ProbSolv_Chapter13

ProbSolv_Chapter13 - CHAPTER 13 - MAGNETICALLY COUPLED...

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Unformatted text preview: 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 INDUCTAN CE Problem 13.1 Given the circuit in Figure 13.1 and k = 1, find I1 and 12. 109 1040° V O 1040° V a 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 M k = =1 L1142 Then, M = 1<./L,L2 = ,/L1L2 221 > Establish a set 0f ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. . 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. Hence, mesh analysis will be used. > ATTEMPT a problem solution. From the circuit in Figure 13.1, we can see that jcoLl : ij2 : j10 _ Thus, 03Ll =coL2 =10 and L1:L2 =L. Hence, M = ,lLle = L and ij = ij =j10. So, the equivalent circuit is y 109 109 1040° V 1040° V Now, using mesh analysis, Loopl: —10+1011+j1011—j1012 =0 (10+j10)11 —j101, =10 (1+j)11-j12 =1 ' 'Loop2: -j101l +j1012 +1012 +10: 0 —j1011+(10+j10)I2 =-10 —jI,+(1+j)I2 2-1 In matrix form, F1+j —j Tllljfl i-j 1+ji12H-1J 01' where A=(1+J')2 —(‘j)2 =(1+j2+j2)_j2 =1+j2- 222 Now, r1+j j _| ~Flll_|1+j2 TIE 11 leJ—[ ‘IJ 1+j2 1+j2 Therefore, 1+j—j 1 140° 11 = _._..__————— ——-—-————= O.4472£-63.43°A 1+j2 "1+j2—$52634? I _j-(1+j)__;L_fl9:_—o4472211657°A 2 1+‘j2 1+j2"J§263.43°‘ ' ' > EVALUATE the solution and check for accuracy. Use KVL to check the solution. The equation produced by KVL of the left loop is -10+1011+j1011—j1012 = 0 The equation produced by KVL of the right loop is 10—j1011+j1012 +1012 = 0 Inserting the values‘for I1 and I2 results in valid equations. Thus, 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. I1 = 0.44724 - 63.43° A 12 = 0.44724116.57° A Problem 13.2 [13. I] For the three coupled coils in Figure 13 .2, calculate the total inductance. 2H Figure 13.2 223 Forcoill, Ll —M12+M,3=6—4+2=4 FOI‘COiIZ, LZ—MZI—M23:8“4_5='1 F0rc0i13, L3 LT =4—1+7=10H or LT =L1+L2 +L3 —2M12 ~2M23 +2M12 LT = 6+8+10—(2)(4)—(2)(5)+(2)(2) LT =6+8+10—8~—10+4= 10H Problem 13.3 For the frequency domain circuit shown in Figure 13.3, determine the value of Vout (t) for vin (t) = lOcos(377t) and a coupling coefficient k = 0.8 . Figure 13.3 Before an equivalent circuit can be drawn, we must determine the value of 03M . Using k = 0.8 , k=M t/LILZ ' Because the circuit is in the frequency domain rather than the time domain, we know the value of COL rather than the value of L. So, transform the equation for k to include 0). Then, 03M _ ,/(mL1)(coL,) ’ coM = k,/(03L1)(ooL2) = (0.8) (5)(20) = 8 We also need to transform the voltage source from the time domain to the frequency domain. Let's assume a reference of Hence, A cos(377t + (1)) . Then, Vin = 10409 224 The circuit can be redrawn as Using the dOt convention, we can draw an equivalent circuit to incorporate the induced voltages from the coupling effects. 