The meter board of a house in India says 220 V. Yet the voltage at the socket is almost never 220 V. It rises to more than 300 V, falls through zero, becomes negative and returns, fifty times each second. It crosses zero a hundred times a second. A tube light and a ceiling fan work on it all the same.
Current that reverses
A cell drives charge round a circuit in one direction only. Its current is a direct current, steady in size and direction. The supply from a power station is different. Its voltage reverses its polarity again and again, so the current it drives reverses too. Such a current is an alternating current, written AC for short.
The alternating voltage of the mains is a sine of time. We write it as \[ v = v_0\sin\omega t. \] Here \(v\) is the instantaneous value of the voltage at time \(t\), and \(v_0\) is its peak value, the largest value it reaches. The quantity \(\omega\) is the angular frequency in rad/s. The voltage repeats after one period \(T\), when \(\omega T = 2\pi\). So \(\omega = 2\pi/T = 2\pi f\), where \(f\) is the frequency in hertz. An alternating current is written in the same way, \(i = i_0\sin\omega t\).
The Indian mains runs at \(f = 50\,\text{Hz}\). So its period is \(T\) \(= 1/50\,\text{s}\) \(= 20\,\text{ms}\), and \(\omega\) \(= 2\pi \times 50\) \(= 100\pi \approx 314\,\text{rad/s}\). Its peak is \(311\,\text{V}\), as the third topic of this lesson shows. In every cycle the voltage passes through zero twice. So it crosses zero \(2 \times 50 = 100\) times each second.
(a) A voltage has a peak of \(10\,\text{V}\) and a period of \(20\,\text{ms}\). Write \(v(t)\) and find \(v\) at \(t = 2.5\,\text{ms}\). (b) A voltage is \(v = 100\sin 314t\) volts. Find its frequency and the first time at which it reaches its peak.
(a) \(f = 1/0.02 = 50\,\text{Hz}\) and \(\omega\) \(= 2\pi \times 50\) \(= 100\pi\,\text{rad/s}\). So \(v = 10\sin 100\pi t\) volts. At \(2.5\,\text{ms}\), \(100\pi \times 0.0025 = \pi/4\), and \(v\) \(= 10\sin 45°\) \(= 7.1\,\text{V}\).
(b) \(\omega = 314 = 2\pi f\) gives \(f = 50\,\text{Hz}\). The first peak comes when \(\omega t = \pi/2\), a quarter period after the start: \(t = T/4 = 5\,\text{ms}\).
Values used: the mains frequency \(50\,\text{Hz}\), \(\omega = 314\,\text{rad/s}\), \(\sqrt2 = 1.414\).
The mains voltage is \(v = 311\sin 314t\) volts. Find \(v\) at \(t = 2.5\,\text{ms}\) after an upward zero crossing, in volts.
The mean is zero
How large is an alternating current? The first idea is its average. The mean value of a current over a time interval is the steady current that carries the same charge in that time: \[ \bar i = \dfrac{1}{T}\int_0^T i\,dt. \] For \(i = i_0\sin\omega t\), the charge that flows forward in the first half cycle flows back in the second half. The two halves are mirror images. So over a full cycle the mean current is zero, however large \(i_0\) may be.
Over the positive half cycle alone the mean is not zero. Integrating from \(0\) to \(T/2\) gives \[ \begin{gathered} \bar i_{half} = \dfrac{2}{T}\int_0^{T/2} i_0\sin\omega t\,dt \\ \bar i_{half} = \dfrac{2i_0}{\pi} \approx 0.64\,i_0. \end{gathered} \] This half-cycle mean is useful when only one half of the wave flows, as in a rectifier.
A moving-coil galvanometer shows this directly. Its coil feels a torque in proportion to the current. On a \(50\,\text{Hz}\) current the torque reverses a hundred times a second. The coil is far too heavy to follow, so it answers only to the mean torque, which is zero. Its needle stays at zero while a large current flows.
A moving-coil galvanometer is connected in series with a lamp on the \(50\,\text{Hz}\) mains. The lamp glows, but the needle stays at zero. Why?
