Chapter 13 solutions_to_exercises (engineering circuit analysis 7th)
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This document contains solutions to chapter thirteen problems from the textbook Engineering Circuit Analysis, 7th Edition. The problems involve calculating voltages, currents, impedances, and power in AC circuits using Kirchhoff's laws. Key steps shown include setting up equations for each mesh using Kirchhoff's voltage law (KVL) and solving the equations simultaneously to find the unknown currents at given frequencies. The maximum power is then calculated for one of the circuits.
![Engineering Circuit Analysis, 7th Edition
6.
8
di1
di
+ 0.4 2 = 5e − t
dt
dt
10 March 2006
[1]
di1
di
+ 8 2 = 3e −2t
dt
dt
Chapter Thirteen Solutions
[2]
0.4
Let i1 = ae −t + be−2t and i2 = ce−t + de−2t
Then from Eq. [1] we have
–8a – 0.4c = 5 [3]
and
–16b – 0.8d = 0 [4]
and
–0.8b – 16d = 3
And from Eq. [2] we have
–0.4a – 8c = 0 [5]
[6]
Solving, we find that a = –0.6266, b = 0.0094, c = 0.03133, and d = –0.1880
(a)
di1 d
⎡
⎤
= ⎣ −0.6266e −t + 0.0094e−2t ⎦ = 0.6266e− t − 0.0188e−2t A/s
dt dt
(b)
di2 d
⎤
= ⎡0.0313e −t − 0.1880e−2t ⎦ = −0.0313e − t + 0.376e −2t A/s
dt dt ⎣
(c) i1 = −0.6266e −t + 0.0094e−2t A
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![Engineering Circuit Analysis, 7th Edition
7.
Chapter Thirteen Solutions
di2 ⎞
⎛ di1
−3
−t
⎜ −2 + 1.5
⎟ ×10 = 2e
dt ⎠
⎝ dt
di1
di2 ⎞
⎛
−3t
⎜ −1.5 + 2
⎟ = 4e
dt
dt ⎠
⎝
10 March 2006
[1]
[2]
Let i1 = ae −t + be−3t and i2 = ce− t + de−3t
Then from Eq. [1] we have
2a – 1.5c = 2×103 [3]
and
6b – 4.5d = 0 [4]
and
4.5b – 6d = 4×103
And from Eq. [2] we have
1.5a – 2c = 0 [5]
[6]
Solving, we find that a = 2286, b = -1143, c = 1714, and d = –1524
(a)
di1 d
= ⎡ 2286e −t − 1143e−3t ⎤ = −2286e − t + 3429e −3t A/s
⎦
dt dt ⎣
(b)
di2 d
= ⎡1714e −t − 1524e−3t ⎤ = −1714e− t + 4572e−3t A/s
⎦
dt dt ⎣
(c) i2 = 1714e− t + 4572e−3t A
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![Engineering Circuit Analysis, 7th Edition
Chapter Thirteen Solutions
10 March 2006
15.
(a)
100 = j5ω (I1 – I2) + j3ωI2 + 6(I1 – I3)
[1]
(4 + j4ω)I2 + j3ω (I1 – I2) + j2ω (I3 – I2) + j6ω (I2 – I3) – j2ω I2 + j5ω (I2 – I1)
[2]
– j3ω I2 = 0
6 (I3 – I1) + j6ω (I3 – I2) + j2ω I2 + 5 I3 = 0
[3]
Collecting terms,
(6 + j5ω) I1 – j2ω I2 – 6 I3 = 100
-j2ω I1 + (4 + j5ω) I2 – j4ω I3 = 0
[2]
-6 I1 - j4ω I2 + (11 + j6ω) I3 = 0
(b)
[1]
[3]
For ω = 2 rad/s, we find
(6 + j10) I1 – j4 I2 – 6 I3 = 100
-j4 I1 + (4 + j10) I2 – j8 I3 = 0
-6 I1 – j8 I2 + (11 + j12) I3 = 0
Solving, I3 = 4.32 ∠ -54.30o A
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![Engineering Circuit Analysis, 7th Edition
29.
Chapter Thirteen Solutions
10 March 2006
Define coil voltages v1 and v2 with the “+” reference at the respective dot.
Also define two clockwise mesh currents i1 and i2. We may then write:
dI1
dI
+M 2
dt
dt
dI
dI
v2 = L2 2 + M 1
dt
dt
v1 = L1
M = k L1 L2
ω = 2π60 rad / s
or, using phasor notation,
V1 = jωL1 I1 + jωM I 2
V2 = jωL2 I 2 + jωM I1
100∠0 = 50 I1 + jωL1 I1 + jωM I 2
−25 I 2 = jωL2 I 2 + jωM I1
Rearrange:
[50 + jωL1 ] I1 + jωMI 2 = 100∠0
jωMI1 [−25 + jωL2 ] I 2 = 0
or
jωM ⎤ ⎡ I1 ⎤ ⎡100∠0 ⎤
⎡50 + jωL1
=
⎢ jωM
⎢
−25 + jωL2 ⎥ ⎣ I 2 ⎥ ⎢ 0 ⎥
⎦
⎣
⎦
⎦ ⎣
We can solve for I2 and V2 = −25I2:
V2 = −
j1.658
k L1L 2 + 1
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![Engineering Circuit Analysis, 7th Edition
Chapter Thirteen Solutions
10 March 2006
43.
