Hayt Engineering Circuit Analysis Solutions Manual

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Hayt Engineering Circuit Analysis 8th Edition Solution Manual

Engineering circuit-analysis-solutions-7ed-hayt. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. Engineering circuit analysis 8th edition hayt solution manual. Solutions Manual for Engineering Circuit Analysis by William H Hayt Jr.

Engineering circuit-analysis-solutions-7ed-hayt. 1. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20061. (a) 12 μs (d) 3.5 Gbits (g) 39 pA (b) 750 mJ (e) 6.5 nm (h) 49 kΩ (c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pAPROPRIETARY MATERIAL.

If you are a student using this Manual, you are using it without permission. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 20062. (a) 1 MW (e) 33 μJ (i) 32 mm (b) 12.35 mm (f) 5.33 nW (c) 47. KW (g) 1 ns (d) 5.46 mA (h) 5.555 MWPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 ⎛ 745.7 W ⎞3. (a) ( 400 Hp ) ⎜ ⎟ = 298.3 kW ⎝ 1 hp ⎠ ⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ (b) 12 ft = (12 ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.658 m ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ (c) 2.54 cm = 25.4 mm ⎛ 1055 J ⎞ (d) ( 67 Btu ) ⎜ ⎟ = 70.69 kJ ⎝ 1 Btu ⎠ (e) 285.4´10-15 s = 285.4 fsPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 20064.

(15 V)(0.1 A) = 1.5 W = 1.5 J/s. 3 hrs running at this power level equates to a transfer of energy equal to (1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 20065. 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 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 20066. The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 400 t (ns) 20Then 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 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 20067. The 1-mJ pulse lasts 75 fs. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 1 t (fs) 75Then 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 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 20068. The power drawn from the battery is (not quite drawn to scale): P (W) 10 6 t (min) 5 7 17 24 (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 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 20069. 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 300 300 s ⎡ 10 ⎤ 10 2 ∫0 ⎢ − 300 t + 10 ⎥ dt = − 600 t + 10t ⎣ ⎦ 0 = 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 kJPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200610.

Total charge q = 18t2 – 2t4 C. (a) q(2 s) = 40 C.

(b) To find the m aximum charge within 0 ≤ t ≤ 3 s, we need to take the f irst and second derivitives: dq/dt = 36t – 8t3 = 0, leading to roots at 0, ± 2.121 s d2q/dt2 = 36 – 24t2 substituting t = 2.121 s into the expression for d2q/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.

Hayt Engineering Circuit Analysis Solutions Manual

(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. 50 70 60 (a) (b) 50 0 40 30 tim e (t) q (C) -50 20 10 0 -100 -10 -20 0 0.5 1 1.5 2 2.5 3 -150 0 0.5 1 1.5 2 2.5 3 tim e (s ) i (A )PROPRIETARY MATERIAL.

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Referring to Fig. 2.6c, ⎧- 2 + 3e −5t A, t0Thus,(a) i1(-0.2) = 6.155 A(b) i1 (0.2) = 3.466 A(c) To determine the instants at which i1 = 0, we must consider t 0 separately:for t 0, -2 + 3e3t = 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.

8 0, we find the energy delivered to be = 18/(200)2 - 1800/(200)3 = 0PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200617. (a) Pabs = (40i)(3e-100t) t = 8 ms = 360 e −100t 2 t = 8 ms = 72.68 W ⎛ di ⎞ (b) Pabs = ⎜ 0.2 ⎟ i = - 180 e −100t 2 t = 8 ms = - 36.34 W ⎝ dt ⎠ ⎜ 0 ⎝ ⎟ ⎠ ( (c) Pabs = ⎛ 30∫ idt + 20 ⎞ 3e −100t t ) t = 8 ms = ⎛ 90e −100t ∫ 3e −100t ′ dt ′ + 60e −100t ⎞ t ⎜ ⎟ = 27.63 W ⎝ 0 ⎠ t = 8 msPROPRIETARY MATERIAL.