5 Q 18 (11-12) in Q With this circuit, we can use mesh analysis to find the answer in the frequency domain. Loop#1: -10+511+j5(Il—12)—-j812=0 Loop #2: j812 + j5(12 —I,)—j8(11 -12)+j2012 +Vout = 0 This is a system of two equations and three unknowns. We need a constraint equation. Due to the open circuit, it is obvious that I2 = 0. Combining like terms and introducing the constraint, the equations become (5+j5)11 =10 and Vout =j1311 Clearly, 10 1040° I = . :_____=\/§4-45° ‘ 5+J5 55449 and V0“t = j13Il =(13490°)(J§4-450)=13fi445o Using the reference assumed above, the voltage converts to the time domain as vow (t) = 13¢? c0s(377t + 45°) v 225 Problem 13.4 Given the circuit in Figure 13.4, find the coupling coefficient, k, and the . voltage across the 1-Q resistor; j100 100 Q Figure 13.4 k=1 vm=0nzmv ENERGY IN A COUPLED CIRCUIT Problem 13.5 Given the circuit in Figure 13.5, V1 = V2 2 10volts, R1 = R2 = 10 ohms, . le = (,)L2 = 10 , and (DM = 5 , find the coupling coefficient, k, the currents in the primary and secondary circuits, I1 and I2 , and the power absorbed. Figure 13.5 The coupling coefficrent 15 k = . ,ILILZ Given values for COM , 03L1 , and COL2 , we need to modify the equation for k to be (0M 5 5 Jamam 10 4— 226 To find the currents, begin by finding an equivalent circuit that takes into account the coupling effects, i.e., the induced voltages. 109 109 Use mesh analysis to find II and 12. Loopl: 10=(10+j10)Il—j512 Loop2: —10=—j511+(10+j10)I2 10+j10 -j5 I1 _ 10 -j5 10+j10 I2 _ -10 where A = (10 + j10)(1 0 + j10)— (-j5)(-j5) = j200 + 25 = (25)(1+ j8) . In matrix form, 10+j10 j5 [11] j5 10+j10[10] 12 (25)(1+ j8) — 10 10+j10 j5 100+j100-j50 [11} [10 ]= 12 J +j -10 j - -j1 (25)(1+j8) (25)(1+j8) (25)(1+j8) l(50)(2+j)] l4+j2l l11l=|(25)(1+j§)|=1 1+j8 ' [12] (sex—21) -4.-j2 i(25)(1+j8)i l1+j8i Thus, I _ fl _ ME ~ 0 55474123 69° A 2 — 1+ jg ‘ 8.0623482.88° ‘4———'—— 227 Now, find the power absorbed in the circuit. Look at the power absorbed by each element. Starting with the primary circuit, pVl : -V I cos0 = -(10)(0.5547) cos(0°—- (-56.3 1°)) 2 -3.0769 W pm =|11|2R1 = (0.5547)2(10)=3.0769 w p1 = -(2.7735)(0.5547) cos(213.69°-— (~56.3 1)°) = 1.5385 cos(270°) = 0 W where p1 is the power absorbed by the induced voltage of L1 . Ending with the secondary circuit, pvz = V I cos0 = (10)(0.5547)cos(0°—123.69°) = -3.0769 W pm =l12|2R2 = (0.5547)2(10) = 3.0769 w p2 = -(2.7735)(0.5547) cos(33.69°— 123.69°) = 1.5385 cos(90°) = 0 W where p2 is the poWer absorbed by the induced voltage of L2 . The volta e sources absorb —3.0769 watts or deliver +3.0769 watts the resistances absorb 3.0769 watts and the induced volta es absorb 0 watts. The inductors do not absorb ower. Problem 13.6 [13.13] Determine the currents Il , 12 , and I 3 in the circuit of Figure 13.6. Find the energy stored in the coupled coils at t = 2 ms. Take 0) = 1000 rad/s. 12 3490° A 2010° V Figure 13.6 Transform the current source to a voltage source as shown below. k = 0.5 12190° A 2040" V 228 k= or M: 03M = k (03L1)(0)L2) = (0.5)(10) : 5 Using mesh Analysis, Meshl, j12= (4+j10-j5)11+j512+j512 =(4+j5)11+j1012 (1) 0: 20+(8+j10—j5)12 +j511+j51l .20=j101,+(8+j5)12 ' (2) Mesh 2, From (1) and (2), lj12 [4+j5 j10ll11l [—20]: le 8+j5ll12 A=107+j60, A1=—60—j296, A2=40_j100 A Ilzx1 = 2.462172.18° A A; A 12: = 0.8784 -97.48° A 13 :11 —12 = 3.329474.89° A i1(t) = 2.462 cos(1000t + 72.18°) A 12 (t) = 0.878 cos(1000t — 97.48°) A At t=2ms, 1000t=2rad=114.6° i,(2 ms) = 2.462cos(114.6°+ 72.18°) = —2.445 12(2 ms) = 0.878 cos(1 14.6°—— 97.48°) = 0.8391 The total energy stored in the coupled coils is w = 0.5Ll i12 + 0.5L2 if + Mi1 i2 Since 03Ll =10 and a) =1000, L1=L2=10mH, M=O.5Ll=5mH w = (0.5)(10)(-2.445)2 + (0.5)(10)(0.8391)2 + (5)(-2.445)(O.8391) w = 23.15 mJ 229 Problem 13.7 Given the circuit in Figure 13.7, VI = V2 = lOvolts, R1 = R2 = 10 ohms, le = 03L2 = 10 , and 03M = 5 , find the coupling coefficient, k, the currents in the primary and secondary circuits, II and 12 , and the power absorbed. As seen in Problem 13.5, To find the currents, begin by finding an equivalent circuit which takes into account the coupling effects, i.e., the induced voltages. 109 109 Use mesh analysis to find I1 and 12 . Loopl: 10=(10+j10)11+j512 -10=j511+(10+j10)12 F10+j10 55 WP] F10] l j5 10+j10_|12 =L-10 where A = (10 + j10)(10 + j10)— (j5)(j5) = j200 + 25 = (25)(1+ j8). Loop 2 : In matrix form, 230 l10+j10 -j5 l . HILL -j5 10+j10li10l llzl‘ <25)(1+j8) l-Iol F10+j1o -j5 l I100+j100+j501 I111: (25)(1+j8) (25)(1+j8)ll10l_i (25)(1+j8) l‘ LBJ—L -j5 10+j10 JLmJ’L-jSOJOO—leOJ (25)(1 + j8) (25)(1 + j8) (25)(1 + j8) I(50)(2+j3)l I4+jél VIII l[(25)(1+j8)I 1 l [12]: "z (50)(2 '3) l 4+j'86l “J J I all (25)(1+ j8) 1 Thus, 4+j6 72111456310 0 1‘ _ 1+ j8 ‘ 8.0623482.88° “W I _ 11E _ W— 0 89444153 43° A 2 — 1+ jg “ 8.0623482.88° _—'~—-—'—- . Now, find the power absorbed in the circuit. Look at the power absorbed by each element. Starting with the primary circuit, pV1 = -V I 0056 = - (10)(O.8944) cos(0° —— (-26.57°)) = -7.9994 W pR1 =|11|2R1 = (0.8944)2(10) = 7.9995 w p1 = (4.4720)(0.8944) cos(243.43°— (-26.57°)) = 3.9998 cos(270°) = O W where p1 is the power absorbed by the induced voltage of L1,. Ending with the secondary circuit, pV2 = V I 0039 = (10)(O.8944) cos(O —153.43°) = —7.9994 W pR2 =|12|2R2 = (0.8944)2(10) = 7.9995 w p2 = (4.4720)(0.8944) cos(63.43°—153.43°) = 3.9998 cos(-90°) = 0 W where p2 is the power absorbed by the induced voltage of L2 . 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. W Problem 13.8 Given the circuit in Figure 13.8, VI = V2 = lOvolts, R1 = R2 = 100hms, 03L1 = coL2 = 10 , and (DM = 5 , find the coupling coefficient, k, the currents in the primary and . secondary circuits, I1 and I2 , and the power absorbed. 231 I1 = 0.55471 - 56.31° A I2 = 0.55471123.69° A The voltage sources absorb ~3.0769 watts, or deliver +3.0769 watts, the resistances absorb 3.0769 watts and the induced volta es absorb 0 watts. The inductors do not absorb ower. Problem 13.9 Given the circuit in Figure 13.9, Vl : V2 =10volts, R1 = R2 = 10 ohms, 1 031,1 = (0L2 = 10 , and (DM = 5 , find the coupling coefficient, k, the currents in the primary and . f secondary circuits, I1 and I2 , and the power absorbed. Figure 13.9 k = .5 I1 = 0.89444 - 26.57° A 2 = 0.89441153.43° A 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. 