Chapter 13, Solution 1.
For coil 1, L1 – M12 + M13 = 6 – 4 + 2 = 4
For coil 2, L2 – M21 – M23 = 8 – 4 – 5 = – 1
For coil 3, L3 + M31 – M32 = 10 + 2 – 5 = 7
LT = 4 – 1 + 7 = 10H
or LT = L1 + L2 + L3 – 2M12 – 2M23 + 2M12
LT = 6 + 8 + 10 = 10H
Chapter 13, Solution 2.
L = L1 + L2 + L3 + 2M12 – 2M23 2M31
= 10 + 12 +8 + 2x6 – 2x6 –2x4
= 22H
Chapter 13, Solution 3.
L1 + L2 + 2M = 250 mH (1)
L1 + L2 – 2M = 150 mH (2)
Adding (1) and (2),
2L1 + 2L2 = 400 mH
But, L1 = 3L2,, or 8L2 + 400, and L2 = 50 mH
L1 = 3L2 = 150 mH
From (2), 150 + 50 – 2M = 150 leads to M = 25 mH
k = M/ L1 L 2 = 2.5 / 50x150 = 0.2887
Chapter 13, Solution 4.
(a) For the series connection shown in Figure (a), the current I enters each coil from
its dotted terminal. Therefore, the mutually induced voltages have the same sign as the
self-induced voltages. Thus,
Leq = L1 + L2 + 2M
Is
L1
I1 I2
+
Vs –
L2
L1 L2
Leq
(a)
(b)
(b) For the parallel coil, consider Figure (b).
Is = I 1 + I2 and Zeq = Vs/Is
Applying KVL to each branch gives,
Vs = jωL1I1 + jωMI2 (1)
Vs = jωMI1 + jω L2I2 (2)
Vs jωL1 jωM I1
or V = jωM jωL 2 I 2
s
∆ = –ω2L1L2 + ω2M2, ∆1 = jωVs(L2 – M), ∆2 = jωVs(L1 – M)
I1 = ∆1/∆, and I2 = ∆2/∆
Is = I1 + I2 = (∆1 + ∆2)/∆ = jω(L1 + L2 – 2M)Vs/( –ω2(L1L2 – M))
Zeq = Vs/Is = jω(L1L2 – M)/[jω(L1 + L2 – 2M)] = jωLeq
i.e., Leq = (L1L2 – M)/(L1 + L2 – 2M)
Chapter 13, Solution 5.
(a) If the coils are connected in series,
L = L1 + L 2 + 2M = 25 + 60 + 2(0.5) 25x 60 = 123.7 mH
(b) If they are connected in parallel,
L1 L 2 − M 2 25x 60 − 19.36 2
L= = mH = 24.31 mH
L1 + L 2 − 2M 25 + 60 − 2x19.36
Chapter 13, Solution 6.
V1 = (R1 + jωL1)I1 – jωMI2
V2 = –jωMI1 + (R2 + jωL2)I2
Chapter 13, Solution 7.
Applying KVL to the loop,
20∠30° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6)
where I is the loop current.
I = 20∠30°/(10 + j6)
Vo = I(j12 + 10 – j4) = I(10 + j8)
= 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V
Chapter 13, Solution 8.
Consider the current as shown below.
j2
1Ω 4Ω
+ +
10 –
I1 j6 j4 I2 -j3 Vo
–
For mesh 1,
10 = (1 + j6)I1 + j2I2 (1)
For mesh 2, 0 = (4 + j4 – j3)I2 + j2I1
0 = j2I1 +(4 + j)I2 (2)
In matrix form,
10 1 + j6 j2 I1
0 = j2 4+ j I 2
∆ = 2 + j25, and ∆2 = –j20
I2 = ∆2/∆ = –j20/(2 + j25)
Vo = –j3I2 = –60/(2 + j25) = 2.392∠94.57°
Chapter 13, Solution 9.
Consider the circuit below.
2Ω 2Ω -j1
+
8∠30o – j4
+
I1 j4 I2 -j2V –
For loop 1,
8∠30° = (2 + j4)I1 – jI2 (1)
For loop 2, ((j4 + 2 – j)I2 – jI1 + (–j2) = 0
or I1 = (3 – j2)i2 – 2 (2)
Substituting (2) into (1), 8∠30° + (2 + j4)2 = (14 + j7)I2
I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12°
Vx = 2I2 = 2.074∠21.12°
Chapter 13, Solution 10.
