Alternating Currents

Alternating currents: period, frequency and peak value

An alternating current is a current that repeatedly reverses its direction, flowing first one way and then the other in a regular cycle. The voltage that drives it, the alternating voltage, does exactly the same: it swings positive and then negative about zero. The sinusoidal case matters most, because the e.m.f. produced by an a.c. generator is very nearly a sine wave, and the whole of the public electricity supply is generated and distributed as a sinusoid.

The period T is the time taken for one complete cycle, from a point on the waveform to the identical point on the next cycle, and it is measured in seconds. The frequency f is the number of complete cycles in one second, measured in hertz (Hz). The two are reciprocals: f = 1 / T and T = 1 / f. The mains supply in the United Kingdom alternates at 50 Hz, so its period is T = 1 / 50 = 0.020 s, which is 20 ms. Each cycle contains two half cycles of 10 ms, one positive and one negative.

The peak value I0 or V0 is the maximum value reached in either direction, and it is also called the amplitude. It is the value that decides whether insulation survives, because at the peak of the cycle the potential difference across the insulation is at its greatest. The peak-to-peak value is twice the peak value, and mixing the two up is a common error: a supply quoted as 230 V has a peak value of 325 V and a peak-to-peak value of 650 V.

The mean value of a sinusoidal current over one complete cycle is zero, because the positive half cycle cancels the negative half cycle exactly. That is why a moving-coil ammeter, which responds to the mean value, reads zero when it is connected to an alternating supply, and why such a meter has to have a rectifier built into it before it can measure alternating current at all.

• Period T: the time for one complete cycle, T = 1 / f, measured in seconds (s).
• Frequency f: the number of complete cycles per second, f = 1 / T, measured in hertz (Hz).
• Peak value I0 or V0: the maximum value reached in either direction, sometimes called the amplitude.
• United Kingdom mains is 50 Hz, so T = 1 / 50 = 0.020 s = 20 ms, with half cycles of 10 ms.
• The mean of a sinusoidal alternating current over a complete cycle is zero.

Writing an alternating quantity as x = x0 sin omega t

Any sinusoidally varying quantity can be written as x = x0 sin omega t, where x is the instantaneous value at time t, x0 is the peak value and omega is the angular frequency. The angular frequency is related to the frequency by omega = 2 pi f, and it is measured in radians per second (rad s-1). Because one complete cycle is 2 pi radians, a frequency of 50 Hz corresponds to an angular frequency of 2 pi x 50 = 314 rad s-1.

The sine form comes from the generator itself. A coil rotating at a constant rate in a uniform magnetic field sweeps through the field so that the flux linkage varies with the cosine of the angle turned through, and by Faraday's law the induced e.m.f. is proportional to the rate of change of that flux linkage, so the e.m.f. varies with the sine of the angle. A constant rotation rate therefore gives a sinusoidal e.m.f. of constant peak value, and the peak value is fixed by the speed of rotation, the strength of the field and the number of turns on the coil.

Worked example. The United Kingdom mains supply is quoted as 230 V, which is a root-mean-square value, so its peak value is 230 x sqrt(2) = 325 V and its angular frequency is 2 pi x 50 = 314 rad s-1. The instantaneous voltage is therefore v = 325 sin(314 t), with v in volts and t in seconds. At t = 2.0 ms, omega t = 314 x 2.0 x 10-3 = 0.63 rad, so v = 325 sin(0.63) = 325 x 0.59 = 190 V.

Check the value at the quarter points of one cycle, because that is the quickest way to show you understand what the equation means. At t = 5.0 ms, which is one quarter of a 20 ms period, sin(pi / 2) = 1 and the voltage is at its peak, 325 V. At t = 10 ms, half way through the cycle, sin(pi) = 0 and the voltage is zero. At t = 15 ms, the voltage is at its negative peak, -325 V, and at 20 ms the cycle begins again.

• x = x0 sin omega t: x is the instantaneous value, x0 the peak value and omega the angular frequency.
• omega = 2 pi f, measured in rad s-1; a 50 Hz supply gives omega = 314 rad s-1.
• v = 325 sin(314 t) describes the United Kingdom mains supply, with v in volts and t in seconds.
• At t = 5.0 ms the voltage is at its peak, at 10 ms it is zero and at 15 ms it is at its negative peak.

