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DRAFT
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How to use this series 00 3.4 Mass and inertia 000 3.5 Moving through fluids 000 How to use this book 00 3.6 Newton’s third law of motion 000 3.7 Understanding SI units 000 Introduction 00 4 Forces 000 1 Kinematics 000 4.1 Combining forces 000 1.1 Speed 000 4.2 Components of vectors 000 1.2 Distance and displacement, scalar 4.3 Centre of gravity 000 and vector 000 4.4 The turning effect of a force 000 1.3 Speed and velocity 000 4.5 The torque of a couple 000 1.4 Displacement–time graphs 000 1.5 Combining displacements 000 5 Work, energy and power 000 1.6 Combining velocities 000 5.1 Doing work, transferring energy 000 1.7 Subtracting vectors 000 5.2 Gravitational potential energy 000 1.8 Other examples of scalar and 5.3 Kinetic energy 000 vector quantities 000 5.4 Gravitational potential to kinetic energy transformations 000 2 Accelerated motion 000 5.5 Down, up, down: energy changes 000 2.1 The meaning of acceleration 000 5.6 Energy transfers 000 2.2 Calculating acceleration 000 5.7 Power 000 2.3 Units of acceleration 000 2.4 Deducing acceleration 000 6 Momentum 000 2.5 Deducing displacement 000 6.1 The idea of momentum 000 2.6 Measuring velocity and acceleration 000 6.2 Modelling collisions 000 2.7 Determining velocity and acceleration 6.3 Understanding collisions 000 in the laboratory 000 6.4 Explosions and crash-landings 000 2.8 The equations of motion 000 6.5 Collisions in two dimensions 000 2.9 Deriving the equations of motion 000 6.6 Momentum and Newton’s laws 000 2.10 Uniform and non-uniform acceleration 000 6.7 Understanding motion 000 2.11 Acceleration caused by gravity 000 2.12 Determining g 000 7 Matter and materials 000 2.13 Motion in two dimensions: projectiles 000 7.1 Density 000 2.14 UnderstandingDRAFT projectiles 000 7.2 Pressure 000 7.3 Archimedes’ principle 000 3 Dynamics 000 7.4 Compressive and tensile forces 000 3.1 Force, mass and acceleration 000 7.5 Stretching materials 000 3.2 Identifying forces 000 7.6 Elastic potential energy 000 3.3 Weight, friction and gravity 000
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8 Electric current 000 13.3 Interference 000 8.1 Circuit symbols and diagrams 000 13.4 The Young double-slit experiment 000 8.2 Electric current 000 13.5 Diffraction gratings 000 8.3 An equation for current 000 14 Stationary waves 000 8.4 The meaning of voltage 000 14.1 From moving to stationary 000 8.5 Electrical resistance 000 14.2 Nodes and antinodes 000 8.6 Electrical power 000 14.3 Formation of stationary waves 000 9 Kirchhoff’s laws 000 14.4 Determining the wavelength and speed of sound 000 9.1 Kirchhoff’s first law 000 9.2 Kirchhoff’s second law 000 15 Atomic structure 000 9.3 Applying Kirchhoff’s laws 000 15.1 Looking inside the atom 000 9.4 Resistor combinations 000 15.2 Alpha-particle scattering and the nucleus 000 10 Resistance and resistivity 000 15.3 A simple model of the atom 000 10.1 The I–V characteristic for a 15.4 Nucleons and electrons 000 metallic conductor 000 15.5 Forces in the nucleus 000 10.2 Ohm’s law 000 15.6 Discovering radioactivity 000 10.3 Resistance and temperature 000 15.7 Radiation from radioactive substances 000 10.4 Resistivity 000 15.8 Energies in a and β decay 000 11 Practical circuits 000 15.9 Equations of radioactive decay 000 11.1 Internal resistance 000 15.10 Fundamental particles 000 11.2 Potential dividers 000 15.11 Families of particles 000 11.3 Sensors 000 15.12 Another look at β decay 000 11.4 Potentiometer circuits 000 15.13 Another nuclear force 000 12 Waves 000 P1 Practical skills for AS 000 12.1 Describing waves 000 P1.1 Practical work in physics 000 12.2 Longitudinal and transverse waves 000 P1.2 Using apparatus and following instructions 000 12.3 Wave energy 000 P1.3 Gathering evidence 000 12.4 Wave speed 000 P1.4 Precision, accuracy, errors and 12.5 The Doppler effect for sound waves 000 uncertainties 000 12.6 Electromagnetic waves 000 P1.5 Finding the value of an uncertainty 000 12.7 Electromagnetic waves 000 P1.6 Percentage uncertainty 000 12.8 Orders of magnitude 000 P1.7 Recording results 000 12.9 The nature of electromagnetic waves 000 P1.8 Analysing results 000 12.10 Polarisation 000 DRAFTP1.9 Testing a relationship 000 13 Superposition of waves 000 P1.10 Combining uncertainties 000 13.1 The principle of superposition P1.11 Identifying limitations in of waves 000 procedures and suggesting improvements 000 13.2 Diffraction of waves 000
