CIRCUIT ANALYSIS II

Will MooreMT 12CIRCUIT ANALYSIS II(AC Circuits)SyllabusComplex impedance, power factor, frequency response of AC networksincluding Bode diagrams, second-order and resonant circuits, dampingand Q factors. Laplace transform methods for transient circuit analysiswith zero initial conditions. Impulse and step responses of second-ordernetworks and resonant circuits. Phasors, mutual inductance and idealtransformers.Learning OutcomesAt the end of this course students should:1.Appreciate the significance and utility of Kirchhoff’s laws.2.Be familiar with current/voltage relationships for resistors, capacitorsand inductors.3.Appreciate the significance of phasor methods in the analysis of ACcircuits.4.Be familiar with use of phasors in node-voltage and loop analysis ofcircuits.5.Be familiar with the use of phasors in deriving Thévenin and Nortonequivalent circuits6.Be familiar with power dissipation and energy storage in circuitelements.

7.Be familiar with methods of describing the frequency response ofAC circuits and in particular8.Be familiar with the Argand diagram and Bode diagram methods9.Be familiar with resonance phenomena in electrical circuits10. Appreciate the significance of the Q factor and damping factor.11. Appreciate the significance of the Q factor in terms of energystorage and energy dissipation.12. Appreciate the significance of magnetic coupling and mutualinductance.13. Appreciate the transformer as a means to transform voltage, currentand impedance.14. Appreciate the importance of transient response of electricalcircuits.15. Be familiar with first order systems16. Be familiar with the use of Laplace transforms in the analysis of thetransient response of electrical networks.17. Appreciate the similarity between the use of Laplace transform andphasor techniques in circuit analysis.2

Circuit Analysis II WRM MT12AC Circuits1. Basic IdeasOur development of the principles of circuit analysis in Circuit Analysis Iwas in terms of DC circuits in which the currents and voltages wereconstant and so did not vary with time. Here in Circuit Analysis II weextend our analysis to consider time varying currents and voltages andthereby we are able to extend our analysis to include capacitors andinductors. In our initial discussions we will limit ourselves to sinusoidalfunctions. We choose this special case because, as you have nowlearnt in P1, it allows us to make use of some very powerful and helpfulmathematical techniques. It is also a common waveform in nature and itis easy to generate in the lab. However as you have also learnt in P1,any waveform can be expressed as a weighted superposition ofsinusoids of different frequencies and hence if we analyse a linear circuitfor sinusoidal functions we can, by appropriate superposition, handleany function of time.Let's begin by considering a sinusoidal variation in voltagev Vm cos t3

in which is the angular frequency and is measured in radians/second.Since the angle t must change by 2 radians in the course of oneperiod, T, it follows that T 2 However the time period T 1where f is the frequency measured infHertz. Thus 2 T 2 fThis is a simple and very important relationship. We naturally measurefrequency in Hz – the mains frequency in the UK is 50Hz – and it is easyto measure the time period, T 1 f from an oscilloscope screen.However as we will soon see, it is mathematically more convenient towork in terms of the angular frequency . Mistakes may be easily madebecause in practice the word frequency is commonly used to refer to4

Circuit Analysis II WRM MT12both and f. It is important in calculations to make sure that if appears, then the correct value for f 50 Hz, say, is 100 rads/sec.A simple point to labour I admit, but if I had a pound for every timesomeone forgets and substitutes 50 . . . . . . . . !!In our example above, v Vm cos t , it was convenient thatv Vm at t 0 . In general this will not be the case and the waveform willhave an arbitrary relationship to the origin t 0 or, equivalently the originmay have been chosen arbitrarily and the voltage, say, may be written interms of a phase angle, , asv Vm cos t TAlternatively, in terms of a different phase angle, , the same waveformcan be written5

v V m sin t where 2 The phase difference between two sinusoids is almost always measuredin angle rather than time and of course one cycle (i.e. one period)corresponds to 2 or 360 . Thus we might say that the waveform aboveis out of phase with the earlier sinusoid by . When 2 we saythat the two sinusoids are said to be in quadrature. When thesinusoids are in opposite phase or in antiphase.2. RMS ValuesWe refer to the maximum value of the sinusoid, Vm, as the “peak” value.On the other hand, if we are looking at the waveform on an oscilloscope,it is usually easier to measure the “peak-to-peak” value 2Vm, i.e. from thebottom to the top. However, you will notice that most meters arecalibrated to measure the root-mean-square or rms value. This is found,as the name suggests, for a particular function, f, by squaring thefunction, averaging over a period and taking the (positive) square root ofthe average. Thus the rms value of any function f(x), over the interval xto x X, where X denotes the period is6

Circuit Analysis II WRM MT12frms 1 x X 2 f y dyX xFor our sinusoidal function v Vm cos tThe average of the square is given by1T 2Vm cos2 t dt T0where the time period T 2 . At this point it's probably easiest tochange variables to t and to write cos2 1 1 cos 2 . Thus the2mean square value becomesVm21 Vm2 2 1 cos2 d 2 2 02The root mean square value, which is simply the positive square root ofthis, may be written asVrms Vm / 2 0.7 Vm.7

