1 CHAPTER 13 ALTERNATING CURRENTS + A B FIGURE XIII.1

Transcription

1 3. Alternating current in an inductance CHAPTE 3 ALTENATNG CUENTS & + A B FGUE X. n the figure we see a current increasing to the right and passing through an inductor. As a consequence of the inductance, a back EMF will be induced, with the signs as indicated. denote the back EMF by = A B. The back EMF is given by = L&. Now suppose that the current is an alternating current given by = ˆ sin ωt. 3.. n that case & = ˆ ωcos ωt, and therefore the back EMF is = L ˆ ωcos ωt, 3.. which can be written = ˆ cos ωt, 3..3 where the peak voltage is ˆ = Lωˆ 3..4 and, of course MS = Lω.(See Section 3..) MS The quantity Lω is called the inductive reactance X L. t is expressed in ohms (check the dimensions), and, the higher the frequency, the greater the reactance. (The frequency ν is ω/(π).) Comparison of equations 3.. and 3..3 shows that the current and voltage are out of phase, and that leads on by 90 o, as shown in figure X.. FGUE X.

2 3. Alternating oltage across a Capacitor + FGUE X.3 At any time, the charge Q on the capacitor is related to the potential difference across it by Q = C. f there is a current in the circuit, then Q is changing, and = C&. Now suppose that an alternating voltage given by = ˆ sin ωt 3.. is applied across the capacitor. n that case the current is = Cωˆ cos ωt, 3.. which can be written = ˆ cos ωt, 3..3 where the peak current is ˆ = Cωˆ 3..4 and, of course MS = Cω. MS The quantity /(Cω) is called the capacitive reactance X C. t is expressed in ohms (check the dimensions), and, the higher the frequency, the smaller the reactance. (The frequency ν is ω/(π).) [When we come to deal with complex numbers, in the next and future sections, we shall incorporate a sign into the reactance. We shall call the reactance of a capacitor /( Cω) rather than merely /( C ω), and the minus sign will indicate to us that lags behind. The reactance of an inductor will remain Lω, since leads on. ] Comparison of equations 3.. and 3..3 shows that the current and voltage are out of phase, and that lags behind by 90 o, as shown in figure X.4. FGUE X.4

3 3.3 Complex Numbers 3 am now going to repeat the analyses of Sections 3. and 3. using the notation of complex numbers. n the context of alternating current theory, the imaginary unit is customarily given the symbol j rather than i, so that the symbol i is available, if need be, for electric currents. am making the assumption that the reader is familiar with the basics of complex numbers; without that background, the reader may have difficulty with much of this chapter. We start with the inductance. f the current is changing, there will be a back EMF given by = L. & f the current is changing as ˆ jωt = e, 3.3. then & ˆ jωt = jωe = jω. Therefore the voltage is given by = jlω The quantity jlω is called the impedance of the inductor, and is j times its reactance. ts reactance is Lω, and, in S units, is expressed in ohms. Equation 3.3. (in particular the operator j on the right hand side) tells us that leads on by 90 o. Now suppose that an alternating voltage is applied across a capacitor. The charge on the capacitor at any time is Q = C, and the current is = C&. f the voltage is changing as ˆ jωt = e, then & ˆ jωt = jωe = jω. Therefore the current is given by That is to say = jcω = j Cω The quantity j / ( Cω) is called the impedance of the capacitor, and is j times its reactance. ts reactance is /( C ω), and, in S units, is expressed in ohms. Equation (in particular the operator j on the right hand side) tells us that lags behind by 90 o. n summary: nductor: eactance = Lω. mpedance = jlω. leads on. Capacitor: eactance = /( C ω) mpedance = j/(cω). lags behind.