Power in a resistive load: why the mean power is half the maximum power

The instantaneous power delivered to a resistor is P = I^2 R, or equally P = V^2 / R. Because the current is squared, the power is positive whether the current is positive or negative, so energy is transferred to the resistor right through the cycle, in both halves of it. The resistor heats up just as it would on a direct current of the same size, which is why a filament lamp on an alternating supply glows steadily instead of flickering.

The power is far from constant. It reaches a maximum of P0 = I0^2 R at the instant the current reaches its peak value, and it falls to zero twice in every cycle, as the current passes through zero. What heats the resistor is the mean power, and for a sine wave the mean value of sin^2(omega t) over a complete cycle is 1/2. The mean power is therefore half the maximum power: mean power = (1/2) I0^2 R.

A numerical check makes the result clear. A current of peak value 4.0 A flows in a resistor of resistance 25 ohm. The maximum power is 4.0^2 x 25 = 400 W. The mean value of the square of the current is 4.0^2 / 2 = 8.0 A2, so the mean power is 8.0 x 25 = 200 W, exactly half of 400 W. Using the r.m.s. current gives the same answer at once: I r.m.s. = 4.0 / sqrt(2) = 2.83 A, and I r.m.s.^2 R = 2.83^2 x 25 = 200 W.

The same reasoning applies to voltage: mean power = (1/2) V0^2 / R = V r.m.s.^2 / R. Notice that the mean power is not found by putting the mean values of current and voltage into P = VI, because those means are both zero. It is the square that is averaged first, and that is the whole point of the root-mean-square value.

• Instantaneous power P = I^2 R, positive in both halves of the cycle because the current is squared.
• Maximum power P0 = I0^2 R, reached only at the peaks of the waveform.
• The mean value of sin^2(omega t) over one cycle is 1/2, so the mean power is (1/2) I0^2 R.
• A peak current of 4.0 A in a 25 ohm resistor gives 400 W at the peaks and 200 W on average.

Root-mean-square values and what they are for

Because the mean of an alternating current is zero, a different kind of average is needed to describe its heating effect. The root-mean-square value of an alternating current is defined as the value of the direct current that would dissipate energy at the same mean rate in the same resistor. The name says how it is calculated: square the current, take the mean of the square, then take the square root. For a sine wave this gives the peak value divided by the square root of 2.

I r.m.s. = I0 / sqrt(2) and V r.m.s. = V0 / sqrt(2). Dividing by the square root of 2 is the same as multiplying by 0.707, so the r.m.s. value is about 71 per cent of the peak value. The United Kingdom mains supply is 230 V r.m.s. at 50 Hz, so the peak voltage is 230 x sqrt(2) = 325 V and the peak-to-peak voltage is 650 V.

Use r.m.s. values in every power and energy question, because that is what they were invented for: mean power = V r.m.s. I r.m.s. = I r.m.s.^2 R = V r.m.s.^2 / R. Use the peak value in questions about insulation, breakdown and peak electric field, because that is the instant at which the material is most highly stressed. Reading the wrong one off a meter is the classic mistake in this topic.

An alternating voltmeter or ammeter is designed to read r.m.s. values for a sine wave, usually by rectifying the current inside the instrument and then scaling the movement. That calibration only holds for a sinusoidal waveform: for a square wave, a triangle wave or a distorted supply, a simple meter gives a reading that has little to do with the true r.m.s. value, which is why true r.m.s. instruments exist. Remember too that the r.m.s. value is not the mean value, which is zero, and it is not half the peak value either.

• Definition: the r.m.s. value of an alternating current is the direct current that would dissipate energy at the same mean rate in the same resistor.
• I r.m.s. = I0 / sqrt(2) and V r.m.s. = V0 / sqrt(2), so the r.m.s. value is 0.707 times the peak value.
• Mean power = V r.m.s. I r.m.s. = I r.m.s.^2 R = V r.m.s.^2 / R.
• 230 V r.m.s. mains has a peak value of 325 V and a peak-to-peak value of 650 V.
• Meters calibrated to read r.m.s. are only correct for a sinusoidal waveform.

Rectification: half-wave and full-wave

Rectification is the conversion of an alternating current into a direct current, meaning a current that flows in one direction only. The component that does the work is the diode, which conducts when it is forward biased with the anode positive with respect to the cathode, and does not conduct when it is reverse biased. A diode is therefore a one-way valve for current, and that single property is the whole basis of rectification.

