Hayt Engineering Circuit Analysis Solutions Manual

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Hayt Engineering Circuit Analysis 8th Solution Manual Pdf
Engineering Circuit Analysis William Hayt Solution Manual 5th Edition

Chapter One: Basic Concepts2 Irwin, Basic Engineering Circuit Analysis, 8/E.

Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 3. (a) ( )⎛ ⎞⎜ ⎟ ⎝ ⎠ 745.7 W400 Hp = 1 hp 298.3 kW (b) 12 ft = 12 in 2.54 cm 1 m(12 ft) 3.658 m 1 ft 1 in 100 cm ⎛ ⎞⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ (c) 2.54 cm = 25.4 mm (d) ( ) 1055 J67 Btu = 1 Btu ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 70.69 kJ (e) 285.4´10-15 s = 285.4 fs 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 5.

Motor power = 175 Hp (a) With 100% efficient mechanical to electrical power conversion, (175 Hp)1 W/ (1/745.7 Hp) = 130.5 kW (b) Running for 3 hours, Energy = (130.5×103 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ (c) A single battery has 430 kW-hr capacity. We require (130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient.

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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 6. The 400-mJ pulse lasts 20 ns.

(a) To compute the peak power, we assume the pulse shape is square: 400 Energy (mJ) t (ns) 20 Then P = 400×10-3/20×10-9 = 20 MW. (b) At 20 pulses per second, the average power is Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 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.

Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 7. The 1-mJ pulse lasts 75 fs. (a) To compute the peak power, we assume the pulse shape is square: 1 Energy (mJ) t (fs) 75 Then P = 1×10-3/75×10-15 = 13.33 GW. (b) At 100 pulses per second, the average power is Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW. 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 8. The power drawn from the battery is (not quite drawn to scale): 5 7 17 24 P (W) 10 6 t (min) (a) Total energy (in J) expended is 6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)60 = 6.9 kJ. (b) The average power in Btu/hr is (6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr. 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.

Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 9. The total energy transferred during the first 8 hr is given by (10 W)(8 hr)(60 min/ hr)(60 s/ min) = 288 kJ The total energy transferred during the last five minutes is given by 300 s 0 10 + 10 300 t dt⎡ ⎤−⎢ ⎥⎣ ⎦∫ = 300 2 0 10 + 10 = 600 t t− 1.5 kJ (a) The total energy transferred is 288 + 1.5 = 289.5 kJ (b) The energy transferred in the last five minutes is 1.5 kJ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

Total charge q = 18 t2 – 2 t4 C. (a) q(2 s) = 40 C. (b) To find the maximum charge within 0 ≤ t ≤ 3 s, we need to take the first and second derivitives: dq/ dt = 36 t – 8 t3 = 0, leading to roots at 0, ± 2.121 s d2 q/ dt2 = 36 – 24 t2 substituting t = 2.121 s into the expression for d2 q/ dt2, we obtain a value of –14.9, so that this root represents a maximum. Thus, we find a maximum charge q = 40.5 C at t = 2.121 s. (c) The rate of charge accumulation at t = 8 s is dq/dt t = 0.8 = 36(0.8) – 8(0.8)3 = 24.7 C/s. 0 0.5 1 1.5 2 2.5 3 -20 -10 0 10 20 30 40 50 60 70 time (s) q (C ) 0 0.5 1 1.5 2 2.5 3 -150 -100 -50 0 50 i (A) tim e (t ) (a) (b) PROPRIETARY MATERIAL.

Hayt Engineering Circuit Analysis 8th Solution Manual Pdf

If you are a student using this Manual, you are using it without permission. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 11.

Referring to Fig. 2.6 c, 0 A, 3 2- 0 A, 3 2- )( 3 5 1 ⎩ ⎨ ⎧ + 0 separately: for t 0, -2 + 3 e3 t = 0 leads to t = (1/3) ln (2/3) = –0.135 s (impossible) Therefore, the current is never negative. (d) The total charge passed left to right in the interval –0. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 12. Referring to Fig.

2.28, (a) The average current over one period (10 s) is iavg = -4(2) + 2(2) + 6(2) + 0(4)/10 = 800 mA (b) The total charge transferred over the interval 1. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 13. (a) VBA = – -19 2 pJ -1.602 10 C× = 12.48 MV (b) VED = -19 0 -1.602 10 C× = 0 (c) VDC = – -19 3 pJ 1.602 10 C× = –18.73 MV (d) It takes – 3 pJ to move +1.602x10–19 C from D to C. It takes 2 pJ to move –1.602x10–19 C from B to C, or –2 pJ to move +1.602x10–19 C from B to C, or +2 pJ to move +1.602x10–19 C from C to B. Thus, it requires –3 pJ + 2 pJ = –1 pJ to move +1.602x10–19 C from D to C to B. Hence, VDB = − × -19 1 pJ 1.602 10 C = –6.242 MV.