4 4 t may be that at this stage you haven't got a very clear idea of the distinction between reactance (symbol X) and impedance (symbol Z) other than that one seems to be j times the other. The next section deals with a slightly more complicated situation, namely a resistor and an inductor in series. (n practice, it may be one piece of equipment, such as a solenoid, that has both resistance and inductance.) Paradoxically, you may find it easier to understand the distinction between impedance and reactance from this more complicated situation. 3.4 esistance and nductance in Series The impedance is just the sum of the resistance of the resistor and the impedance of the inductor: Z = + jlω Thus the impedance is a complex number, whose real part is the resistance and whose imaginary part Lω is the reactance. For a pure resistance, the impedance is real, and and are in phase. For a pure inductance, the impedance is imaginary (reactive), and there is a 90 o phase difference between and. The voltage and current are related by = Z = ( + jlω ) Those who are familiar with complex numbers will see that this means that leads on, not by 90 o, but by the argument of the complex impedance, namely tan ( Lω / ). Further the ratio of the peak (or MS) voltage to the peak (or MS) current is equal to the modulus of the impedance, namely + L ω. 3.5 esistance and Capacitance in Series Likewise the impedance of a resistance and a capacitance in series is Z = j / ( Cω ) The voltage and current are related, as usual, by = Z. Equation 3.5. shows that the voltage lags behind the current by tan [ / ( Cω )], and that ˆ ˆ / = + /( Cω). 3.6 Admittance n general, the impedance of a circuit is partly resistive and partly reactive: Z = + jx. 3.6.

5 5 The real part is the resistance, and the imaginary part is the reactance. The relation between and is = Z. f the circuit is purely resistive, and are in phase. f is it purely reactive, and differ in phase by 90 o. The reactance may be partly inductive and partly capacitive, so that Z = + j( X C ). L + X 3.6. ndeed we shall describe such a system in detail in the next section. Note that X C is negative. [The equation is sometimes written Z = + j X L X ), in which X C is intended to represent the ( C unsigned quantity /( C ω). n these notes X C is intended to represent /( C ω).] The reciprocal of the impedance Z is the admittance, Y. Thus Y = = Z + jx And of course, since = Z, = Y. Whenever we see a complex (or a purely imaginary) number in the denominator of an expression, we always immediately multiply top and bottom by the complex conjugate, so equation becomes Z * jx Y = = Z + X This can be written where the real part, G, is the conductance: Y = G + jb, G = + X, and the imaginary part, B, is the susceptance: B X = + X The S unit for admittance, conductance and susceptance is the siemens (or the "mho" in informal talk). leave it to the reader to show that

6 = 6 G G + B and X B = G + B Example What is the impedance of the circuit below to alternating current of frequency 000/π Hz (ω = 4000 rad s )? A C 5 Ω 40 Ω 5 µf.5 mh B D think that the following will be readily agreed. (emember, the admittance is the reciprocal of the impedance; and, whenever you see a complex number in a denominator, immediately multiply top and bottom by the conjugate.) mpedance of AB = (5 50j) Ω mpedance of CD = (40 + 0j) Ω + j Admittance of AB = S = 5 4 j Admittance of CD = S = j 750 S 0 5 j 750 S

7 Admittance of the circuit = j S mpedance of the circuit = 30(6 7 j) 9 Ω The current leads on the voltage by 5º Another example A B D 3 = e ˆ jωt C F E G Three resistors and a capacitor are connected to an AC voltage source as shown. The point E is grounded (earthed), and its potential can be taken as zero. Calculate the three currents, and the potential at B. We can do this by using Kirchhoff s rules in the usual way. When did this found the algebra to be slightly heavy going, and found that it was much simplified by writing a C ω =, where a is j a dimensionless number. Then, instead of writing the impedance of the section BDG as, C ω write it as ( ja). Kirchhoff s rules, applied to two circuits and the point B, are = ja) ( 3 = 3 = These equations are to be solved for the three currents,, 3. These will all be complex numbers, representing alternating currents. Solution could proceed, for example, by eliminating 3 from equations (0) and (), and then eliminating 3 from equations (0) and (). This results in two equations in and. We can eliminate from these to obtain. make it =, (3 aj) but then we immediately multiply top and bottom by 3+ aj to obtain