Half-wave rectification uses one diode in series with the load. The diode conducts during the half cycle in which it is forward biased, so current flows through the load; during the other half cycle the diode is reverse biased and no current flows at all. The output is unidirectional, but it arrives in separate pulses, one for each cycle of the supply, with a gap at zero volts lasting a whole half cycle between one pulse and the next. The mean output is low, and the gaps make the output hard to smooth.

Full-wave rectification uses four diodes connected as a bridge rectifier. During one half cycle two of the diodes are forward biased and conduct, carrying current through the load in one direction; during the other half cycle the other two conduct and push current through the load in the same direction again. The load therefore receives a pulse in every half cycle, so a 50 Hz supply produces 100 pulses per second, and the mean power delivered to the load is twice what the same supply gives through a half-wave rectifier.

On an oscilloscope the two outputs are unmistakably different. The half-wave trace shows a rounded pulse, then a flat section at zero for the same length of time, then the next pulse: two humps in the time of two supply cycles. The full-wave trace shows humps one after another with no gaps, and at twice the supply frequency: two humps in the time of one supply cycle. Both outputs have the same peak value as each other, a little below the peak of the input because of the forward voltage drop of the conducting diodes: about 0.7 V is lost in a single-diode circuit and about 1.4 V in a bridge, where two diodes carry the current in series.

Smoothing: the effect of a capacitor across the load

A rectified output is unidirectional but nowhere near steady: it rises and falls, and the variation about the mean is called the ripple. Connecting a single capacitor in parallel with the load resistor smooths the output, turning a series of pulses into something much closer to a steady direct voltage.

The capacitor works in two stages. While the rectifier output is rising towards its peak, the capacitor charges and the voltage across it follows the output, while the current that charges it also flows through the load. As soon as the rectifier output begins to fall, the voltage on the capacitor is higher than the output voltage, the diodes stop conducting, and the capacitor discharges through the load resistor instead. The load therefore keeps drawing current while the rectifier is, in effect, doing nothing, and the output voltage falls only a little before the next pulse arrives and recharges the capacitor.

How well this works depends on the time constant RC of the capacitor and the load, compared with the time between pulses. If RC is much greater than the interval between pulses, the capacitor discharges only a little in each interval and the output is nearly steady; if RC is small, the voltage sags a long way and the ripple is large. After full-wave rectification of a 50 Hz supply the pulses arrive every 10 ms, so a capacitor of 4700 microfarads (4.7 x 10-3 F) with a load of 100 ohm gives RC = 0.47 s, nearly fifty times the interval between pulses, and the output is very steady. Increasing the capacitance, or increasing the load resistance, reduces the ripple; reducing either one makes it worse.

The capacitor also raises the average output, because the load is supplied at a voltage near the peak for most of each cycle instead of at the mean of the pulses, so an unregulated supply with a large smoothing capacitor gives a steady output close to the peak value of the rectified waveform. The price is a large surge of current through the diodes every time the capacitor recharges, together with a large electrolytic capacitor that takes up space and must be connected the right way round.

• The capacitor charges to the peak of the rectified output and then discharges through the load resistor when the output falls.
• Smoothing works because the capacitor supplies the load during the gaps between pulses.
• The ripple is small when the time constant RC is much greater than the time between pulses.
• Increasing the capacitance or increasing the load resistance increases RC and reduces the ripple.
• 4700 microfarads with a 100 ohm load gives RC = 0.47 s, far longer than the 10 ms between pulses of a full-wave 50 Hz supply.

Transformers: turns ratio, current ratio and efficiency

A transformer changes the size of an alternating voltage. It consists of two coils, the primary and the secondary, wound on a laminated soft iron core. An alternating current in the primary sets up a changing magnetic flux in the core, the core carries almost all of that flux to the secondary, and the changing flux linkage induces an alternating e.m.f. in the secondary by Faraday's law. The core is laminated, made of thin sheets insulated from one another, to break up the eddy currents that would otherwise circulate in it and waste energy.

Because the same flux passes through every turn of both coils, the e.m.f. induced in each turn is the same, so the voltages are in the ratio of the numbers of turns: Vp / Vs = Np / Ns. This ratio of turns is the turns ratio. If the secondary has more turns than the primary the transformer is a step-up transformer and the voltage rises; if it has fewer turns it is a step-down transformer and the voltage falls. The frequency is unchanged, because a transformer cannot alter the frequency of a supply.