If you are a student using this Manual, you are using it without permission. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 200618. (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. (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 th e maximum current corres ponds to zero voltage, and likewise, the maximum voltage occu rs at zero curren t. The m aximum 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 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200619. (a) P first 2 hours = ( 5 V )( 0.001 A ) = 5 mW P next 30 minutes = (? V )( 0 A ) = 0 mW P last 2 hours = ( 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 JPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200620. Note that in the table below, only th e –4-A sou rce 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) -4W 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 powe r quantities sum to –4 – 16 + 40 – 50 + 30 = 0, as de manded from conservation of energy.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200621. 20 A 32 V 8V 40 V –16 A –12 A 40 V 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: ∑ supplied power = 64 + 256 – 640 + 800 – 480 = 0 (ok)PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200622. We are told that V x = 1 V, and from Fig. 2.33 w e see that the current flowing through the dependent source (and hence th rough each element of t he circuit) is 5V x = 5 A.

We will co mpute absorbed power by using the current flowing into the po sitive reference terminal of the a ppropriate voltage (p assive sign conven tion), and we will compute supplied power by using the current flowing out of the pos itive 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 su m of the power supplied = the sum of the power absorbed, as we expect from the principle of conservation of energy.PROPRIETARY MATERIAL.

If you are a student using this Manual, you are using it without permission. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 200623. We are asked to determine the voltage vs, which is identical to the voltage labeled v1.

The only remaining reference to v1 is in the expression for the current flowing through the dependent source, 5v1. This current is equal to –i2. Thus, 5 v1 = -i2 = - 5 mA Therefore v1 = -1 mV and so vs = v1 = -1 mVPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 200624. The voltage across the dependent source = v2 = –2ix = –2(–0.001) = 2 mV.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200625. 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 APROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200626. The supply voltage is 110 V, and the m aximum dissipated power is 500 W. The fuse s are specified in te rms of current, so we need to determine the m aximum current th at can flow through the fuse.

P=VI therefore Imax = Pmax/V = (500 W)/(110 V) = 4.545 A If we choose the 5-A fuse, it will allow up to (110 V)(5 A) = 550 W of power to be delivered to the application (we must assume here that the fuse absorbs zero power, a reasonable assumption in practice). This exceeds the specified maximum power. If we choose the 4.5-A fuse instead, we will hav e a m aximum current of 4.5 A. This leads to a maximum power of (110)(4.5) = 495 W delivered to the application. Although 495 W is less than the m aximum po wer allowed, this fuse will provid e adequate protection for the app lication circuitry. If a faul t occurs and the application circuitry attempts to draw too m uch power, 1000 W for example, the fuse will blow, no current will f low, and the application c ircuitry will be p rotected.

However, if the application circuitry tries to draw its maximum rated power (500 W ), the fuse will also blow. In practice, m ost equipm ent will not draw its m aximum rated powe r continuously—although to be safe, we typically assume that it will.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200627.

(a) imax = 5/ 900 = 5.556 mA imin = 5/ 1100 = 4.545 mA(b) p = v2 / R so pmin = 25/ 1100 = 22.73 mW pmax = 25/ 900 = 27.78 mWPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200628.

P = i2 R, so pmin = (0.002)2 (446.5) = 1.786 mW and (more relevant to our discussion) pmax = (0.002)2 (493.5) = 1.974 mW 1.974 mW would be a correct answer, although power ratings are typically expressed as integers, so 2 mW might be more appropriate.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200629. (a) Pabs = i2R = 20e-12t 2 (1200) μW = 20e-1.2 2 (1200) μW= 43.54 mW(b) Pabs = v2/R = 40 cos 20t 2 / 1200 W keep in mind we 2= 40 cos 2 / 1200 W are using radians= 230.9 mW(c) Pabs = v i= 8t 1.5 W= 253.0 mWPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

It’s probably best to begin this problem by sketching the voltage waveform: v (V) +10 40 60 t (ms) 20 -10(a) vmax = +10 V(b) vavg = (+10)(20×10-3) + (-10)(20×10-3)/(40×10-3) = 0(c) iavg = vavg /R = 0 2 vmax(d) pabs max = = (10)2 / 50 = 2 W R 1 ⎡ (+10) 2 (−10) 2 ⎤(e) pabs = ⎢ ⋅ 20 + ⋅ 20⎥ = 2 W avg 40 ⎣ R R ⎦PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200631. Since we are inform ed that the same current must flow through each com ponent, we begin by defining a current I flowing out of the positive reference term inal of thevoltage source.