8 8 3 + aj a = t is then straightforward to return to the original equations to obtain 6 + a + aj a = and a aj a = For example, suppose that the frequency and the capacitance were such that a =, then 8 + j = j = and 5 j = Thus l leads on by 7º.; leads on by 33º.7; and 3 lags behind by º.3. The vector (phasor) diagram for these three currents is shown below, in which the phasor representing the alternating voltage is directed along the real axis. 3

9 Bearing in mind that the potential at E is zero, we see that the potential at B is just 9 3 and is in phase with 3. There is another method of finding, which we now try. f we get the same answer by both methods, this will be a nice check for possible mistakes in the algebra. ll re-draw the circuit diagram as follows: = e ˆ jωt C To calculate we have to calculate the admittance Y of the circuit, and then we have immediately = Y. The impedance of and C in series is ja and so its admittance is. The admittance of the rectangle is therefore + =. ja. The ja ja ja impedance of the rectangle is. ja, and the impedance of the whole circuit is plus this, ja which is. 3 ja. The admittance of the whole circuit is. ja. ja 3 ja 6 bottom by the conjugate of the denominator to obtain. + a + ja a. 6 + a + ja =, which is what we obtained by the Kirchhoff method a Hence Multiply top and

10 0 f you want to invent some similar problems, either as a student for practice, or as an instructor looking for homework or examination questions, you could generalize the above problem as follows. A B Z 3 D = e ˆ jωt Z 3 Z F E G Each of the three impedances in the circuit could be various combinations of capacitors and inductors in series or in parallel, but, whatever the configuration, each could be written in the form + jx. Three Kirchhoff equations could be constructed as follows = Z 3Z3 Z Z 3 f have done my algebra correctly, make the solutions = = ( Z + Z ) 3 = 3.6. ZZ3 + Z3Z + ZZ Z = 3 ZZ3 + Z3Z + ZZ Z = ZZ3 + Z3Z + ZZ Each must eventually be written in the form e + j m, or ( G + jb). For example, suppose that the three impedances, in ohms, are Z = 3 j, Z = 4 + j, Z3 = 3 3 j. n that case, believe (let me know if m wrong, jtatum at uvic.ca) that equation becomes, after j 49 manipulation, 3 =. This means that 3 leads on by 64º.0, and that the peak value of 3 is 0.8 A, where is in. The potential at B is 3 Z 3. Both of these are complex numbers and the potential at B us not in phase with 3 unless Z 3 is purely resistive.

11 n composing a problem, you probably want all resistances and reactances to be of comparable magnitudes, say a few ohms each. As a guide, if you choose the frequency to be 500/π Hz, so that ω = 0 3 rad s, and if you choose inductances to be about 0 mh and capacitances about 00 µf, your reactances will each be about 0 Ω. You can probably also compose problems with various bridge circuits, such as Z Z Z 3 Z 6 Z 4 Z 5 There are six independent impedances, so you ll need six equations. Three for Kirchhoff s second rule, to cover the complete circuit once; and three of Kirchhoff s first rule, at three points. Good luck in solving them. emember that, in an equation involving complex numbers, the real and imaginary parts are separately equal. And remember, as soon as a complex number appears in a denominator, multiply top and bottom by the conjugate. Alternatively, and easier, we could do what we did with a similar problem with direct currents in Section 4., using a delta-star transformation. We ll try an example in Section 3.9, subsection 3.9.4, and we make some further remarks on the delta-star transform with alternating currents in Section The LC Series Acceptor Circuit A resistance, inductance and a capacitance in series is called an "acceptor" circuit, presumably because, for some combination of the parameters, the magnitude of the inductance is a minimum, and so current is accepted most readily. We see in figure X.5 an alternating voltage jωt = e ˆ applied across such an, L and C. = e ˆ jωt L C FGUE X.5