Z in = Z11 +
ω2 M 2
R22 + jX 22
1
1
= 31.83 ⇒ ω =
= 314 rad / s
31.83 × C
ωC
ie. a 50Hz system
Z in = 20 + jω100 × 10−3 +
ω2 k 2 L1 L2
2 − j 31.83
ω2 k 2 L1 L2 2 jω2 k 2 L1 L2 31.83
−
22 + 31.832
22 + 31.832
7840 ⎤ 2
⎡ 493
= 20 + j 31.4 + ⎢
−j
k
1020 ⎥
⎣1020
⎦
= 20 + j 31.4 + [0.483 − j 7.69]k 2
Ω
(a) Z in (k = 0) = 20 + j 31.4
(b) Z in (k = 0.5) = 20.2 + j 27.6 Ω
(c) Z in (k = 0.9) = 20.4 + j 24.5 Ω
(d) Z (k = 1.0) = 20.5 + j 23.7 Ω
Z in = 20 + jω100 × 10−3 +
in
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Chapter 13 solutions_to_exercises (engineering circuit analysis 7th)
- 1. Engineering Circuit Analysis, 7th Edition 1. v2(t) = M21 di1 ( t ) dt Chapter Thirteen Solutions 10 March 2006 = − M 21 (400)(120π ) sin(120π t ) Taking peak values and noting sign is irrelevant, 100 = M21(400)(120π). Thus, M21 = 663.1 μH PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 2. v1 = M 12 i2 = di2 dt therefore 1 1 ⎛ 115 2 ⎞ o v1dt = ⎜ ∫ ⎜ 120π ⎟ sin 120π t − 16 ⎟ M 12 M 12 ⎝ ⎠ Equating peak values, M 12 = ( ) 1 ⎛ 115 2 ⎞ ⎜ ⎟ = 9.59 mH 45 ⎜ 120π ⎟ ⎝ ⎠ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. Engineering Circuit Analysis, 7th Edition 3. Chapter Thirteen Solutions 10 March 2006 1 and 3, 2 and 4 1 and 4, 2 and 3 3 and 1, 2 and 4 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. Engineering Circuit Analysis, 7th Edition 4. (a) v1 = − L1 Chapter Thirteen Solutions 10 March 2006 di1 di +M 2 dt dt Substituting in i1 = 30 sin 80t and i2 = 30 cos 80t, we find that v1 = –2400 cos 80t – 1200 sin 80t 1200 ⎞ ⎛ = – 24002 + 12002 cos ⎜ 80t − tan −1 ⎟ 2400 ⎠ ⎝ = –2683 cos (80t – 26.57o) V (b) v2 = − L2 di2 di +M 1 dt dt Substituting in i1 = 30 sin 80t and i2 = 30 cos 80t, we find that v2 = 7200 sin 80t + 1200 cos 80t = 7200 ⎞ ⎛ 72002 + 12002 cos ⎜ 80t − tan −1 ⎟ 2400 ⎠ ⎝ = 7299 cos (80t – 80.54o) V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. Engineering Circuit Analysis, 7th Edition 5. Chapter Thirteen Solutions 10 March 2006 di ⎞ ⎛ di (a) v1 = − ⎜ L1 1 + M 2 ⎟ dt ⎠ ⎝ dt Substituting in i1 = 3 cos 800t nA and i2 = 2 cos 800t nA,we find that v1 = − ⎡ −(22 × 10−6 )(3)(800) × 10−9 sin 800t − (5 × 10−6 )(2)(800) × 10−9 sin 800t ⎤ ⎣ ⎦ = 60.8 sin 800t pV di ⎞ ⎛ di (b) v2 = + ⎜ L2 2 + M 1 ⎟ dt ⎠ ⎝ dt Substituting in i1 = 3 cos 800t nA and i2 = 2 cos 800t nA,we find that v1 = −(15 × 10−6 )(2)(800) ×10−9 sin 800t − (5 × 10−6 )(3)(800) ×10−9 sin 800t = 36 sin 800t pV PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. Engineering Circuit Analysis, 7th Edition 6. 8 di1 di + 0.4 2 = 5e − t dt dt 10 March 2006 [1] di1 di + 8 2 = 3e −2t dt dt Chapter Thirteen Solutions [2] 0.4 Let i1 = ae −t + be−2t and i2 = ce−t + de−2t Then from Eq. [1] we have –8a – 0.4c = 5 [3] and –16b – 0.8d = 0 [4] and –0.8b – 16d = 3 And from Eq. [2] we have –0.4a – 8c = 0 [5] [6] Solving, we find that a = –0.6266, b = 0.0094, c = 0.03133, and d = –0.1880 (a) di1 d ⎡ ⎤ = ⎣ −0.6266e −t + 0.0094e−2t ⎦ = 0.6266e− t − 0.0188e−2t A/s dt dt (b) di2 d ⎤ = ⎡0.0313e −t − 0.1880e−2t ⎦ = −0.0313e − t + 0.376e −2t A/s dt dt ⎣ (c) i1 = −0.6266e −t + 0.0094e−2t A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. Engineering Circuit Analysis, 7th Edition 7. Chapter Thirteen Solutions di2 ⎞ ⎛ di1 −3 −t ⎜ −2 + 1.5 ⎟ ×10 = 2e dt ⎠ ⎝ dt di1 di2 ⎞ ⎛ −3t ⎜ −1.5 + 2 ⎟ = 4e dt dt ⎠ ⎝ 10 March 2006 [1] [2] Let i1 = ae −t + be−3t and i2 = ce− t + de−3t Then from Eq. [1] we have 2a – 1.5c = 2×103 [3] and 6b – 4.5d = 0 [4] and 4.5b – 6d = 4×103 And from Eq. [2] we have 1.5a – 2c = 0 [5] [6] Solving, we find that a = 2286, b = -1143, c = 1714, and d = –1524 (a) di1 d = ⎡ 2286e −t − 1143e−3t ⎤ = −2286e − t + 3429e −3t A/s ⎦ dt dt ⎣ (b) di2 d = ⎡1714e −t − 1524e−3t ⎤ = −1714e− t + 4572e−3t A/s ⎦ dt dt ⎣ (c) i2 = 1714e− t + 4572e−3t A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 8. (a) −V2 = jω 0.4 1∠0 V2 = − j100π× 0.4 × 1∠0 = 126 ∠ 90o V Thus, v(t) = 126 cos (100πt + 90o) V (b) Define V2 across the 2-H inductor with + reference at the dot, and a clockwise currents I1 and I2, respectively, in each mesh. Then, V = -V2 and we may also write V2 = jωL2 I2 + jωMI1 or -V = jωL2 V + jωM 10 Solving for V, − ( j100π )(0.4) 125.7∠ − 90o 125.7∠ - 90o = = 2.000 ∠ - 179.1o = 1 + ( j100π )(2 ) 1 + j 62.83 62.84∠89.09o Thus, v(t) = 2 cos (100πt – 179.1o) V. V= (c) Define V1 across the left inductor, and V2 across the right inductor, with the “+” reference at the respective dot; also define two clockwise mesh currents I1 and I2. Then, V1 = jωL1 I1 + jω M I 2 V2 = jωL2 I 2 + jω M I1 Now I1 = 1∠0 − V1 and Vout = −V2 4 V and I 2 = out 10 Vout ⎡1∠0 − V1 ⎤ ⇒ V1 = jωL1 ⎢ ⎥ + jωM 10 EQN 1 4 ⎣ ⎦ V ⎡1∠0 − V1 ⎤ −Vout = jωL2 out + jωM ⎢ ⎥ EQN 2 10 4 ⎣ ⎦ − j ωM ⎤ ⎡ jωL1 ⎡ jωL1 1∠0 ⎤ ⎢1 − 4 ⎥ ⎡ V1 ⎤ ⎢ ⎥ 10 4 ⎢ ⎥⎢ ⎥ = ⎢ jωM 1∠0 ⎥ jωL2 ⎥ ⎣Vout ⎦ ⎢ ⎢ jωM ⎥ −1 + ⎢ 4 ⎢ ⎥ ⎣ ⎦ 10 ⎥ 4 ⎣ ⎦ − j12.6 ⎤ ⎡ V1 ⎤ ⎡ 39.3 j ⎤ ⎡1 − j 39 ⎢ j 31.4 −1 + j 62.8⎥ ⎢V ⎥ = ⎢31.4 j ⎥ ⎦ ⎣ ⎦ ⎣ out ⎦ ⎣ Solving, we find that Vout (= V) = 1.20 ∠ -2.108o V and hence v(t) = 1.2 cos (100πt – 2.108o) V. