Bst Handout e03

ROYAL SCHOOL OF ARTILLERYBASIC SCIENCE & TECHNOLOGY SECTION

GUNNERY CAREERS COURSES

Alternating Current Theory

DWR Alternating Current Theory

E.03-1 E03_AC_Electricity.QXD

INTRODUCTION

The vast majority of power generation and distributionsystems operate using alternating current. This

includes military systems, such as Rapier, where smallpetrol or diesel engines are used to turn a machinecalled an ‘alternator’ to convert mechanical energy intoelectrical energy. This handout covers the basic proper-ties of alternating current systems.

DIRECT AND ALTERNATING CURRENT

Direct Current is produced by batteries and by gen-erators that are turned by steam turbine (power sta-

tion), piston engine (portable) or muscle power. TheEMF acts in one direction, only, and the current flow isfairly steady, in one direction. (Like the motion of thechain of a bicycle.)

Alternating current (AC) is also made by a rotatingelectrical machine, but here, the EMF reverses directionfor each half-turn of the machine. (Like the motion ofyour feet as you pedal a bicycle.) The ‘waveform’ of analternating current is a graph showing the variation involtage or current with time. The standard waveform ofthe European mains supply is shown in Figure One. It iscalled alternating current because the direction of theEMF and the flow of current alternates between positiveand negative values.

The European mains supply takes 20 milli-seconds(20 ms) to complete one cycle of the waveform and thisis called the ‘period’ or ‘periodic time’ of the alternatingcurrent. The symbol for the period is ‘τ’, pronounced ‘tor’.

8 Oct 02

There are 1 000 milli-seconds in one second - if eachcycle takes 20 ms then there must be 1000 ÷ 20 = 50cycles of the waveform in each second. This is calledthe ‘frequency’ of the alternating current, symbol ‘f’. Thefrequency and the period are calculated using the fol-lowing formulae:

f = 1 and τ = 1τ f

The unit of frequency is the Hertz, symbol ‘Hz’,although you might see, in some older books, frequencymeasured in cycles per second, cps or c/s.

AC MEASUREMENTS

Any attempt to measure the voltage or current of anac supply must take into account its constantly

changing values. The waveform in Figure One wouldgive a voltage of zero if it were measured at a time of 10ms, + 325 V (positive peak value) if it were measured attime 5 ms and – 325 V (negative peak value) if meas-ured at time 15 ms. It is no use taking an average –because the average is zero as there are equal amountsof positive and negative voltage.

The solution to the measurement of ac voltage orcurrent is to calculate the number of volts or amps ofdirect current that would produce the same power (in aheater or light bulb, for example). This figure is calledthe ‘root mean square’ of the alternating current or volt-age and is calculated using the following formulae:

Vrms = Vpeak

√2Ιrms = Ιpeak

√2

This means that a 12 V light bulbwill shine with the same brightness on12 V AC as it does on 12 V DC. Therms value is also called the ‘dc-equiva-lent’ value of the alternating voltage orcurrent. For European mains, the peakis 325 V and the rms is 230 V. (For aquick calculation, the rms is about 70%of the peak.) All AC measurements areassumed to be rms unless otherwisestated.

The mains supply in the USA is110 V and 60 Hz. The period is 1/60

sec or 16.7 ms. Industrial supplies inFigure 1: The Standard Waveform of the AC Mains

ms

V

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the UK are 400 V 50 Hz whilst many military systemsuse 400 Hz.

AC IN A RESISTIVE CIRCUIT

When alternating current is flowing in a resistor,there are two waveforms: voltage and current.

The two waveforms have an identical shape and theyrise and fall in step. At all times, the current is governedby Ohm’s Law and is equal to EMF ÷ Resistance. Thisis illustrated in Figures Two (a) and Two (b) for a 10 VAC supply connected to a 5 W resistor.

Using Ohm’s Law, Ι = V/R, the current will be 2 A. Ofcourse, the current will be 2 A rms because we used 10V rms to calculate it. The current rises as high as 2.83 A(√2 × 2 A) when the voltage is 14.14 V.

The power in an electrical circuit is found by multi-plying Volts by Amps. In the case of AC - this must becalculated over one cycle and the graphical result isshown in Figure 2 (c) where each point on the graph hasbeen obtained by multiplying the corresponding valuesof Voltage (Figure 2 a) and Current (Figure 2 b).

You will see that the power is always positivebecause when the voltage is negative the current is alsonegative and, when multiplied as two negative numbers,their product is positive.

This means that, for example, during the first half-cycle of the supply a lamp would receive electricalpower and emit light. During the second half-cycle, thevoltage and current both reverse and the lamp againemits light. If this were not so then, during the secondhalf-cycle, the lamp would have to suck back the lightand change it back into electricity again - obviously thisdoes not happen with ordinary light bulbs.

The average power over the whole cycle of the waveis 20 W (half the peak) - this can be deduced fromexamining the graph of Figure Two (c) which is symmet-rical about the 20 W line. This average of 20 W is exact-ly what you would have got by using the DC formula forpower:

Power = Volts × Amps

provided that the rms values of 10 V and 2 A are usedin the calculation. Remember that the rms values are theeffective values of the alternating voltage and current.