The power supplied by the voltage source is Vs I. The power absorbed by resistor R1 is I2R1. The power absorbed by resistor R2 is I2R2. Since we know that the total power supplied is equal to the total power absorbed, we may write: Vs I = I2R1 + I2R2 or Vs = I R1 + I R2 Vs = I (R1 + R2) By Ohm’s law, I = VR2 / R2 so that VR2 Vs = (R1 + R2 ) R2 Solving for VR2 we find R2 VR2 = Vs Q.E.D. (R1 + R2 )PROPRIETARY MATERIAL.

© 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200632. (a) 6 5 4 3 2 c urrent (m A ) 1 0 -1 -2 -3 -4 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 voltage (V ) (b) W e see from our answer to part (a ) that this device has a reasonably linear characteristic (a not unreasonable degree of e xperimental error is evident in the data).

Thus, we choose to estimate the resistance using the two extreme points: Reff = (2.5 – (-1.5)/5.23 – (-3.19) kΩ = 475 Ω Using the last two points instead, we find Reff = 469 Ω, so that we can state with some certainty at least that a reasonable estimate of the resistance is approximately 470 Ω.(c) 2 1.5 1 c urrent (m A ) 0.5 0 -0.5 -1 -1.5 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 voltage (V )PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200633. Top Left Circuit: I = (5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW Top Right Circuit: I = -(5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW Bottom Left Circuit: I = (-5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW Bottom Right Circuit: I = -(-5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mWPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 200634. The voltage vout is given by vout = -10-3 vπ (1000) = - vπSince vπ = vs = 0.01 cos 1000t V, we find that vout = - vπ = -0.01 cos 1000t VPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 200635. Vout = -vπ = -vS = -2sin 5t V vout (t = 0) = 0 V vout (t = 0.324 s) = -2sin (1.57) = -2 V (use care to em ploy r adian m ode on your calculator or convert 1.57 radians to degrees)PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200636. 18 AWG wire has a resistance of 6.39 Ω / 1000 ft. Thus, we require 1000 (53) / 6.39 = 8294 ft of wire. (Or 1.57 miles.

Or, 2.53 km).PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200637. We need to create a 47 0-Ω resistor from 28 AWG wire, knowing that the am bient tem perature is 108oF, or 42.22oC.

Referring to Table 2.3, 28 AWG wire is 65.3 m Ω/ft at 20 oC, and using the equation provided we compute R2/R1 = (234.5 + T2)/(234.5 + T1) = (234.5 + 42.22)/(234.5 + 20) = 1.087 We thus find that 28 AWG wire is (1.087)(65.3) = 71.0 mΩ/ft. Thus, to repair the transmitter we will need (470 Ω)/(71.0 × 10-3 Ω/ft) = 6620 ft (1.25 miles, or 2.02 km). Note: This seems like a lot of wire to be washing up on shore.

We may find we don’t have enough. In that case, perhaps we shoul d take our cue from Eq.

6, and try to squash a piece of the wire flat so that it has a very small cross-sectional area.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200638. We are given that the conductivity σ of copper is 5.8×107 S/m. (a) 50 ft of #18 (18 AWG) copper wire, which has a diameter of 1.024 mm, will have a resistance of l/(σ A) ohms, where A = the cross-sectional area and l = 50 ft. Converting the dimensional quantities to meters, l = (50 ft)(12 in/ft)(2.54 cm/in)(1 m/100 cm) = 15.24 m and r = 0.5(1.024 mm)(1 m/1000 mm) = 5.12×10-4 m so that A = π r2 = π (5.12×10-4 m)2 = 8.236×10-7 m2 Thus, R = (15.24 m)/( 5.8×107)( 8.236×10-7) = 319.0 mΩ (b) We assume that the conductivity value specified also holds true at 50oC. The cross-sectional area of the foil is A = (33 μm)(500 μm)(1 m/106 μm)( 1 m/106 μm) = 1.65×10-8 m2 So that R = (15 cm)(1 m/100 cm)/( 5.8×107)( 1.65×10-8) = 156.7 mΩ A 3-A current flowing through this copper in the direction specified would lead to the dissipation of I2R = (3)2 (156.7) mW = 1.410 WPROPRIETARY MATERIAL.