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16 Circular motion 000 20.5 Changing temperature 000 16.1 Describing circular motion 000 20.6 Ideal gas equation 000 16.2 Angles in radians 000 20.7 Modelling gases: the kinetic model 000 16.3 Steady speed, changing velocity 000 20.8 Temperature and molecular kinetic energy 000 16.4 Angular speed 000 16.5 Centripetal forces 000 21 Uniform electric fields 000 16.6 Calculating acceleration and force 000 21.1 Attraction and repulsion 000 16.7 The origins of centripetal forces 000 21.2 The concept of an electric field 000 17 Gravitational fields 000 21.3 Electric field strength 000 21.4 Force on a charge 000 17.1 Representing a gravitational field 000 17.2 Gravitational field strengthg 000 22 Coulomb’s law 000 17.3 Energy in a gravitational field 000 22.1 Electric fields 000 17.4 Gravitational potential 000 22.2 Coulomb’s law 000 17.5 Orbiting under gravity 000 22.3 Electric field strength for a radial field 000 17.6 The orbital period 000 22.4 Electric potential 000 17.7 Orbiting the Earth 000 22.5 Gravitational and electric fields 000 18 Oscillations 23 Capacitance 000 18.1 Free and forced oscillations 000 23.1 Capacitors in use 000 18.2 Observing oscillations 000 23.2 Energy stored in a capacitor 000 18.3 Describing oscillations 000 23.3 Capacitors in parallel 000 18.4 Simple harmonic motion 000 23.4 Capacitors in series 000 18.5 Representing s.h.m. graphically 000 23.5 Comparing capacitors and resistors 000 18.6 Frequency and angular frequency 000 23.6 Capacitor networks 000 18.7 Equations of s.h.m. 000 23.7 Charge and discharge of capacitors 000 18.8 Energy changes in s.h.m. 000 18.9 Damped oscillations 000 24 Magnetic fields and 18.10 Resonance 000 electromagnetism 000 24.1 Producing and representing magnetic fields 19 Thermal physics 000 000 19.1 Changes of state 000 24.2 Magnetic force 000 19.2 Energy changes 000 24.3 Magnetic flux density 000 19.3 Internal energy 000 24.4 Measuring magnetic flux density 000 19.4 The meaning of temperature 000 24.5 Currents crossing fields 000 19.5 Thermometers 000 24.6 Forces between currents 000 19.6 Calculating energy changes 000 24.7 Relating SI units 000 DRAFT24.8 Comparing forces in magnetic, electric 20 Ideal gases 000 and gravitational fields 000 20.1 Particles of a gas 000 20.2 Explaining pressure 000 25 Motion of charged particles 000 20.3 Measuring gases 000 25.1 Observing the force 000 20.4 Boyle's law 000 25.2 Orbiting charged particles 000
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25.3 Electric and magnetic fields 000 30 Medical imaging 000 25.4 The Hall effect 000 30.1 The nature and production of X-rays 000 25.5 Discovering the electron 000 30.2 X-ray attenuation 000 26 Electromagnetic induction 000 30.3 Improving X-ray images 000 30.4 Computerised axial tomography 000 26.1 Observing induction 000 30.5 Using ultrasound in medicine 000 26.2 Explaining electromagnetic induction 000 30.6 Echo sounding 000 26.3 Faraday’s law of electromagnetic induction 000 30.7 Ultrasound scanning 000 26.4 Lenz’s law 000 30.8 Positron Emission Tomography 000 26.5 Everyday examples of electromagnetic 31 Astronomy and cosmology 000 induction 000 31.1 Standard candles 000 27 Alternating currents 000 31.2 Luminosity and radiant flux intensity 000 27.1 Sinusoidal current 000 31.3 Stellar radii 000 27.2 Alternating voltages 000 31.4 The expanding Universe 000 27.3 Power and alternating current 000 P2 Planning, analysis 27.4 Rectification 000 and evaluation 000 28 Quantum physics 000 P2.1 Planning and analysis 000 28.1 Modelling with particles and waves 000 P2.2 Planning 000 28.2 Particulate nature of light 000 P2.3 Analysis of the data 000 28.3 The photoelectric effect 000 P2.4 Treatment of uncertainties 000 28.4 Threshold frequency and wavelength 000 P2.5 Conclusions and evaluation 28.5 Photons have momentum too 000 of results 000 28.6 Line spectra 000 Glossary 000 28.7 Explaining the origin of line spectra 000 28.8 Photon energies 000 Index 000 28.9 The nature of light: waves or particles? 000 28.10 Electron waves 000 28.11 Revisiting photons 000 29 Nuclear physics 000 29.1 Balanced equations 000 29.2 Mass and energy 000 29.3 Energy released in radioactive decay 000 29.4 Binding energy and stability 000 29.5 Randomness and radioactive decay 000 29.6 The mathematicsDRAFT of radioactive decay 000 29.7 Decay graphs and equations 000
29.8 Decay constant λ and half-life t1 000 2
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Chapter 27 Alternating currents
LEARNING OUTCOMES
In this chapter you will learn how to: • 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
I0 • distinguish between root-mean-square (r.m.s.) and peak values and recall and use Ir.m.s. = and V0 2 Vr.m.s. = for a sinusoidal alternating current 2 • distinguish graphicallyDRAFT 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 value of capacitance and the load resistance.