Example:

1. A supply of 24 V AC is connected to a lamp of resist-ance 10 W. Calculate the current and power.

Solution: all values are rms so the DC formulae canbe used.

Ι = V/R = 24 ÷ 10 = 2.4 A

Pwr = V × I = 24 × 2.4 = 57.6 W.

2. What is the peak value of a 12 V AC supply?

Solution: use the standard formula:

Vpeak = Vrms × √2= 12 × 1.414= 17 V

[Note: all graphs in this handout have been produced usingExcel. You may request a file copy from BST Staff.]

[Try self-test questions 1 - 5 now.]


8 Oct 02E03 AC Electricity.QXD

ms

ms

ms

V

A

W

Figure 2a: The Waveform of 10 V, AC

Figure 2b: The Waveform of 2 A, AC

Figure 2c: The Waveform of 20 W, AC

There are no negative values!

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AC IN AN INDUCTIVE CIRCUIT

When an inductor is used with direct current, theeffect of the induced EMF is to delay the rise of

current when the circuit is switched on and to try tomaintain the current when the circuit is switched off.(This is covered in another handout.)

When an inductor is used in a circuit with an alternat-ing current then, since the current is constantly chang-ing, the induced EMF will be present all the time - notjust at the instant of switching on or off.

The induced EMF is always opposing the changethat is causing it. For this reason, it is often called a‘back’ EMF. The effects of this back EMF on both DCand AC include:

• If the current is increasing then the induced EMF willtry to prevent that increase. As this occurs, the mag-netic field gets stronger and takes energy from theEMF - leaving less to make the current flow.

• If the current is decreasing then the induced EMFwill try to keep it flowing. As this occurs, the magnet-ic field gets weaker as its energy is released to main-tain current flow.

There are two main effects of the back EMF on AC,where it continually opposes the changing current.These are:

• to oppose the flow of current.• to cause the current to have a phase lag of 90° com-

pared to the voltage.

The amount of opposition to current flow increaseswith frequency because the changes that cause it occurat a faster rate. Unlike the opposition to current flow thatoccurs in a resistor - which is the same at all frequen-cies, this opposition varies with frequency. Although sim-ilar to resistance and measured in Ohms, it is differentand is called ‘Reactance’, symbol ‘X’.

The reactance of an inductor can be calculatedusing the formula:

XL = 2 π f L

Where ‘f’ is the frequency and L the inductance inHenries. For example, an inductor of 10 mH at a fre-quency of 50 Hz will have a reactance of 3.14 Ω.

The set of diagrams in Figure Three shows the volt-age, current and power values for an alternating supplyof 10 V connected to an inductor of reactance 5 Ω. Thefollowing points should be noted:

• The graph of voltage, Figure Three (a), is identical tothat of Figure Two (a) used for a resistor.

• The graph of current, Figure Three (b), shows aphase lag of 90° to the voltage but is otherwise iden-tical to that of Figure Two (b).

• The graph of power, Figure Three (c), is very differ-ent from its equivalent graph for the resistor. Thephase angle between current and voltage has re-

aligned the values so that some of the power valuesare negative. The average power is now zero.

The power in an inductor is zero, when averaged overone cycle because the inductor can both store energy in,and release energy from,its magnetic field. Energy stor-age is is indicated by the positive parts of the graph inFigure Three (c). Energy release is indicated by the neg-ative parts of the graph. Taken over a cycle, it is true tosay that any energy that comes from the supply is stored


E03_AC_Electricity.QXD8 Oct 02

ms

ms

ms

V

A

W


Figure 3b: The Waveform of 2 A, AC, 90° lagging

Figure 3c: Waveform of Inductive Reactive Power

Equal amounts of positive & negative values!

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in the inductor’s magnetic field and then returned. Thus,the power is zero and no heat is produced.

This is another difference between reactance andresistance: the resistor converts energy from electricalto thermal energy (heat) and this is a one-way process.The inductor converts electrical energy into magneticenergy and back again - this is a two-way process andthe energy is not converted to heat. (In practice, theinductor is not perfectly efficient and a small proportionof the energy is wasted as heat.)

Calculation of Reactance: The reactance of aninductor can be calculated using the standard formula:

XL = 2 π f L

where ‘f’ is the frequency and ‘L’ the inductance, inHenries (H). For example, an inductor of 10 mH wouldhave a reactance of 3.14 Ω at 50 Hz (2 × 3.14 × 50 ×0.01) and a reactance of 6.28 Ω at 100 Hz, and so on.(Actually, the many turns of copper wire that make upthe coil of the inductor usually have some resistance,but this can be ignored in our simplified view of AC cir-cuits.)

Variation of Reactance with Frequency: The for-mula for reactance (see above) matches the pattern ofequations that produces a straight-line when plotted asa graph. The reactance is proportional to the frequency.This means that, for example, if the frequency is dou-bled then the reactance will double, etc. The graph ofreactance versus frequency, between Zero and 5 kHz,for an inductor of 10 mH is shown in Figure Four.