Since R = ρ l / A, it follows that ρ = R A/ l. From Table 2.4, we see that 28 AWG soft copper wire (cross-sectional area = 0.080 4 mm2) is 65.3 Ω per 1000 ft. Thus, R = 65.3 Ω. L = (1000 ft)(12 in/ft)(2.54 cm/in)(10 mm/cm) = 304,800 mm. A = 0.0804 mm2. Thus, ρ = (65.3)(0.0804)/304800 = 17.23 μΩ/ mm or ρ = 1.723 μΩ.cm which is in fact consistent with the representative data for copper in Table 2.3.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 200640. (a) From the text, (1) Zener diodes, (2) Fuses, and (3) Incandescent (as opposed to fluorescent) light bulbs This last on e requires a few facts to be put together. We have st ated that temperature can affect resistance—in other words, if the temperature changes during operation, the resistance will not rem ain constant and hence nonlinear behavior will be observed. Most discrete resistors are rated for up to a sp ecific power in o rder to ensure th at temperature variation during operation will no t significantly change th e res istance value. Light bulbs, however, become rather warm when operating and can experience a significant change in resistance.

(b) The energy is dissipated by the resistor, converted to heat which is transferred to the air surrounding the resistor. The resistor is unable to store the energy itself.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 200641. The quoted resistivity ρ of B33 copper is 1.7654 μΩ.cm. A = πr2 = π(10–3)2 = 10–6π m2. Thus, R = ρ l / A = (1.7654 ×10 −6 ) Ωcm (1 m/100 cm )(100 m ) = 0.5619 Ω 10−6 π And P = I2R = (1.5)2 (0.5619) = 1.264 W.PROPRIETARY MATERIAL.

We know that for any wire of cross-sectional area A and length l, the resistan ce is given by R = ρ l / A. If we keep ρ fixed by choosing a material, and A fixed by choosing a wire gauge (e.g. 28 AWG), changing l will change the resistance of our “device.” A simple variable resistor concept, then: Copper wire Rotating sh ort Leads to wire de ter- connect to mines length circuit of long wire used in circuit.

But this is s omewhat impractical, as the l eads may turn out to have alm ost the sam e resistance unless we have a very long wi re, which can also be im practical. One improvement would be to replace the coppe r wire shown with a coil of insulated copper wire. A s mall amount of insulation would then need to be removed from where the moveable wire touches the coil so that electrical connection could be made.PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200643.

(a) We need to plot the negative and positive voltage ranges separately, as the positive voltage range is, after all, exponential! -6 x 10 -4 16 10 14 12 10 -5 10 c urrent (A ) c urrent (A) 8 -6 10 6 4 -7 10 2 0 -8 -2 10 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 voltage (V ) voltage (V) (b) To dete rmine the resist ance of the device at V = 550 mV, we com pute the corresponding current: I = 10-9 e39(0.55) – 1 = 2.068 AThus, R(0.55 V) = 0.55/2.068 = 266 mΩ (c) R = 1 Ω corresponds to V = I. Thus, we need to solve the transcendental equation I = 10-9 e39I – 1 Using a scientific calculator or the tried-and-true trial and error approach, we find that I = 514.3 mAPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200644. We require a 10- Ω resistor, and are told it is f or a portable application, implying that size, weight or both would be im portant to consider when selecting a wire gauge. W e have 10,000 ft of each of the gaug es listed in Table 2.3 w ith which to work. Quick inspection o f the values listed e liminates 2, 4 a nd 6 AW G w ire as the ir respec tive resistances are too low for only 10,000 ft of wire.

Using 12-AWG wire would require (10 Ω) / (1.59 mΩ/ft) = 6290 ft. Using 28-AWG wire, the narrowest available, would require (10 Ω) / (65.3 mΩ/ft) = 153 ft. Would the 28-AW G wire weight less? Again referring to T able 2.3, we see that the cross-sectional area of 28-AWG wire is 0.0804 mm 2, and that of 12-AWG wire is 3.31 mm2. The volume of 12-AWG wire required is therefore 6345900 mm 3, and that of 28-AWG wire required is only 3750 mm3.