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BEFORE YOU START
• In pairs, try to recall and explain the relationship for power dissipation in terms of current, potential difference and resistance from Chapter 8. • The physics of alternating currents has similarities with simple harmonic motion (see Chapter 18). Discuss what you remember about period, frequency and angular frequency. • Write down what you know about the behaviour of diodes in circuits. What’s the most important property of a diode? • Discuss the discharge of a capacitor through a resistor. Can you remember the factors that affect the time constant of a circuit?
DESCRIBING ALTERNATING CURRENT
In many countries, mains electricity is a supply of alternating current (a.c.). The first mains electricity supplies were developed towards the end of the 19th century; at that time, a great number of different voltages and frequencies were used in different places. In some places, the supply was direct current (d.c.). Nowadays, this has been standardised across much of the world, with standard voltages of 110 V or 230 V (or similar), and frequencies of 50 Hz or 60 Hz. Mains electricity is transported along many kilometres of high-voltage power lines (cables). Transformers are used for stepping-up and stepping- down alternating voltages between the power Figure 27.1: This engineer is working on a transformer stations and the consumers (Figure 27.1). From your used for increasing (stepping-up) the size of the prior knowledge of transformers and transmission alternating voltage to help with the transportation of electrical energy, can you remember why it is of electrical energy. necessary for power lines to use high voltage?
18), and it can be interpreted in the same way. In a wire 27.1 Sinusoidal current with a.c., the free electrons within the wire move back and forth with s.h.m. The variation of the current with An alternating current can be represented by a graph such time is a sine curve, so it is described as sinusoidal. (In as that shown in Figure 27.2. This shows that the current principle, any current whose direction changes between varies regularly. During half of the cycle, the current is positive and negative can be described as alternating, positive, and in the other half it is negative. This means but we will only be concerned with those which have a that the direction of the current reverses every half cycle. regular, sinusoidal pattern.) Whenever you use a mains appliance, the charges (free electrons) within the wire andDRAFT appliance flow backwards and forwards. At any instant in time, the current has a KEY WORD particular magnitude and direction given by the graph. sinusoidal: having a magnitude that varies as a The graph has the same shape as the graphs used to sine curve. represent simple harmonic motion (s.h.m.) (see Chapter
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b Determine the time the current next has the / A 3 same value, but negative. 2 c State the time T for one complete cycle (the 1 period of the a.c). d Determine the frequency of this alternating 0 t / ms 5 10 15 20 25 30 35 current. –1 2 The following questions relate to the graph in Figure 27.2. –2 a Determine the values of I0 and ω. –3 b Write an equation to represent this alternating current. Figure 27.2: A graph to represent a sinusoidal alternating 3 An alternating current, measured in ampere (A), is current. represented by the equation: I = 5.0 sin (120πt)
a Determine the values of I0, ω, f and T. An equation for a.c. b Sketch a graph to represent the current. As well as drawing a graph, we can write an equation to represent alternating current. This equation gives us the value of the current I at any time t: 27.2 Alternating voltages
I = I0 sin ωt Alternating current is produced in power stations by large generators like those shown in Figure 27.3. where I is the current at time t, I0 is the peak value of the alternating current, and ω is the angular frequency of the supply, measured in rad s−1 (radians per second). The peak value is the maximum magnitude of the current. It’s very much like the ‘amplitude’ of the alternating current, except the unit is that of current. This is related to the frequency f in the same way as for s.h.m.: ω = 2πf and the frequency and period are related by: 1 f = T
KEY EQUATION
I = I0 sin ωt Figure 27.3: Generators in the generating hall of a large Remember that your calculator must be in the power station. radian mode when using this equation. As you have already seen in Chapter 26, a generator DRAFTconsists of a coil rotating in a magnetic field. An e.m.f. is induced in the coil according to Faraday’s and Lenz’s Questions laws of electromagnetic induction. 1 The following questions relate to the graph in This e.m.f. V varies sinusoidally, and so we can write an Figure 27.2. equation to represent it which has the same form as the a State the value of the current I and its equation for alternating current: direction when time t = 5 ms. V = V0 sin ωt
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where V0 is the peak value of the voltage. We can also a Determine the values of V0, ω and f for this represent this graphically, as shown in Figure 27.4. alternating voltage. b Calculate V when t = 0.002 s. (Remember V that 100πt is in radians when you calculate V0 this.) c Sketch a graph to show two complete cycles of this voltage. 0 T T 3T t 2 2 Measuring frequency and –V0 voltage An oscilloscope can be used to measure the frequency Figure 27.4: An alternating voltage. and voltage of an alternating current. Practical 27.1 explains how to do this. There are two types of oscilloscope. The traditional cathode-ray oscilloscope Question (CRO) uses an electron beam. The alternative is a 4 An alternating voltage V, in volt (V), is represented digital oscilloscope, which is likely to be much more by the equation: compact and which can store data and display the V = 300 sin (100πt) traces later.