12 The impedance is Z = + j Lω Cω We can see that the voltage leads on the current if the reactance is positive; that is, if the inductive reactance is greater than the capacitive reactance; that is, if ω > / LC. (ecall that the frequency, ν, is ω/(π)). f ω < / LC, the voltage lags behind the current. And if ω = / LC, the circuit is purely resistive, and voltage and current are in phase. The magnitude of the impedance (which is equal to ˆ / ˆ ) is ( Lω /( ω) ), Z = + C 3.7. and this is least (and hence the current is greatest) when ω = / LC, the resonant frequency, which shall denote by ω 0. t is of interest to draw a graph of how the magnitude of the impedance varies with frequency for various values of the circuit parameters. can reduce the number of parameters by defining the dimensionless quantities Ω = ω / ω and Q z = L C Z = You should verify that Q is indeed dimensionless. We shall see that the sharpness of the resonance depends on Q, which is known as the quality factor (hence the symbol Q). n terms of the dimensionless parameters, equation 3.7. becomes z = + Q ( Ω / Ω ) This is shown in figure X.6, in which it can be seen that the higher the quality factor, the sharper the resonance.

13 3 8 FGUE X.6 6 mpedance z Q = Angular frequency Ω n particular, it is easy to show that the frequencies at which the impedance is twice its minimum value are given by the positive solutions of Ω Q Ω + = f denote the smaller and larger of these solutions by Ω and Ω +, then Ω + Ω will serve as a useful description of the width of the resonance, and this is shown as a function of quality factor in figure X.7.

14 4 FGUE X Width Q 3.8 The LC Parallel ejector Circuit n the circuit below, the magnitude of the admittance is least for certain values of the parameters. When you tune a radio set, you are changing the overlap area (and hence the capacitance) of the plates of a variable air-spaced capacitor so that the admittance is a minimum for a given frequency, so as to ensure the highest potential difference across the circuit. This resonance, as we shall see, does not occur for an angular frequency of exactly / LC, but at an angular frequency that is approximately this if the resistance is small. = e ˆ jωt L C

15 5 The admittance is Y = jcω jlω After some routine algebra (multiply top and bottom by the conjugate; then collect real and imaginary parts), this becomes + jω( L Cω + C L) Y = L ω The magnitude of the admittance is least when the susceptance is zero, which occurs at an angular frequency of ω 0 = LC L f << L / C, this is approximately / LC. 3.9 AC Bridges We have already met, in Chapter 4, Section 4., the Wheatstone bridge, which is a DC (direct current) bridge for comparing resistances, or for "measuring" an unknown resistance if it is compared with a known resistance. n the Wheatstone bridge (figure.9), balance is achieved when 3 =. Likewise in a AC (alternating current) bridge, in which the power supply is an 4 AC generator, and there are impedances (combinations of, L and C ) in each arm (figure X.8), i Z Z Z 3 Z 4 FGUE X.8

16 balance is achieved when 6 Z Z Z3 = 3.9. Z 4 or, of course, Z Z =. This means not only that the MS potentials on both sides of the detector Z3 Z4 must be equal, but they must be in phase, so that the potentials are the same at all times. ( have drawn the "detector" as though it were a galvanometer, simply because that is easiest for me to draw. n practice, it might be a pair of earphones or an oscilloscope.) Each side of equation 3.9. is a complex number, and two complex numbers are equal if and only if their real and imaginary parts are separately equal. Thus equation 3.9. really represents two equations which are necessary in order to satisfy the two conditions that the potentials on either side of the detector are equal in magnitude and in phase. We shall look at three examples of AC bridges. t is not recommended that these be committed to memory. They are described only as examples of how to do the calculation The Owen Bridge i L =? C 4 C 3 4 FGUE X.9

