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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)11j12 =1 ' 'Loop2: j101l +j1012 +1012 +10: 0
—j1011+(10+j10)I2 =10
—jI,+(1+j)I2 21 In matrix form, F1+j —j Tllljﬂ
ij 1+ji12H1J 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 (1112) 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 = . :_____=\/§445°
‘ 5+J5 55449
and V0“t = j13Il =(13490°)(J§4450)=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 1Q 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+j100j50
[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 =112R1 = (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 =l122R2 = (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+j10j5)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 =L10 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) lIol 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’LjSOJOO—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 =112R1 = (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 =122R2 = (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 IIequivalent 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 '
L1L2M2 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=—1011412+(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)(1j)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. THREEPHASE TRANSFORMERS Problem 13.13 [13. 53] In order to meet an emergency, three singlephase transformers with 12,470/ 7200 V rms are connected in A — Y to form a threephase transformer which is
fed by a 12,470V 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 109 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=1122 R: (112)2(10): 1254 w 241 ...
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