Will Moore MT & MT CIRCUIT ANALYSIS II (AC Circuits) Syllabus Complex impedance, power factor, frequency response of AC networks including Bode diagrams, second-order and resonant circuits, damping and Q factors. Laplace transform methods for transient circuit analysis with zero initial conditions. Impulse and step responses of second-order networks and resonant circuits. Phasors, mutual inductance and ideal transformers. Learning Outcomes At the end of this course students should:. Appreciate the significance and utility of Kirchhoff s laws.. Be familiar with current/voltage relationships for resistors, capacitors and inductors. 3. Appreciate the significance of phasor methods in the analysis of AC circuits. 4. Be familiar with use of phasors in node-voltage and loop analysis of circuits. 5. Be familiar with the use of phasors in deriving Thévenin and Norton equivalent circuits 6. Be familiar with power dissipation and energy storage in circuit elements.
7. Be familiar with methods of describing the frequency response of AC circuits and in particular 8. Be familiar with the Argand diagram and Bode diagram methods 9. Be familiar with resonance phenomena in electrical circuits 0. Appreciate the significance of the Q factor and damping factor.. Appreciate the significance of the Q factor in terms of energy storage and energy dissipation.. Appreciate the significance of magnetic coupling and mutual inductance. 3. Appreciate the transformer as a means to transform voltage, current and impedance. 4. Appreciate the importance of transient response of electrical circuits. 5. Be familiar with first order systems 6. Be familiar with the use of Laplace transforms in the analysis of the transient response of electrical networks. 7. Appreciate the similarity between the use of Laplace transform and phasor techniques in circuit analysis.
Circuit Analysis II WRM MT AC Circuits. Basic Ideas Our development of the principles of circuit analysis in Circuit Analysis I was in terms of DC circuits in which the currents and voltages were constant and so did not vary with time. We will now extend this analysis to consider time varying currents and voltages. In our initial discussions we will limit ourselves to sinusoidal functions. We choose this special case because, as you have now learnt in P, it allows us to make use of some very powerful and helpful mathematical techniques. It is a common waveform in nature and it is easy to generate in the lab. However as you have also learnt in P, any waveform can be expressed as a weighted superposition of sinusoids of different frequencies and hence if we analyse a linear circuit for sinusoidal functions we can, by appropriate superposition, handle any function of time. Let's begin by considering a sinusoidal variation in voltage v = Vm cos ωt 3
in which ω is the angular frequency and is measured in radians/second. Since the angle ωt must change by π radians in the course of one period, T, it follows that ωt = π However the time period Hertz. Thus T = where f is the frequency measured in f π ω = = πf T This is a simple and very important relationship. We naturally measure frequency in Hz the mains frequency in the UK is 50Hz and it is easy to measure the time period, T ( = f ) from an oscilloscope screen. However as we will soon see, it is mathematically more convenient to work in terms of the angular frequency ω. Mistakes may be easily made because in practice the word frequency is commonly used to refer to both ω and f. It is important in calculations to make sure that if ω appears, then the correct value for f = 50 Hz, say, is ω = 00π rads/sec. A simple point to labour I admit, but if I had a pound for every time someone forgets and substitutes ω = 50........!! 4
Circuit Analysis II WRM MT In our example above, v = Vm cos ωt, it was convenient that v = Vm at t = 0. In general this will not be the case and the waveform will have an arbitrary relationship to the origin t = 0 or, equivalently the origin may have been chosen arbitrarily and the voltage, say, may be written in terms of a phase angle, φ, as v = Vm cos t ( ω φ) T Alternatively, in terms of a different phase angle, ψ, the same waveform can be written v = V m sin t ( ω +ψ ) where ψ = π φ 5
The phase difference between two sinusoids is almost always measured in angle rather than time and of course one cycle (i.e. one period) corresponds to π or 360. Thus we might say that the waveform above is out of phase with the earlier sinusoid by φ. When φ = ± π we say that the two sinusoids are said to be in quadrature. When sinusoids are in opposite phase or in antiphase. φ = π the. RMS Values We refer to the maximum value of the sinusoid, V m, 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 V m, i.e. from the bottom to the top. However, you will notice that most meters are calibrated to measure the root-mean-square or rms value. This is found, as the name suggests, for a particular function, f, by squaring the function, averaging over a period and taking the (positive) square root of the average. Thus the rms value of any function f(x), over the interval x to x+x, where X denotes the period is f rms = x + X x X f ( y)dy For our sinusoidal function v = Vm cos ωt 6
