9478 / 2027
Currents overview

Lesson 6 of 7

Mean power and rms values

  • I can define and use period, frequency, peak and rms values for alternating current or voltage.Syllabus

    Syllabus 9478, 15(f). Define and use period, frequency, peak and rms values for alternating current or voltage.

  • I can deduce that mean power in a resistive load is half the peak for sinusoidal alternating current.Syllabus

    Syllabus 9478, 15(h). Deduce that mean power in a resistive load is half the peak for sinusoidal alternating current.

  • I can distinguish rms from peak values and use both I_rms = I0/sqrt(2) and V_rms = V0/sqrt(2) for sinusoids.Syllabus

    Syllabus 9478, 15(i). Distinguish rms from peak values and use both I_rms = I0/sqrt(2) and V_rms = V0/sqrt(2) for sinusoids.

Opposite current directions can produce the same heating in a resistor. Mean signed current therefore does not tell you its mean heating power.

Use a constant resistive load R and a sinusoidal current i = I0 sin(ωt). With the passive reference v = iR, instantaneous power is:

p = vi = i2R
= I02R sin2(ωt)

Voltage and current reverse together, so their product stays non-negative. Over a complete cycle the positive and negative current contributions cancel, giving zero mean signed current. Squaring them removes that sign cancellation.

Deduce the mean power

The identity sin2(θ) + cos2(θ) = 1 holds at every angle. The two squared functions have the same full-cycle mean because one is a quarter-cycle shift of the other. Their means must therefore each be 1/2.

Mean power = I02R × mean[sin2(ωt)]
= ½I02R
= ½Ppeak

This deduction uses the full sinusoidal variation. It is not an instruction to average only the maximum and minimum readings for any waveform.

Worked heating comparison

Two power peaks in one current cycle

Let i = 4.00 sin(100πt) A through 6.00 Ω. The current period is 20.0 ms, while:

p = 96.0 sin2(100πt) W
Ppeak = 4.002 × 6.00 = 96.0 W
Pmean = 48.0 W

Power peaks at both 5.00 and 15.0 ms, when the current has its positive and negative extrema. Its repeat period is 10.0 ms, half the current period.

Equal full-period areas mean equal transferred energy

Both panels use the same 6.00 Ω resistor and the same time and power scales. The shaded area covers one complete 20.0 ms current period. Squaring the sinusoidal current produces two power maxima in that interval.

Sinusoidal current: two power maxima

The squared-sinusoidal power transfers 0.960 joule in twenty millisecondsTime in milliseconds is horizontal and power in watts vertical. Both comparison plots use x equals sixty plus twelve times time in milliseconds and y equals 320 minus twice power in watts. The smooth power curve is 96 times sine squared of one hundred pi t, with t in seconds. It is zero at zero, ten and twenty milliseconds and reaches 96 watts at five and fifteen milliseconds. A dashed horizontal guide marks the full-period mean forty-eight watts, rather than an instantaneous constant power. The full shaded area is 0.960 joule in either case. The steady d.c. comparison uses rms current, two square root two amperes, in the same resistor. Zero signed mean alternating current does not imply zero energy transfer.0510152004896p / WMean 48.0 W (dashed)t / msFull-period area: 0.960 J

Equivalent d.c.: constant power

A constant forty-eight-watt input transfers 0.960 joule in twenty millisecondsTime in milliseconds is horizontal and power in watts vertical. Both comparison plots use x equals sixty plus twelve times time in milliseconds and y equals 320 minus twice power in watts. Power is forty-eight watts throughout, so its shaded rectangle has height forty-eight watts and width 0.0200 second. The full shaded area is 0.960 joule in either case. The steady d.c. comparison uses rms current, two square root two amperes, in the same resistor. Zero signed mean alternating current does not imply zero energy transfer.0510152004896p / WConstant 48.0 Wt / msFull-period area: 0.960 J

The equivalent steady current is Irms = 2√2 A, giving the same 48.0 W mean power. The comparison uses the whole period and the same resistor; it is not an average of only the peak and zero readings.

The sinusoidal-current power curve and the constant 48.0 W comparison use the same time and power scales. Their shaded areas over the full 20.0 ms current cycle are equal, each representing 0.960 J.
Energy in one current cycle
= PmeanT = 48.0 × 0.0200 = 0.960 J

A signed mean current of zero is therefore compatible with a positive energy transfer. The constant-power rectangle compares energies over the same duration.

The equivalent steady current

The root-mean-square current is the steady d.c. current producing the same mean heating power in the same resistor. It is the square root of the mean of the squared instantaneous current:

Irms = √[mean(i2)]
Irms2R = Pmean

For the specified sinusoid, substitute Pmean = I02R/2 and take the non-negative square root. The voltage relation follows in the same way from p = v2/R:

Irms = I0/√2
Vrms = V0/√2

In the example, Irms = 2√2 A ≈ 2.83 A and Vrms = 12√2 V ≈ 17.0 V. For this in-phase, purely resistive load:

Pmean = Irms2R
= Vrms2/R
= VrmsIrms = 48.0 W

The rms value belongs to a full-cycle squared average. The fact that the sinusoidal current happens to equal +2.83 A at 2.50 ms does not define rms by that instant.

Keep instantaneous value, signed mean, peak, peak-to-peak and rms distinct. Dividing a peak by √2 applies to a sinusoid; the general rms definition does not give every periodic waveform that ratio. The product VrmsIrms above also uses the stated in-phase resistive condition.

Try it step by step

Work it out in steps: r.m.s. current I0 / sqrt 2 in A, then Mean power Irms2 R in W.

  1. Problem 1. A sinusoidal current of peak value 4.0 A flows through a 10 ohm resistor. Find the r.m.s. current and the mean power.
  2. Problem 2. A sinusoidal current of peak value 3.0 A flows through a 20 ohm resistor. Find the mean power.
  3. Problem 3. A sinusoidal current of peak value 0.50 A flows through a 100 ohm resistor. Find the r.m.s. current and the mean power.
Show the answer

Problem 1.

r.m.s. current I0 / sqrt 2 in A: Irms = 4.0 / 1.414 = 2.83 A

Mean power Irms2 R in W: Mean power = 2.832 x 10 = 80 W

Problem 2.

r.m.s. current I0 / sqrt 2 in A: Irms = 3.0 / 1.414 = 2.12 A

Mean power Irms2 R in W: Mean power = 2.122 x 20 = 90 W

Problem 3.

r.m.s. current I0 / sqrt 2 in A: Irms = 0.50 / 1.414 = 0.354 A

Mean power Irms2 R in W: Mean power = 0.3542 x 100 = 12.5 W

Check your understanding

The current i = 4.00 sin(100 pi t) A flows through a constant 6.00 ohm resistor. Over a complete current cycle, which comparison is correct?

  1. Mean current is zero, mean power is 48.0 W, rms current about 2.83 A.
  2. Mean heating power is 96.0 W and rms current is 4.00 A.
  3. Mean current and mean heating power are both zero.
Show the answer

Mean current is zero, mean power is 48.0 W, rms current about 2.83 A.

The signed halves cancel in the current mean, but i2 R stays non-negative. Mean sin2 is 1/2, giving 4.002 x 6.00/2 = 48.0 W and I_rms = 4.00/sqrt(2).

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define the root-mean-square (r.m.s.) value of an alternating voltage. A sinusoidal alternating voltage of peak value 12.0 V is connected across an 8.0 Ω resistor. Calculate the r.m.s. voltage and the mean power dissipated. [3]

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