Alternating currents

Both subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

Syllabus points

21.1 Characteristics of alternating currents

• Understand and use the terms period, frequency and peak value as applied to an alternating current or voltage.
• Use equations of the form x = x₀ sin ωt representing a sinusoidally alternating current or voltage.
• Recall and use the fact that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current.
• Distinguish between root-mean-square (r.m.s.) and peak values and recall and use I_r.m.s. = I₀/√2 and V_r.m.s. = V₀/√2 for a sinusoidal alternating current.

21.2 Rectification and smoothing

• Distinguish graphically between half-wave and full-wave rectification.
• Explain the use of a single diode for the half-wave rectification of an alternating current.
• Explain the use of four diodes (bridge rectifier) for the full-wave rectification of an alternating current.
• Analyse the effect of a single capacitor in smoothing, including the effect of the values of capacitance and the load resistance.

Describing an alternating current

An alternating current reverses direction periodically. For the sinusoidal case:

I = I₀ sin ωt

V = V₀ sin ωt

• Peak value I₀ or V₀: the maximum value reached in either direction. Read from a graph as the height above the axis, not the height from trough to crest, which is the peak-to-peak value and is twice as large.
• Period T: the time for one complete cycle, including both the positive and the negative half.
• Frequency f = 1/T, in hertz. Mains supplies run at 50 Hz or 60 Hz.
• Angular frequency ω = 2πf = 2π/T, in rad s⁻¹. The ω in the equation is angular frequency, not frequency, and substituting f for ω is a frequent slip.

Worked example. A supply is described by V = 325 sin 314t, with t in seconds.

Comparing with V = V₀ sin ωt, the peak voltage is 325 V and ω is 314 rad s⁻¹.

f = 314 / (2 × 3.142) = 49.97

so the frequency is 50 Hz, and

T = 1 / 50 = 0.02

so the period is 20 ms.

Power, and why the mean is half the peak

The instantaneous power dissipated in a resistor is

P = I²R

Because the current is squared, the power is positive throughout the cycle, even during the negative half when the current is reversed. A resistor does not care which way the current flows; it heats up either way.

Squaring a sine wave gives a curve that oscillates between zero and the peak power, at twice the frequency of the current, and symmetrically about its own midpoint. So the mean power is half the peak power:

⟨P⟩ = ½P₀ = ½I₀²R = ½I₀V₀

That factor of a half is the origin of everything else in this section.

Root-mean-square values

The r.m.s. value of an alternating current is the value of the direct current that would dissipate energy in a resistor at the same mean rate. It is the number that makes an a.c. supply comparable with a d.c. one, which is why every stated supply voltage is an r.m.s. value.

Setting the d.c. power equal to the mean a.c. power:

I_r.m.s.²R = ½I₀²R

so

I_r.m.s. = I₀ / √2

and identically

V_r.m.s. = V₀ / √2

Since √2 is 1.414, the r.m.s. value is about 0.707 of the peak, and the peak is about 1.414 times the r.m.s.

The name says how it is calculated, in the same order as for molecular speeds in topic 15: square the values, take the mean, take the root. The mean of the current itself over a whole cycle is zero, which is precisely why a simple mean is useless here and the r.m.s. is needed.

Worked example. A 230 V mains supply is quoted as an r.m.s. value. What is the peak voltage?

V₀ = 230 × 1.414 = 325.2

so the peak is about 325 V. That is the number in the equation above, and it is why insulation in mains equipment has to withstand 325 V rather than 230 V.

Worked example. A heater of resistance 46 Ω runs from that supply. Find the mean power.

Use r.m.s. values with the ordinary d.c. formulae, which is the whole point of r.m.s.:

P = 230² / 46 = 1150

so the mean power is 1150 W. The peak power is twice that:

1150 × 2 = 2300

that is 2300 W.

The rule to carry away: r.m.s. values behave exactly like d.c. values in P = VI, P = I²R and P = V²/R. Peak values do not, and putting a peak value into those formulae gives the peak power, which is twice the useful answer.

Rectification

Rectification converts alternating current into current that flows in one direction only. It relies on the diode, which conducts in forward bias and effectively does not conduct in reverse bias.

Half-wave rectification

A single diode in series with the load. During the half-cycle when the diode is forward biased, current flows through the load. During the other half-cycle the diode is reverse biased, no current flows, and the output is zero.

The output graph shows the positive humps of the input, with flat gaps where the negative humps were.

The drawback follows straight from the picture: half the power is thrown away, and the mean power delivered is half that of full-wave rectification of the same supply.

Full-wave rectification with a bridge rectifier

Four diodes arranged in a bridge, with the supply across one diagonal and the load across the other.

On each half-cycle, two of the four diodes are forward biased and conduct, while the other two are reverse biased and block. The pair that conducts swaps over when the supply reverses, but they are arranged so that the current through the load is in the same direction both times.

The output graph shows the negative humps flipped up into positive ones, so there are twice as many humps as with half-wave and no gaps.

Being able to trace the current path through the bridge for each half-cycle, and say which two diodes conduct, is a standard question. Work it out from the rule that current can only pass through a diode in the direction of its arrow.

Smoothing

Rectified output is unidirectional but far from steady. A capacitor connected in parallel with the load smooths it.

The mechanism, in the order the marks are awarded:

1. As the rectified voltage rises, the capacitor charges up, and the output follows the input.
2. As the rectified voltage falls past the peak, the capacitor is at a higher potential than the supply, so it discharges through the load, maintaining the current when the supply cannot.
3. The output therefore falls slowly, along an exponential discharge curve with time constant RC, until the next peak arrives and recharges it.

The result is a nearly steady voltage with a small ripple.

The size of the ripple depends on the time constant RC compared with the time between peaks:

• A larger capacitance stores more charge and gives a longer time constant, so the voltage falls less between peaks and the ripple is smaller.
• A larger load resistance draws a smaller current, so again the capacitor discharges more slowly and the ripple is smaller. A heavily loaded supply, meaning a small R, has a large ripple.
• Full-wave rectification gives peaks twice as often as half-wave, so there is less time to discharge between them and the ripple is smaller for the same capacitor.

Note that smoothing does not remove the ripple entirely; it reduces it. And a larger capacitor smooths better but draws a larger surge of current when the supply is first switched on, which is why a real design is a compromise.

Common mistakes

• Reading the peak value off a graph as the peak-to-peak distance.
• Substituting f for ω in x = x₀ sin ωt.
• Quoting a mains supply voltage as a peak value. Quoted supply voltages are r.m.s.
• Putting peak values into P = V²/R and reporting the answer as the mean power. It is the peak power, twice too large.
• Dividing the peak by 2 instead of by √2.
• Saying the mean current over a cycle is the r.m.s. current. The mean current is zero.
• Saying the mean power is zero because the current reverses. Power depends on the square of the current and is positive throughout.
• Drawing half-wave rectification with the negative humps inverted, which is full-wave.
• Saying all four diodes in a bridge conduct at once. Two conduct on each half-cycle.
• Placing the smoothing capacitor in series with the load rather than in parallel with it.
• Saying a larger load resistance increases the ripple. A larger R means a smaller current and a smaller ripple.
• Saying a capacitor removes the ripple completely.