Circuit Analysis II WRM MT The average of the square is given by T T 0 V m cos ωt dt where the time period T = π ω. At this point it's probably easiest to change variables to mean square value becomes θ = ωt and to write cos θ = ( + cosθ). Thus the m V m π V π 0 ( + cosθ) dθ = The root mean square value, which is simply the positive square root of this, may be written as V rms = V m / 0.7 V m. Since we nearly always use rms values in our AC analysis, we assume rms quantities unless told otherwise so by convention we just call it V as in: 7
V = V m /. So, for example, when we say that the UK mains voltage is 30V what we are really saying is that the rms value 30V. Its peak or maximum value is actually 30 35 V. To see the real importance of the rms value let's calculate the power dissipated in a resistor. Here the current is given by i = v R Vm i = cos ωt = Im cosωt R where Im = Vm R and the power, p = vi, is given by Vm p = cos ωt R 8
Circuit Analysis II WRM MT If we want to calculate the average power dissipated over a cycle we must integrate from ωt = 0 to ωt = ωt = π. If we again introduce θ = ωt, the average power dissipated, P, is given by = π Vm P π R. cos 0 θdθ = π V P m. π R 0 ( + cos θ) dθ P = Vm R If we now introduce the rms value of the voltage V = V m then the average power dissipated may be written as P = V R Indeed if the rms value of the current I = I m is also introduced then P = V R = I R which is exactly the same form of expression we derived for the DC case. Therefore if we use rms values we can use the same formula for the average power dissipation irrespective of whether the signals are AC or DC. 9
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Circuit Analysis II WRM MT 3. Circuit analysis with sinusoids Let us begin by considering the following circuit and try to find an expression for the current, i, after the switch is closed. The Kirchhoff voltage law permits us to write di L + Ri = Vm cos ωt dt This is a linear differential equation, which you know how to solve. We begin by finding the complementary function, from the homogeneous equation: which yields the solution: di L dt + Ri =0 i = A exp ( Rt L)
We now need to find the particular integral which, for the sinusoidal "forcing function" V m the full solution is given by cos ωt, will take the form B cos ωt + C sin ωt. Thus i ( t) = A exp ( Rt L) + B cos ωt + C sin ωt We see that the current consists of a "transient" term, A ( Rt L) exp, which eventually decays and becomes negligible in comparison with the "steady state" response. The transient response arises because of the sudden opening or closing of a switch but we will concentrate here on the final sinusoidal steady state response. How long do we have to wait for the steady state? If for example R =00Ω and L = 5mH then 3 R L = 4 0 sec and so after only ms exp Rt L = exp ( 4) = 0. 08 and so any measurements we are likely to make on this circuit will be truly 'steady state' measurements. Thus our solution of interest reduces to
Circuit Analysis II WRM MT i = B cos ωt + C sin ωt In order to find B and C we need to substitute this expression back into the governing differential equation to give { C cos ωt B sin ωt} + R{ B cos ωt + C sin ωt} = V cos t ωl m ω It is now a simple matter to compare coefficients of cosωt and sinωt to obtain expressions for B and C which lead, after a little algebra, to V i = m ω R + sin ( ) { R cosω t + ω L t} ωl If we now introduce the inductive reactance X L ( = ωl) we can write this equation as V m R XL i = cosωt + sinωt R + XL R + XL R + XL The expression in curly brackets is of the form cos φcos ωt + sin φ sin ωt = cos ( ωt φ) and hence i = V m R + X L cos( ωt ϕ) where 3
" ϕ =tan $ # X L R % ' & Thus we see that the effect of the inductor has been to introduce a phase lag φ between the current flowing in the circuit and the voltage source. Similarly the ratio of the maximum voltage to the maximum current is given by X L R + which since it is a combination of resistance and reactance is given the new name of impedance. It is apparent that we could solve all networks containing combinations of resistors, inductors and capacitors in this way. We would end up with a series of simultaneous equations to solve just as we did when analysing DC circuits the problem is that they would be simultaneous differential equations which, given the effort we went through to solve one equation in the simple example above, would be very tedious and therefore rather error-prone. Fortunately there is an easier way. We are saved because the differential equations we have to solve are linear and hence the principle of superposition applies. This tells us that if a forcing function v (t) produces current i (t) and a forcing function v ( t ) provides current i ( t ) then v ( t) + v ( t) produces i ( t) + i( t). The trick then is to choose a more general forcing function ( t) v ( t) v ( t) which, say, ( t) v + v corresponds to cos ωt and which made the differential equation easy to solve. We achieve this with complex algebra. V m = in You should know that 4