Inductor in Series with a Lamp: A series circuitwith a lamp and inductor will show a variation of currentwith frequency because the inductive reactance varieswith frequency. Using the same lamp that we had in theDC Handout (6 V, 40 mA, 150 Ω) and using a 10 mHinductor whose variation of reactance is illustrated inFigure Four, the effects of different frequencies (whichyou will see as a demonstration in the classroom) are asfollows:

• At DC (Zero Hz) the reactance of any inductor isalways zero and the resistance of the circuit is dueto the lamp, 150 Ω. The resistor dominates the cir-cuit. The lamp will be bright.

• At frequencies higher than about 5 kHz, the induc-tor’s reactance is several hundreds of Ohms and the150 Ω of the resistor doesn’t make much difference- the inductor dominates the circuit because it hasmost of the Ohms. The current is small and the lampwill be out because the number of Ohms is large.

• At frequencies around 2.5 kHz, the inductive reac-tance and the resistance are similar and the lamp isnoticeably dimmer. As the frequency increasesabove 2.5 kHz, the lamp gets progressively lessbright.

Because the reactance of an inductor increases withfrequency, it can be used as a ‘filter’ to allow low fre-quencies to pass whilst blocking high frequencies. Twosimple applications of this are:

• in two-way loudspeakers, the ‘woofer’, or largerloudspeaker, has an inductor wired in series with itso that only the lower frequencies (the bass notes)can pass through to reach the loudspeaker.

• many computer data cables (e.g. key-board to PC, PC to monitor) have a lump ofmagnetic material (ferrite) clamped aroundthem. This increases the inductance of thecable, even though it has only one turn ofwire, sufficiently to block the passage ofhigh frequencies from nearby radios, cell-phones, etc., that might interfere with thedata in the wire.




Figure 4: Variation of Inductive Reactance with Frequency.10 mH Inductor from Zero to 5 000 Hz.

ΩΩ

Hz

10 mH6 V

150 Ohms

Figure 5: A Lamp in Series with an Inductor

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AC IN A CAPACITIVE CIRCUIT

When a capacitor is used with direct current, theeffect of the capacitor is to delay the rise of volt-

age (PD) across the capacitor when the circuit isswitched on and to try to maintain the voltage (becausethe capacitor is charged) when the circuit is switched off.The current flows only temporarily, whenever the circuitis switched on or off. (This is covered in another hand-out.)

When a capacitor is used in a circuit with alternatingcurrent then, since the voltage is constantly changing,the capacitor will be always charging or discharging -not just when switching on or off.

The charge in the capacitor causes a displacementof the electric charge in its dielectric - a bit like stretch-ing an elastic band. This produces a voltage that oppos-es the changes that are taking place in the alternatingsupply. The effects of this, on both DC and AC includethe following:

• If the EMF is increasing then the displacementincreases and absorbs energy from it. This delaysthe rise of voltage (PD) across the capacitor.

• If the EMF is decreasing then the displacementdecreases and releases energy to delay the fall ofvoltage across the capacitor.

There are two main effects of the storage andrelease of energy by the displacement current in analternating current circuit:

• it allows alternating current to flow apparentlythrough an insulator. In fact, it does not flow through- but it does displace from side to side, between theplates of the capacitor. Since charge is moving thena current must be flowing.

• it causes the voltage to have a phase lag of 90° com-pared to the current.

Reactance: although a capacitor allows some alter-nating current to flow though it, there is some oppositionbecause the displacement of charge requires energy.The energy is returned later and so, just as with theinductor, the opposition is called a reactance. It is meas-ured in Ohms and given the symbol ‘XC’. The reactanceof a capacitor varies with frequency in the opposite wayto that of an inductor.

The set of diagrams in Figure Six shows the voltage,current and power values for an alternating supply of10 V connected to a capacitor of reactance 5 Ω. Thesegraphs should be compared to those of Figures Two andThree and the following points noted:

• The top graph, the EMF, is the same for all - as thisis the supply voltage.

• The graph of current, Figure Six (b), shows a lead of90° compared to the voltage and is the inverse (180°out of phase) of the inductor’s current.

• The graph of power again shows an average over acycle of zero. It is also the inverse of the correspon-

ding graph for the inductor. (In other words: when theinductor releases energy then the capacitor absorbsenergy, and vice versa.)

The power in a capacitor is zero, when averagedover a cycle of the alternating current because thecapacitor can both store energy in, and release energyfrom, the displaced charge in its dielectric. Energy stor-age is indicated by the positive parts of the graph inFigure Six (c). Energy release is indicated by the nega-tive parts of the graph. Taken over a cycle, it is true to




Figure 6b: The Waveform 2A, AC, 90° Lead

Figure 6c: Waveform of Capacitive Reactive Power

V

A

W

ms

ms

ms

Equal amounts of positive & negative values!

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say that any energy that comes from the supply is storedin the capacitor and then returned to the supply. Thus,the power is zero and no heat is produced. (As with theinductor, the conversion process is not perfectly efficientand a small proportion of the energy is wasted as heat.)