Consider the circuit below.
jωM
jωL jωL
Io
I1 I2
Iin∠0o
1/jωC
M = k L1 L 2 = L2 = L, I1 = Iin∠0°, I2 = Io
Io(jωL + R + 1/(jωC)) – jωLIin – (1/(jωC))Iin = 0
Io = j Iin(ωL – 1/(ωC)) /(R + jωL + 1/(jωC))
Chapter 13, Solution 11.
Consider the circuit below.
R2 V2
+–
I3
R1 jωL1
jωM
1/jωC
V1
+ I1 jωL2
–
I2
For mesh 1, V1 = I1(R1 + 1/(jωC)) – I2(1/jωC)) –R1I3
For mesh 2,
0 = –I1(1/(jωC)) + (jωL1 + jωL2 + (1/(jωC)) – j2ωM)I2 – jωL1I3 + jωMI3
For mesh 3, –V2 = –R1I1 – jω(L1 – M)I2 + (R1 + R2 + jωL1)I3
or V2 = R1I1 + jω(L1 – M)I2 – (R1 + R2 + jωL1)I3
Chapter 13, Solution 12.
Let ω = 1. j4
j2
•
+ j6 j8 j10
1V
- I1 I2
•
Applying KVL to the loops,
1 = j8 I 1 + j 4 I 2 (1)
0 = j 4 I 1 + j18 I 2 (2)
Solving (1) and (2) gives I1 = -j0.1406. Thus
1 1
Z= = jLeq
→ Leq = = 7.111 H
I1 jI 1
We can also use the equivalent T-section for the transform to find the equivalent
inductance.
Chapter 13, Solution 13.
We replace the coupled inductance with an equivalent T-section and use series and
parallel combinations to calculate Z. Assuming that ω = 1,
La = L1 − M = 18 − 10 = 8, Lb = L2 − M = 20 − 10 = 10, Lc = M = 10
The equivalent circuit is shown below:
12 Ω j8 Ω j10 Ω 2Ω
j10 Ω
-j6 Ω
Z
j4 Ω
Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244Ω
Chapter 13, Solution 14.
To obtain VTh, convert the current source to a voltage source as shown below.
j2
5Ω j6 Ω j8 Ω -j3 Ω 2Ω
a
+ I +
+ 8V –
j10 V – VTh
–
b
Note that the two coils are connected series aiding.
ωL = ωL1 + ωL2 – 2ωM
jωL = j6 + j8 – j4 = j10
Thus, –j10 + (5 + j10 – j3 + 2)I + 8 = 0
I = (– 8 + j10)/ (7 + j7)
But, –j10 + (5 + j6)I – j2I + VTh = 0
VTh = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7)
VTh = 5.349∠34.11°
To obtain ZTh, we set all the sources to zero and insert a 1-A current source at the terminals
a–b as shown below.
j2
5Ω j6 Ω j8 Ω -j3 Ω 2Ω
a
+
I1 1A Vo I2
–
b
Clearly, we now have only a super mesh to analyze.
(5 + j6)I1 – j2I2 + (2 + j8 – j3)I2 – j2I1 = 0
(5 + j4)I1 + (2 + j3)I2 = 0 (1)
But, I2 – I1 = 1 or I2 = I1 – 1 (2)
Substituting (2) into (1), (5 + j4)I1 +(2 + j3)(1 + I1) = 0
I1 = –(2 + j3)/(7 + j7)
Now, ((5 + j6)I1 – j2I1 + Vo = 0
Vo = –(5 + j4)I1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.332∠50°
ZTh = Vo/1 = 2.332∠50° ohms
Chapter 13, Solution 15.
To obtain IN, short-circuit a–b as shown in Figure (a).