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. Engineering Circuit Analysis, 7th Edition 9. (a) Chapter Thirteen Solutions 10 March 2006 100 = (50 + j 200) I1 + j300 I2 , (2000 + j 500) I2 + j 300 I1 = 0 ∴ I2 = ⎛ − j3 900 ⎞ , 100 = ⎜ 50 + j 200 + ⎟ I1 20 + j 5 20 + j 5 ⎠ ⎝ ∴100 = ∴ PS ,abS 900 + j 4250 I1 ∴ I1 = 0.47451 ∠ − 64.01° A 20 + j 5 1 = − × 100 × 0.4745cos 64.01° = −10.399 W 2 2 (b) 1 1 − j3 = 4.769 W P50 = × 50 × 0.47452 = 5.630 W, P2000 = × 2000 × 0.47452 × 20 + j 5 2 2 (c) 0 each (d) 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10. iS 1 = 4t A, iS 2 = 10t A (a) v AG = 20 × 4 + 4 ×10 = 120 V (b) vCG = −4 × 6 = −24 V (c) 10 March 2006 vBG = 3 × 10 + 4 × 4 − 6 × 4 = 30 + 16 − 24 = 22 V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 11. (a) Vab ,oc = 100 (− j 300) = 145.52∠ − 165.96° V 50 + j 200 100 = (50 + j 200) I1 + j 300 I2SC , j 500 I2SC + j 300 I1 = 0 ⎡ ⎤ 5 ⎛ 5⎞ ∴ I1 = − I2 SC , 100 = ⎢ (50 + j 200) ⎜ − ⎟ + j 300 ⎥ I2 SC ∴ I2 SC = 1.1142∠158.199° A 3 ⎝ 3⎠ ⎣ ⎦ ∴ Zth = Vab ,bc / I2 SC = (b) 145.52∠ − 165.96° = 130.60∠35.84° = 105.88 + j 76.47 Ω 1.1142∠158.199° Z L = 105.88 − j 76.47 Ω ∴ IL = ∴ PL max = 145.52 = 0.6872 A 2 × 105.88 1 × 0.68722 × 105.88 = 25.00 W 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 12. KVL Loop 1 100 ∠0 = 2(I1 – I2) + jω3 (I1 – I3) + jω2 (I2 – I3) KVL Loop 2 2(I2 – I1) + 10I2 + jω4 (I2 – I3) + jω2 (I1 – I3) = 0 KVL Loop 3 5I3 + jω3 (I3 – I1) + jω2 (I3 – I2) + jω4 (I3 – I2) + jω2 (I3 – I1) = 0 ∴LINEAR EQUATIONS ⎡ 2 + jω 3 − 2 + jω 2 − jω 5 ⎤ ⎡ I1 ⎤ ⎡100∠0⎤ ⎢− 2 + jω 2 12 + jω 4 − jω 6 ⎥ ⎢I ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢ − jω 5 ⎢ 0 ⎥ 5 + j11⎥ ⎢I 3 ⎥ jω 2 ⎣ ⎦⎣ ⎦ ⎣ ⎦ Since ω = 2πf = 2π(50) = 314.2 rad/s, the matrix becomes ⎡ 2 + j 942.6 − 2 + j 628.4 − j1571 ⎤ ⎡ I1 ⎤ ⎡100∠0⎤ ⎢− 2 + j 628.4 12 + j1257 ⎥ ⎢I ⎥ = ⎢ 0 ⎥ − j1885 ⎥ ⎢ 2 ⎥ ⎢ ⎢ ⎥ ⎢ − j1571 ⎢ 0 ⎥ 5 + j 3456⎥ ⎢I 3 ⎥ j 628.4 ⎣ ⎦⎣ ⎦ ⎣ ⎦ Solving using a scientific calculator or MATLAB, we find that I1 = 278.5 ∠ -89.65o mA, I2 = 39.78 ∠ -89.43o mA, I3 = 119.4 ∠ -89.58o mA. Returning to the time domain, we thus find that i1(t) = 278.5 cos (100πt – 89.65o) mA, i2(t) = 39.78 cos (100πt – 89.43o) mA, and i3(t) = 119.4 cos (100πt – 89.58o) mA. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 13. 10t 2u (t ) 1000t 2 u (t ) = 0.01i′S ∴ i′S = 2 t 2 + 0.01 t + 0.01 15t 2 1500t 2 ′S = 2 vx = 0.015i u (t ), 100vx = 2 u (t ) t + 0.01 t + 0.01 ⎞ d ⎛ 15t 2 (t 2 + 0.01)2t − t 2 × 2t u (t ) ⎟ = 15 × 10−4 u (t ) ∴ iC = 100 × 10−6 v′x = 10−4 ⎜ 2 dt ⎝ t + 0.01 (t 2 + 0.01) 2 ⎠ vs = ∴ iC = 15 × 10−4 0.02t 30t ∴ iC (t ) = 2 μA, 2 (t + 0.01) (t + 0.01) 2 2 t>0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. Engineering Circuit Analysis, 7th Edition 14. Chapter Thirteen Solutions (a) ′ ′ v A (t ) = L1i′ − Mi′2, vB (t ) = L1i1 − Mi′2 + L 2i′2 − Mi1 1 (b) 10 March 2006 V1(jω) = jωL1 IA + jωM(IB + IA) V2(jω) = jωL2 (IB + IA) + jωMIA PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 15. (a) 100 = j5ω (I1 – I2) + j3ωI2 + 6(I1 – I3) [1] (4 + j4ω)I2 + j3ω (I1 – I2) + j2ω (I3 – I2) + j6ω (I2 – I3) – j2ω I2 + j5ω (I2 – I1) [2] – j3ω I2 = 0 6 (I3 – I1) + j6ω (I3 – I2) + j2ω I2 + 5 I3 = 0 [3] Collecting terms, (6 + j5ω) I1 – j2ω I2 – 6 I3 = 100 -j2ω I1 + (4 + j5ω) I2 – j4ω I3 = 0 [2] -6 I1 - j4ω I2 + (11 + j6ω) I3 = 0 (b) [1] [3] For ω = 2 rad/s, we find (6 + j10) I1 – j4 I2 – 6 I3 = 100 -j4 I1 + (4 + j10) I2 – j8 I3 = 0 -6 I1 – j8 I2 + (11 + j12) I3 = 0 Solving, I3 = 4.32 ∠ -54.30o A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. Engineering Circuit Analysis, 7th Edition 16. (a) Chapter Thirteen Solutions Va = jωL1 I a + jωM I b I a = I1 Vb = jωL2 I b + jωM I a 10 March 2006 Ib = − I 2 V1 = I1 R1 + Va = I1 R1 + jω L1 I a + jωM I b = I1 R1 + jω L1 I1 − jωM I 2 V2 = I 2 R2 − Vb = I 2 R2 − jω L2 I b − jωM I a = I 2 R2 + jω L2 I 2 − jωM I1 (b) Assuming that the systems connecting the transformer are fully isolated. Va = jωL1 I a + jωMI b I a = − I1 Vb = jωL2 I b + jωMI a Ib = − I 2 V1 = I1 R − Va = I1 R − jωL1 I a − jωM I b = I1 R + jωL1 I1 + jωM I 2 V2 = Vb + I b R2 = − I 2 R2 + jω L2 I b + jωM I a = − I 2 R2 − jω L2 I 2 − jωM I1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 17. (a) ω2 (0.2) 2 Z = 2 + jω0.1 + 5 + jω 0.5 5ω2 (0.2) 2 jω0.5 ω2 (0.2) 2 = 2 + jω 0.1 + 2 − 2 5 + (ω0.5) 2 5 + (ω0.5) 2 = 2+ ⎡ 0.2ω2 0.02ω2 ⎤ + jω ⎢0.1 − 25 + 0.25ω2 25 + 0.25ω2 ⎥ ⎣ ⎦ (b) (c) Zin(jω) at ω = 50 is equal to 2 + 0.769 + j(50)(0.023) = 2.77 + j1.15 Ω. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 18. Z in = Z11 + ω2 M 2 Z 22 = jω50 ×10−3 + ω2 M 2 8 + jω10 × 10−3 ⇒ Z in = jω50 ×10−3 + ω2 M 2 8 jω10 ×10−3 ω2 M − 2 82 + (ω10 × 10−3 ) 2 8 + (ω10 × 10−3 ) 2 ⎡ 10 ×10−3 ω2 M 2 ⎤ ω2 M 2 8 −3 = 2 + jω ⎢50 × 10 − 2 ⎥ 8 + (ω10 × 10−3 ) 2 8 + (ω10 ×10−3 ) 2 ⎦ ⎣ In this circuit the real power delivered by the source is all consumed at the speaker, so 1 2 2 2 2 V ⎛ 20 ⎞ × 2 ω M 8−3 2 P = rms ⇒ 3.2 = ⎜ ⎟ R ⎝ 2 ⎠ 8 (ω10 ×10 ) ω2 M 2 8 202 ⇒ 2 = 8 + (ω10 × 10−3 ) 2 2 × 3.2 = 62.5 W PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 19. iS 1 = 2 cos10t A, iS 2 = 1.2 cos10t A (a) 10 March 2006 v1 = 0.6(−20sin10t ) − 0.2(−12sin10t ) + 0.5(−32sin10t ) + 9.6 cos10t ∴ v1 = 9.6 cos10t − 25.6sin10t = 27.34 cos (10t + 69.44°) V (b) v2 = 0.8(−12sin10t ) − 0.2(−20sin10t ) − 16sin10t + 9.6 cos10t ∴ v2 = 9.6 cos10t − 21.6sin10t = 23.64 cos (10t + 66.04°) V (c) 1 1 PS 1 = × 27.34 × 2 cos 69.44° = 9.601 W, PS 2 = × 23.64 × 1.2 cos 66.04° = 5.760 W 2 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. Engineering Circuit Analysis, 7th Edition 20. Chapter Thirteen Solutions 10 March 2006 Va = jω8 I a + jω4 I b * Vb = jω10 I b + jω4 I a = jω10 I b + jω5 I c Vc = jω6 I c + jω5 I b Also I = − I a = − I b = I c Now examine equation *. − jω10 I − jω4 I = − jω10 I + jω5 I c ∴ the only solution to this circuit is I = and hence v(t ) = 120 cos ωt V. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 21. 100 = j10 I1 − j15 I2 0 = j 200 I2 − j15 I1 − j15 IL 0 = (5 + j10) IL − j15 I2 ∴ I2 = ⎛ 1+ j2 ⎞ 5 + j10 1+ j2 IL = IL ∴ 0 = j 200 ⎜ − j15 ⎟ IL − j15 I1 j15 j3 ⎝ j3 ⎠ 200 ⎞ j118.33 + 66.67 ⎛ 400 ∴0 = ⎜ j − j15 + IL ⎟ IL − j15 I1 ∴ I1 = 3 ⎠ j15 ⎝ 3 ⎡2 ⎤ ∴100 = ⎢ (66.67 + j118.33) − 5 − j10 ⎥ IL = (39.44 + j 68.89) IL ⎣3 ⎦ ∴ IL = 1.2597∠ − 60.21° A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 22. is = 2 cos10t A, t = 0 (a) 10 March 2006 1 1 a − b O.C. ∴ w(0) = × 5 × 22 + × 4 × 22 = 10 + 8 = 18 J 2 2 (b) 1 12 = 3 H 2 j 20 3 = 1.1390∠9.462°A ∴ i2 = 1.1390 cos (10t + 9.462°) A ( j 30 + 5) I2 − j10 3 × 2, ∴ I2 = 5 + j 30 1 ∴ i2 (0) = 1.1235− ∴ w(0) = 10 + 8 − 3 × 2 × 1.1235 + × 3 × 1.12352 = 16.001 J 2 a − b S.C. ω = 10, IS = 2∠0° A, M = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 23. Vs = 12∠0° V rms, ω = 100 rad/s 12 = (6 + j 20) I1 + j100(0.4K) I2 , (24 + j80) I2 + j 40K I1 = 0 ∴ I1 = ⎡ ⎤ 3 + j10 3 + j10 + j 40K ⎥ I2 I2 ∴12 = ⎢(6 + j 20) − j 5K − j 5K ⎣ ⎦ ∴12 = − j 60K 18 − 200 + j 60 + j 60 + 200K 2 I2 ∴ I2 = − j 5K −182 + 200K 2 + j120 ∴ P24 = 86, 400 K 2 2.16K 2 602 K 2 24 = = 4 W (200K 2 − 182) 2 + 1202 40, 000K 4 − 72,800K 2 + 47,524 K − 1.82K 2 + 1.1881 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. Engineering Circuit Analysis, 7th Edition 24. Chapter Thirteen Solutions 10 March 2006 M • • 2Ω k= Zin → M L1 L2 ω = 250k rad / s j10 Ω M = L1 L2 = 2 × 80 × 10−6 = 12.6μH Zin n = Z11 + ω2 M 2 R22 − jM 2 ω2 X 22 + 2 2 2 2 R22 + X 22 R22 + X 22 Z11 = j × 250 ×103 × 2 ×10−6 R22 = 2Ω X 22 = (250 × 103 ) (80 ×10−6 ) = j 0.5 = 20 Thus, Zin = j0.5 + 19.8/404 – j198/ 404 = 0.049 + j0.010 Ω. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 25. ω = 100 rad/s (a) 10 March 2006 K1 → j 50Ω, K 2 → j 20Ω, 1H → j100 Ω 100 = j200 I1 − j 50 I2 − j 20 I3 0 = (10 + j100) I2 − j 50 I1 0 = (20 + j100) I3 − j 20 I1 ∴ I3 = ⎡ j2 j5 j5 j2 ⎤ I1 , I2 = I1 ∴10 = ⎢ j 20 − j 5 I1 − j2 2 + j10 1 + j10 1 + j10 2 + j10 ⎥ ⎣ ⎦ ⎛ 25 4 ⎞ ∴10 = ⎜ j 20 + + ⎟ I1 ∴ I1 = 0.5833 ∠ − 88.92° A, I2 = 0.2902∠ − 83.20° A, 1 + j10 2 + j10 ⎠ ⎝ I3 = 0.11440 ∠ − 77.61° A ∴ P10Ω = 0.29022 ×10 = 0.8422 W (b) P20 = 0.11442 × 20 = 0.2617 W (c) Pgen = 100 × 0.5833cos88.92° = 1.1039 W PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. Engineering Circuit Analysis, 7th Edition 26. (a) k= Chapter Thirteen Solutions 10 March 2006 M L1 L2 ⇒ M = 0.4 5 × 1.8 = 1.2H (b) I1 + I 2 = I 3 ⇒ I 2 = I 3 − I1 −t 5 = 5 × 10 − 4 × 10 (c) −t 10 The total energy stored at t = 0. I1 = 4 A I 2 = 1A 1 1 2 L1 I12 + L2 I 2 + M 12 I1 I 2 2 2 1 1 = × 5 × 16 + × 1.8 × 1 − 1.2 × 4 × 1 2 2 = 40 + 0.9 − 4.8 = 36.1J W total = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 27. K → j1000K L1L 2 , L1 → j1000L1 , L 2 → j1000L 2 ∴ Vs = (2 + j1000L1 ) I1 − j1000K L1L 2 I2 0 = − j1000K L1L 2 I1 + (40 + j1000L 2 ) I2 ω = 1000 rad/s ∴ I1 = 40 + j1000L 2 I2 j1000K L1L 2 ∴ Vs = (2 + j1000L1 )(40 + j1000L 2 ) + 106 K 2 L1L 2 I2 j1000K L1L 2 ∴ I2 = j1000K L1L 2 80 + j 40, 000L1 + j 2000L 2 − 106 L1L 2 (1 − K 2 ) ∴ j 40, 000K L1L 2 V2 = 6 Vs 80 − 10 L1L 2 (1 − K 2 ) + j (40, 000L1 + 2000L 2 ) (a) L1 = 10−3 , L2 = 25 ×10−3 , K = 1 ∴ (b) L1 = 1, L 2 = 25, K = 0.99 ∴ ∴ (c) V2 j 40 × 5 j 200 = = = 1.6609∠41.63° Vs 80 − 0 + j (40 + 50) 80 + j 90 V2 j 40, 000 × 0.99 × 5 = 6 V3 80 − 25 ×10 (1 − 0.992 ) + j (40, 000 + 50, 000) V2 j198, 000 = = 0.3917∠ − 79.74° VS 80 − 497,500 + j 90, 000 L1 = 1, L 2 = 25, K = 1 ∴ V2 j 40, 000 × 5 j 200, 000 = = = 2.222∠0.05093° Vs 80 − 0 + j 90, 000 80 + j 90, 000 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. Engineering Circuit Analysis, 7th Edition 28. (a) Chapter Thirteen Solutions 10 March 2006 L AB ,CDOC = 10 mH, LCD , ABOC = 5 mH L AB ,CDSC = 8 mH ∴ L1 = 10 mH, L 2 = 5 mH, 8 = 10 − M + M (5 − M) (mH) ∴ 8 = 10 − M + ∴K = (b) M(5 − M) , ∴ 5M = (10 − 8)5 + 5M − M 2 ∴ M = 3.162 mH (= 10) 5 3.162 ∴ K = 0.4472 50 Dots at A and D, i1 = 5 A, wtot = 100 mJ 1 1 2 ∴100 ×10−3 = × 10 × 10−3 × 25 + × 5 × 10−3 i2 − 10 × 5i2 × 10−3 2 2 2 10 ± 40 − 40 2 2 = 10 100 = 125 + 2.5i2 − 5 10 i2 ∴ i2 − 2 10 i2 + 10 = 0, i2 = 2 ∴ i2 = 3.162 A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. Engineering Circuit Analysis, 7th Edition 29. Chapter Thirteen Solutions 10 March 2006 Define coil voltages v1 and v2 with the “+” reference at the respective dot. Also define two clockwise mesh currents i1 and i2. We may then write: dI1 dI +M 2 dt dt dI dI v2 = L2 2 + M 1 dt dt v1 = L1 M = k L1 L2 ω = 2π60 rad / s or, using phasor notation, V1 = jωL1 I1 + jωM I 2 V2 = jωL2 I 2 + jωM I1 100∠0 = 50 I1 + jωL1 I1 + jωM I 2 −25 I 2 = jωL2 I 2 + jωM I1 Rearrange: [50 + jωL1 ] I1 + jωMI 2 = 100∠0 jωMI1 [−25 + jωL2 ] I 2 = 0 or jωM ⎤ ⎡ I1 ⎤ ⎡100∠0 ⎤ ⎡50 + jωL1 = ⎢ jωM ⎢ −25 + jωL2 ⎥ ⎣ I 2 ⎥ ⎢ 0 ⎥ ⎦ ⎣ ⎦ ⎦ ⎣ We can solve for I2 and V2 = −25I2: V2 = − j1.658 k L1L 2 + 1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 30. i1 = 2 cos 500t A Wmax at t = 0 1 1 1 ∴ wmax = × 4 × 22 + × 6 × 22 + × 5 × 22 + 3 × 22 2 2 2 = 8 + 12 + 10 + 12 = 42 J PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. Engineering Circuit Analysis, 7th Edition 31. (a) Reflected impedance = Chapter Thirteen Solutions ω 2M 2 Z 22 10 March 2006 . Z 22 = 2 + 7∠320 + jω10−2 where ω = 100π Thus, the reflected impedance is 4.56 – j3.94 nΩ (essentially zero). (b) Zin = Z11 + reflected impedance = 10 + jω(20×10–2) + (4.56 – j3.94)×10–9 = 10 + j62.84 Ω (essentially Z11 due to small reflected impedance) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. Engineering Circuit Analysis, 7th Edition 32. Reflected impedance = ω 2M 2 Z 22 Chapter Thirteen Solutions = 10 March 2006 ω 2M 2 . 3.5 + j (ω L2 + X L ) We therefore require 1 + jω ( 3 ×10−3 ) = ω 210−6 . 3.5 + j (10−3 ω + X L ) Thus, ⎡ ⎤ ω 210−6 XL = − j ⎢ − 3.5 − j10−3 ω ⎥ = −0.448 + j 3.438 . ⎢1 + jω 3 × 10−3 ⎥ ⎣ ⎦ ( ) This is physically impossible; to be built, XL must be a real number. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. Engineering Circuit Analysis, 7th Edition 33. Chapter Thirteen Solutions 10 March 2006 M = 5 H. L1 – M = 4 H, therefore L1 = 9 H L2 – M = 6 H, therefore L2 = 11 H. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 34. Engineering Circuit Analysis, 7th Edition 34. Chapter Thirteen Solutions 10 March 2006 Lz = L1 – M = 300 – 200 = 100 mH Ly = L2 – M = 500 – 200 = 300 mH Lx = M = 200 mH PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 35. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 35. (a) All DC: L1− 2 = 2 − 1 = 1 H (b) AB SC: L1− 2 = −1 + 2 8 = 0.6 H (c) BC SC: L1− 2 = 2 + ( −1) 9 = 2 − 9 / 8 = 0.875 H (d) 10 March 2006 AC SC: L1− 2 = (2 − 1) (1 + 2) = 1 3 = 0.750 H PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 36. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 36. (a) IL = VS = (b) 1 j 2ω (20 + jω ) 15 + j 3ω + 20 + j 3ω ⎛ j 2ω ⎞ ⎜ ⎟ ⎜ 20 + j 3ω ⎟ ⎠ ⎝ j 2ω 300 − 11ω 2 + j145ω −145 ± 1452 − 13, 200 = −2.570, − 10.612 22 + Be −10.61t , ∴ 0 = A + B vs (t ) = 100u (t ), is (0) = 0, iL (0) = 0, s1,2 = iL = iLf + iLn , iLf = 0, ∴ iL = Ae −2.57 t 100 = 15is + 5i′s − 2i′L , 0 = 20iL + 3i′L − 2i′s At t = 0 + : 100 = 0 + 5i′s (0 + ) − 2i′L (0 + ) and 0 = 0 + 3i′L (0+ ) − 2i′s (0+ ) ∴ i′s (0+ ) = 1.5i′L (0+ ) ∴100 = 7.5i′L (0+ ) − 2i′L (0+ ) = 5.5i′L (0+ ) ∴ i′L (0+ ) = 18.182 A/s ∴18.182 = −2.57A − 10.61B = −2.57A + 10.61A = 8.042A ∴ A = 2.261, B = −2.261, iL (t ) = 2.261(e −2.57 t − e−10.612t ) A, t > 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 37. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 37. (a) Open-Circuit T Z oc× A = jω4 M Ω T Z oc× B = jω4 M Ω (b) Short-Circuit T T Z SS× A = Z SS× B = − jω4 M Ω + jω8 jω10 M Ω (c) If the secondary is connected in parallel with the primary T Z in× A = − jω4 − jω10 + jω8 M Ω T Z in×B = jω26 jω12 − jω8 M Ω PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 38. Engineering Circuit Analysis, 7th Edition 38. Chapter Thirteen Solutions 10 March 2006 Define three clockwise mesh currents I1, I2, and I3 beginning with the left-most mesh. Vs = j8ω I1 – j4ω I2 0 = -4jω I1 + (5 + j6ω) I2 – j2ω I3 0 = -j2ω I2 + (3 + jω) I3 Solving, I3 = jω / (15 + j17ω). Since Vo = 3 I3, Vo j 3ω = VS 15 + j17ω PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 39. Engineering Circuit Analysis, 7th Edition 39. Chapter Thirteen Solutions 10 March 2006 Leq = 2/ 3 + 1 + 2 + 6/5 = 4.867 H Z(jω) = 10 jω (4.867)/ (10 + jω4.867) = j4.867ω/ (1 + j0.4867ω) Ω. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 40. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 40. ω = 100 rad/s Vs = 100∠0° V rms (a) Zina −b = 20 + j 600 + j 400(10 − j 200) 80, 000 + j 4, 000 = 20 + j 600 + 10 + j 200 10 + j 200 = 210.7∠73.48o V and Voc = 0. (b) 100( j 400) = 39.99∠1.146° V rms 20 + j1000 −240, 000 + j8, 000 j 400(20 + j 600) = − j 200 + = 40.19∠85.44°Ω Zincd , VS = 0 = − j 200 + 20 + j1000 20 + j1, 000 VOC ,cd = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 41. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 41. L1 = 1 H, L 2 = 4 H, K = 1, ω = 1000 rad/s (a) Z L = 1000 Ω ∴ Zin = j1000 + (b) 4 × 106 Z L = j1000 × 0.1 ∴ Zin = j1000 + = j 24.39 Ω j 4000 + j100 (c) ZL = − j100 ∴ Zin = j1000 + 10 March 2006 106 × 1× 4 = 24.98 + j 0.6246 Ω j 4000 + 100 4 × 106 = − j 25.46 Ω j 4000 − j100 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 42. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 42. L1 = 6 H, L 2 = 12 H, M = 5 H #1, LinAB ,CDOC = 6 H #2, LinCD , ABOC = 12 H #3, LinAB ,CDSC = 1 + 7 5 = 3.917 H #4, LinCD , ABSC = 7 + 5 1 = 7.833 H #5, LinAC , BDSC = 7 + 1 = 8 H #6, LinAB , ACSC , BDSC = 7 1 + 5 = 5.875 H #7, LinAD , BCSC = 11 + 17 = 28 H #8, LinAB , ADSC = −5 + 11/17 = 1.6786 H PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 43. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 43. Z in = Z11 + ω2 M 2 R22 + jX 22 1 1 = 31.83 ⇒ ω = = 314 rad / s 31.83 × C ωC ie. a 50Hz system Z in = 20 + jω100 × 10−3 + ω2 k 2 L1 L2 2 − j 31.83 ω2 k 2 L1 L2 2 jω2 k 2 L1 L2 31.83 − 22 + 31.832 22 + 31.832 7840 ⎤ 2 ⎡ 493 = 20 + j 31.4 + ⎢ −j k 1020 ⎥ ⎣1020 ⎦ = 20 + j 31.4 + [0.483 − j 7.69]k 2 Ω (a) Z in (k = 0) = 20 + j 31.4 (b) Z in (k = 0.5) = 20.2 + j 27.6 Ω (c) Z in (k = 0.9) = 20.4 + j 24.5 Ω (d) Z (k = 1.0) = 20.5 + j 23.7 Ω Z in = 20 + jω100 × 10−3 + in PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 44. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 44. ↑ L1 → 125 H, L 2 → 20 H, K = 1, ∴ M = 2500 = 50 H, jωM = j 5000 Ω (a) Zina −b = 20 + j 7500 + = 20 + j 7500 + j 5000(10 − j 3000) 10 + j 2000 15 × 106 + j 50, 000 = 82.499∠0.2170° Ω 10 + j 2000 = 82.498 + j 0.3125− Ω VOC = 0 (b) 100( j 5000) = 39.99995∠0.09167° V rms 20 + j12,500 j 5000(20 + j 7500) Zincd , VS = 0 = − j 3000 + = 3.19999 + j 0.00512 Ω 20 + j12,500 VOC ,cd = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 45. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 45. 280 × 2 = 0.438A 1280 1000 × 2 = 1.56A Ib = 1280 ∴ Ia = ∴ I1 = 1.56A ⇒ I 2 = 5 × 1.56 = 7.8A ⇒ I 3 = 1.5 × 7.8 A = 11.7A 2 ⇒ P(1k ) = I a R = 0.4382 × 1× 103 = 192W ⇒ P(30Ω) = I12 R = (1.56) 2 × 30 = 73W ⇒ P(1Ω) = I R = 7.82 ×1 2 2 = 60.8W ⇒ P(4Ω) = I 32 R = 11.7 2 × 4 = 548W PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 46. Engineering Circuit Analysis, 7th Edition 46. (a) Chapter Thirteen Solutions 10 March 2006 R L sees 10 × 42 = 160 Ω ∴ use R L = 160 Ω 2 PL max (b) ⎛ 100 ⎞ =⎜ ⎟ ×10 = 250 W ⎝ 20 ⎠ R L = 100 Ω V2 − V1 3V1 = 40 40 I 3V 4V ⎛ 3V ⎞ ∴100 = 10 ⎜ I1 1 ⎟ + V1 , 1 = 1 + 1 4 40 100 ⎝ 40 ⎠ I 2 = I1 / 4, V2 = 4 V1 ∴ I X = ∴ I1 = 0.46V1 ∴100 = 10(0.46V1 − 0.075V1 ) + V1 = 4.85 V1 ∴ V1 = ∴ V2 = 4V1 = 100 4.85 400 82.47 2 = 82.47 V ∴ PL = = 68.02 W 4.85 100 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 47. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 47. V2 V ∴ I1 = 2 , V1 = 5V2 8 40 ∴100 = 300(C + 0.025) V2 + 5V2 I2 = ∴ V2 = 100 12.5 + 300C (a) 82 C = 0 ∴ V2 = 8 V ∴ PL = = 8 W 8 (b) 100 ⎛ 100 ⎞ 1 C = 0.04 ∴ V2 = ∴ PL = ⎜ ⎟ = 2.082 W (neg. fdbk ) 24.5 ⎝ 24.5 ⎠ 8 (c) C = −0.04 ∴ V2 = 2 100 2002 = 200 V ∴ PL = = 5000 W (pos. fdbk ) 0.5 8 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 48. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 48. Apply Vab = 1 V ∴ Ix = 0.05 A, V2 = 4 V 4 −1 = 0.05 A 60 ∴ I1 = 0.2 A ∴ Iin = 0.25 A ∴ R th = 4 Ω, Vth = 0 ∴ 4 = 60 I2 + 20 × 0.05 ∴ I2 = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 49. Engineering Circuit Analysis, 7th Edition 49. Chapter Thirteen Solutions 10 March 2006 Pgen = 1000 W, P100 = 500 W 500 = 5 A, VL = 100 5 V 100 1000 IS = = 10 A ∴ V1 = 100 − 40 = 60 V 100 ∴ IL = Now, P25 = 1000 − 500 − 102 × 4 = 100W ∴ I X = Ix = b 5 = 2, b = 100 = 2 A; also 25 2 = 0.8944 5 Around center mesh: 60a = 2 × 25 + 100 5 1 300 ∴a = =5 0.8944 60 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 50. Engineering Circuit Analysis, 7th Edition 50. (a) Chapter Thirteen Solutions 10 March 2006 2 16 22 22 2 ⎛ 4 ⎞ 16 Ω, +2= Ω, 3× ⎜ ⎟ = (3) = 66 Ω 3 3 3 3 ⎝ 3⎠ 100 = 1.0989∠0° A = I1 66 + 25 = 91Ω 91 (b) I2 = 3I1 = 3.297∠0° A (c) 4 I3 = − × 3.297 = 4.396∠180° A 3 (d) P25 = 25 ×1.09892 = 30.19 W (e) P2 = 3.297 2 × 2 = 21.74 W (f) P3 = 4.3962 × 3 = 57.96 W PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 51. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 51. V1 = 2.5 V2 , I1 = 0.4 I2 , I50 = I2 + 0.1 V2 60 + 2.5 V2 16 Also, 60 = 50 ( I2 + 0.1 V2 ) + V2 = 50 I2 + 6V2 ∴ 60 = 40(0.4 I2 ) − 2.5V2 ∴ I2 = ⎛ 60 + 2.5 V2 ⎞ ∴ 60 = 50 ⎜ ⎟ + 6 V2 = 187.5 + (7.8125 + 6) V2 16 ⎝ ⎠ 60 − 187.5 ∴ V2 = = −9.231 V 13.8125 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 52. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 52. 400 = 16 Ω, 16 48 = 12Ω, 12 + 4 = 16 Ω 52 16 10 = 4 Ω ∴ Is = = 2 A ∴ P1 = 4 W 2 2 4 +1 2 = 1 A ∴ P4 = 4 W, 10 − 2 × 1 = 8 V 2 8 × 2 = 16 V, 16 − 4 × 1 = 12 V, 122 / 48 = 3 W = P48 , 12 × 5 = 60 V P400 = 602 =9 W 400 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 53. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 53. I1 = 2I 2 , 2I 2 = I s + I x ∴ I x + I s − 2I 2 = 0 1 100 = 3I s + (4I 2 + 20I 2 − 20I x ) 2 ∴10I x − 3I s − 12I 2 = −100 100 = 3 I s − 5I x + 20I 2 − 20I x ∴ 25I x − 3I s − 20I 2 = −100 0 1 −2 −100 −3 −12 ∴IX = −100 −3 −20 −800 0 + 100(−26) − 100(−18) = = = 4.819 A 1 1 −2 1(60 − 36) − 10(−20 − 6) + 25(−12 − 6) −166 10 −3 −12 25 −3 −20 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 54. Engineering Circuit Analysis, 7th Edition Chapter Thirteen Solutions 10 March 2006 54. (a) 50 10 = 25 25 100 V Ω ∴ VAB = 1× 4 × = 3 3 3 2 ⎛ 100 ⎞ 1 1000 ∴ P10AB = ⎜ = = 111.11 W ⎟ 9 ⎝ 3 ⎠ 10 25 252 VCD = 1× 3 × = 62.5 W = 25 V, P10CD = 3 10 (b) Specify 3 A and 4 A in secondaries I AB = I f + 4 25 25 (I f + 4) = (−I f − 3) 3 3 ∴ 2I f = −7, I f = −3.5 A ICD = − Ib − 3 ∴ ∴ VAB = VCD = 25 25 (−3.5 + 4) = V 3 6 ∴ P10 AB = P10CD ⎛ 25 ⎞ 1 =⎜ ⎟ = 1.7361 W ⎝ 6 ⎠ 10 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 55. Engineering Circuit Analysis, 7th Edition 55. Chapter Thirteen Solutions 10 March 2006 Corrections required to the problem text: both speakers that comprise the load are 4-Ω devices. We desire a circuit that will connect the signal generator (whose Thévenin resistance is 4 Ω) to the individual speakers such that one speaker receives twice the power delivered to the other. One possible solution of many: We can see from analysing the above circuit that the voltage across the right-most 1.732 speaker will be or 2 times that across the left speaker. Since power is 1.225 proportional to voltage squared, twice as much power is delivered to the right speaker. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 56. Engineering Circuit Analysis, 7th Edition 56. Chapter Thirteen Solutions 10 March 2006 (a) We assume Vsecondary = 230∠0o V as a phasor reference. Then, Iunity PF load = I0.8 PF load = 8000 o ∠0 = 34.8∠0o A 230 ( and ) 15000 ∠ − cos −1 0.8 = 65.2∠ − 36.9o A 230 Thus, Iprimary = ( 230 34.8∠0o + 65.2∠ - 36.9o 2300 ) = 0.1 (86.9 – j39.1) = 9.5 ∠-24.3o A (b) The magnitude of the secondary current is limited to 25×103/230 = 109 A. If we include a new load operating at 0.95 PF lagging, whose current is I0.95 PF load = | I0.95 PF load | ∠ (-cos-1 0.95) = | I0.95 PF load | ∠ -18.2o A, then the new total secondary current is 86.9 – j39.1 + | I0.95 PF load | cos 18.2o – j | I0.95 PF load | sin 18.2o A. Thus, we may equate this to the maximum rated current of the secondary: 109 = (86.9 + | I 0.95 PF load | cos 18.2o ) 2 + (39.1 + | I 0.95 PF load | sin 18.2o ) 2 Solving, we find | I 0.95 PF load |2 = - 189 ± 189 2 + (4)(2800) 2 So, |I0.95 PF load | = 13.8 A (or –203 A, which is nonsense). This transformer, then, can deliver to the additional load a power of 13.8×0.95×230 = 3 kW. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 57. Engineering Circuit Analysis, 7th Edition 57. Chapter Thirteen Solutions 10 March 2006 After careful examination of the circuit diagram, we (fortunately or unfortunately) determine that the meter determines individual IQ based on age alone. A simplified version of the circuit then, is simply a 120 V ac source, a 28.8-kΩ resistor and a (242)RA resistor all connected in series. The IQ result is equal to the power (W) dissipated in resistor RA divided by 1000. 2 ⎛ ⎞ 120 P= ⎜ ⎜ 28.8 × 103 + 576R ⎟ × 576R A ⎟ A ⎠ ⎝ 2 ⎞ 1 ⎛ 120 ⎜ Thus, IQ = ⎜ 28.8 × 103 + 576 × Age ⎟ × 576 × Age ⎟ 1000 ⎝ ⎠ (a) Implementation of the above equation with a given age will yield the “measured” IQ. (b) The maximum IQ is achieved when maximum power is delivered to resistor RA, which will occur when 576RA = 28.8×103, or the person’s age is 50 years. (c) Well, now, this arguably depends on your answer to part (a), and your own sense of ethics. Hopefully you’ll do the right thing, and simply write to the Better Business Bureau. And watch less television. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 58. Engineering Circuit Analysis, 7th Edition 58. Chapter Thirteen Solutions 10 March 2006 We require a transformer that converts 240 V ac to 120 V ac, so that a turns ratio of 2:1 is needed. We attach a male european plug to the primary coil, and a female US plug to the secondary coil. Unfortunately, we are not given the current requirements of the CD writer, so that we will have to over-rate the transformer to ensure that it doesn’t overheat. Checking specifications on the web for an example CD writer, we find that the power supply provides a dual DC output: 1.2 A at 5 V, and 0.8 A at 12 V. This corresponds to a total DC power delivery of 15.6 W. Assuming a moderately efficient ac to DC converter is being used (e.g. 80% efficient), the unit will draw approximately 15.6/0.8 or 20 W from the wall socket. Thus, the secondary coil should be rated for at least that (let’s go for 40 W, corresponding to a peak current draw of about 333 mA). Thus, we include a 300-mA fuse in series with the secondary coil and the US plug for safety. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 59. Engineering Circuit Analysis, 7th Edition 59. Chapter Thirteen Solutions 10 March 2006 You need to purchase (and wire in) a three-phase transformer rated at 3 (208)(10) = 3.6 kVA. The turns ratio for each phase needs to be 400:208 or 1.923. ( ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 60. Engineering Circuit Analysis, 7th Edition 60. Chapter Thirteen Solutions 10 March 2006 (a) The input to the left of the unit will have the shape: and the output voltage will be: We need to reduce the magnitude from 115-V (rms) to a peak voltage of 5 V. The corresponding peak voltage at the input will be 115 2 = 162.6 V, so we require a transformer with a turns ratio of 162.6:5 or about 32.5:1, connected as shown: 115 V rms ac ± a = 1/ 32.5 (b) If we wish to reduce the “ripple” in the output voltage, we can connect a capacitor in parallel with the output terminals. The necessary size will depend on the maximum allowable ripple voltage and the minimum anticipated load resistance. When the input voltage swings negative and the output voltage tries to reduce to follow, current will flow out of the capacitor to reduce the amount of voltage drop that would otherwise occur. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.