The best (but not the only) choice for a portable application is clear: 28-AWG wire!PROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y toteachers 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 200645. Our target is a 100-Ω resistor.

We see from the plot that at ND = 1015 cm-3, μn 2x103 2 cm /V-s, yielding a resistivity of 3.121 Ω-cm. At ND = 1018 cm-3, μn 230 cm2/ V-s, yielding a resistivity of 0.02714 Ω-cm.

Thus, we s ee that the lower doping level cl early provides m aterial with higher resistivity, requiring less of the available area on the silicon wafer. Since R = ρL/A, where we know R = 100 Ω and ρ = 3.121 Ω-cm for a phosphorus concentration of 1015 cm-3, we need only define the resistor geom etry to complete the design. We choose a geometry as shown in the figure; our contact area is arbitrarily chosen as 100 μm by 250 μm, so that only the length L remains to be specified. Solving, R (100 Ω)(100 μ )(250 μ ) m m L= A= = 80.1 μ m ρ ( ( 3.121 Ω cm ) 10 μ m/cm 4 ) Design summary (one possibility): ND = 1015 cm-3 L = 80.1 μm Contact width 100 μm = (Note: this is som ewhat atypical; in the semiconductor industry contacts are typically made to the top and/o r bottom surface of a wafer. So, there’s more than one solu tion based on geometry as well as doping level.) 100 μm Wafer surface 250 μm contact 80.1 μm contactPROPRIETARY MATERIAL. © 2007 Th e McGraw-Hill Companies, Inc.

Lim ited distr ibution perm itted onl y toteachers 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 Three Solutions 10 March 20061.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 20062. (a) six nodes; (b) nine branches.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 20063. (a) Four nodes; (b) five branches; (c) path, yes – loop, no.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

Limited distribution permitted only toteachers 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 Three Solutions 10 March 20064. (a) Five nodes; (b) seven branches; (c) path, yes – loop, no.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

(a) The number of nodes remains the same – four (4). (b) The number of nodes is increased by one – to five (5). (c) i) YES ii) NO – does not return to starting point iii) YES iv) NO – does not return to starting point v) NO – point B is crossed twicePROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 20066. (a) By KCL at the bottom node: 2 – 3 + iZ – 5 – 3 = 0 So iZ = 9 A. (b) If the left-most resistor has a value of 1 Ω, then 3 V appears across the parallel network (the ‘+’ reference terminal being the bottom node) Thus, the value of the other resistor is given by 3 R= = 600 mΩ. −(−5)PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

Limited distribution permitted only toteachers 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 Three Solutions 10 March 20067. (a) 3 A; (b) –3 A; (c) 0.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 20068.

Engineering Circuit Analysis Pdf

By KCL, we may write: 5 + iy + iz = 3 + ix (a) ix = 2 + iy + iz = 2 + 2 + 0 = 4 A (b) iy = 3 + ix – 5 – iz iy = –2 + 2 – 2 iy Thus, we find that iy = 0. (c) 5 + iy + iz = 3 + ix → 5 + ix + ix = 3 + ix so ix = 3 – 5 = -2A.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 20069. Focusing our attention on the bottom left node, we see that ix = 1 A.

Focusing our attention next on the top right node, we see that iy = 5 A.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200610.

We obtain the current each bulb draws by dividing its power rating by the operating voltage (115 V): I100W = 100/115 = 896.6 mA I60W = 60/115 = 521.7 mA I40W = 347.8 mA Thus, the total current draw is 1.739 A.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200611. The DMM is connected in parallel with the 3 load resistors, across which develops the voltage we wish to measure.

If the DMM appears as a short, then all 5 A flows through the DMM, and none through the resistors, resulting in a (false) reading of 0 V for the circuit undergoing testing. If, instead, the DMM has an infinite internal resistance, then no current is shunted away from the load resistors of the circuit, and a true voltage reading results.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200612. In either case, a bulb failure will adversely affect the sign.