Heating does not cancel
A resistor heats up whichever way the current flows through it. The heat is produced at the rate \(i^2R\), and \(i^2\) is never negative. So heating is a fair measure of the size of an alternating current. We compare the alternating current with a steady current that produces the same heat in the same resistor in the same time.
The rms value of an alternating current is the steady direct current that produces the same heat in a resistor in the same time. It is the square root of the mean of the square of the current: \(I_{rms} = \sqrt{\langle i^2\rangle}\). The letters rms stand for root mean square.
\[ i^2 = i_0^2\sin^2\omega t \]
Why. square the current at every instant; the square is positive in both halves of the cycle.
\[ \sin^2\omega t = \dfrac{1 - \cos 2\omega t}{2} \]
Why. the double-angle identity \(\cos 2\theta = 1 - 2\sin^2\theta\).
\[ \langle\sin^2\omega t\rangle = \dfrac12 \]
Why. over a full cycle \(\cos 2\omega t\) runs through two complete cycles and averages to zero; only the constant \(\tfrac12\) remains.
\[ \langle i^2\rangle = \dfrac{i_0^2}{2} \]
Why. \(i_0^2\) is a constant, so it comes outside the mean.
\[ I_{rms} = \sqrt{\langle i^2\rangle} = \dfrac{i_0}{\sqrt2} \approx 0.707\,i_0 \]
Read it. take the square root, as the definition of the rms value requires.
Every alternating voltage or current quoted in daily life is an rms value. That includes the \(220\,\text{V}\) of the mains, the rating on an appliance and the reading of an AC meter. When a problem says "a \(220\,\text{V}\) supply" it means \(V_{rms} = 220\,\text{V}\) unless it says otherwise. The peak is then \(v_0\) \(= \sqrt2 \times 220\) \(= 311\,\text{V}\).
A bulb is rated \(100\,\text{W}\), \(220\,\text{V}\). Find the rms current, the peak current and the resistance of its filament.
The rating is in rms values: \(P = 100\,\text{W}\) at \(V_{rms} = 220\,\text{V}\). The filament is a resistor.
\(I_{rms} = P/V_{rms}\), \(i_0 = \sqrt2\,I_{rms}\), \(R = V_{rms}^2/P\).
\(I_{rms}\) \(= 100/220\) \(= 0.45\,\text{A}\). \(i_0\) \(= 1.414 \times 0.455\) \(= 0.64\,\text{A}\). \(R\) \(= 220^2/100\) \(= 484\,\Omega\).
The filament carries \(0.45\,\text{A}\) rms, swinging to \(0.64\,\text{A}\) at each peak. Its resistance when hot is \(484\,\Omega\).
Now compare the two currents of equal peak. A steady \(2\,\text{A}\) heats a resistor at \(4R\) watts. An alternating current of peak \(2\,\text{A}\) has \(I_{rms} = 1.41\,\text{A}\). So it heats the same resistor at \(1.41^2 R = 2R\) watts, half as much. The alternating current spends most of each cycle below its peak.
Back to the meter board. The \(220\,\text{V}\) there is neither the peak, \(311\,\text{V}\), nor the mean, zero. It is the rms value. A heater on the mains gets exactly as hot as it would on a steady \(220\,\text{V}\). That is the sense in which the mains "is" \(220\,\text{V}\).
An AC ammeter in series with a room heater reads \(2\,\text{A}\). Find the peak current, in amperes.
A steady current of \(3\,\text{A}\) and an alternating current of peak \(3\,\text{A}\) flow in turn through the same resistor for the same time. How does the heat from the alternating current compare?
Other waveforms
The factor \(1/\sqrt2\) belongs to the sine alone. For any other waveform the rms value comes from its definition, \(I_{rms} = \sqrt{\langle i^2\rangle}\). Square the current, take the mean of the square over one period, and take the root. The mean value comes from its own definition, the mean of \(i\) itself.