Calculation of Reactance: The reactance of acapacitor can be calculated using the formula:

XC = 12 π f C

Where ‘f’ is the frequency and ‘C’ the capacitance inFarads. For example, a capacitor of 1 μF at a frequencyof 50 Hz will have a reactance of 3.18 kΩ. The samecapacitor would have a reactance of 1.09 kΩ at 100 Hzand so on.

Variation of Reactance with Frequency: The for-mula for reactance (see above) matches the pattern ofequations that produces a hyperbola when plotted as agraph. The reactance is inversely proportional to the fre-quency. This means that, for example, if the frequencyis doubled then the reactance will halve, etc. The graphof reactance versus frequency, between Zero and5 kHz, for an capacitor of 1 μH is shown in FigureSeven.

Capacitor in Series with a Lamp: A series circuitwith a lamp and capacitor will show a variation of currentwith frequency because the capacitive reactance varieswith frequency. Using the circuit of Figure Eight with thesame lamp that we had in the DC Handout (6 V, 40 mA,150 Ω) and using a 1 μF capacitor whose variation ofreactance is illustrated in Figure Seven, the effects ofdifferent frequencies (which you will see as a demon-stration in the classroom) are as follows:

• At DC (Zero Hz) the reactance of any capacitor isalways infinite and the resistance of the circuit isinfinity. The 150 Ω of the lamp makes no differenceas the reactance of the capacitor dominates the cir-cuit. The lamp will be off.

• At frequencies higher than about 5 kHz, the capaci-tor’s reactance is several tens of Ohms and the150 Ω of the resistor now dominates the circuitbecause it has most of the Ohms. The current ismaximum and the lamp will be brightly lit.

• At frequencies around 1 kHz, the capacitive reac-tance and the resistance of the lamp are similar andthe lamp is noticeably dimmer. As the frequencyincreases above 1 kHz, the lamp gets progressivelybrighter.

Because the reactance of an capacitor decreaseswith frequency, it can be used as a ‘filter’ to allow highfrequencies to pass whilst blocking low frequencies.Two simple applications of this are:

• in two-way loudspeakers, the ‘tweeter’, or smallerloudspeaker, has an capacitor wired in series with itso that only the higher frequencies (the treble notes)can pass through to reach the loudspeaker.

• to suppress radio interference in a car’s electricalcircuits, capacitors are connected between the ‘live’wire and earth. At high frequencies, the reactance isvery low and the interference is conducted to earth.The car’s main electrics run from direct currentwhich, apart from the instant of switching on, cannot

pass through the capacitor.

With all real components, there will beminor imperfections that cause deviationsfrom the theoretically perfect situation. Forexample, to reduce weight, the conductingplates of real capacitors are made from verythin metal foil. This thin conductor has asmall amount of resistance which will act inseries with the capacitor’s reactance. Minorimperfections in the dielectric will allow asmall current to leak through - effectivelyplacing a very high resistance in parallelwith the reactance. These points areignored in this simple treatment of capaci-tive circuits.

MEASURING REACTANCE

The reactance of a capacitor or an induc-tor can be found by measuring the rms

values of voltage and current and using the



Figure 7: Variation of Capacitive Reactance with Frequency.1 μF Capacitor from Zero to 5 000 Hz.

Ω

Hz

1 Micro-Farad

6 V

150 Ohms

Figure 8: A Lamp in Series with a Capacitor

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reactive version of Ohm’s Law to calculate the numberof Ohms:

X = VrmsIrms

Here, the symbol ‘X’ is used, without a subscript ‘L’or ‘C’, to show that it applies to either an inductor or acapacitor.

COMBINATIONS OF COMPONENTS

When a series circuit contains some resistance(phase angle zero degrees) and some inductance

(phase angle 90° current lagging) or capacitance (phaseanlge 90° current leading) then, since the current mustbe the same all around the circuit, some balance mustbe reached. In such cases, the phase angle betweensupply voltage and current will be somewhere between0° and 90° depending on the circumstances. In thiscase, the total opposition to current flow is called‘Impedance’ (symbol ‘Ζ’) and may be made up from anycombination of resistance, capacitance and inductance.Impedance generally applies to practical circuits whereseveral elements are present. It is measured in Ohmsand calculated for a circuit from measurements of sup-ply volts and supply current, using Ohm’s Law:

Ζ = Supply VrmsSupply Irms

The analysis of these circuits is beyond the scope ofthis course - apart from one special case called ‘reso-nance’, which is covered in the next section.

Most real circuits contain various amounts of resist-ance, inductance and capacitance - so it is common torefer to the ‘Impedance’ of a circuit rather than its‘Resistance’. The value of the impedance will usuallyvary with frequency and it is usually quoted at the cen-tre of the range of operating frequencies for a circuit ordevice.

RESONANCE

Previous sections of this handout have includedgraphs showing the current and power in an induc-

tor and a capacitor. By comparing the graphs for aninductor with those for a capacitor, you should see thatthey are opposites - in anti-phase. This feature allowsthe creation of a family of circuits where, over a narrowrange of frequencies called the ‘bandwidth’, the effectsof the inductance in the circuit are cancelled by theeffects of the capacitance in the same circuit. This iscalled ‘resonance’.