Circuit Analysis II WRM MT exp j ωt =cosωt + j sinωt. where j = (-), [electrical engineers like to use i for current] so let s solve the differential equation with the general forcing function v ( t )=V m exp j ωt =V m cosωt + j V m sinωt where v ( t )=Re{ v ( t )}=Re{ V m exp j ωt }=V m cosωt. The solution will be of the form i (t ) =I exp j ωt where I = I exp j will, in general, be a complex number. Then in order to find that part of the full solution corresponding to the real part of the forcing function, i ( t). Thus V m exp j ωt we merely need to find the real part of { ( )} i ( t )=Re{ I exp jωt }=Re I exp j ωt ϕ ( ) = I cos ωt ϕ Let's illustrate this by returning to our previous example where we tried to solve: 5
di L + Ri = Vm cos ωt dt Now, instead, we solve the more general case: di L + Ri = Vm exp j ωt dt and take the real part of the solution. As suggested above an appropriate particular integral is i = I exp j ωt which leads to j ωli exp j ωt + RI exp j ωt = Vm exp j ωt The factor exp j ωt is common and hence ( R + j ωl) I = Vm in which R + j ωl may be regarded as a complex impedance. The complex current I is now given by = Vm Vm I = exp φ R + j ωl j R + ( ωl) with φ= ( ωl R) tan and hence Vm i( t) = Re{ I exp j ωt} = cos t R + ( ωl) ( ω φ) 6
Circuit Analysis II WRM MT which, thankfully, is the same solution as before but arrived at with considerably greater ease. Let us be clear about the approach. We have (i) introduced a complex forcing function exp j ωt knowing that in reality the voltage source must be real i.e. { exp j ωt} V m Re. V m (ii) We solved the equations working with complex voltages and complex currents, V exp j ωt and I exp j ωt (or rather V and I since the time dependence exp jωt cancelled out). (iii) Since the actual voltage is given by { exp j ωt} Re the actual current is given by { I exp j ωt} = Re{ I exp j ψ exp j ωt} = I cos ( ωt + ψ) V m Re. (iv) Since the differential of exp jωt = exp jωt. exp j is simply jω. exp jωt. exp j and since we always take exp jωt out as a common factor, you may see now that our differential equations turn into polynomial equations in jω (and you knew how to solve these at GCSE!) This is a very powerful approach that will permit us to solve AC circuit problems very easily. 7
Example Let's now do an example to show, formally, how we can solve AC problems. Let's imagine we want to find the steady state current, i, flowing through the capacitor in the following example The two KVL loop equations may be written and Ri + Ldi +R ( i dt i )=E m cos( ωt +α) R ( i i )+ C i dt =0 Replacing E m cos( ω t +α) by j ( ωt + α) = E exp j ωt E m exp where E = Em exp jα and further introducing I and I via i = I exp j ωt and i = I exp j ωt we obtain 8
Circuit Analysis II WRM MT ( R + j ωl) I R + I j ωc RI RI = E = 0 and, after a little algebra I = R + L CR E + j ωl ωc Em exp j α = M + j N where M = R + L CR and N = ωl ωc. We note that this may be written, introducing tan θ = N M Em I = exp j M + N ( α θ) and hence the actual current i = ( I exp j ωt ) Re may be written as i m ( t) = cos( ωt + α θ) M E + N 9
4. Phasors We have just introduced a very powerful method of circuit analysis. In essence we have introduced the use of complex quantities to represent sinusoidal functions of time. The complex number A exp jφ (often written A φ) when used in this context to represent Acos ( ωt + φ) is called a phasor. Since the phase angle φ must be measured relative to some reference we may call the phasor A 0 the reference phasor. Since the phasor, A exp j φ, is complex it may be represented in Cartesian form x + jy just like any other complex quantity and thus x = Acosφ y = A sin φ A= x + y Further since the phasor is a complex quantity it is very easy to display it on an Argand diagram (also in this context called a phasor diagram). Thus the phasor A exp j φ is drawn as a line of length A at an angle φ to the real axis. 0
Circuit Analysis II WRM MT We emphasise that this is a graphical representation of an actual A cos ω t +φ. The rules for addition, subtraction and sinusoid ( ) multiplication of phasors are identical to those for complex numbers. Thus addition: For multiplication it is easiest to multiply the magnitudes and add the phases. Consider the effect of multiplying a phasor by j j Aexp j φ=exp j π Aexp j φ= Aexp j ( ) φ+ π which causes the phasor to be rotated by 90 o.
