A2 · Paper 4 practice · Alternating currents

Peaks, r.m.s., and ripple.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Alternating currents: peak and r.m.s. values, mean power, and half-wave, full-wave and smoothed rectification. Several questions carry the characteristics of a.c. into the rectifier that follows. Each is linked to its lessons; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Use Iₕₘₛ = I₀/√2, Vₕₘₛ = V₀/√2, mean power = ½ peak power, and τ = RC where needed.
Keep these straight

Values, power, ripple.

01
Paper 4
[10 marks]

A sinusoidal current is described by i = I₀ sinωt, with a peak value of 8.0 A and a frequency of 50 Hz.

(a) State what is meant by the peak value. [1]

(b) Calculate the r.m.s. value of the current. [2]

(c) Calculate the period of the current. [1]

(d) Calculate the angular frequency ω. [2]

(e) Write the equation for the current with these numerical values. [1]

(f) Calculate the current at t = 5.0 ms. [2]

(g) State the time at which the current first returns to zero after t = 0. [1]

  • (a)The maximum value of the current (the amplitude) ✓
  • (b)Iₕₘₛ = I₀/√2 = 8.0/1.414 ✓ = 5.7 A ✓
  • (c)T = 1/f = 1/50 = 0.020 s (20 ms) ✓
  • (d)ω = 2πf = 2π × 50 ✓ = 314 rad s⁻¹ ✓
  • (e)i = 8.0 sin(314t) ✓
  • (f)ωt = 314 × 0.0050 = 1.57 rad (½π) ✓; i = 8.0 sin(½π) = 8.0 A ✓
  • (g)At ωt = π, t = T/2 = 10 ms ✓
02
Paper 4
[10 marks]

A heating element of resistance 46 Ω is connected to the 230 V (r.m.s.), 50 Hz mains supply.

(a) Calculate the peak voltage of the supply. [2]

(b) Calculate the r.m.s. current in the element. [2]

(c) Calculate the mean power dissipated. [2]

(d) Calculate the peak power dissipated. [2]

(e) State the ratio of mean power to peak power. [1]

(f) State why r.m.s. values, rather than the simple average, are quoted for a.c. [1]

  • (a)V₀ = Vₕₘₛ × √2 = 230 × 1.414 ✓ = 325 V ✓
  • (b)Iₕₘₛ = Vₕₘₛ/R = 230/46 ✓ = 5.0 A ✓
  • (c)Mean power = Vₕₘₛ²/R = 230²/46 ✓ = 1150 W (1.15 kW) ✓
  • (d)Peak power = V₀²/R = 325²/46 ✓ = 2300 W ✓
  • (e)Mean : peak = 1 : 2 (mean is half the peak) ✓
  • (f)The simple average of a sinusoid over a cycle is zero, so it is useless; the r.m.s. gives the equivalent steady (d.c.) value for power ✓
03
Paper 4
[9 marks]

A resistor of resistance 12 Ω carries a sinusoidal current of peak value 3.0 A.

(a) Calculate the r.m.s. current. [2]

(b) Calculate the mean power dissipated in the resistor. [2]

(c) Calculate the peak (maximum instantaneous) power. [2]

(d) State the relationship between the mean power and the peak power. [1]

(e) State the value of the steady direct current that would dissipate the same mean power. [2]

  • (a)Iₕₘₛ = 3.0/√2 ✓ = 2.1 A ✓
  • (b)Mean power = Iₕₘₛ²R = 2.12² × 12 ✓ = 54 W (= ½ × 3.0² × 12) ✓
  • (c)Peak power = I₀²R = 3.0² × 12 ✓ = 108 W ✓
  • (d)The mean power is half the peak power ✓
  • (e)2.1 A ✓, which is the r.m.s. current (the d.c. that gives the same mean power) ✓
04
Paper 4
[9 marks]

Fig. 4.1 shows how an alternating current varies with time.

i / A t / ms 2.0 20 40
Fig. 4.1

(a) State the peak value of the current. [1]

(b) Read the period from the graph and hence calculate the frequency. [2]

(c) Calculate the angular frequency ω. [2]

(d) Write the equation of the current in the form i = I₀ sinωt. [1]

(e) Calculate the r.m.s. value of the current. [2]

(f) State the time at which the current first reaches its peak. [1]

  • (a)2.0 A ✓
  • (b)T = 40 ms = 0.040 s ✓; f = 1/T = 25 Hz ✓
  • (c)ω = 2π/T = 2π/0.040 ✓ = 157 rad s⁻¹ ✓
  • (d)i = 2.0 sin(157t) ✓
  • (e)Iₕₘₛ = 2.0/√2 ✓ = 1.4 A ✓
  • (f)At a quarter period, t = 10 ms ✓
05
Paper 4
[8 marks]

A single diode is connected in series with a resistor and a sinusoidal a.c. supply.