For resonance to occur, the effects of the capacitorand inductor must be not only opposite but also equal.This can only occur at one frequency - which can bededuced from the graphs of their variation of reactancewith frequency. The resonant frequency is the crossingpoint of the two lines, where the reactance of the capac-itor equals the reactance of the inductor. These graphsare shown in Figure Nine.

The resonant frequency is, from the graph, about1.6 kHz, and the effects of resonance are a maximum atthis frequency. At higher and lower frequencies, thereactance values diverge and the resonance effects arerapidly reduced as the frequency moves away from theresonant frequency. The common symbol for a resonantfrequency is fo.

Calculation of the Resonant Frequency: the reso-nant frequency can be found from a graph or it can becalculated using the formula:

fo = 12 π√(L C )

Example: the resonant frequency of a combination of a10 mH inductor and a 1 μF capacitor is:

fo = 12 π √(10 m × 1 μ )

Using a VPAM calculator press buttons asfollows:

2 × π × √( 10 Shift 5 × 1 Shift 4 ) = x–1

The answer is 1.591 kHz, quite close to the1.6 kHz that we estimated from the crossingpoint of the lines on the graph.




Figure 9: Variation of Reactance with Frequency.10 mH Inductor and 1 μF Capacitor from Zero to 5 000 Hz.

Ω

Hz

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SERIES RESONANCE

In any series circuit, the current is the same all aroundthe circuit. Therefore, the current in the capacitor must

be the same as the current in the inductor - consequent-ly, their currents cannot cancel. It is their voltages (PDs)that cancel when resonance occurs. Since their voltages(back-EMF and EMF of displacement charge) provideopposition to current flow then, when that opposition iscancelled, there will be an increase of current at reso-nance. Thus, the behaviour of a resonant circuit differsgreatly at, or near, its resonant frequency from itsbehaviour at other frequencies.

Resonance is easily demonstrated using the circuitof Figure Ten - which contains a small lamp (6 V, 40 mAand 150 Ω). The action of the circuit can be developedby referring to the reactance values shown in FigureNine over the range of frequencies shown:

• At low frequencies (e.g. 500 Hz, the capacitor hasabout 300 Ω of reactance and this is hardly can-celled by the 30 Ω of the inductor. The result is thatthe lamp is dimly lit at this frequency because thecurrent is too small.

• At high frequencies (e.g. 5 kHz), the inductor hasabout 300 Ω of reactance and, again, this is hardlycancelled by the 50 Ω of the capacitor. The result isthat the lamp is dimly lit at this frequency.

• At frequencies below about 500 Hz, there will beeven less current because the reactance of thecapacitor increases even more.

• At frequencies above about 5 kHz, there will be evenless current because the reactance of the inductorincreases even more.

• At 1.6 kHz, the inductive and capacitive reactancescancel, leaving just the 150 Ω of the lamp. The lampshines with maximum brightness.

A graph of current versus frequency, for the circuit ofFigure Ten is shown in Figure Eleven. This graph has

been plotted for frequencies between 1 Hz and 10 MHzto show the full extent of the resonance curve.

This graph has a logarithmic (log) scale for frequen-cy. A log scale is not evenly spaced like the linearscales that have been used in graphs such as FigureSeven. The log scale is used when an axis on the graphcovers a very wide range of numbers (in this case fromone to ten million). The log scale expands the smallernumbers and compresses the larger numbers and it iseasily recognised by the pattern of gridlines as they arenot equally spaced. It is very common to show a fre-quency range using a log scale. (MS Excel has anoption to use a log scale - right-click an axis, choose‘Format Axis’ and it is under the ‘Scale’ tab.)



1 10 100 1 k 10 k 100 k 1 M Hz 10 M

mA

Figure 11: Graph of Current versus Frequency for the SeriesResonant Circuit of Figure Ten. (Log Scale of Frequency)

10 mH6 V

150 Ohms

1 micro-Farad

Figure 10: A Series Resonant Circuit, withResistor (lamp) , Capacitor and Inductor

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PARALLEL RESONANCE

For any components in parallel, the voltage (PD) isthe same for each. Therefore, in the circuit of Figure

Twelve, the PD across the capacitor must be the sameas the PD across the inductor - consequently, their volt-ages cannot cancel. It is their currents that cancel whenresonance occurs. Since their currents are cancelled,there will be a decrease of current through the lamp atresonance because the total current flows through thelamp.

Parallel resonance is easily demonstrated using thecircuit of Figure Twelve - which contains a small lamp (6V, 40 mA and 150 Ω). The action of the circuit can bedeveloped by referring to the reactance values shown inFigure Nine over the range of frequencies shown:

• At low frequencies (e.g. 500 Hz, the capacitor hasabout 300 Ω of reactance and takes a little current.However, the inductor, with 30 Ω, offers little opposi-tion to the current. The result is that the lamp isbrightly lit at this frequency because the current cantake the easy route through the inductor.

• At high frequencies (e.g. 5 kHz), the inductor hasabout 300 Ω of reactance and blocks the current onits side of the parallel circuit but, again, the currentcan flow through the 50 Ω of the capacitor. The resultis that the lamp is brightly lit at this frequency.