17 7 This bridge can be used for measuring inductance. Note that the unknown inductance is the only inductance in the bridge. eactance is supplied by the capacitors. Equation 3.9. in this case becomes + jl ω = j / ( C3ω) j / ( C ω) L That is, 4 j = j C ω C C ω On equating real and imaginary parts separately, we obtain L = C and C4 = C The Schering Bridge This bridge can be used for measuring capacitance. i C =? C 4 C 3 4 FGUE X.0

18 8 The admittance of the fourth arm is + jc4ω, and its impedance is the reciprocal of this. leave 4 the reader to balance the bridge and to show that and The Wien Bridge C C4 = C 3 C34 = i 3 C 3 4 C 4 FGUE X. This bridge can be used for measuring frequency. The reader will, think, be able to show that

19 4 3 9 C3 + = C 4 and ω = C 3 C Bridge Solution by Delta-Star Transform n the above examples, we have calculated the condition that there is no current in the detector -. i.e. that the bridge is balanced. Such calculations are relatively easy. But what if the bridge is not balanced? Can we calculate the impedance of the circuit? Can we calculate the currents in each branch, or the potentials at any points? This is evidently a little harder. We should be able to do it. Kirchhoff s rules and the delta-star transform still apply for alternating currents, the complication being that all impedances, currents and potentials are complex numbers. Let us start by trying the following problem: ω = 500 rad s i 0 mh Ω A 5 Ω 3Ω 400 µf 6 Ω 500 µf 6 mh 00 µf 4 Ω

20 0 That is to say: ω = 500 rad s i B B Z =5+5j 3 Z 4 = 5j The impedances are indicated in ohms. A Z 3 =3 4j 4 C 5 Z =6 0j Z 5 =4+3j D Our question is: What is the impedance of the circuit at a frequency of ω = 500 rad s? We refer now to Chapter 4, Section. We are going to replace the left hand delta with its equivalent star. ecall equations With alternating currents, We can use the same equations with alternating currents, provided that we replace the 3GG G3 in the delta and the r r r3 ggg3 in the star with Z ZZ3Y YY3 in the delta and zz z3 y y y3 in the star. That is, we replace the resistances with impedances, and conductances with admittances. This is going to need a little bit of calculation, and familiarity with complex numbers. used a computer - hand ZZ3 calculation was too tedious and prone to mistakes. This is what got, using z = Z + Z + Z and 3 cyclic variations:

21 C ω = 500 rad s i B = i = 4 4 z Z 4 = 5j The impedances are indicated in ohms. A z 3 P C Potential here taken to be zero. i = 5 5 z Z 5 =4+3j z = j D z = j z 3 = j The impedance of PBC is j Ω. The impedance of PDC is j Ω. The admittance of PBC is j S. The admittance of PDC is j S. The admittance of PC is j S. The impedance of PC is j Ω. The impedance of AC is j Ω. The admittance of AC is j S.

22 Suppose the applied voltage is = = ˆ A C cos ωt, with ω = 500 rad s and ˆ = 4. Can we find the currents in each of Z, Z, Z3, Z4, Z5, and the potential differences between the several points? This should be fun. m sitting in front of my computer. Among other things, have trained it instantly to multiply two complex numbers and also to calculate the reciprocal of a complex number. f ask it for ( j)(.9 3.4j) it will instantly tell me j. f ask it for /(0.5 +.j) it will instantly tell me j. can instantaneously convert between impedance and admittance. The current through any element depends on the potential difference across it. We can take any point to have zero potential, and determine the potentials at other points relative to that point. choose to take the potential at C to be zero, and have indicated this by means of a ground (earth) symbol at C. We are going to try to find the potentials at various other points relative to that at C. n drawings B and C have indicated the currents with arrows. Since the currents are alternating, they should, perhaps, be drawn as double-headed arrows. However, have drawn them in the direction that think they should be at some instant when the potential at A is greater than the potential at C. f any of my guesses are wrong, ll get a negative answer in the usual way. The total current is times the admittance of the circuit. That is: = 4 ( j) = ( j) A. The peak current will be A (because the modulus of the admittance is S), and the current lags behind the voltage by 3º.9. hope the following two equations are obvious from drawing C. From this, = ( z + Z4) = 5( z + Z5) 4 = z + z z + Z + Z Z 5 = j j ( j) = j A 4 leads on by 8º.7 ˆ4 = A. z + Z4 5 = = j z z Z4 Z A 5 lags behind by º. ˆ5 = A