Similarly, dividing by j leads to Aexp j φ = j exp ( j π ) A exp j φ = Aexp j ( φ π ) i.e. a rotation of 90 o. We finally note that it is usual to use rms values for the magnitude of phasors.
Circuit Analysis II WRM MT 5. Phasor relations in passive elements Consider now a voltage V m cos ωt applied to a capacitor. As we have indicated we elect to use the complex form V m exp j ωt and so omit the "real part" as we calculate the current via i = C dv dt = C d dt ( V exp j ωt) = j ωcv exp j ωt m m If we now drop the exp j ωt notation and write the voltage phasor V m as V and the current phasor as I we have I = j ωcv or V = I j ωc In terms of a phasor diagram, taking the voltage V as the reference from which we confirm two things we already knew 3
(i) the ratio of the voltage to the current is ωc - the reactance. (ii) the current leads the voltage by 90. The pre-multiplying factor j describes this. For the inductance an analogous procedure leads to V = j ωl I Where the reactance is now jωl and, if we now take, say, the current as the reference phasor we have and here the current lags the voltage by 90. 4
Circuit Analysis II WRM MT [It is important to get these relationships the right way around and as a check we may use the memory aid CIVIL in a capacitor, the current leads the voltage CIVIL and in an inductor, the current lags the voltage CIVIL.] Finally for a resistor we know that the current and voltage are in phase and hence, in phasor terms V = I R 5
6. Phasors in circuit analysis We are now in a position to summarise the method of analysis of AC circuits. (i) We include all reactances as imaginary quantities j L ( = j ) an inductor and j ωc ( = j ) for a capacitor. X c ω for X L (ii) All voltages and currents are represented by phasors, which usually have rms magnitude, and one is chosen as a reference with zero phase angle. (iii) All calculations are carried out in complex notation. (iv) The magnitude and phase of, say, the current is obtained as I exp j φ. This can, if necessary, be converted back into a time varying I cos ω t +φ. expression ( ) 6
Circuit Analysis II WRM MT Suppose we wish to find the current flowing through the inductor in the circuit below The reactances have been calculated and marked on the diagram. The left hand voltage source has been chosen as reference and provides 0V rms. The right hand source produces 5V rms but at a phase angle of 37 with respect to the 0V source. If we introduce phasor loop currents I and I as shown then we may write KVL loop equations as 0 = 5I 5exp j 37 + j 0 o ( I I ) = 4 + j 3 = ( I I ) j 0 ( j 5) I where we have noted that 5exp j 37 =5cos37 + j 5sin37 =4+ j3. It is routine to solve these simultaneous equations to give o o o I ( 7 j ) = and.5 I 6.5+ j 8 =.5 and hence the current I = I I becomes - 7
0.5 j9 o I = = 0.7 86.8 = 0.7 exp j 86. 8.5 o Since rms values are involved, if we want to convert this into a function of time we must multiply by to obtain the peak value. Thus o ( t) =.0cos( ωt 86. ) i 8 In our example we do not know the value of ω but it was accounted for in the value of the reactances. Since everything is linear and the sources are independent it would be a good exercise for you to check this result by using the principle of superposition. We have used mesh or loop analysis in our examples so far. It is, of course equally appropriate to use node-voltage analysis if that looks like an easier way to solve the problem. As an example let's suppose we would like to find the voltage V in the circuit below where the reactances have been calculated corresponding to the frequency, ω, of the source 8
Circuit Analysis II WRM MT It's probably as easy as anything to introduce two phasor node voltages V and V. The two node voltage relationships may be written as V 0 V V V 0 + + 0 j5 j5 = 0 and V V j5 V 0 V 0 + + 5 j0 5 + j0 = 0 We note that in writing these equations no thought was given to whether currents flowing into or out of the nodes were being considered. As in the DC case it is merely necessary to be consistent. It is now straightforward to solve these two simultaneous equations to yield o V = 0 7. 6 or, if the time domain result is required, remembering o that the voltage supply is 0V rms then v ( t) 0 cos( ωt 7. 6 ) =. 9