PRACTICAL 27.1
Measurements using an oscilloscope may be many controls on a CRO, even more than those shown on the CRO illustrated in Figure 27.6. A CRO is an electron beam tube, as shown in Figure 25.4, but with an extra set of parallel plates to anode produce a horizontal electric field at right angles to vacuum the beam (Figure 27.5). X 2 Y The principles of a cathode-ray 2 X1 oscilloscope (CRO) Y1 The signal into the CRO is a repetitively varying heated cathode voltage. This is applied to the y-input, which deflects electron beam the beam up and down using the parallel plates electron gun screen Y1 and Y2 shown in Figure 27.5. The time-base produces a p.d. across the other set of parallel Figure 27.5: The construction of a cathode-ray plates X and X to move the beam from left to right 1 2 oscilloscope. Cathode rays (beams of electrons) are across the screen. produced in the electron gun and then deflected by When the beam hits the screen of the CRO, it electric fields before they strike the screen. produces a small spot of light. If you look at the screen and slow the movement down, you can see the spot The controls move from left to right, while the applied signal moves DRAFTThe X-shift and the Y-shift controls move the the spot up and down. When the spot reaches the right side of the screen, it flies back very quickly and whole trace in the x-direction and the y-direction, waits for the next cycle of the signal to start before respectively. The two controls that you must moving to the right once again. In this way, the signal know about are the time-base and the Y-gain, or is displayed as a stationary trace on the screen. There Y-sensitivity.
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CONTINUED Determining frequency and amplitude (peak value of voltage) If you look at the CRO trace shown in Figure 27.7, you can see that the amplitude of the waveform, or the peak value of the voltage, is equivalent brightness focus to 2 cm and the period of the trace is equivalent to 4 cm. time-base Y-gain If the Y-gain or Y-sensitivity setting is 2 V/cm, then the peak voltage is 2 × 2 = 4 V. If the time-base setting is 5 ms/cm, then the period is X-shi Y-shi 4 × 5 = 20 ms. on Y input In the example, since: o 1 frequency = Figure 27.6: The controls of a typical CRO. period 1 = You can see in Figure 27.6 that the time-base 00. 2 control has units marked alongside. Let us suppose = 50Hz that this reads 5 ms/cm, although it might be 5 ms/ division. This shows that 1 cm (or 1 division) on the x-axis represents 5 ms. Varying the time-base control alters the speed with which the spot moves across the screen. If the time-base is changed to 1 ms/cm, then the spot moves faster and each centimetre represents a smaller time. The Y-gain control has a unit marked in volts/cm, or sometimes volts/division. If the actual marking is 5 V/ cm, then each centimetre on the y-axis represents 5 V in the applied signal. 1 cm It is important to remember that on the CRO screen, Figure 27.7: A typical trace on the screen of a CRO. the x-axis represents time and the y-axis represents voltage.