20 Ω j20 Ω 20 Ω j20 Ω
a a
j5 j5
I1 j10 Ω 1
+ j10 Ω +
– IN –
60∠30o I1 I2
I2
b b
(a) (b)
For mesh 1,
60∠30° = (20 + j10)I1 + j5I2 – j10I2
or 12∠30° = (4 + j2)I1 – jI2 (1)
For mesh 2,
0 = (j20 + j10)I2 – j5I1 – j10I1
or I1 = 2I2 (2)
Substituting (2) into (1), 12∠30° = (8 + j3)I2
IN = I2 = 12∠30°/(8 + j3) = 1.404∠9.44° A
To find ZN, we set all the sources to zero and insert a 1-volt voltage source at terminals a–
b as shown in Figure (b).
For mesh 1, 1 = I1(j10 + j20 – j5x2) + j5I2
1 = j20I1 + j5I2 (3)
For mesh 2, 0 = (20 + j10)I2 + j5I1 – j10I1 = (4 + j2)I2 – jI1
or I2 = jI1/(4 + j2) (4)
Substituting (4) into (3), 1 = j20I1 + j(j5)I1/(4 + j2) = (–1 + j20.5)I1
I1 = 1/(–1 + j20.5)
ZN = 1/I1 = (–1 + j20.5) ohms
Chapter 13, Solution 16.
To find IN, we short-circuit a-b.
jΩ
8Ω -j2 Ω
a
• •
+ j4 Ω j6 Ω I2 IN
80∠0 V o
I1
-
b
− 80 + (8 − j 2 + j 4) I 1 − jI 2 = 0 → (8 + j 2) I 1 − jI 2 = 80 (1)
j 6 I 2 − jI 1 = 0
→ I1 = 6I 2 (2)
Solving (1) and (2) leads to
80
IN = I2 = = 1.584 − j 0.362 = 1.6246∠ − 12.91o A
48 + j11
To find ZN, insert a 1-A current source at terminals a-b. Transforming the current source
to voltage source gives the circuit below.
jΩ
8Ω -j2 Ω 2Ω
a
• • +
j4 Ω j6 Ω I2 2V
I1
-
b
jI 2
0 = (8 + j 2) I 1 − jI 2
→ I1 = (3)
8 + j2
2 + (2 + j 6) I 2 − jI 1 = 0 (4)
Solving (3) and (4) leads to I2 = -0.1055 +j0.2975, Vab=-j6I2 = 1.7853 +0.6332
Vab
ZN = = 1.894∠19.53o Ω
1
Chapter 13, Solution 17.
Z = -j6 // Zo
where
144
Z o = j20 + = 0.5213 + j15.7
j30 − j2 + j5 + 4
− j6 xZ o
Z= = 0.1989 − j9.7Ω
− j6 + Z o
Chapter 13, Solution 18.
Let ω = 1. L1 = 5, L2 = 20, M = k L1 L2 = 0.5 x10 = 5
We replace the transformer by its equivalent T-section.
La = L1 − (− M ) = 5 + 5 = 10, Lb = L1 + M = 20 + 5 = 25, Lc = − M = −5
We find ZTh using the circuit below.
-j4 j10 j25 j2
-j5
ZTh
4+j6
j 6(4 + j )
Z Th = j 27 + (4 + j ) //( j 6) = j 27 + = 2.215 + j 29.12Ω
4 + j7
We find VTh by looking at the circuit below.
-j4 j10 j25 j2
+
-j5
+
VTh
o
120Chapter 13, Solution 20.
Transform the current source to a voltage source as shown below.
k=0.5
4Ω j10 j10 8Ω
I3
+
– j12 I1 -j5 I2 +
20∠0o –
k = M/ L1 L 2 or M = k L1 L 2
ωM = k ωL1ωL 2 = 0.5(10) = 5
For mesh 1, j12 = (4 + j10 – j5)I1 + j5I2 + j5I2 = (4 + j5)I1 + j10I2 (1)
For mesh 2, 0 = 20 + (8 + j10 – j5)I2 + j5I1 + j5I1
–20 = +j10I1 + (8 + j5)I2 (2)
j12 4 + j5 + j10 I1
From (1) and (2), 20 = + j10 8 + j5 I
2
∆ = 107 + j60, ∆1 = –60 –j296, ∆2 = 40 – j100
I1 = ∆1/∆ = 2.462∠72.18° A
I2 = ∆2/∆ = 0.878∠–97.48° A
I3 = I1 – I2 = 3.329∠74.89° A
i1 = 2.462 cos(1000t + 72.18°) A
i2 = 0.878 cos(1000t – 97.48°) A
At t = 2 ms, 1000t = 2 rad = 114.6°
i1 = 0.9736cos(114.6° + 143.09°) = –2.445
i2 = 2.53cos(114.6° + 153.61°) = –0.8391
The total energy stored in the coupled coils is
w = 0.5L1i12 + 0.5L2i22 – Mi1i2
Since ωL1 = 10 and ω = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH
w = 0.5(10)(–2.445)2 + 0.5(10)(–0.8391)2 – 5(–2.445)(–0.8391)
w = 43.67 mJ
Chapter 13, Solution 21.