Play the sine in Fig 1.2. The grey mean line rises during the first half cycle, then falls back to zero by the end. The brown line of the lower graph never falls to zero, because \(i^2\) is never negative. It settles at \(12.5\,\text{A}^2\), so the rms value is \(\sqrt{12.5} = 3.54\,\text{A}\). The square wave is always \(5\,\text{A}\) in size, so its square is always \(25\,\text{A}^2\) and its rms value is \(5\,\text{A}\), while its mean is zero.
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. In (b), \(i = 3 + 4\sin\omega t\) amperes. In (c), a half-wave rectifier is fed with \(i = 4\sin\omega t\) and passes only the positive halves.
(a) The current rises uniformly, so its mean is \(3\,\text{A}\). With \(i = 6t/T\), \(\langle i^2\rangle\) \(= \dfrac{1}{T}\int_0^T \dfrac{36t^2}{T^2}\,dt\) \(= 12\), so \(I_{rms}\) \(= \sqrt{12}\) \(= 3.5\,\text{A}\).
(b) The sine averages to zero, so the mean is \(3\,\text{A}\). \(i^2 = 9 + 24\sin\omega t + 16\sin^2\omega t\) averages to \(9 + 0 + 8 = 17\), so \(I_{rms}\) \(= \sqrt{17}\) \(= 4.1\,\text{A}\).
(c) One half of each cycle carries a mean of \(2i_0/\pi\) and the other half nothing, so \(\bar i\) \(= i_0/\pi\) \(= 4/\pi\) \(= 1.3\,\text{A}\). \(\langle i^2\rangle\) is half that of the full sine, \(i_0^2/4\), so \(I_{rms}\) \(= i_0/2\) \(= 2\,\text{A}\).
A sum of a sine and a cosine of the same frequency is again one sine. For example, \(2\sin\omega t + 2\cos\omega t = 2\sqrt2\sin(\omega t + 45°)\). Its peak is \(2\sqrt2 = 2.8\,\text{A}\), and its rms value is \(2.8/\sqrt2 = 2\,\text{A}\). The next lesson shows how to see such a sum at a glance, with phasors.
A current is \(i = 4 + 3\sin\omega t\) amperes. Find its rms value, in amperes.
Put these currents in order of their rms values, smallest first.
An electric iron is rated \(1000\,\text{W}\) on the \(220\,\text{V}\) mains. Estimate the peak current through its element, in amperes.
An AC ammeter must read the same whichever way the current flows. A moving-coil meter cannot do this, as the second topic showed. The old hot-wire ammeter measured how much a thin wire stretched as the current heated it. Heating goes as \(i^2\), so its scale read rms values directly. A modern digital multimeter samples the waveform and computes the rms value. A cheaper one assumes a sine and can misread a square wave.
Edison's first power stations in New York supplied direct current at about \(110\,\text{V}\). At that low voltage the heat lost in the wires limited each station to customers within about a mile. George Westinghouse backed alternating current, whose voltage a transformer can raise for the journey and lower again at the house. Edison fought back with public warnings about the danger of AC. In 1893 Westinghouse lit the Chicago World's Fair with alternating current. In 1896 power from Niagara Falls reached Buffalo, about \(40\,\text{km}\) away, as AC. Alternating current has run the world's grids ever since.
- Using the peak value where a problem gives "a \(220\,\text{V}\) supply".Supply voltages, ratings and meter readings are rms values. The peak is \(\sqrt2\) times larger.
- Writing \(I_{rms} = i_0/\sqrt2\) for every waveform.The factor \(1/\sqrt2\) is for a sine only. For any other waveform, work out \(\sqrt{\langle i^2\rangle}\).
- Taking the mean current as the measure of an alternating current.The mean of a sine over a cycle is zero. Heating goes as \(i^2\), so the rms value measures the size.
- Using \(\omega = 50\,\text{rad/s}\) for the mains.\(50\,\text{Hz}\) is the frequency \(f\). The angular frequency is \(\omega\) \(= 2\pi f\) \(= 314\,\text{rad/s}\).