(a) Explain how the diode produces half-wave rectification. [2]

(b) Describe the output voltage across the resistor over one cycle. [2]

(c) State the fraction of each cycle during which current flows in the resistor. [1]

(d) Explain why the mean power delivered is less than for full-wave rectification of the same supply. [2]

(e) State one disadvantage of half-wave rectification. [1]

  • (a)The diode conducts only when it is forward biased ✓, so current passes during only one half of each cycle ✓
  • (b)Positive half-cycles appear across the resistor; during the other half the output is zero ✓✓
  • (c)One half of each cycle ✓
  • (d)Energy is delivered for only half of each cycle ✓; full-wave uses both halves, so its mean power is greater ✓
  • (e)Half the available power is wasted (or the output has a large ripple) ✓
06
Paper 4
[8 marks]

Fig. 6.1 shows the output of a full-wave rectifier (no smoothing) supplied from 50 Hz a.c.

V t
Fig. 6.1

(a) State the number of diodes in a bridge rectifier. [1]

(b) Explain how the bridge produces full-wave rectification. [3]

(c) State the frequency of the output shown in Fig. 6.1, and explain your answer. [2]

(d) State one advantage of full-wave over half-wave rectification. [2]

  • (a)Four ✓
  • (b)On each half-cycle two of the four diodes are forward biased ✓; they route the current through the load in the same direction both times ✓, so both halves of every cycle are used ✓
  • (c)100 Hz ✓; there are two output bumps per input cycle, so the frequency is twice the 50 Hz supply ✓
  • (d)It uses both halves of each cycle, so it delivers more power ✓ and is easier to smooth (smaller ripple) ✓
07
Paper 4
[8 marks]

A capacitor is added to the output of a rectifier to smooth it.

(a) State how the smoothing capacitor is connected relative to the load. [1]

(b) Explain how the capacitor smooths the output. [3]

(c) State and explain the effect on the ripple of using a larger capacitance. [2]

(d) State the effect on the ripple of increasing the load resistance. [1]

(e) State what the output would be if the load were disconnected entirely. [1]

  • (a)In parallel with the load ✓
  • (b)The capacitor charges up to the peak voltage ✓; when the rectified voltage falls, the capacitor discharges through the load ✓, holding the voltage up and filling in the dips ✓
  • (c)The ripple is smaller ✓, because a larger C gives a longer time constant RC, so the capacitor discharges less between peaks ✓
  • (d)The ripple is smaller (larger R also lengthens RC) ✓
  • (e)The capacitor stays charged at the peak voltage, giving a constant output (no ripple) ✓
08
Paper 4
[8 marks]

An a.c. supply of r.m.s. value 12 V and frequency 50 Hz is connected to a single-diode (half-wave) rectifier and a load. Parts (a) and (b) use the a.c. characteristics; the rest use rectification.

(a) Calculate the peak voltage of the supply. [2]

(b) Stating any assumption, give the peak output voltage across the load. [1]

(c) For how long in each cycle does the diode conduct? [2]

(d) A smoothing capacitor is now connected across the load. Describe the output. [2]

(e) State the ripple frequency of this smoothed half-wave output. [1]

  • (a)V₀ = 12 × √2 ✓ = 17 V ✓
  • (b)About 17 V ✓, assuming the diode has no voltage drop (ideal diode)
  • (c)One half of the 20 ms cycle, so 10 ms ✓✓
  • (d)A nearly steady d.c. near the peak voltage, with a small ripple as the capacitor discharges between peaks ✓✓
  • (e)50 Hz (one charging pulse per cycle for half-wave) ✓
09
Paper 4
[8 marks]

A full-wave rectifier supplied from 50 Hz a.c. feeds a load of resistance 2.2 kΩ with a smoothing capacitor of 100 μF in parallel.

(a) Calculate the time constant of the load and capacitor. [2]

(b) State the frequency and period of the ripple before smoothing. [2]

(c) By comparing the time constant with the ripple period, explain whether the smoothing is effective. [2]

(d) State and explain the effect of replacing the capacitor with a 470 μF one. [2]

  • (a)τ = RC = 2200 × 100 × 10⁻⁶ ✓ = 0.22 s ✓
  • (b)Full-wave ripple frequency = 100 Hz ✓; period = 1/100 = 10 ms ✓
  • (c)τ (220 ms) is much greater than the ripple period (10 ms) ✓, so the capacitor barely discharges between peaks: the smoothing is effective and the ripple is small ✓
  • (d)The ripple is smaller still ✓; a larger C gives a longer time constant, so even less discharge between peaks ✓
10
Paper 4
[11 marks]

This question follows a power supply from end to end. A transformer secondary provides 9.0 V (r.m.s.) at 50 Hz to a bridge rectifier, whose output is smoothed and connected to a load.

(a) Calculate the peak voltage at the secondary. [2]

(b) State the type of rectification and the number of diodes used. [2]

(c) State the ripple frequency of the rectified output. [2]

(d) A smoothing capacitor is added across the load. Describe the output voltage. [2]

(e) The load is changed to draw a larger current (a smaller resistance). State and explain the effect on the ripple. [3]

  • (a)V₀ = 9.0 × √2 ✓ = 12.7 V ✓
  • (b)Full-wave rectification ✓, using four diodes (a bridge) ✓
  • (c)Twice the supply frequency = 100 Hz ✓✓
  • (d)A nearly steady d.c. close to the peak (about 12.7 V) with a small ripple ✓✓
  • (e)The ripple becomes larger ✓; the smaller load resistance gives a shorter time constant RC ✓, so the capacitor discharges more between peaks ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.