Questions 27.3 Power and 5 The Y-sensitivity and time-base settings are 5 V/cm and 10 ms/cm. The trace seen on the CRO screen is alternating current the one shown in Figure 27.7. We use mains electricity to supply us with energy. If Determine the amplitude, period and frequency of the current and voltage are varying all the time, does the signal applied to theDRAFT Y-input of the CRO. this mean that the power is varying all the time too? 6 Sketch the CRO trace for a sinusoidal voltage of The answer to this is yes. You may have noticed that frequency 100 Hz and amplitude 10 V, when the some fluorescent lamps flicker continuously, especially time-base is 10 ms/cm and the Y-sensitivity if you observe them out of the corner of your eye or is 10 V/cm. when you move your head quickly from one side to the
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other. A tungsten filament lamp would flicker too, but The lamps in Practical 27.2 are the ‘resistive loads’. A the frequency of the mains has been chosen so that the full analysis, which we will come to shortly, shows that
filament does not have time to cool down noticeably Ir.m.s. is related to I0 by: between peaks in the supply. I0 Ir.m.s. = 2 ≈×0 707. I Root-mean-square (r.m.s.) 0 This is where the factor of 70% comes from. Note values that this factor only applies to sinusoidal alternating There is a mathematical relationship between the currents. peak value V0 of the alternating voltage and a direct We also have r.m.s. voltage V across the resistive load. voltage which delivers the same average electrical r.m.s. Vr.m.s is related to the peak voltage V0 by: power. The direct voltage is about 70% of V0. (You might have expected it to be about half, but it is more V0 Vr.m.s. = than this, because of the shape of the sine graph.) This 2 steady direct voltage is known as the root-mean-square (r.m.s.) value of the alternating voltage. In the same KEY EQUATIONS way, we can think of the root-mean-square value of an
alternating current, Ir.m.s. I0 Ir.m.s. = 2
KEY DEFINITION ≈×0.707 I0
root-mean-square (r.m.s.) value: the r.m.s. value where I0 is the peak (maximum) current. of an alternating current is that steady current V0 which delivers the same average power as the Vr.m.s. = a.c. to a resistive load. 2 ≈×0.707 V0
The r.m.s. value of an alternating current is that steady where V0 is the peak (maximum) voltage. current which delivers the same average power as the a.c. to a resistive load.
PRACTICAL 27.2
Comparing alternating current (a.c.) and direct (on the right) and the other to a d.c. supply (the current (d.c.) batteries on the left). The a.c. supply is adjusted so that the two lamps are equally bright, indicating Because the power supplied by an alternating that the two supplies are providing energy at the current is varying all the time, we need to have some same average rate. The output voltages are then way of describing the average power which is being compared on the double-beam oscilloscope. supplied. To do this, we compare an alternating current with a direct current, and try to find the A typical trace is shown in Figure 27.9. This shows direct current that supplies the same average power that the a.c. trace sometimes rises above the steady as the alternating current. d.c. trace, and sometimes falls below it. This makes sense: sometimes the a.c. is delivering more power Figure 27.8 shows how this can be done in practice. DRAFTthan the d.c., and sometimes less, but the average Two filament lamps (our resistive loads) are placed power is the same for both. side by side; one is connected to an a.c. supply
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CONTINUED
Figure 27.9: The oscilloscope trace from the experiment shown in Figure 27.8.
Figure 27.8: Comparing direct and alternating currents that supply the same power. The lamps are equally bright.
calculate the average power dissipated in a resistor, we Questions can use the usual formulae for power: 7 The alternating current (in ampere, A) in a resistor V 2 is represented by the equation: I = 2.5 sin (100πt) PI=2RI = V = R Calculate the r.m.s. value for this alternating current. Remember that it is essential to use the r.m.s. 8 The mains supply to domestic consumers in many values of I and V, as in Worked example 1. If you European countries has an r.m.s. value of 230 V use peak values, your answer will be too great by a for the alternating voltage. (Note that it is the factor of 2. r.m.s. value which is generally quoted, not the peak Where does this factor of 2 come from? Recall that r.m.s. value.) and peak values are related by: Calculate the peak value of the alternating II= 2 voltage. 0 r.m.s 2 So, if you calculate I R using I0 instead of Ir.m.s., you will 2 Calculating power introduce a factor of ()22≡ . The same is true if you calculate power using V0 instead of Vr.m.s.. It follows that, The importance of r.m.s. values is that they allow us for a sinusoidal alternating current, peak power is twice to apply equations from our study of direct current average power. to situations where the current is alternating. So, to
WORKED EXAMPLE
1 A 20 Ω resistor is connected to an alternating Step 1 Calculate the r.m.s. value of the voltage. supply. The voltage acrossDRAFT the resistor has peak V0 value 25 V. Vr.m.s = 2 Calculate the average power dissipated in the 25 = resistor. 2 = 17 7. V
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CONTINUED
Step 2 Now calculate the average power Note that, if we had used V0 rather than V r.m.s., dissipated. (Remember you must use the we would have found: r.m.s. value, and not the peak value.) 252 P = V 2 P = 20 R = 31.W 3 17. 72 = 20 which is double the correct answer. = 15.W 6
Questions 0 9 Calculate the average power dissipated in a resistor of resistance 100 Ω when a sinusoidal alternating 0 t current has a peak value of 3.0 A. 10 The sinusoidal voltage across a 1.0 kΩ resistor has a peak value 325 V. – 0 2 a Calculate the r.m.s. value of the alternating 0 2 2 voltage. < 2> b Use V IR to calculate the r.m.s. current in the = 2 resistor. 0 2 c Calculate the average power dissipated in the resistor. 0 t d Calculate the peak power dissipated in the resistor. Figure 27.10: An alternating current I is alternately positive and negative, while I2 is always positive.