• At frequencies below about 500 Hz,there will be plenty of current becausethe reactance of the inductor decreas-es towards zero.

• At frequencies above about 5 kHz,there will be plenty of current becausethe reactance of the capacitordecreases towards zero.

• At 1.6 kHz, the inductive and capaci-tive currents cancel, and the supplycurrent is zero. (In fact exact cancel-lation is difficult to achieve and thereis usually a small current.) The lampis out as there is no current flowing.

A graph of current versus frequency,for the circuit of Figure Twelve is shown inFigure Thirteen. There are minor differ-ences between the series and parallelresonance circuits, but these are beyondthe scope of this course.

One unusual feature of the parallel resonance circuitbecomes visible when additional lamps are placed in theinductive and capacitive branches of the circuit of FigureTwelve. This is shown in Figure Fourteen. When demon-strated, this circuit shows that Lamps Two and Threecan be brightly lit whilst Lamp One is quite dimly lit. Thisis because the current in Lamp One is the DIFFER-ENCE between the currents on Lamps Two and Three.This is similar to the resonance that occurs when a par-ent pushes a child on a swing - at each push the move-ment of the swing increases. The movement of theswing can easily be ten times the movement of the par-ent’s hand because the swing stores energy from previ-ous pushes. A similar effect happens in this circuit,where energy is stored in an inductor and a capacitor.The current in the parallel branch is bigger than the cur-rent from the supply because the parallel branch is stor-ing energy from many previous cycles. The supply cur-rent represents the energy in just one cycle.

One use of this circuit is in radio and radar receivers.The weak signal from the antenna is connected to a res-onant circuit whose resonant frequency is the same asthe frequency of the transmission that you are trying to



10 mH6 V

150 Ohms

1 micro-Farad

Figure 12: A Parallel Resonant Circuit, withResistor (lamp) , Capacitor and Inductor

1 10 100 1 k 10 k 100 k 1 M Hz 10 M

mA

Figure 13: Graph of Current versus Frequency for the ParallelResonant Circuit of Figure Twelve. (Log Scale of Frequency)

10 mH

6 V

Lamp1

Lamp2 Lamp3

1 micro-Farad

Figure 14: AC Parallel Circuit with Lamps to

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receive. Resonance effects can increase the amount ofsignal received by a factor of several hundred. This ismost useful since an receiving antenna will normallyhave a very small signal that can be conveniently boost-ed using a resonant circuit. An additional advantage isthat other frequencies - from nearby transmitters - arenot increased by the resonant circuit since they do notmatch the resonant frequency.

Inductors and capacitors form part of the TUAM orTURF units that are mounted at the base of whip anten-nae in HF radios. The antennae function much more effi-ciently with these units in place and correctly adjustedfor resonance.

BANDWIDTH

Since the frequency response graph for a resonantcircuit is a curve, there is no clearly defined point of

cut-off. The resonance diminishes gradually. In electron-ics, there is a standard technique that is used to deter-mine the point of cut-off and it is defined as follows:

• For graphs showing Voltage or Current, the cut-offpoint is when the voltage or current level is 1/√2 or71% of the peak height. In other words, the currenthas reduced to 71% of the maximum value.

• For graphs showing Power or deci-Bels, the cut-offpoint is when the power is half the maximum or thenumber of deci-Bels is three less than the maximum.

In all cases, the cut-off point represents half-power.It is often called the half-power point, the -3 dB pint orthe 3 dB down point. There are two cut-off points on theresonance curve and these are indicated by thick, greylines in Figure Fourteen. The bandwidth is the frequen-cy difference between the two cut-off points and theseare about 3 100 Hz and 800 Hz, giving a bandwidth of2 300 Hz or 2.3 kHz.

Bandwidth is an important feature of a frequencyselective circuit such as the resonant circuit. Wheneverinformation is transmitted electronically from one place

to another then bandwidth is required -and the amount needed depends on therate at which information must be trans-ferred.

For the best performance from aresonance circuit, the amount of ‘extra’resistance must be minimised. This isachieved by techniques such as wind-ing the coils of the inductor with thethickest copper wire that can be accom-modated and by using a good conduc-tor for the plates of the capacitor. If theextraneous resistance in the circuit isexcessive then then resonance curvebecomes wider and the height of thepeak is reduced. This widens the band-width and reduces the effectiveness ofthe circuit. In the case of a radio, wherethe resonant frequency is used to selectthe station that you want to receive,having too wide a bandwidth might

cause your radio to receive several stations simultane-ously or give poor performance by admitting noise andinterference (these both increase as the bandwidthincreases).