23 3 Check: = show below graphs of the potential difference between A and C ( = A C ) and the currents, 4 and 5. The origin for the horizontal scale is such that the potential at A is zero at t = 0. The vertical scale is in volts for C, and is five times the current in amps for the three currents ωt deg We don t really need to know i, i, i3, but we do want to know,, 3. Let s first see if we can find some potentials relative to the point C. From drawing C we see that = = Z, which results in = ( j) B C B 4 4 B B lags behind by rad = 49º.5 ˆ = B From drawing C we see that

24 D C = D = Z, which results in = ( j) 5 5 D 4 D leads on by rad = 4º.7 ˆ = B B D We can now calculate 3 (see drawing B) from 3 = = ( B D ) Y3. find Z 3 = j A 3 3 lags behind by rad = 60º.0. ˆ3 = A. A B can now be found from = Z = ( A B ) Y. The real part of A is 4, and (since we have grounded C), its imaginary part is zero. f in doubt about this verify that Z = ( j)( j) = (4 + 0 j). find = j A lags behind by rad = 5º.4. ˆ = A. n a similar manner, find = j A leads on by rad = 49º.4. ˆ =.6689 A. Summary: = j = j = j 3 = j 4 = j 5 = j A A A A A A These may be checked by verification of Kirchhoff s first rule at each of the points A, B, C, D.

25 Further remarks on the delta-star transform with alternating currents are given in Section The Transformer We met the transformer briefly in Section 0.9. There we pointed out that the EMF induced in the secondary coil is equal to the number of turns in the secondary coil times the rate of change of magnetic flux; and the flux is proportional to the EMF applied to the primary times the number of turns in the primary. Hence we deduced the well known relation N = 3.0. N relating the primary and secondary voltages to the number of turns in each. We now look at the transformer in more detail; in particular, we look at what happens when we connect the secondary coil to a circuit and take power from it. = e ˆ jωt i L L M FGUE X. jωt n figure X., we apply an AC EMF = e ˆ to the primary circuit. The self inductance of the primary coil is L l, and an alternating current flows in the primary circuit. The self inductance of the secondary coil is L, and the mutual inductance of the two coils is M. f the coupling between the two coils is very tight, then M = L L ; otherwise it is less than this. am supposing that the resistance of the primary circuit is much smaller than the reactance, so am going to neglect it.

26 6 The secondary coil is connected to a resistance. An alternating current flows in the secondary circuit. Let us apply Ohm's law (or Kirchhoff's second rule) to each of the two circuits. n the primary circuit, the applied EMF is opposed by two back EMF's: = L & + & M That is to say = jωl + jωm Similarly for the secondary circuit: 0 = j M + j L + ω ω These are two simultaneous equations for the currents, and we can (with a small effort) solve them for and : L ωl L L + j ωm = j M M M ωm ωm M and + j L. ω = L L This would be easier to understand if we were to do the necessary algebra to write these in the forms = ( a + jb) and = ( c + jd). We could then easily see the phase relationships between the current and as well as the peak values of the currents. There is no reason why we should not try this, but am going to be a bit lazy before do it, and am going to assume that we have a well designed transformer in which the secondary coil is really tightly wound around the primary, and M L L. f you wish, you may carry on with a less efficient transformer, with = M = k L L, where k is a coupling coefficient less than, but 'm going to stick with M = L L. n that case, equations and 6 eventually take the forms = L L j L ω = N N j L ω and L L N = = N These equations will tell us, on examination, the magnitudes of the currents, and their phases relative to.

