Baixe sadiku 5ª edi...icos e extras - chapter 13 magnetically coupled circuits e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! CHAPTER 13 - MAGNETICALLY COUPLED CIRCUITS List of topics for this chapter : Mutual Inductance Energy in a Coupled Circuit Linear Transformers Ideal Transformers Three-Phase Transformers PSpice Analysis of Magnetically Coupled Circuits Applications MUTUAL INDUCTANCE Problem 13,1 Given the circuit in Figure 13.1 and k =1, find 1, and I,. joM 102 109 MAN MA e . 1020º Y C L ) qo E jo o (1) (O) 1020º V Figure 13.1 » Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. » PRESENT everything you know about the problem. We know all values of the independent sources. We also know the values of all the elements. In order to find the equivalent circuit containing the induced voltages, we need to know the mutual inductance, M. We know that the coupling coefficient is ko = Jul, Then, M=RLT; = JD 221 > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. e The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. The goal of the problem is to find two currents. Ience, mesh analysis will be used. > ATTEMPT a problem solution. From the circuit in Figure 13.1, we can see that joL, = joL, = jt0. Thus, oL =oL,=10andL =L,=L. Hence, M=yLL, =L and joM= joL=jl0. So, the equivalent circuit is | 109 100 AN MAs F jo j10 0 1020ºV O) K L O 1020ºV j10L, j101 Now, using mesh analysis, Loop !: -10+101 +jt01, -jlOI, = 0 (10+j10)X, —j101, =10 d+DL-jh=l Loop2: -jlOI, +j101, +10L, +10=0 -j101, +(10+j10)I, =-10 “5 +0+DL =-1 In matrix form, [1+j -5 E! [a Li 1+j 11, or [5 dj 1 [ti qt) where A=(+)º CP =(+pP+))-P=i+p. 222 The circuit can be redrawn as 1020º V O k Using the dot convention, we can draw an equivalent circuit to incorporate the induced voltages from the coupling effects. 59 Bh) joao ] With this circuit, we can use mesh analysis to find the answer in the frequency domain. Loop 41 : -10+51, +51, —1,)- 581, =0 Loop *2: i8L, +50, —1))-58(1, = 1,)+ 520 + Voa =0 This is a system of two equations and three unknowns. We noed a constraint equation. Due to the open circuit, it is obvious that T, = 0. Combining like terms and introducing the constraint, the equations become +91 =10 and Vou = II, Clearly, 10 10209 L=—.=— = 24-45º "545 5/2248º and V out = j131, = (13/00)422-45) = 13/2245º Using the reference assumed above, the voltage converts to the time domain as Vou (1) = 1342 cos(377t + 45º) V 225 Problem 13.4 Given the circuit in Figure 13.4, find the coupling coefficient, k, and the 0 voltage across the 1-0) resistor. j100 100 9 1020º Y (1) Figure 13.4 k=1 Via =0,120ºV ENERGY IN À COUPLED CIRCUIT Problem 13.5 Given the circuit in Figure 13.5, V, = V, = 10 volts, R, = R, = 10ohms, e oL, =oL, = 10, and oM =5, find the coupling coefficient, k, the currents in the primary and secondary circuits, 1, and I,, and the power absorbed. M Ri R; AM My . . “O (1) vg Bu (1) Or Figure 13.5 The coupling coefficient is k = . LL, Given values for 0M, QL,, and wL, , we need to modify the equation for k to be o oM VOL XOL,) 5 5 k= Jaodo) 10 =45 226 'To find the currents, begin by finding an equivalent circuit that takes into account the coupling e effects, i.e., the induced voltages. 