Physics › Magnetic fields › Alternating currents

Alternating currents

Alternating only means reversing, and the generator's version of it is a sine wave, so ac needs its own vocabulary. Peak, peak-to-peak, and the rms value, which is the steady dc equivalent for power in the same resistor. The oscilloscope is the instrument that lays a waveform out to be measured.

Year 13AQA 3.7.5.5CIE 21.1.1, 21.1.2, 21.1.3, 21.1.4OCR A 4.4.1(b)(ii)EDEXCEL 128, 129

Builds on Electromagnetic induction: Faraday and Lenz and Current, charge and the direction problem.

IN THIS TOPIC

  • Define alternating current, and read peak, peak-to-peak and rms values from sinusoidal waveforms.
  • Use the rms relations and apply them to mains electricity.
  • Use an oscilloscope to measure voltages, time intervals and frequencies.

COMMON MISCONCEPTION

Mains is 230 volts at its peak.

Three sizes of one sine

Three sizes for one sine wave: the peak, the peak-to-peak, and the rms level that matches dc for powerpeak V₀rms = 0.707 V₀peak-to-peak = 2V₀
FIG. 1Peak, peak-to-peak, and the rms level at 0.707 of the peak: three measurements of the same wave.

An alternating current is one that reverses direction periodically, and that is the whole of the definition. Square waves, triangular waves and the battered waveforms coming off a cheap inverter are every bit as alternating as the mains. The sine is simply the shape a coil rotating steadily in a uniform field produces, so it is the one A-level works with, and one equation, x = x0 sin ωt, describes current and pd alike, with ω = 2πf setting the frequency. A sinusoidal supply swings between +V0 and −V0. The peak value V0 is the height of the swing, while the peak-to-peak value is the full 2V0 span, which is the easiest thing to read off a screen.

Neither is a fair average. The supply spends most of its time below the peak, and its plain average over a cycle is zero. The useful measure is the root mean square, the steady dc value that would produce the same mean power in the same resistor. Square the waveform, average it over a cycle, root it again. For a sinusoid, and for a sinusoid only, that gives

Irms=I02I_{rms} = \frac{I_{0}}{\sqrt{2}}ON THE AQA DATA SHEET
Vrms=V02V_{rms} = \frac{V_{0}}{\sqrt{2}}ON THE AQA DATA SHEET

about 0.707 of the peak. Feed the same resistor a square wave of the same height and its rms value is the full peak, since the wave never sits anywhere else; a triangular wave comes in at 0.577 of its peak. The factor belongs to the shape, not to ac in general, so a question that never mentions the waveform has told you it is a sine.

Quoted ac values are rms unless stated otherwise. UK mains is 230 V rms, so its peak is 230 × 1.414 ≈ 325 V and its peak-to-peak span is about 650 V. Insulation gets engineered for the peak, never for the label. CIE likes one further statement on its own line. Since each rms value is the peak over √2, the mean power in a resistor is exactly half the peak power.

WORKED EXAMPLE

What the rms label means

A 100 Ω heater runs from the 230 V rms mains. Find its average power, and say what dc supply would match it.

The rms value exists for exactly this calculation, so P = Vrms2/R = 2302/100 = 530 W.

A steady dc supply of 230 V would heat the same resistor identically, and that equivalence is the definition of rms, not a coincidence.

Using the 325 V peak instead would give 1060 W, double the correct value, because the sine spends most of its time below its peak. The factor of two is the square of the missing √2, hiding in plain sight.

GUIDED PRACTICE

An unfamiliar mains

In the United States the mains is quoted as 120 V rms. Find the peak and peak-to-peak voltages.

Show the working

V0 = 120 × 1.414 = 170 V.

Peak-to-peak is double that, 340 V. The quoted label is always the dc-equivalent rms, and the insulation, as ever, has to be built for the peak.

The oscilloscope

The oscilloscope as a voltmeter and clock: a dc input draws a level line, an ac input draws its waveform against the squared screendc: a steady lineac: the waveform itselfvolts per square on the y-gainseconds per square on the timebaseheight reads the voltageone cycle's width reads the period: f = 1/T
FIG. 2Two screens: dc draws a level line, ac draws its waveform, and the grid converts squares to volts and seconds.

The oscilloscope plots voltage against time on a squared screen, and two settings translate those squares. The y-gain gives the volts per division vertically, and the timebase gives the seconds per division horizontally. A dc input draws a level line whose height times the y-gain is the voltage, which is the scope working as a dc voltmeter. An ac input draws its actual waveform. Amplitude times y-gain gives the peak voltage, the width of one full cycle times the timebase gives the period, and f = 1/T finishes the job. Only its use is examined, never the instrument's internal workings.

ASSESSMENT FOCUS

  • State what rms means, not only the formula. It is the dc value that would deliver the same power in the same resistor, and that sentence is the mark.
  • The root-two relations hold for sinusoidal waveforms only, and saying so takes one clause. Mains numbers are the stock application, so learn the trio of 230 V rms, 325 V peak and 650 V peak-to-peak.
  • Scope arithmetic is two multiplications. Divisions times y-gain for volts, divisions times timebase for seconds. Write the readings down in divisions first, then convert.
  • Frequency comes only through the period. Read the width of one full cycle, convert it to seconds, then take f = 1/T. Never try to scale a frequency straight off the screen.

CHECK YOURSELF

An oscilloscope shows a sine wave of amplitude 3.0 divisions with one cycle spanning 4.0 divisions. The y-gain is 5.0 V per division and the timebase 2.0 ms per division. Find the peak voltage, the rms voltage, and the frequency.

Show a hint

Squares to volts and seconds first; the physics is two multiplications and a reciprocal.

Show the answer

Peak: V0 = 3.0 × 5.0 = 15 V.

Vrms=V0/2V_{rms} = V_{0}/\sqrt{2}: 15 / 1.414 = 10.6 V.

Period T = 4.0 × 2.0 ms = 8.0 ms, so f = 1/T = 125 Hz.

Alternating means reversing, and only a sinusoid puts rms at 0.707 of the peak.

Quoted ac is rms, the dc equivalent for power in the same resistor.

On the scope, squares become volts and seconds; f comes from 1/T.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 59 KB · print edition · white background

Question workbook

PDF · 60 KB · screen edition · navy

Answer book

PDF · 56 KB · fully worked · navy
18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the alternating currents questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

  • Define alternating current, and read peak, peak-to-peak and rms values from sinusoidal waveforms.
  • Use the rms relations and apply them to mains electricity.
  • Use an oscilloscope to measure voltages, time intervals and frequencies.

Open the full revision checklist to track your progress across the whole unit.