27 Now look at the circuit shown in figure X.3. 7 = e ˆ jωt i L N N FGUE X.3 n figure X.3 we have a resistance ( N / N ) in parallel with an inductance L. The admittances of these two elements are, respectively, ( N / N) / and j / ( Lω ), so the total N admittance is j N L ω. Thus, as far as the relationship between current and voltage is concerned, the primary circuit of the transformer is precisely equivalent to the circuit drawn in figure X.3. To see the relationship between and, we need look no further than figure X.3. Likewise, equation shows us that the relationship between and is exactly as if we had an AC generator of EMF N / N connected across, as in figure X.4. N N jωt e ˆ i FGUE X.4 Note that, if the secondary is short-circuited (i.e. if = 0 and if the resistance of the secondary coil is literally zero) both the primary and secondary current become infinite. f the secondary circuit is left open (i.e. = ), the secondary current is zero (as expected), and the primary current, also as expected, is not zero but is j/ ( Lω ); That is to say, the current is of magnitude / ( L ) ω and it lags behind the voltage by 90 o, just as if the secondary circuit were not there.

28 8 3. oot-mean-square values, power and impedance matching We have been dealing with alternating currents of the form jωt = e ˆ. have been using the notation Î to denote the peak (i.e. maximum) value of the current. Of course, in the notation of complex numbers, this is synonymous with the modulus of. That is to say ˆ = mod =. shall use one or other notation wherever it is convenient. This will often mean using ^ when describing time-varying quantities, and when describing constant (but perhaps frequency dependent) quantities, such as impedances. Suppose we have a current that varies with time as = ˆ sin ωt. During a complete period ( P = π / ω) the average or mean current is zero. The mean of the square of the current, however, is not zero. The mean square current,, is defined such that P 0 P = dt. 3.. With ˆ = sin ωt, this gives = ˆ. The square root of this is the root-mean-square current, or the MS value of the current: = ˆ = 0.707ˆ. 3.. MS When we are told that an alternating current is so many amps, or an alternating voltage is so many volts, it is usually the MS value that is meant, though we cannot be sure of this unless the speaker or writer explicitly says so. f you wish to be understood and not misunderstood in your own writings, you will always make it explicitly clear what meaning you intend. f an alternating current is flowing through a resistor, at some instant when the current is, the instantaneous rate of dissipation of energy in the resistor is. The mean rate of dissipation of energy during a complete cycle is MS. This is one obvious reason why the concept of MS current is important. Now cast your mind back to Chapter 4, Section 4.8. There we imagined that we connected a resistance across a battery of EMF E and internal resistance r. We calculated that the power E delivered to the resistance was P = =, and that this was greatest (and equal to ( + r) E / r ), when the external resistance was equal to the internal resistance of the battery. 4