109 1090 Ay | 190 oo 1V O h (15) (O 10V sk ist, Use mesh analysis to find 1, and 1,. Loop 1: 10= (10+)10)1, — j51, Loop2: -10= -j51, +(10+310)1, IO+j0 -j5 L] [io “5 10+j0||L] [-10 where A = (10+j10)(10+ 10) (-]5X-55) = 1200 +25 = (251 + 98). in matrix form, Io+jlo | j5 e L (ESA + j8) -10 10+jt0 ES 100+ j100- 50 Ir] | Cod+i) Cars || 10 |) +) LI j 1o+j0 ||.10| | j50-100-j00 (25X1+58) (25X1+ 8) (25X1+ 8) [ Goc+d ] [4+;2 ] [1] ese) | E Li, | ges D | [é (25X1+ 58) le Thus, A+i2 AMTIL2657º CR Sc gos 31 A L=Trijg T 80623282.880 155474 SÓSIA -4-.2 44702-153.43º = dE TSE = 0,5547/123.69º A L=Trjg” 8.0623282.88º = 0.55472123.690 A 227 Problem 13.7 Given the circuit in Figure 13.7, V, = V, =lOvolts, R,=R, =10ohms, ' oL, =qL, =10, and 0M =5, find the coupling coefficient, k, the currents in the primary and secondary circuits, 1, and I,, and the power absorbed. M i R: R: MA MA . “O (1) “EB 1) Os e Figure 13.7 As seen in Problem 13.5, 5 oM 5 “JOL VImão “107 LS To find the currents, begin by finding an equivalent circuit which takes into account the coupling effects, i.e., the induced voltages. 1092 100 0 My MA jo jo 10V h k 10V 5h Sh Use mesh analysis to find 1 and I,. Loop 1: 10=(10+j10)1, + j51, Loop2: -10=;51, +(10+310)1, In matrix form, [rodo js E [o 5 1o+j10 ) 1, =[.10 where À = (10 + jLOXLO + LO) — (555) = 200 +25 = (25)(1+ 58). 230 [io+po js | e [1,1] [ - 1o+j1o 10] L, [FT exEm -10 T 10+j0 -5 7 [ 100+ 100 :+350 ] K] | Cd+) C5d+;) [to | e)dri) LI -j5 10+jl0 [10 “| tt Les) sue | (25X1+ 8) ] E (502 +43) | k (5X + j8) |- E Pee 5) 1 ENTRO [ Pelé jo | | | ; | Thus, í 2 4+j6 7211256310 agr. MST A ! 1+;8 8.0623282.88º Toi o -4-j6 T2N12-123.69º . L=rig goias — MSMMALISSAS A e Now, find the power absorbed in the circuit. Look at the power absorbed by each element. Starting with the primary circuit, vi =-VIcos6 = -(100.8944) cos(0º — (-26.57º)) = -7.9994 W RS | HER, = (0.8944)? (10) = 7.9995 W bp; = (4.4720X0.8944) cos(243.43º — (-26.57º)) = 3.9998cos(2709) = 0 W where p, is the power absorbed by the induced voltage of L, . Ending with the secondary circuit, py; = VIcosO = (10)(0.8944) cos(0 —153.43º) = -7.9994 W wo =| LR, = (0.89447 (10) = 7.9995 W = (4.4720)(0.8944) c0s(63.43º — 153.43º) = 3.9998cos(-909) = O W where p, is the power absorbed by the induced voltage of L,. The voltage sources absorb —7.9994 watts, or deliver +7.9994 watts, the resistances absorb 7.9995 watts, and the induced voltages absorb 0 watts. The inductors do not absorb power. Problem 13.8 Given the circuit in Figure 13.8, V, = V, = 10 volts, R, = R, = 10 ohms, oL, =oL, =10,and oM =, find the coupling coefficient, k, the currents in the primary and ) secondary circuits, I, and É, and the power absorbed. 231 R; M GY by AM Vi h L; L;, (1) A . . Figure 13.8 k= 6.5 1,=0.55472-56,31º A 1, = 0.5547/123.69º A The voltage sources absorb —3.0769 watts, or deliver +3.0769 watts, the resistances absorb 3.0769 watts, and the induced voltages absorb 0 watts. The inductors do not absorb power. Problem 13.9 Given the circuit in Figure 13.9, V, = V, =lOvolts, R|,=R, = 10ohms, | oL, =ÓL, =10, and oM =5, find the coupling coefficient, k, the currents in the primary and secondary circuits, 1, and L,, and the power absorbed. Ri M Ay ay e MN R Yi K L; L; (1) Vo É Figure 13.9 K=05 1, = 0.89442..26.57º A T, = 0.89442153.43º A The voltage sources absorb -7.9994 watts, or deliver +7.9994 watts, the resistances absorb 7.9995 watts, and the induced voltages absorb O watts. The inductors do not absorb power. 