1 10 100 1 k 10 k 100 k 1 M Hz 10 M

mA

Figure 15: Frequency Response of a Resonant Circuit with theHalf-Power Points Indicated. (Log Scale of Frequency)

E.03-11



FORMULA SUMMARY

Frequency = 1 Period = 1Period Frequency

Peak Value = RMS Value × √2

XL = 2 π f L (Inductive Reactance)

XC = 1 (Capacitive Reactance)2 π f C

X = Vrms (Ohm’s Law for ‘L’ or ‘C’)Irms

Ζ = Supply Vrms (Ohm’s Law for ac)Supply Irms

fo = 1 (Resonance)2 π √(L C )

At resonance, the reactance of the inductor is equal andoposite to the reactance of the capacitor:

E.03-12



SELF-TEST QUESTIONS

1. The period of an alternating current of frequency400 Hz is:

a. 400 msb. 40 msc. 25 msd. 2.5 ms

2. An alternating voltage has a period of 200 μs. Its fre-quency is:

a. 5 kHzb. 10 kHzc. 20 kHzd. 200 kHz

3. An alternating supply is rated at 48 V. This means that48 V is the:

a. maximum voltage.b. minimum voltagec. rms voltaged. average voltage

4. An alternating current of 4 A has a maximum, or peakvalue of:

a. 230 Vb. 4 Ac. 5.66 Ad. 2.83 A

5. When alternating current flows through a resistor, thecurrent is:

a. lagging the PD by 90°b. in phase with the PDc. leading the PD by 90°d. in anti-phase with the PD

.6. The opposition to the flow of current in an induc-tor is called:

a. resistanceb. impedancec. conductanced. reactance

7. When alternating current flows though an inductor,the current is:

a. lagging the PD by 90°b. in phase with the PDc. leading the PD by 90°d. in anti-phase with the PD.

8. When an inductor is connected to an alternating sup-ply the power in the circuit is:

a. equal to the rms Volts multiplied by the rms Ampsb. zero at all timesc. an average of zero over each cycled. greater than the power in a resistance of the same

number of Ohms.

9. When the frequency of alternating current through aninductor is increased then this will increase its:

a. resonant frequencyb. reactancec. bandwidthd. current

10. An inductor of 47 mH when used at 100 Hz has a reac-tance of approximately:

a. 2128 Ω.b. 100 Ω.c. 47 Ω.d. 29.5 Ω.

1. ττ= 1/f=1/400= 2.5 ms(d)2. f= 1/ττ=1/200 μs= 5 kHz(a)3. AC is always quoted in rms(c)4. Vpk= Vrms×√√2= 4 √√2 = 5.66 V(c)5. Resistive current is in phase with PD(b)

Answers

6. Reactance = opposition of an Inductor(d)7. Inductive current lags Volts by 90°(a)8. Power flows back & forth, average = Zero(c)9. Reactance is proportional to frequency(b)10. ΞΞL= 2 ππφφΛΛ= 2 × ππ×100 × 47 mH = 29.5 Ω(d)

Answers

E.03-13



SELF-TEST QUESTIONS

11. The opposition to the flow of current in a capacitor iscalled:

a. resistanceb. impedancec. conductanced. reactance

12. When alternating current flows though a capacitor,the current is:

a. leading the PD by 90°b. in phase with the PDc. lagging the PD by 90°d. in anti-phase with the PD.

13. When a capacitor is connected to an alternating sup-ply the power in the circuit:

a. equal to the rms Volts multiplied by the rms Ampsb. zero at all timesc. an average of zero over each cycled. greater than the power in a resistance of the same

number of Ohms.

14. When the frequency of alternating current through ancapacitor is increased then this will cause:

a. a decrease of currentb. a decrease of reactancec. an increase of reactanced. an increase of bandwidth

15. An capacitor of 20 μF when used at 300 Hz has a reac-tance of approximately:

a. 15 Ω.b. 300 Ω.c. 320 Ω.d. 26.5 Ω.

16. When a capacitor is in series with a lamp and the fre-quency of the alternating supply is varied then the lampwill:

a. shine brightly only at low frequenciesb. shine brightly only at high frequenciesc. go dim at the resonant frequencyd. go bright at the resonant frequency.

17. A capacitor has a reactance of 150 Ω at 2 kHz. WHenthe frequency is reduced to 1 kHz then the reactancewould be:

a. 300 Ω.b. 150 Ω.c. 75 Ω.d. 1 kΩ.

18. A capacitor is connected to a 10 V AC supply and acurrent of 0.2 A flows. From these figures it can be calcu-lated that:

a. 2 Watts of heat is produced.b. the frequency was 50 Hz.c. the reactance of the capacitor was 50 Ω.d. the capacitor is operating at resonance.

19. When a resonant circuit is operating at the resonantfrequency, the reactance of the inductor will be:

a. zerob. infinitec. equal to the resistance of the lampd. equal to the reactance of the capacitor

20. When a capacitor of 50 μF is used in a resonant cir-cuit with an inductor of 25 mH then the resonant frequen-cy will be approximately:

a. 142 Hzb. 500 Hzc. 1250 Hzd. 2000 Hz

11. Reactance = opposition of a Capacitor(d)12. Capacitive current leads Volts by 90°(a)13. Power flows back & forth, average = Zero(c)14. Reactance is inversely proportional to the

frequency(b)15. ΞΞc= 1/(2 ππφφΧΧ))= 1 ÷ (2 ππ× 300 × 20 μF)= 26.5 ΩΩ(d)16. Capacitors let through high freqs(b)17. Freq down - reactance up(a)18. ΞΞ= ςς/ΙΙ= 10 ÷ 0.2 = 50 ΩΩ(c)19. Equal (&opposite) reactances at resonance(d)20. Standard formula gives 142 Hz(a)

Answers

E.03-14



SELF-TEST QUESTIONS

The graph in Figure STQ1 shows the variation of currentwith frequency for a resonant circuit. Questions 21 - 25are about this graph and its corresponding circuit.