29 9 What is the corresponding situation with alternating current? Suppose we have a box (a j t source ) that delivers an alternating voltage (which is represented by a complex number ˆ e ω ), and that this box has an internal impedance z = r + jx. f we connect across the box a device (a load ) that has an impedance Z = + jx, what will be the power delivered to the load, and can we match the external impedance of the load to the internal impedance of the box in such a manner that the power delivered to the load is greatest? The second question is quite easy to answer. eactance can be either positive (inductive) or negative (capacitive), and so it is quite possible for the total reactance of the entire circuit to be zero. Thus for a start, we want to ensure that X = x. That is, the external reactance should be equal in magnitude but opposite in sign to the internal reactance. The circuit is then purely resistive, and the power delivered to the circuit is just what it was in the direct current case, namely E P = =, where the current and EMF in this equation are now MS values. And, as ( + r) in the direct current case, this is greatest if = r. The conclusion is that, for maximum power transfer, + jx should equal r jx. That is, for the external and internal impedances to be matched for maximum power transfer, the source impedance. * Z = z. The load impedance should equal the conjugate of What is the power delivered to the load when the impedances are not matched? n other words, when z = r + jx and Z = + jx. t is P = ˆ. The current is given by the equation E = ( Z + z). (These are all complex numbers - i.e. they are all periodic functions with different phases. E and vary with time.) Now if w and w are two complex numbers, it is well known (from courses in complex numbers) that w w = w w. We apply this now to E = ( Z + z). [ shall use ^ for the peak of the time-varying quantities, and for the modulus of the impedances] We obtain E ˆ = ˆ Z + z. ˆ MS Thus ˆ E E P = = = 3..3 Z + z ( + r) + ( X + x) 3. Some emarks on the Star-Delta Transform We pointed out in Section that we can use the star-delta transform with alternating currents, provided that we replace the several, G, r, g quantities in equations with the corresponding Z, Y, z, y quantities of alternating current theory. Because Z, Y, z, y are all complex, the calculation of the transforms may be tedious, and it requires care and patience even though it is in principle straightforward. The calculation is eased somewhat by understanding that if two complex numbers are equal, their real and imaginary parts are separately equal, and by remembering what to do if a complex number appears in the denominator of an expression. We successfully performed such a calculation in Section

30 30 A warning may be in order. t may sometimes happen that, in applying the delta-star transform to a circuit with an alternating current, in the transformed geometry the real part of the impedance of an arm (i.e. the resistance) turns out to be negative. This may be disconcerting when first encountered. Nevertheless, even in such cases, the delta-star transform may still be a useful computational device in solving a circuit problem, even though the transformed circuit is a physical impossibility. A negative didn t happen in our example in Section 3.9.4, but here s a simple example where it does happen. What is the equivalent delta of the star shown below, assuming that the angular 4 frequency of the current is ω = 0 rad s (frequency ν = 0 4 /π Hz)? 0. Ω 00 µf 0.06 mh Let s relabel the drawing showing the impedances in ohms. z = 0. z = 0.5j z 3 =.j Using the transforms the delta, in ohms, are: Z = j Z = j Z 3 = 0.583& 0.5j Z z z3 + z3z + zz =, etc., we find that the impedances of the sides of z

31 ecall that inductive reactance is Lω and capacitive reactance is /(Cω). With this in mind we see that: We could represent Z as a resistance of.50 Ω in series with an inductance of mh. We could represent Z as a resistance of 0.76 Ω in series with an inductance of 0.06 mh. We could represent Z 3 as a resistance of 0.583& Ω in series with a capacitance of 00 µf, 3 n a typical problem we may be given the elements of a star, and, for a given frequency, be asked to determine the elements of the corresponding star, or vice versa. However, since the reactance of every element is frequency-dependent, this gives the opportunity of devising a problem in which the resistances, capacitances and inductances are given in all arms of both the star and the corresponding delta, and asking what the frequency must be. For example, suppose the star has First arm: A resistance of 0. Ω Second arm: A capacitance of 00 µf Third arm: An inductance of 0.06 mh and the corresponding delta has First side: A resistance of.50 Ω in series with an inductance of mh Second side: A resistance of 0.76 Ω in series with an inductance of 0.06 mh Third side: A resistance of 0.583& Ω in series with a capacitance of 00 µf. What is the frequency? The impedances in ohms are: n the star z = 0. z z 3 = 5000 j / ω = 6 0 n the delta: Z Z Z 3 = 5 jω jω = jω = 0.583& 5000 j / ω f we now use Z z z3 + z3z + zz 4 = we find, with some care, that ω = 0 rad s z

32 3 f one were to start with the elements of the delta and the corresponding star in Section 3.9.4, one could presumably recover the angular frequency of 500 rad s. don t think can summon up the energy myself.