Still, in the parallel-connected case, at least 10 (up to 11) of the other characters will be lit, so the sign could be read and customers will know the restaurant is open for business.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

Limited distribution permitted only toteachers 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 Three Solutions 10 March 200613. (a) vy = 1(3vx + iz) vx = 5 V and given that iz = –3 A, we find that vy = 3(5) – 3 = 12 V (b) vy = 1(3vx + iz) = –6 = 3vx + 0.5 Solving, we find that vx = (–6 – 0.5)/3 = –2.167 V.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200614. (a) ix = v1/10 + v1/10 = 5 2v1 = 50 so v1 = 25 V.

By Ohm’s law, we see that iy = v2/10 also, using Ohm’s law in combination with KCL, we may write ix = v2/10 + v2/10 = iy + iy = 5 A Thus, iy = 2.5 A. (b) From part (a), ix = 2 v1/ 10.

Substituting the new value for v1, we find that ix = 6/10 = 600 mA. Since we have found that iy = 0.5 ix, iy = 300 mA. (c) no value – this is impossible.PROPRIETARY MATERIAL.

Hayt Engineering Circuit Analysis 8th Edition Solution Manual

© 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200615. We begin by making use of the information given regarding the power generated by the 5-A and the 40-V sources.

The 5-A source supplies 100 W, so it must therefore have a terminal voltage of 20 V. The 40-V source supplies 500 W, so it must therefore provide a current IX of 12.5 A. (1) By KVL, –40 + (–110) + R(5) – 20 = 0 Thus, R = 34 Ω. (2) By KVL, -VG – (-110) + 40 = 0 So VG = 150 V Now that we know the voltage across the unknown conductance G, we need only to find the current flowing through it to find its value by making use of Ohm’s law.

KCL provides us with the means to find this current: The current flowing into the “+” terminal of the –110-V source is 12.5 + 6 = 18.5 A. Then, Ix = 18.5 – 5 = 13.5 A By Ohm’s law, I x = G VG So G = 13.5/ 150 or G = 90 mSPROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200616. (a) -1 + 2 + 10i – 3.5 + 10i = 0 Solving, i = 125 mA (b) +10 + 1i - 2 + 2i + 2 – 6 + i = 0 Solving, we find that 4i = -4 or i = - 1 A.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200617.

Starting at the bottom node and proceeding clockwise, we can write the KVL equation +7 – 5 – 2 – 1(i) = 0 Which results in i = 0. Again starting with the bottom node and proceeding in a clockwise direction, we write the KVL equation -9 +4i + 4i = 0 (no current flows through either the -3 V source or the 2 Ω resistor) Solving, we find that i = 9/8 A = 1.125 A.PROPRIETARY MATERIAL.

© 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200618. Begin by defining a clockwise current i.vS + v1 + v2 = 0 so vS = v1 + v2 = i(R1 + R2) vS and hence i =.

R1 + R2 R1 R2 Thus, v1 = R1i = vS and v2 = R2i = vS. R1 + R2 R1 + R2 Q.E.D.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200619.

Given: (1) Vd = 0 and (2) no current flows into either terminal of Vd. Calculate Vout by writing two KVL equations. Begin by defining current i1 flowing right through the 100 Ω resistor, and i2 flowing right through the 470 Ω resistor.5 + 100i1 + Vd = 0 1 -5 + 100i1 + 470i2 + Vout = 0 2 Making use of the fact that in this case Vd = 0, we find that i1 = 5/100 A. Making use of the fact that no current flows into the input terminals of the op amp, i1 = i2. 2 reduces to -5 + 570(5/100) + Vout = 0 or Vout = -23.5 V (hence, the circuit is acting as a voltage amplifier.)PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only toteachers 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 Three Solutions 10 March 200620.

(a) By KVL, -2 + vx + 8 = 0 so that vx = -6 V. (b) By KCL at the top left node, iin = 1 + IS + vx/4 – 6 or iin = 23 A (c) By KCL at the top right node, IS + 4 vx = 4 - vx/4 So IS = 29.5 A. (d) The power provided by the dependent source is 8(4vx) = -192 W.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

Book Name: Engineering Circuit Analysis Author Name: William H. Kemmerly Steven M. Durbin Edition: 6th Edition Type: Solution Manual Download Solution Manual here. Its in PDF format if you don't have Adobe Reader then please 1st download it.