Since we nearly always use rms values in our AC analysis, we assumerms quantities unless told otherwise so by convention we just call it V asin:V Vm / 2.So, for example, when we say that the UK mains voltage is 230V whatwe are really saying is that the rms value 230V. Its peak or maximumvalue is actually 230 2 325 V.To see the real importance of the rms value let's calculate the powerdissipated in a resistor.Here the current is given by i v Ri Vmcos t I m cos tR8

Circuit Analysis II WRM MT12where Im Vm R and the power, p vi , is given byVm2p cos2 tRIf we want to calculate the average power dissipated over a cycle wemust integrate from t 0 to t T 2 . If we again introduce t ,the average power dissipated, P, is given byP 1 Vm2 2 .cos2 d 2 R 01 Vm21 2 P . 1 cos 2 d 2 R20P Vm2 2RIf we now introduce the rms value of the voltage V Vm2 then theaverage power dissipated may be written asP V 2 RIndeed if the rms value of the current I ImP V 2 R I 2 R92 is also introduced then

which is exactly the same form of expression we derived for the DCcase.Therefore if we use rms values we can use the same formula for theaverage power dissipation irrespective of whether the signals are AC orDC.3. Circuit analysis with sinusoidsLet us begin by considering the following circuit and try to find anexpression for the current, i, after the switch is closed.10

Circuit Analysis II WRM MT12The Kirchhoff voltage law permits us to writeLdi Ri Vm cos tdtThis is a linear differential equation, which you know how to solve.We begin by finding the complementary function, from the homogeneousequation:Ldi Ri 0dtwhich yields the solution:i A exp Rt L We now need to find the particular integral which, for the sinusoidal"forcing function" Vm cos t , will take the form B cos t C sin t . Thusthe full solution is given byi t A exp Rt L B cos t C sin t11

We see that the current consists of a "transient" term, A exp Rt L ,which eventually decays and becomes negligible in comparison with the"steady state" response. The transient response arises because of thesudden opening or closing of a switch but we will concentrate here onthe final sinusoidal steady state response. How long do we have to waitfor the steady state? If for example R 100 and L 25mH thenR L 4 10 3 sec 1 and so after only 1ms exp Rt L exp 4 0.018and so any measurements we are likely to make on this circuit will betruly 'steady state' measurements. Thus our solution of interest reducestoi B cos t C sin tIn order to find B and C we need to substitute this expression back intothe governing differential equation to give L C cos t B sin t R B cos t C sin t Vm cos t12

Circuit Analysis II WRM MT12It is now a simple matter to compare coefficients of cos t and sin t toobtain expressions for B and C which lead, after a little algebra, toi VmR 2 L 2 R cos t L sin t If we now introduce the inductive reactance X L L we can write thisequation asi XLR cos t sin t 2222R 2 X L2 R XR X LL VmThe expression in curly brackets is of the formcos cos t sin sin t cos t and hencei VmR X22L(cos wt - j)whereæX öL èR øj tan-1 çThus we see that the effect of the inductor has been to introduce aphase lag between the current flowing in the circuit and the voltagesource. Similarly the ratio of the maximum voltage to the maximum13

current is given byR 2 X L2 which since it is a combination ofresistance and reactance is given the new name of impedance.It is apparent that we could solve all networks containing combinationsof resistors, inductors and capacitors in this way. We would end up witha series of simultaneous equations to solve – just as we did whenanalysing DC circuits – the problem is that they would be simultaneousdifferential equations which, given the effort we went through to solveone equation in the simple example above, would be very tedious andtherefore rather error-prone. Fortunately there is an easier way.We are saved because the differential equations we have to solve arelinear and hence the principle of superposition applies. This tells usthat if a forcing function v1(t) produces current i1(t) and a forcing functionv 2 t provides current i 2 t then v1 t v 2 t produces i1 t i 2 t . Thetrick then is to choose a more general forcing function v t v1 t v 2 t inwhich, say, v1 t corresponds to Vm cos t and which made thedifferential equation easy to solve. We achieve this with complexalgebra.You should know thatexp j wt coswt j sinwt .where j (-1),[electrical engineers like to use i for current]so let’s solve the differential equation with the general forcing function14

Circuit Analysis II WRM MT12()v t Vm exp j wt Vm coswt j Vm sinwtwhere(){ ( )}v 1 t Re v t Re {Vm exp j wt } Vm coswt.The solution will be of the formi (t ) I exp j wtwherewill, in general, be a complex number. Then inorder to find that part of the full solution corresponding to the real part ofthe forcing function, Vm exp j t we merely need to find the real part ofi t . Thus(){(i 1 t Re {I exp jwt } Re I exp j wt - j( I cos wt - j))}Let's illustrate this by returning to our previous example where we triedto solve:Ldi Ri Vm cos tdtNow, instead, we solve the more general case:Ldi Ri Vm exp j tdt15

and take the real part of the solution. As suggested above anappropriate particular integral is i I exp j t which leads toj L I exp j t RI exp j t Vm exp j tThe factor exp j t is common and hence R j L I Vmin which R j L may be regarded as a complex impedance. Thecomplex current I is now given byI VmVm exp j 22R j LR L with tan 1 L R and hencei t Re I exp j t VmR L 22cos t which, thankfully, is the same solution as before but arrived at withconsiderably greater ease.Let us be clear about the approach. We have16