Physics › Magnetic fields › Alternating currents

Alternating currents

An alternating current reverses direction periodically, and for a sinusoid the rms value is the peak divided by root two, or 0.707 of it. That rms value is the steady dc delivering the same mean power in the same resistor, and mains is 230 V rms with a 325 V peak. On an oscilloscope, the y-gain and time base convert divisions to volts and seconds.

Alternating current

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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 potential difference.

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.

230 V is the rms value of the mains, not the peak: the peak voltage is √2 times larger, about 325 V.

Peak, peak-to-peak and rms values

One sine wave with three measurements marked: the peak height above the axis, the double-headed peak-to-peak span, and an amber dashed rms line at 0.707 of the peak, the dc-equivalent level for power.
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. Square waves, triangular waves and the distorted waveforms from a cheap inverter are all alternating. The sine is the shape produced by a coil rotating steadily in a uniform field, so it is the one used at A-level. One equation, x = x0 sin ωt, describes current and pd alike, with ω = 2πf setting the frequency. A sinusoidal supply varies between +V0 and −V0. The peak value V0 is the amplitude, and the peak-to-peak value is the full 2V0 range, which is the easiest quantity to read from a screen.

Neither value gives the mean power. The supply is below its peak for most of the cycle, and its mean over a whole cycle is zero. The useful measure is the root mean square value, the steady dc value that would deliver the same mean power in the same resistor: square the waveform, average it over a cycle, and take the square root. For a sinusoid, and only for a sinusoid,

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

about 0.707 of the peak. A square wave of the same amplitude has an rms value equal to its peak, since its magnitude is constant; a triangular wave gives 0.577 of its peak. The factor depends on the waveform. A question that does not name the waveform is describing a sinusoid.

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 value is about 650 V. Insulation must be rated for the peak, not the quoted value. CIE requires one further statement: since each rms value is the peak divided by √2, the mean power in a resistor is 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 is defined for exactly this calculation, so P = Vrms2/R = 2302/100 = 530 W.

A steady dc supply of 230 V would deliver the same power to the same resistor, which is the definition of the rms value.

Using the 325 V peak instead gives 1060 W, twice the correct value, because the sine is below its peak for most of the cycle. The factor of two is the square of the √2 that was omitted.

GUIDED PRACTICE

Peak values of a 120 V supply

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 twice that, 340 V. The quoted value is the rms, the dc equivalent for power, and the insulation must be rated for the peak.

The oscilloscope

Two oscilloscope screens with centimetre squares: a dc input drawing a flat line whose height reads the voltage, and an ac input drawing a sine whose amplitude spans one and a half divisions and whose cycle spans four, ready to convert through the y-gain and timebase settings.
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 convert those squares. The y-gain gives volts per division vertically, and the timebase gives seconds per division horizontally. A dc input gives a horizontal line whose height multiplied by the y-gain is the voltage, which is the oscilloscope used as a dc voltmeter. An ac input displays the waveform: amplitude × y-gain gives the peak voltage, the width of one full cycle × the timebase gives the period, and f = 1/T gives the frequency. Only the use of the instrument is examined, not its internal circuitry.

ASSESSMENT FOCUS

  • State what rms means, not only the formula: it is the dc value that would deliver the same mean power in the same resistor. That statement carries the mark.
  • The root-two relations hold for sinusoidal waveforms only, so state that condition. Mains values are the standard application: 230 V rms, 325 V peak and 650 V peak-to-peak.
  • Oscilloscope calculations are two multiplications: divisions × y-gain for volts, divisions × timebase for seconds. Record the readings in divisions first, then convert.
  • Obtain the frequency through the period: read the width of one full cycle, convert it to seconds, then use f = 1/T. The frequency cannot be read directly from 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

Convert divisions to volts and seconds first, then take the reciprocal of the period.

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.

An alternating current reverses direction periodically; the rms value is 0.707 of the peak for a sinusoid only.

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

On an oscilloscope, divisions convert to volts and seconds, and f = 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

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Question workbook

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Answer book

PDF · 73 KB · fully worked · navy
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  • 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.

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