For mesh 1, 36∠30° = (7 + j6)I1 – (2 + j)I2 (1)
For mesh 2, 0 = (6 + j3 – j4)I2 – 2I1 jI1 = –(2 + j)I1 + (6 – j)I2 (2)
36∠30° 7 + j6 − 2 − j I1
Placing (1) and (2) into matrix form, 0 = − 2 − j 6 − j I
2
∆ = 48 + j35 = 59.41∠36.1°, ∆1 = (6 – j)36∠30° = 219∠20.54°
∆2 = (2 + j)36∠30° = 80.5∠56.56°, I1 = ∆1/∆ = 3.69∠–15.56°, I2 = ∆2/∆ = 1.355∠20.46°
Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts
Chapter 13, Solution 22.
With more complex mutually coupled circuits, it may be easier to show the effects of the
coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure
13.85 then becomes,
-j50
Io
I3
j20Ic j10Ib j60
j40
+ − + − − + − +
Ia Ix
− j30Ib j20Ia
j30Ic +
50∠0° V
+ j80 100 Ω
− I1 I2
Ib
−
+ j10Ia
Note the following,
Ia = I 1 – I3
Ib = I2 – I1
Ic = I 3 – I2
and Io = I 3
Now all we need to do is to write the mesh equations and to solve for Io.
Loop # 1,
-50 + j20(I3 – I2) j 40(I1 – I3) + j10(I2 – I1) – j30(I3 – I2) + j80(I1 – I2) – j10(I1 – I3) = 0
j100I1 – j60I2 – j40I3 = 50
Multiplying everything by (1/j10) yields 10I1 – 6I2 – 4I3 = - j5 (1)
Loop # 2,
j10(I1 – I3) + j80(I2–I1) + j30(I3–I2) – j30(I2 – I1) + j60(I2 – I3) – j20(I1 – I3) + 100I2 = 0
-j60I1 + (100 + j80)I2 – j20I3 = 0 (2)
Loop # 3,
-j50I3 +j20(I1 –I3) +j60(I3 –I2) +j30(I2 –I1) –j10(I2 –I1) +j40(I3 –I1) –j20(I3 –I2) = 0
-j40I1 – j20I2 + j10I3 = 0
Multiplying by (1/j10) yields, -4I1 – 2I2 + I3 = 0 (3)
Multiplying (2) by (1/j20) yields -3I1 + (4 – j5)I2 – I3 = 0 (4)
Multiplying (3) by (1/4) yields -I1 – 0.5I2 – 0.25I3 = 0 (5)
Multiplying (4) by (-1/3) yields I1 – ((4/3) – j(5/3))I2 + (1/3)I3 = -j0.5 (7)
Multiplying [(6)+(5)] by 12 yields (-22 + j20)I2 + 7I3 = 0 (8)
Multiplying [(5)+(7)] by 20 yields -22I2 – 3I3 = -j10 (9)
(8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3 (10)
(9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces,
I3 = j3.333 + (-1.2273 – j1.1623)I3
or I3 = Io = 1.3040∠63o amp.
Chapter 13, Solution 23.