232 LINEAR TRANSFORMERS Problem 13.10 [13.23] For the circuit in Figure 13.10, find : (a) the T—equivalent circuit, (b) the II-equivalent circuit. 5H m 15H 20H Figure 13.10 (a) La=L1—M=10H Lb=L2—-M=15H . LC=M=5H (b) L1 L2 ~M2 =300—25=275 L1L2 — M2 275 LA=——I:‘_—M—=-l—5—=18.33H L M_Z7SH B— Ll—M ' L1L2--M2 275 LC— M — 5 —55H 233 IDEAL TRANSFORMERS Problem 13.11 Given the ideal transformer circuit in Figure 13.11, find V109. '9 Q 1 : 10 + 1040° V I 2 Vlog 10 Q Figure 13.11 Carefully DEFINE the problem. Each component is labeled completely. The problem is clear. PRESENT everything you know about the problem. We know the values of the independent source. We also know the values of all the elements. Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success. The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. Using either nodal or mesh analysis will produce two equations with four unknowns. In either case, we need to find two constraint equations. So, let's use mesh analysis for the initial attempt to find a solution. Then, nodal analysis will be used to check the solution. ATTEMPT a problem solution. We need to make some assumptions. First, assume that a positive voltage for the inductor in the primary circuit yields a positive voltage for the inductor in the secondary circuit. Second, assume that a positive (or clockwise) current in the primary circuit yields a positive (or clockwise) current in the secondary circuit. The assumptions are shown in the following circuit. 0.9 Q 234 1040° V O Mesh analysis yields, Loop 1 : 10 = 0911+ Vl Loop2: V2 =lOI2 This is a set of two equations and four unknowns. Two constraint equations are needed. From the ideal transformer, as shown, we know that V2 = nV1 and I1 = n12 . V2 =10Vl and I1 =1012 which are the two constraint equations. This implies that There are many ways to find the values of V1, V2 , I1 , and 12. Let's find I1 . To do this, find V1 in terms of I1 and substitute into the equation for loop 1. Vi = 0.1V2 = (0.1)(1012) = I2 = 0.111 and the equation for loop 1 becomes 10 = 0.911 +0.111 10 =11 Hence, I1 =10 A V1 =1V I2 =lA V2 =10V Therefore, . V100 = V2 =10 V EVALUATE the solution and check for accuracy. Using nodal analysis, V1 ~10 At node 1, 0.9 +11 = 0 or V1—10+0.9Il = 0 (1) V2 —0 Atnode2, 10 +(-IZ)=0 or VZ—IOI2 =0 (2) 235 Again, this is a set of two equations and four unknowns. Two constraint equations are needed. From the ideal transformer, we have V2 =10Vl (3) I1 = 1012 (4) From (2) and (4), V2 =1012 = (10)(1/10)11 =11 (5) From (1), (3), and (5), V1—10+0.9V2 =V1—10+(0.9)(10)V1=(1+9)V1—10=0 10Vl=10 ——> V1=1V Then, V2 =10V1=10V 11=V2 =10A and 12 =(1/10)11_=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. V100 = 10 V Problem 13.12 [13.33] For the circuit in Figure 13.12, find Vo . Switch the dot on the secondary side and find V0 again. 