Every plug point in an Indian home is rated at \(220\,\text{V}\). Yet the voltage between its two holes is never steady. It rises to about \(311\,\text{V}\), falls through zero, reaches \(-311\,\text{V}\) and returns, fifty times each second. Over one full cycle its average is exactly zero. A heater on this supply still becomes hot, so the zero average cannot be the whole story. And \(220\) is neither the peak nor the average. So which number is it?
The mean value of a changing quantity
A cyclist's speed changes all through a ride. The average velocity is the displacement divided by the time taken. The displacement is \(\int v\,dt\), the area under the v-t graph. So the average is that area divided by the width of the interval. The same rule gives the mean of any quantity \(f\) that changes with \(x\). It is the height of a rectangle with the same width and the same area as the region under the curve.
The mean value of \(f(x)\) from \(x = a\) to \(x = b\) is \(\bar f = \frac{1}{b - a}\int_a^b f(x)\,dx\). It is the constant height that encloses the same area as \(f\) over the same interval.
For a sine over its positive half cycle, the area is \(\int_0^\pi \sin\theta\,d\theta = \left[-\cos\theta\right]_0^\pi = 2\). The width is \(\pi\), so the mean is \(2/\pi \approx 0.637\). Over a full cycle, the negative half has the same area with a minus sign. So the mean of a sine over a full cycle is zero. This is why the average of the mains voltage is zero: every positive half cancels the negative half that follows it.
A current \(i = 5\sin\omega t\) amperes passes through a diode that lets only the positive halves through. What is the mean current over one positive half cycle, in amperes?
Why \(\sin^2\) averages to one half
The power in a heater depends on the square of the current. So the mean that matters is the mean of \(\sin^2\), not of \(\sin\). The square is never negative, so nothing cancels. The derivation uses the identity \(\sin^2\theta = \tfrac12(1 - \cos 2\theta)\), which changes a square into a cosine that averages to zero.
\[ \sin^2\theta = \tfrac12\left(1 - \cos 2\theta\right) \]
Why. the double-angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\), rearranged.
\[ \int_0^{2\pi} \sin^2\theta\,d\theta = \tfrac12\int_0^{2\pi} d\theta - \tfrac12\int_0^{2\pi} \cos 2\theta\,d\theta \]
Why. integrate term by term over one full cycle, from \(0\) to \(2\pi\).
\[ = \pi - \tfrac14\Big[\sin 2\theta\Big]_0^{2\pi} = \pi \]
Why. the first integral is \(\tfrac12 \times 2\pi\). The cosine completes two whole cycles, so its integral is zero.
\[ \overline{\sin^2\theta} = \dfrac{\pi}{2\pi} = \tfrac12 \]
Read it. divide the area by the width of the interval, \(2\pi\).
There is a second way to see the answer. The graphs of \(\sin^2\theta\) and \(\cos^2\theta\) have the same shape, one shifted a quarter cycle from the other. So over a full cycle their means are equal. Their sum is \(\sin^2\theta + \cos^2\theta = 1\) at every instant. Two equal means that add to 1 must each be one half.
The derivation used exactly one full cycle. Over a window that is not a whole number of cycles, the mean of \(\sin^2\) is not exactly one half. The figure below lets the window grow from the left. For a short window the mean swings well above and below one half. As the window grows, each part-cycle at its end matters less, and the mean settles on one half.
The root mean square
The name of the root mean square lists its steps backwards. Square the quantity at every instant. Take the mean of the square over one period. Then take the square root, so the unit returns to the unit of the quantity. For a sine of peak \(x_0\), the mean of the square is \(x_0^2 \times \tfrac12\), so the rms value is \(x_0/\sqrt2 \approx 0.707\,x_0\).
A heater makes heat at the rate \(i^2R\). Its mean heating is therefore \(\overline{i^2}R\), which equals \(I_{rms}^2R\). So an alternating current of rms value \(I_{rms}\) heats a resistor exactly as a steady current \(I_{rms}\) does. This is what the \(220\,\text{V}\) on a plug point means. It is the rms value of the supply, and the peak is \(220 \times \sqrt2 = 311\,\text{V}\).
These are parts (a) and (b) of the worked example "Three waveforms" of Alternating Current and the LCR Circuit, lesson 1, done with the tool. Find the mean and the rms value of each current. In (a), a sawtooth current rises uniformly from \(0\) to \(6\,\text{A}\) in each period \(T\). In (b), \(i = 3 + 4\sin\omega t\) amperes.
(a) In one period, \(i = 6t/T\) from \(t = 0\) to \(T\). Over a whole period in (b), the mean of \(\sin\) is zero and the mean of \(\sin^2\) is one half.
(a) \(\bar i = \frac1T\int_0^T \frac{6t}{T}\,dt\) and \(\overline{i^2} = \frac1T\int_0^T \frac{36t^2}{T^2}\,dt\). (b) \(i^2 = 9 + 24\sin\omega t + 16\sin^2\omega t\).
(a) \(\bar i = 3\,\text{A}\) and \(\overline{i^2} = \frac{36}{T^3} \cdot \frac{T^3}{3} = 12\,\text{A}^2\). (b) The mean of \(i\) is \(3\,\text{A}\), and the mean of \(i^2\) is \(9 + 0 + 8 = 17\,\text{A}^2\).
(a) \(I_{rms} = \sqrt{12} = 3.5\,\text{A}\). (b) \(I_{rms} = \sqrt{17} = 4.1\,\text{A}\). These match the physics chapter. In both, the rms value is larger than the mean.
A bulb rated \(100\,\text{W}\), \(220\,\text{V}\) runs on the mains. Its rms current is \(100/220 = 0.455\,\text{A}\). What is the peak current, in amperes?
In a tiny model of a gas, four molecules have speeds of \(1\), \(2\), \(3\) and \(4\,\text{km/s}\). Find their root mean square speed, in km/s.
A square-wave current jumps between \(+5\,\text{A}\) and \(-5\,\text{A}\), spending equal times at each. What are its mean and its rms value over a period?
The rule "rms equals peak divided by \(\sqrt2\)" holds only for a pure sine. A square wave, a sawtooth or a sine with a steady part added needs \(\sqrt{\overline{f^2}}\) calculated afresh, as in the example. Two more slips are common. The mean of a square is not the square of the mean. And a mean over part of a cycle is not the mean over a whole cycle, as Fig 5.2 shows.
The plug point is now explained. The mains voltage is a sine of peak \(311\,\text{V}\). Its mean is zero over every cycle, so the mean says nothing about heating. The mean of its square is \(311^2/2\), and the square root of that is \(220\,\text{V}\). A heater on the mains therefore makes the same heat as on a steady \(220\,\text{V}\) supply.
Means and rms values appear wherever a quantity swings or spreads. In Alternating Current and the LCR Circuit the rms values of current and voltage give the heating and the mean power. In Kinetic Theory of Gases the pressure of a gas depends on the mean of \(v^2\), and the rms speed follows. In the chapters on waves, the mean energy and the mean power of a harmonic wave come from the mean of \(\sin^2\). The same mean gives the light that passes a polaroid and the output of a full-wave rectifier.
- Using peak divided by \(\sqrt2\) for a waveform that is not a sine.Square the waveform, take the mean over one period, then the square root.
- Taking the square of the mean instead of the mean of the square.Square first at every instant; average afterwards.
- Averaging a sine over part of a cycle and expecting zero.The mean of \(\sin\) is zero and of \(\sin^2\) is one half only over whole cycles.