21. From the graph in Figure STQ1. it can be deducedthat the type of circuit used was:

a. parallel inductor and capacitor.b. inductor in series with a lampc. capacitor in series with a lampd. capacitor in series with an inductor.

22. The resonant frequency of the circuit used for FigureSTQ1 is approximately:

a. 10 Hzb. 1.7 kHzc. 4.5 kHzd. 1 MHz

23. The bandwidth of the circuit used for Figure STQ1 isapproximately:


24. In a series resonant circuit, when the resonant fre-quency is reached then:

a. the supply current is a maximum.b. the inductive reactance has its maximum valuec. The capacitive reactance is zerod. the supply current is a minimum

25. One effect on the performance of a resonant circuitof using a better quality inductor (i.e. lower resistance inits coils) would be:

a. wider bandwidth.b. less current.c. narrower bandwidthd. higher resonant frequency.

21. It is a series LC circuit, maximum current(d)22. Use standard formula, 1.7 kHz(b)23. FInd 71% points, 57 mA, width = 4500 Hz(c)24. Series resonance gives max current(a)25. Less resistance = higher & narrower resonance(d)

Answers

Figure STQ1

Hz

mA

E.03-15



SELF-TEST QUESTIONS

The graph in Figure STQ2 shows the variation of currentwith frequency for a resonant circuit. Questions 26 - 30are about this graph and its corresponding circuit.

26. From the graph in Figure STQ2. it can be deducedthat the type of circuit used was:

a. inductor in series with a lampb. parallel inductor and capacitorc. capacitor in series with a lampd. capacitor in series with an inductor.

27. The resonant frequency of the circuit used for FigureSTQ1 is approximately:


28. The bandwidth of the circuit used for Figure STQ1 isapproximately:


29. In a parallel resonant circuit, when the resonant fre-quency is reached then:

a. the supply current is a minimum.b. the inductive reactance has its maximum valuec. The capacitive reactance is zerod. the supply current is a maximum

30. In the parallel resonant circuit of Figure STQ2, at afrequency of 10 kHz:

a. the inductor has the lower reactanceb. the capacitor has the higher reactancec. the inductor and capacitor have the same reac-

tanced. the capacitor has the lower reactance

Figure STQ2

Hz

mA

26. It is a parallel LC circuit, min current(b)27. Lowest point on graph, approx. 1.7 kHz(c)28. FInd 71% points, 85 mA, width = 2200 Hz(c)29. Parallel resonance gives min current(a)30. Above resonance the capacitor has lower Xc(d)

Answers

E.03-16



Teaching Objectives Comments

E.03.01: Describe an Alternating Current Using Standard Terminology

E.03.01.01 Sketch the waveform of an AC supply given its parameters and vice -versa.

Includes rms, peak, period, frequency and phase.

E.03.01.02 State that t he average value of an alternating current or voltage is zero.

E.03.01.03 Sketch the waveform of the power in an AC circuit. Resistive circuit.

E.03.01.04 Use rms values to calculate the power in a resistive circuit.

E.03.02: Describe the Effects of A lternating Current in a Pure Inductive Circuit.

E.03.02.01 Describe how the induced emf opposes the flow of alternating current.

E.03.02.03 Use the reactance formula to calculate the reactance of an inductor.

E.03.02.03 Sketch a graph showing the vari ation of reactance with frequency.

Demo with sig gen & lamp.

E.03.02.04 State that the phase angle is 90°, current lagging.

E.03.02.05 Sketch the waveform of the power in an inductive circuit

E.03.02.06 Describe how the energy flows back and forth in an inductive circuit.

E.03.03: Describe the effects of Alternating Current in a Pure Capacitive Circuit.

E.03.03.01 Describe how the charge in a capacitor opposes the flow of alternating current.

E.03.03.03 Use the reactance formula to calculate the reactance of a capacitor.

E.03.03.03 Sketch a graph showing the variation of reactance with frequency.

Demo with sig gen & lamp.

E.03.03.04 State that the phase angle is 90°, current leading.

E.03.03.05 Sketch the waveform of the power in a capacitive circuit

E.03.03.06 Describe how the energy flows back and forth in a capacitive circuit.

E.03.04: Describe the Basic Properties of Resonant Circuits.

E.03.04.01 Describe resonance as the ability to show a pronounced response over a relatively narrow range of frequencies.

E.03.04.02 State that resonance occurs when stored energy is accumulated over many cycles.

E.03.03.03 Identify series and parallel resonant circuits and sketch their graphs of current versus frequency.

Indicate bandwidth and effe cts of resistance.

E.03.03.04 Calculate the resonant frequency.

Bst Handout e03

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