232 Mesh analysis yiclds, Loopl: 10=0.91,+YV, Loop2: V,=10l, This is a set of two equations and four unknowns. Two constraint equations are needed. From the ideal transformer, as shown, we know that V, =nV, and 1, =nl,. l:m “OO EB (1) [)u This implies that Vv,=10V, and 1 =10, which are the two constraint equations. There are many ways to find the values of V,, V,, 1, and L,. Let's find I,. To do this, find V, in terms of 1, and substitute into the equation for loop 1. V, =0.1V, = (0.)0L)=1, =0.1L, and the equation for loop 1 becomes 10=0,91,+0.11, 10=1, Hence, IL =10A V5=1V L=1A V, =10V Therefore, Vig = V, =10V EVALUATE the solution and check for accuracy. Using nodal analysis, V,-10 Atnode 1, 09 +1,=0 or v,-10+0.91, = 0 <1) v,-0 Atnode 2, 10 +(1,)=0 or V,-101,=0 2) 235 Again, this is a set of two equations and four unknowns. Two constraint equations are nceded. From the ideal transformer, we have e V, =10V, G) 1 =101, (4) From (2) and (4), V, = 101, = (10)(1/10)1, =], (5) From (1), (3), and (5), V,-10+0.9V, =V,-10+(0.9)10)V, = (1+9)V, -10=0 10V =10 —> V=1V Then, V,=10V, =10V 1L=V,=10A and L=(/0)L=1A Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If'so, present the solution; if not, then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. Voo = 10V Problem 13.12 [13.33] For the circuit in Figure 13.12, find V,. Switch the dot on the secondary side and find V, again. 20 mF I£ R 10 cos(5t) V Figure 13.12 1 1 C=0.02 F becomes -— = em = jlO joc” 0.02) 236 We apply mesh analysis to the circuit shown below. 0 100 Je K TO no 3:1 +. + + noso v(t)( du ly | v (1) 20 Ev, Using mesh analysis, For mesh 1, 10=101, -101, +Y, 1) For mesh 2, V,=2L=V, (2) For mesh 3, 0=(10-310)1, -101,+V, —V, (3) At the terminais, V,=nV=V, /3 (4) L=nl,=1,/3 (0) From (2) and (4), e V,=61, (6) Substituting this into (1), 10=101, -101, (D Substituting (4) and (6) into (3) yields 0=-101, -41, +(10)0- DL, (8) From (5), (7), and (8), [1 -0333 O úlfo 1º 6 10 |L ! -10 -4 10-j10 (] o | 1, = 22. 1090-100 | 49) 200 20 A -20-j9333 : V,=21, =2.963232.9º V Switching the dot on the secondary side affects only equations (4) and (5). 0 Vo =-Vi/3 (9) 237 (eee PSPICE ANALYSIS OF MAGNETICALLY COUPLED CIRCUITS Problem 13.14 [13.63] | Use PSpiceto find V,, V,, and I, in the circuit in Figure 13.13. 16092 892 + « Elê» mary 40260º V Figure 13.13 The schematic is shown below. o ACMAG=40V Ty 2 0.25 aves 30 v2 ACMAG=30V AcpHaSE-SO(AS) COUPLING=0.899 mo) ACPHASE=0 Lt TURNS=400000 a t2 TURNS=200000 break % In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes FREQ IMLPRINTD IP(VPRINTD 1552 EO 1.955 E401 8.332 E+0I FREQ IM(V PRINT TP(V PRINTS) 1592 Ei 6.847 E+01 4640 EXOJ FREQ IM(V PRINTS) , TP PRINT 1552 Edi 4434 EI 9.260 E+01 0 240 Thus, V, — 19.55283,32º V V, = 68.47246.4º V 1, = 443.47 -92,6º mA -——— 4 APPLICATIONS Problem 13.15 [13.73] A 4800-V rms transmission line feeds a distribution transformer with 1200 turns on the primary and 28 turns on the secondary. When a 10-9) load is connected across the secondary, find : (a) the secondary voltage, (b) the primary and secondary currents, (c) the power supplied to the load. Va N, a —=*=n (a 1 N, N, (2) - v= x, Vi =| 7500 (4800) = 112 V vo 112 (b) = RE To JA l=nl,, where n = 28/1200 I (a 2) = 261.3 mA 1=L 7300 /11-D = 261.8 mA (o p=|LfPR=012200)=1254W 241