Explaining root-mean-square Now, if we consider
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Summarising this process: to find the r.m.s. value of the This type of rectification is known as half-wave current, we find the root of the mean of the square of rectification. For one-half of the time the voltage is zero, the current–hence r.m.s. and this means that the power available from a half-wave rectified supply is reduced.
27.4 Rectification The bridge rectifier Many electrical appliances work with alternating To overcome this problem of reduced power, a bridge current. Some, like electrical heaters, will work rectifier circuit is used. This consists of four diodes equally well with d.c. or a.c. However, there are many connected across the input alternating voltage, as shown appliances, such as electronic equipment, which in Figure 27.12. The output voltage Vout is taken across require d.c. For these, the alternating mains voltage the load resistor R. The resulting output voltage across must be converted to direct voltage by the process of the load resistor R is full-wave rectified. rectification. The way in which this works is shown in Figure 27.13. KEY WORD • During the first half of the positive cycle of the input alternating voltage, the terminal A rectification: the process of converting is positive. Current flows through diode 2, alternating voltage into direct voltage. downwards through the load resistor R, through diode 3 and back the supply via terminal B. In this half of the cycle, current cannot flow through diode diodes 1 or 4 because they are pointing the wrong way (reverse bias). • In the second half of the negative cycle of load the input alternating voltage, the terminal B a.c. supply ~ resistor Vout R is positive. Current flows through diode 4, downwards through the load resistor R, through diode 1 and back to the supply via terminal A. Diodes 2 and 3 do not conduct because they are pointing the wrong way. Vout
t 1 2 A Figure 27.11: Half-wave rectification of a.c. requires a a.c. supply ~ single diode. B
A simple way to do this is to use a diode, which is a 3 4 component that will only allow current in only one R Vout direction. (You have already met diodes in Chapter 10.) Figure 27.11 shows a circuit for doing this. An alternating input voltage is applied to a circuit with a diode and a resistor in series. The diode will only V conduct during the positive cycles of the input voltage. out Hence, there will be a currentDRAFT in the load resistor only during these positive cycles. The output voltage Vout t across the resistor will fluctuate as shown in theV out against time t graph. This graph is identical to the input Figure 27.12: Full-wave rectification of a.c. using a diode alternating voltage, except the negative cycles have been bridge. ‘chopped-off’.
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a necessary in the circuit. This capacitor, of capacitance C, is in parallel with the load resistor of resistance 2 R. This is shown in Figure 27.14. The idea is that the + A capacitor charges up and maintains the voltage at a high level. It discharges gradually when the rectified voltage – B 3 drops, but the voltage soon rises again and the capacitor + charges up again. The result is an output voltage with ‘ripple’. R
b Vin ~ C R Vout 1 – A
+ B + 4 C charging C discharging R V out
t
Figure 27.13: Direction of current during full-wave Figure 27.14: A smoothing capacitor is connected across rectification a for positive cycles and b for negative cycles. (in parallel with) the load resistor.
Note that in both halves of the cycle, current direction The amount of ripple can be controlled by carefully in the load resistor R is always the same (downwards). choosing the capacitance C of the capacitor and the This means that the top end of R must always be resistance R of the load resistor. A capacitor with a positive. large capacitance value discharges more slowly than a capacitor with a small capacitance value, so will give You can construct a bridge rectifier using light-emitting a smaller ripple. Similarly, if the resistance R of the diodes (LEDs) which light up when current flows resistor is increased, then this too leads to a slower through them. By connecting this bridge to a slow a.c. discharge of the capacitor. You may have already met supply (for instance 1 Hz from a signal generator), you the physics of discharging capacitors in Chapter 23. can see the sequence in which the diodes conduct during So, the size of the ripple can be reduced by increasing rectification. the time constant CR of the capacitor–resistor circuit. Ideally, though this is definitely not a general rule,CR must be much greater than the time interval between Question the adjacent peaks of the output signal–you want the 11 Explain why, when terminal B in Figure 27.13 is capacitor to be still discharging between the ‘gaps’ positive (during the negative cycle), the current between the positive cycles. This is illustrated in Worked flows through diodes 1 and 4, but not through example 2. diodes 2 and 3. Note that, in Figures 27.11 to 27.14, we have represented DRAFTthe load on the supply by a resistor. This represents any components that are connected to the supply. For Smoothing example, a rectifier circuit can be used to charge the In order to produce steady d.c. from the ‘bumpy’ d.c. battery of a mobile phone or provide a direct voltage that results from rectification, a smoothing capacitor is supply for small radio.