20 mF 10 cos(5t) V Figure 13.12 1 1 ' C=0.02Fb ——=————=—'10 mm“ jcoC j(5)(0.02) J 236 We apply mesh analysis to the circuit shown below. . —j10 o 1040° V C Using mesh analysis, Formesh 1, 10:1011—1013+Vl (1) For mesh 2, V2 = 212 = Vo (2) Formesh3, 0=(10—le)I3—IOII+V2—Vl (3) At the terminals, V2 = nV1 = V1 / 3 (4) I1 =n12=Iz/3 (5) From (2) and (4), . V1 = 612 (6) Substituting this into (1 ), 10:1011—1013 (7) Substituting (4) and (6) into (3) yields 0=—1011--412+(10)(l—j)13 (8) _. lll 10333 0 llllol l l1 10 6 -10 I2 = 10 [—10 —4 10—j10[13J i0] From (5), (7), and (8), Vo = 212 = 2.963432.9° V Switching the dot on the secondary side affects only equations (4) and (5). . V2 = —V1/3 (9) 237 11 = 42/3 (10) . From (2) and (9), V1 = -6I2 Substituting this into (1), ~ 10:1011—1013—612 =(23—j5)I1 (11) Substituting (9) and (10) into (3), 0=-IOII+4IZ+(10)(1--j)I3 (12) From (10) to (12), we get li10.333 o T11l i] [.1105 '46 1011310 Jiijfli? J I ~31—M—1482z 1471‘ A 2” A “-2o+j93.33"' ‘ ' » Vo = 212 = 2.9634 -147.1° v Wm. THREE-PHASE TRANSFORMERS Problem 13.13 [13. 53] In order to meet an emergency, three single-phase transformers with 12,470/ 7200 V rms are connected in A — Y to form a three-phase transformer which is fed by a 12,470-V transmission line. If the transformer supplies 60 MVA to a load, find : (a) the turns ratio for each transformer, (b) the currents in the primary and secondary windings of the transformer, (c) the incoming and outgoing transmission line currents. (a) Consider just one phase at a time. 238 (b) (c) n_.__VL__L299___ 1 fiVLp 1247045 _3_ The load carried by each transformer is 60/3 = 20 MVA 20 MVA Hence, ILP = m = 1,604 A I _ M5 _ 2 778 A LS ‘ 7.2kV ' ’ The current in each incoming line a, b, c is J3er = J3‘x 1603.85 = 2,778 A The current in each outgoing line A, B, C is 2778 = 4,812 A m5 239 .m, .3 PSPICE ANALYSIS OF MAGNETICALLY COUPLED CIRCUITS Problem 13.14 [13.63] 169 jso 40460° V 3040° V The schematic is shown below. R2 L1 IPRINT ACMAG=40V 0.08333 20 V2 ACMAG=30V COUPLING=0.999 ACPHASE=0 ACPHASE=60 a L1 TURNS=400000 L2:TURNS=200000 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 Use PSpice to find V1, V2 , and I0 in the circuit in Figure 13.13. mm— IM(V_PR1NT1)‘ IPWJRINTI) 1.592 E—Ol 1.955 E+01 8.332 E+01 FREQ IM(V_PRINT2) IP(V_PR1NT2) m 6.847E+m 4.640E+01 —fl§_ IM<v_PRINT3> \ IP<V.PRINT3> 1.592 E—Ol 4.434 E—Ol —9.260 E+01 240 Thus, Vl = 1955483320 V V2 = 68.47446.4° V I0 = 443.44 - 92.6° mA ¥gam5ggfiwwfimwmgfififififi=aawfifi=am§ 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. n_&_n (a) V1 — N] — V —-I-\I—1V —[—2§—)(4800)—112V 2 1 1 1200 __ V 112 (b) 12=-E2-=T0—=11.2A I1 =nIZ, where n=28/1200 28 I1 = (fija 1.2) = 261.3 mA (0) p=112|2 R: (112)2(10): 1254 w 241 ...
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ProbSolv_Chapter13 - CHAPTER 13 - MAGNETICALLY COUPLED...

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