7. Combining impedances As we have seen before the ratio of the voltage to the current phasors is in general a complex quantity, Z, which generalises Ohms law, in terms of phasors, to V = Z I where Z in general takes the form Z = R + j e X e where the overall effect is equivalent to a resistance, R e, in series with a reactance X e. If X e is positive the effective reactance is inductive whereas negative values suggest that the effective reactance is capacitative. Consider the circuit below 30
Circuit Analysis II WRM MT V = R + j ωl + I j ωc Thus the combined impedance Z is given by Z = R + j ωl = R + ωc j X This may be visualised on an Argand or phasor diagram We note that the reactance may be positive or negative according to the relative values of ω L and ωc. Indeed at a frequency = LC ω we see that X = 0 and that the impedance is purely resistive. We will return to this point later. 3
It is straightforward to show, and hopefully intuitive, that all the DC rules for combining resistances in series and parallel carry over to impedances. Thus if we have n elements in series, Z, Z, Z 3 Z n Where Z eff = Z + Z + Z + Z 3 n = n i = Z i and similarly for parallel elements Z eq = Z + Z + Z 3 + = N i= Z i We note that the inverse of impedance, Z, is known as admittance, Y. Thus, as in the DC case it is sometimes more convenient to write n i= Y eq = Y i 3
Circuit Analysis II WRM MT and finally since Y is also a complex number it may be written Y = G + j B where G is a conductance and B is known as the susceptance. 33
34 Example Find the equivalent impedance of the circuit below Using the usual combination rules gives ( ) C j L j R C j L j R Z Z Z Z Z ω ω ω ω + + + = + = Which we can simplify to C R j LC L j R Z ω + ω ω + =
Circuit Analysis II WRM MT 8. Operations on phasors We have just introduced a method of analysing AC circuits in terms of complex currents and voltages. This method inevitably involves the manipulation of complex phasor quantities and so we list below the results for manipulating these quantities which are, course, simply the standard rules for complex numbers. Sometimes it is easier to use the a+jb notation and sometimes the r exp jθ = r θ notation is easiest. We summarise below the important relationships. Addition and Subtraction If then I = a+ jb and I = c + jd ( ) I I = a ± c + j b ± d ± where the real and imaginary parts add/subtract 35
Multiplication Here it is easiest by far to use the r θ rotation. If I = r θ = r exp jθ and I = r exp jθ I I = r r exp j θ + θ = r r θ + θ then ( ) when we see the amplitudes multiply and the arguments add For division we have I I = = exp r r exp j θ exp j θ r r j r r ( θ θ ) = θ θ the amplitudes divide and the arguments subtract. 36
Circuit Analysis II WRM MT Complex conjugates also appear. If I = a + j b = r exp j θ = r θ then the complex conjugate I*, is given by I * = a j b = r exp j θ= r θ from which we see I + I * = Re * {} I ; I I = j Im{} I where Re{ } denotes the real part the Im { } denotes the imaginary part. 37
Rationalising We are often confronted with expressions of the form a+ j b c + j d And sometimes we wish to rationalise them. We do this in one of two ways. The first is to multiply top and bottom by c j d. This gives a + j b a + j b = c + j d c + j d c c j d j d ac + bd = + c + d bc ad j c + d Alternatively, we can write a + jb as r exp j θ with a + r = b and tan θ =b a. Similarly c + jd may be written as r exp j θ and hence a + jb r = c + jd r = exp exp jθ exp jθ r r j r r ( θ θ ) = θ θ where the amplitudes divide and the arguments subtract. There is no golden rule as to which approach to take it is determined by the problem at hand. However, if you do not have a pressing need to rationalise the expression, we will see some quite good reasons why we may often prefer to stick with the factorised form and not rationalise at all. 38
