose vertical projection is the sinusoid; same-frequency sinusoids add as vectors.
$I_{\text{rms}} = \sqrt{\langle i^2\rangle}$ ; for a sinusoid $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ . Mean over a cycle = 0; mean over a half-cycle = $\frac{2I_0}{\pi}$ .
AC meters and ratings give rms values; insulation must survive the peak.

Sources and credits

15.2: AC Sources, from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Rwanda's supply (facts quoted): The Rwanda Grid Code, Rwanda Utilities Regulatory Authority (RURA), August 2013, published by REG; Reticulation Standards for Electricity Distribution Planning, Rwanda Energy Group (REG), August 2020.
Video: What is RMS value? (& why should we care?), Khan Academy India - English, embedded from YouTube.

The experiment, the six diagrams, the worked examples and the practice exercises are written by Ganzaa.

1What "230 V ~ 50 Hz" really means

2Describing a sinusoidal quantity

3Phasors: a rotating picture of a sinusoid

4Mean and root mean square values

Deriving the rms value

The mean over half a cycle

5Experiment 1.1: reading an AC signal on an oscilloscope

What "230 V ~ 50 Hz" really means

Describing a sinusoidal quantity

Phasors: a rotating picture of a sinusoid

Mean and root mean square values

Experiment 1.1: reading an AC signal on an oscilloscope