ω = 10
0.5 H converts to jωL1 = j5 ohms
1 H converts to jωL2 = j10 ohms
0.2 H converts to jωM = j2 ohms
25 mF converts to 1/(jωC) = 1/(10x25x10-3) = –j4 ohms
The frequency-domain equivalent circuit is shown below.
j2
j5 j10
+ I1 –j4 I2
12∠0° −
5Ω
For mesh 1, 12 = (j5 – j4)I1 + j2I2 – (–j4)I2
–j2 = I1 + 6I2 (1)
For mesh 2, 0 = (5 + j10)I2 + j2I1 –(–j4)I1
0 = (5 + j10)I2 + j6I1 (2)
From (1), I1 = –j12 – 6I2
Substituting this into (2) produces,
I2 = 72/(–5 + j26) = 2.7194∠–100.89°
I1 = –j12 – 6 I2 = –j12 – 163.17∠–100.89 = 5.068∠52.54°
Hence, i1 = 5.068cos(10t + 52.54°) A, i2 = 2.719cos(10t – 100.89°) A.
At t = 15 ms, 10t = 10x15x10-3 0.15 rad = 8.59°
i1 = 5.068cos(61.13°) = 2.446
i2 = 2.719cos(–92.3°) = –0.1089
w = 0.5(5)(2.446)2 + 0.5(1)(–0.1089)2 – (0.2)(2.446)(–0.1089) = 15.02 J
Chapter 13, Solution 24.
(a) k = M/ L1 L 2 = 1/ 4 x 2 = 0.3535
(b) ω = 4
1/4 F leads to 1/(jωC) = –j/(4x0.25) = –j
1||(–j) = –j/(1 – j) = 0.5(1 – j)
1 H produces jωM = j4
4 H produces j16
2 H becomes j8
j4
2Ω
j8
+ I1 I2 0.5(1–j)
12∠0° −
j16
12 = (2 + j16)I1 + j4I2
or 6 = (1 + j8)I1 + j2I2 (1)
0 = (j8 + 0.5 – j0.5)I2 + j4I1 or I1 = (0.5 + j7.5)I2/(–j4) (2)
Substituting (2) into (1),
24 = (–11.5 – j51.5)I2 or I2 = –24/(11.5 + j51.5) = –0.455∠–77.41°
Vo = I2(0.5)(1 – j) = 0.3217∠57.59°
vo = 321.7cos(4t + 57.6°) mV
(c) From (2), I1 = (0.5 + j7.5)I2/(–j4) = 0.855∠–81.21°
i1 = 0.885cos(4t – 81.21°) A, i2 = –0.455cos(4t – 77.41°) A
At t = 2s,
4t = 8 rad = 98.37°
i1 = 0.885cos(98.37° – 81.21°) = 0.8169
i2 = –0.455cos(98.37° – 77.41°) = –0.4249
w = 0.5L1i12 + 0.5L2i22 + Mi1i2
= 0.5(4)(0.8169)2 + 0.5(2)(–.4249)2 + (1)(0.1869)(–0.4249) = 1.168 J
Chapter 13, Solution 25.
m = k L1 L 2 = 0.5 H
We transform the circuit to frequency domain as shown below.
12sin2t converts to 12∠0°, ω = 2
0.5 F converts to 1/(jωC) = –j
2 H becomes jωL = j4
j1
Io 4 Ω a 1Ω 2Ω
–j1
+ j2 j2
12∠0° − j4
10 Ω
b
Applying the concept of reflected impedance,
Zab = (2 – j)||(1 + j2 + (1)2/(j2 + 3 + j4))
= (2 – j)||(1 + j2 + (3/45) – j6/45)
= (2 – j)||(1 + j2 + (3/45) – j6/45)
= (2 – j)||(1.0667 + j1.8667)
=(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085∠17.9° ohms
Io = 12∠0°/(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2∠–4.88°
io = 2.2sin(2t – 4.88°) A
Chapter 13, Solution 26.
M = k L1L 2
ωM = k ωL1ωL 2 = 0.6 20x 40 = 17
The frequency-domain equivalent circuit is shown below.
j17
50 Ω –j30
Io
+ I1 j20 j40
200∠60° I2 10 Ω
−
For mesh 1,
200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2 (1)
For mesh 2,
0 = (10 + j40)I2 + j17I1 (2)
200∠60° 50 − j10 j17 I1
In matrix form, 0 = j17 10 + j40 I 2
∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30°
I2 = ∆2/∆ = 3.755∠–36.34°
Io = I2 = 3.755∠–36.34° A
Switching the dot on the winding on the right only reverses the direction of Io.
This can be seen by looking at the resulting value of ∆2 which now becomes
3400∠150°. Thus,
Io = 3.755∠143.66° A
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