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WORKED EXAMPLE 2 Figure 27.15 shows the output voltage from a Step 2 Compare the time constant with the time half-wave rectifier. The load resistor has resistance interval between the adjacent peaks of 1.2 kΩ. A student wishes to smooth the output the output signal. voltage by placing a capacitor across the load resistor. The time constant of 0.012 ms is very small compared with time interval of V 40 ms between the adjacent peaks of the output. If this capacitor were to be used, 0 it would discharge far too quickly. There t/ms 0 50 100 would be no smoothing of the output Figure 27.15: Output from a half-wave rectifier. voltages–the 10 pF capacitor is not suitable.
With the help of a calculation, suggest if a 10 pF Step 3 Repeat the steps for the 500 µF capacitor. capacitor or a 500 µF capacitor would be suitable 6 for this task. Time constant = CR = 500 × 10− × 1.2 × 103 = 0.60 s (= 600 ms) Step 1 Calculate the time constant with the 10 pF capacitor. Now, the time constant of 600 ms is much larger than 40 ms. This capacitor will not Time constant = CR = 10 × 10−9 × 1.2 × 103 discharge completely between the positive = 1.2 × 10−5 s (= 0.012 ms) cycles of the half-wave rectified signal. The 500 µF capacitor would be adequate for the smoothing task.
Questions 1 2 12 Sketch the following voltage patterns: A V a a sinusoidal alternating voltage in ~ B b the same voltage as part a, but half-wave 3 rectified 4 R Vout c the same voltage as part b, but smoothed d the same voltage as part a, but full-wave rectified e the same voltage as part d, but smoothed. Figure 27.16: A bridge rectifier circuit that is wired 13 A student wires a bridge rectifier incorrectly as incorrectly. For Question 13. shown in Figure 27.16. Explain what you would expect to observe when an oscilloscope is connected across the load resistor R. 14 A bridge rectifier circuit is used to rectify an alternating current through a resistor. A smoothing capacitor is connected DRAFTacross the resistor. Figure t 27.17 shows how the current varies. Use sketches to show the changes you would expect: Figure 27.17: A smoothed, rectified current. For a if the resistance R of the resistor is increased Question 14. b if the capacitance C of the capacitor is decreased.
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REFLECTION
Without looking at your textbook, summarise all the key equations from this chapter. Make a list of mains operated devices in your laboratory. For each device, determine the power, r.m.s. current and r.m.s. voltage. Give yourself and a classmate one minute to draw a circuit diagram for a full-wave rectifier circuit. Compare your circuit diagrams. Which diagram was more accurate? How would you make this diagram more accurate if you were to draw it in the future?
DRAFT
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EXAM-STYLE QUESTIONS 1 The maximum power dissipated in a resistor carrying an alternating current is 10 W. What is the mean power dissipated in the resistor? [1] A 5.0 W B 7.1 W C 10 W D 14 W 2 The alternating current I in ampere (A) in a filament lamp is represented by the equation: I = 1.5 sin (40t). Which of the following is correct? [1] A The angular frequency of the alternating current is 40 rad s−1. B The frequency of alternating current is 40 Hz. C The maximum current is 3.0 A. D The peak voltage is 1.5 V. 3 Write down a general expression for the sinusoidal variation with time t of: a an alternating voltage V [1] b an alternating current I (you may assume that I and V are in phase) [1] c the power P dissipated due to this current and voltage. [1] 4 The alternating current I in ampere (A) in a circuit is represented by the equation: I = 2.0 sin (50πt). a State the peak value of the current. [1] b Calculate the frequency of the alternating current. [2] c Sketch a graph to show two cycles of the variation of current with time. Mark the axes with suitable values. [2]
d Calculate Ir.m.s., the r.m.s. value of current, and mark this on your graph in part c. [1]
e Determine two values of time t at which the current I = Ir.m.s.. [3] 5 A heater of resistance 6.0 Ω is connected to an alternating current supply. The output voltage from the supply is 20 V r.m.s. Calculate: a the average power dissipated in the heater [2] b the maximumDRAFT power dissipated in the heater [1] c the energy dissipated by the heater in 5.0 minutes. [2]
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CONTINUED 6 An oscilloscope is used to display the variation of voltage across a 200 Ω resistor with time. The trace is shown. The time-base of the oscilloscope is set at 5 ms div−1 and the Y-gain at 0.5 V div−1.
1 division
Determine: a the period and hence the frequency of the alternating voltage [2] b the peak voltage and hence the r.m.s. voltage [2] c the r.m.s. current in the resistor [1] d the mean power dissipated in the resistor. [2]
7 a State the relationship between the peak current I0 and the r.m.s. current Irms for a sinusoidally varying current. [1] b The current in a resistor connected to a steady d.c. supply is 2.0 A. When the same resistor is connected to an a.c. supply, the current in it has a peak value of 2.0 A. The heating effects of the two currents in the resistor are different. i Explain why the heating effects are different and state which heating effect is the greater. [2] ii Calculate the ratio of the power dissipated in the resistor by the d.c. current to the power dissipated in the resistor by the a.c. current. [2] 8 A sinusoidal voltage of 6.0 V r.m.s. and frequency 50 Hz is connected to a diode and a resistor R of resistance 400 Ω as shown in the diagram.