(a) Pabs = (+3.2 V)(-2 mA) = –6.4 mW (or +6.4 mW supplied) (b) Pabs = (+6 V)(-20 A) = –120 W (or +120 W supplied) (d) Pabs = (+6 V)(2 ix) = (+6 V)(2)(5 A) = +60 W (e) Pabs = (4 sin 1000 t V)(-8 cos 1000 t mA) t = 2 ms = +12.11 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 16. I = 3 te-100 t mA and v = 6 – 600 t e-100 t mV (a) The power absorbed at t = 5 ms is Pabs = ( ) W 3 60 μ msttt teet =−− ⋅− = 0.01655 μW = 16.55 nW (b) The energy delivered over the interval 0 0, we find the energy delivered to be = 18/(200)2 - 1800/(200)3 = 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.

Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 17. (a) Pabs = (40 i)(3 e-100 t) t = 8 ms = 2 8100 360 mstte =− = 72.68 W (b) Pabs = W36.34- 180- 2.0 2 8100 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − mst tei dt di (c) Pabs = ( ) mst tt eidt 8 100 0 3 20 30 = −⎟ ⎠ ⎞⎜ ⎝ ⎛ +∫ = mst tt tt etdee 8 100 0 100100 60 3 90 = −′−− ⎟ ⎠ ⎞⎜ ⎝ ⎛ +′∫ = 27.63 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 18. (a) The short-circuit current is the value of the current at V = 0. Reading from the graph, this corresponds to approximately 3.0 A.

Hayt engineering circuit analysis 8th edition solution manual free download

(b) The open-circuit voltage is the value of the voltage at I = 0. Reading from the graph, this corresponds to roughly 0.4875 V, estimating the curve as hitting the x-axis 1 mm behind the 0.5 V mark.

(c) We see that the maximum current corresponds to zero voltage, and likewise, the maximum voltage occurs at zero current. The maximum power point, therefore, occurs somewhere between these two points. By trial and error, Pmax is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 19. (a) ( )( ) =first 2 hoursP = 5 V 0.001 A 5 mW ( )( ) =next 30 minutesP =? V 0 A 0 mW ( )( )− = −last 2 hoursP = 2 V 0.001 A 2 mW (b) Energy = (5 V)(0.001 A)(2 hr)(60 min/ hr)(60 s/ min) = 36 J (c) 36 – (2)(0.001)(60)(60) = 21.6 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 20. Note that in the table below, only the –4-A source and the –3-A source are actually “absorbing” power; the remaining sources are supplying power to the circuit. Source Absorbed Power Absorbed Power 2 V source (2 V)(-2 A) - 4 W 8 V source (8 V)(-2 A) - 16 W -4 A source (10 V)-(-4 A) 40 W 10 V source (10 V)(-5 A) - 50 W -3 A source (10 V)-(-3 A) 30 W The 5 power quantities sum to –4 – 16 + 40 – 50 + 30 = 0, as demanded from conservation of energy. 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 21. 40 V 40 V 32 V 8 V 20 A –12 A –16 A P8V supplied = (8)(8) = 64 W (source of energy) P32V supplied = (32)(8) = 256 W (source of energy) P–16A supplied = (40)(–16) = –640 W P40V supplied = (40)(20) = 800 W (source of energy) P–12A supplied = (40)( –12) = –480 W Check: = 64 + 256 – 640 + 800 – 480 = 0 (ok) supplied power∑ 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 22. We are told that Vx = 1 V, and from Fig. 2.33 we see that the current flowing through the dependent source (and hence through each element of the circuit) is 5Vx = 5 A. We will compute absorbed power by using the current flowing into the positive reference terminal of the appropriate voltage (passive sign convention), and we will compute supplied power by using the current flowing out of the positive reference terminal of the appropriate voltage. (a) The power absorbed by element “A” = (9 V)(5 A) = 45 W (b) The power supplied by the 1-V source = (1 V)(5 A) = 5 W, and the power supplied by the dependent source = (8 V)(5 A) = 40 W (c) The sum of the supplied power = 5 + 40 = 45 W The sum of the absorbed power is 45 W, so yes, the sum of the power supplied = the sum of the power absorbed, as we expect from the principle of conservation of energy. PROPRIETARY MATERIAL.

Engineering Circuit Analysis William Hayt Solution Manual 5th Edition

This current is equal to – i2. Thus, 5 v1 = - i2 = - 5 mA Therefore v1 = -1 mV and so vs = v1 = -1 mV PROPRIETARY MATERIAL.

Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 25. The battery delivers an energy of 460.8 W-hr over a period of 8 hrs. (a) The power delivered to the headlight is therefore (460.8 W-hr)/(8 hr) = 57.6 W (b) The current through the headlight is equal to the power it absorbs from the battery divided by the voltage at which the power is supplied, or I = (57.6 W)/(12 V) = 4.8 A PROPRIETARY MATERIAL.