Circuit Analysis II WRM MT 9. Phasor diagrams Although direct calculation is easily carried out using phasors it is sometimes useful to use a phasor (Argand) diagram to show the relationships between say a voltage and current phasor graphically. In this way it is easy to see their relative amplitudes and phases and hence gain quick insight into how the circuit operates. As a simple example consider the circuit below Since the current, I, flows through both elements it is sensible to choose this as the reference phasor. Having made this choice the voltage drop across the resistor, V R = IR, whereas that across the capacitor, V c = I jωc = j ωc I. The sum of these voltages must equal V. The phasor diagram is easily drawn as 39
from which it is clear that the voltage lags the current by an angle φ. The angle φ may be obtained from the diagram as ( ) φ=tan ωcr. Let's consider another example We could obtain the relationship between I and V by using the equivalent impedance derived earlier. However we will use a phasor diagram to show the various currents and voltages which appear across the various components. Since the voltage, V, is the same across each arm it is sensible to choose this as the reference phasor. The relationships are I + = jωcv V = I R + jωli and I = I I The two phasor diagrams are 40
Circuit Analysis II WRM MT or, combining onto a single diagram where φ denotes the phase angle between V and I. In the diagram above I leads V by φ. 4
0. Thévenin and Norton equivalent circuits The Thévenin and Norton theorems apply equally well in the AC case. Here we replace any arbitrarily complicated circuit containing resistors, capacitors, inductors by a circuit whose behaviour, as far as the outside world is concerned, is entirely equivalent. The two choices are the Thévenin equivalent and the Norton equivalent The methods for determining V, I and Z are identical to those used in the DC case. In general (i) Calculate the open circuit voltage, V oc (ii) Calculate the short circuit current, I sc 4
Circuit Analysis II WRM MT From which V V = Voc, I = Isc and Z = I oc sc We note, of course that in the absence of dependent sources it is often easier to "set the sources to zero" and simply calculate the terminal impedance Z ab. We emphasise again that when a voltage source is "set to zero" it is replaced by a short circuit whereas when a current source is "set to zero", no current flows, and hence the source is replaced by an open circuit. Finally we note that the equivalent impedance Z is also frequently referred to as either (i) internal impedance or (ii) output impedance. 43
Example Find the Norton and Thévenin equivalents of Noting that there are no dependent sources it is easiest to calculate Z ab directly with the voltage source replaced by a short circuit. This is easy since the circuit then reduces to an inductor in parallel with a resistor. Thus j 40.0 = Z = =6+ j 8 0+ j40 Z ab We now need to calculate the open circuit voltage between a and b. This is easy since the circuit is essentially a voltage divider V oc 0 = 0 + j 40 = 50 exp j 0 5 exp j 63.4 =.3 53.4 505 exp j 0 =.3 exp V j 53.4 Thus the Thévenin equivalent circuit takes the form 44
Circuit Analysis II WRM MT In order to find the Norton equivalent we need to find the current flowing between the terminals a and b when they are shorted together. In this case the circuit becomes and I sc 50 exp j 0 50 exp j 0 = = =.5 exp j 80 j 40 40 exp j 90 =.5 80 and hence the Norton equivalent becomes 45
We have elected to find the Norton equivalent directly. However it is equally possible to transform between Thévenin and Norton equivalents directly as we did in the DC case. It is left as an exercise to confirm that Hence we could have worked out the Norton current source in our example directly from the Thévenin equivalent as V I = Z oc.36 exp j 53.04 = 6 + j 8.36 exp j 53.4 = 7.89 exp j 6.6 =.5 exp j 80 =.5 80 which, of course, is the value we previously calculated. 46