6.0 V 400 Ω R DRAFTr.m.s.
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CONTINUED a Sketch a graph showing the variation with time of both the supply waveform (use a dotted line) and the voltage across R (use a solid line). Put numerical scales on both the voltage and time axes. [4] b An uncharged capacitor C is connected across R. When the 6.0 V r.m.s. supply is switched on, the capacitor charges fully during the first quarter of a cycle. You may assume that the p.d. across the diode is zero when it conducts. For the next three-quarters of the first cycle, the diode stops conducting and the p.d. across R falls to one-half of the peak value. During this time the mean p.d. across R is 5.7 V. For the last three-quarters of the first cycle, calculate: i the time taken [1] ii the mean current in R [2] iii the charge flowing through R [2] iv the capacitance of C. [2] c Explain why the diode stops conducting during part of each cycle in part b. [2] 9 The rectified output from a circuit is connected to a resistor R of resistance 1000 Ω. Graph A shows the variation with time t of the p.d. V across the resistor. Graph B shows the variation of V when a capacitor is placed across R to smooth the output.
V / V V / V
1.0 1.0
0.8 0.8
0.6 0.6
0.4 0.4
0.2 0.2
0 0 0 0.01 0.02 0.03 t / s 0 0.01 0.02 0.03 0.04 t / s Graph A Graph B
Explain how the rectification is achieved. Draw a circuit diagram to show the components involved. [6] b Explain the action of the capacitor in smoothing the output. [3] c Using graphDRAFT B between t = 0.005 and t = 0.015 s, determine: i the time during which the capacitor is charging [1] ii the mean value of the p.d. across R [1] iii the average power dissipated in R. [2]
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CONTINUED 10 Electrical energy is supplied by a high-voltage power line which has a total resistance of 4.0 Ω. At the input to the line, the root-mean-square (r.m.s.) voltage has a value of 400 kV and the input power is 500 MW. a i Explain what is meant by root-mean-square voltage. [2] ii Calculate the minimum voltage that the insulators which support the line must withstand without breakdown. [2] b i Calculate the value of the r.m.s. current in the power line. [2] ii Calculate the power loss on the line. [2] iii Suggest why it is an advantage to transmit the power at a high voltage. [2] 11 A student has designed a full-wave rectifier circuit. The output voltage for this circuit is taken across a resistor of resistance 120 Ω. The variation of the output voltage with time is shown.
V/V 100 50 0 t/ms 0 10 20 30 40
A capacitor is now connected across the resistor. The graph shows the new variation of the output voltage with time.
V/V B 100 A C 50 0 t/ms 0 10 20 30 40
a Explain the variation of the output variation between points: i AB [1] ii BC. [1] b Use the second graph to determine the value of the capacitance C. [3] t − (You may use the equation V =Ve CR from Chapter 23.) DRAFT0
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SUMMARY
A sinusoidal alternating current can be represented by I = I0 sin ωt, where I0 is the peak value of the current. The root-mean-square (r.m.s.) value of an alternating current is that steady current which delivers the same average power as the a.c. to a resistive load; for a sinusoidal a.c.:
I0 Ir.m.s. = 2
≈×0 707. I0
The relationship between root-mean-square (r.m.s.) voltage Vr.m.s and peak voltage V0 is:
V0 Vr.m.s. = 2
The power P dissipated in a resistor can be calculated using the equations: V 2 P = VI, P = I 2R and P = R where V and I are the r.m.s. values of the voltage and current respectively. A single diode is used for the half-wave rectification of an alternating current. Four diodes (bridge rectifier) are used for the full-wave rectification of an alternating current. A capacitor placed in parallel with a resistive load will smooth the rectified alternating voltage. The greater the time constant CR of the capacitor-resistor network, the smaller is the size of the ripple.
SELF-EVALUATION
Needs Almost Ready to I can more work there move on understand the terms period, frequency and peak value as applied to an alternating current or voltage
use the equations I = I0 sin ωt and V = V0 sin ωt for sinusoidally alternating current and voltage respectively understand that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current understand root-mean-square (r.m.s.) and peak values recall and use: DRAFT I0 V0 Ir.m.s. = and Vr.m.s. = 2 2
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CONTINUED
Needs Almost Ready to I can more work there move on understand half-wave and full-wave rectification explain how a single diode produces half-wave rectification explain how four diodes (bridge rectifier) produce full- wave rectification understand smoothing capacitors, and understand how smoothing effects are governed by capacitance of the smoothing capacitor and the resistance of the load resistor.
DRAFT
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