For an ideal transformer, which wastes no energy, the power delivered to the secondary equals the power taken from the primary: Vp Ip = Vs Is. Combining this with the turns ratio gives Np / Ns = Is / Ip, so a transformer that steps the voltage up steps the current down in the same proportion, and the other way round. A real transformer is less than 100 per cent efficient, so the secondary current is a little below the ideal value, and the efficiency is calculated as efficiency = (Vs Is / Vp Ip) x 100%. The losses are resistive heating in the windings, eddy currents and hysteresis in the core, and a little flux leakage.

Worked example. A step-down transformer for a 12 V lighting circuit has 1150 turns on the primary and is connected to the 230 V mains. The number of turns on the secondary is Ns = Np Vs / Vp = 1150 x 12 / 230 = 60 turns. If the secondary supplies 4.0 A, an ideal transformer would draw Ip = Is Ns / Np = 4.0 x 60 / 1150 = 0.21 A from the mains, and a real transformer with an efficiency of 95 per cent draws 0.22 A. Note that the transformer works only on alternating current: a steady direct current in the primary gives a steady flux, no changing flux linkage and therefore no induced e.m.f. in the secondary at all.

• A transformer is two coils on a laminated soft iron core, working by mutual induction.
• Turns ratio: Vp / Vs = Np / Ns. More turns on the secondary means a step-up transformer.
• Ideal transformer: Vp Ip = Vs Is, so Np / Ns = Is / Ip.
• Efficiency = (Vs Is / Vp Ip) x 100%, reduced by resistance, eddy currents, hysteresis and flux leakage.
• A transformer cannot work on steady direct current, because there is no changing flux.

The National Grid: why power is transmitted at high voltage

The National Grid carries electrical power from power stations to the places where it is used. Power stations in the United Kingdom generate at about 25 kV, the grid transmits at 400 kV and 275 kV, and the voltage is stepped down in substations to 132 kV, 33 kV and 11 kV before it finally reaches homes and offices at 230 V. Transformers at every stage make this possible, and that is one reason the supply is alternating.

The reason for transmitting at a high voltage is the power wasted heating the cables. The power carried is P = VI, and the power wasted in the cables is I^2 R, where R is the total resistance of the conductors on the route. For a fixed transmitted power, raising the transmission voltage lowers the current in proportion, and the wasted power falls with the square of the current. Transmitting 200 MW through cables of total resistance 1.0 ohm needs 8000 A at 25 kV, which wastes 8000^2 x 1.0 = 64 MW, nearly a third of the power generated. At 400 kV the current is only 500 A and the loss is 500^2 x 1.0 = 250000 W, which is 0.25 MW. Raising the voltage by a factor of 16 has cut the waste by a factor of 256.

There is a second, economic advantage. For the same current density a smaller current needs a smaller cross-sectional area of conductor, so the cables can be thinner and lighter and less aluminium is needed, which means cheaper pylons and conductors. The saving in conductor material is a large part of the reason the grid is designed the way it is.

The voltage cannot be raised indefinitely, and evaluating the compromise is a standard examination task. Higher voltages need thicker insulation and greater clearance between the conductors and from the ground and buildings, because air breaks down and corona discharge begins when the electric field around a conductor becomes strong enough. The transformers and switchgear needed at very high voltage are themselves expensive, and the pylons must be taller. In practice 400 kV is close to the economic optimum for overhead lines on land.

For very long submarine cables, high-voltage direct current (h.v.d.c.) is sometimes used instead, for example on the cross-Channel link between England and France. A long alternating cable behaves as a capacitor, and the charging current that flows because of that cable capacitance is a loss that grows with the length of the cable; with direct current there is no charging current, so the cable can carry more useful power. At each end of such a link the alternating supply is rectified to direct current and then converted back to alternating current by power electronic converters.

• Power stations generate at about 25 kV, the grid transmits at 400 kV and 275 kV, and substations step the voltage down to 230 V for use.
• P = VI, so for a fixed power a higher voltage means a smaller current.
• Cable losses are I^2 R, so raising the transmission voltage cuts the wasted power by the square of the voltage rise.
• 200 MW at 25 kV wastes 64 MW in 1.0 ohm of cable, while the same power at 400 kV wastes only 0.25 MW.
• A smaller current needs thinner, cheaper conductors, and the voltage is limited by insulation, clearance and cost.
• Very long submarine cables use h.v.d.c. instead, because an alternating cable draws a large charging current.