Solution Manual-Engineering Circuit Analysis

Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 1. (a) 12 ? s (b) 750 mJ (c) 1. 13 k? (d) 3. 5 Gbits (e) 6. 5 nm (f) 13. 56 MHz (g) 39 pA (h) 49 k? (i) 11. 73 pA 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 2. (a) 1 MW (b) 12. 35 mm (c) 47. kW (d) 5. 46 mA (e) 33 ? J (f) 5. 33 nW (g) 1 ns (h) 5. 555 MW (i) 32 mm 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 3. (a) ( 400 Hp ) ? ? 745. 7 W ? ? = ? 1 hp ? 298. 3 kW ? 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 = ? 1055 J ? (d) ( 67 Btu ) ? ? = ? 1 Btu ? (e) 285. 4? 10-15 s = 25. 4 mm 70. 69 kJ 285. 4 fs PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

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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 4. (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 kJ 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: Energy (mJ) 400 20 t (ns) 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: Energy (mJ) 1 75 t (fs) 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 . 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 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 2 ? 10 ? ? ? 300 t + 10 ? dt = ? 600 t + 10t ? ? 300 = 1. 5 kJ 0 (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. 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 10. Total charge q = 18t2 – 2t4 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 = 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. (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. d) See Fig. (a) and (b). 50 70 60 50 40 30 20 10 0 -10 -20 0 0. 5 1 1. 5 tim e (s ) 2 2. 5 3 (a) 0 (b) tim e (t) q (C) -50 -100 -150 0 0. 5 1 1. 5 i (A ) 2 2. 5 3 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 11. Referring to Fig. 2. 6c, t0 ?- 2 + 3e ? 5t A, i1 (t ) = ? 3t ? – 2 + 3e A, Thus, (a) i1(-0. 2) = 6. 155 A (b) i1 (0. 2) = 3. 66 A (c) To determine the instants at which i1 = 0, we must consider t < 0 and t > 0 separately: for t < 0, – 2 + 3e-5t = 0 leads to t = -0. 2 ln (2/3) = +0. 0811 s (impossible) 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 < t < 0. 1 s is q(t) = = ? 0. 1 ?0. 8 1 i (t )dt ?5 t ?? 0. 8 ? ?2 + 3e ? dt ? ? 0 + ? 0. 1 0 ? ? 2 + 3e3t ? dt ? ? 3t 0. 1 0 3 ? ? = ? ?2t ? e? 5t ? 5 ? ? -0. 8 = 33. 91 C 0 + ( ? 2t + e ) 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 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 < t < 12 s is qtotal = ? 12 1 i (t )dt = -4(1) + 2(2) + 6(2) + 0(4) – 4(2) = 4 C (c) See Fig. below q (C) 8 16 2 4 6 8 10 12 14 16 t(s) -8 -16

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 13. (a) VBA = – 2 pJ -1. 602 ? 10-19 C = 12. 48 MV (b) VED = 0 -1. 602 ? 10-19 C 3 pJ 1. 602 ? 10-19 C = 0 (c) VDC = – = –18. 73 MV (d) It takes – 3 pJ to move +1. 602×10–19 C from D to C. It takes 2 pJ to move –1. 602×10–19 C from B to C, or –2 pJ to move +1. 602×10–19 C from B to C, or +2 pJ to move +1. 02×10–19 C from C to B. Thus, it requires –3 pJ + 2 pJ = –1 pJ to move +1. 602×10–19 C from D to C to B. Hence, VDB = ?1 pJ = –6. 242 MV. 1. 602 ? 10-19 C 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 14. + V1 – + Voltmeter – – V2 + Voltmeter + From the diagram, we see that V2 = –V1 = +2. 86 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 15. (a) Pabs = (+3. 2 V)(-2 mA) = –6. 4 mW (b) Pabs = (+6 V)(-20 A) = –120 W (or +6. 4 mW supplied) (or +120 W supplied) (d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W (e) Pabs = (4 sin 1000t V)(-8 cos 1000t 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 = 3te-100t mA and v = [6 – 600t] e-100t mV (a) The power absorbed at t = 5 ms is Pabs = (6 ? 600t ) e ? 100t ? 3te ? 100t [ ] t = 5 ms ?W = 0. 01655 ? W = 16. 55 nW (b) The energy delivered over the interval 0 < t < ? is ? ? 0 Pabs dt = ? ? 0 3t (6 ? 600t ) e ? 200t dt ?J Making use of the relationship ? ? 0 x n e ? ax dx = n! a n +1 where n is a positive integer and a > 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 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. (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 he 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) 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 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 2 V source 8 V source -4 A source 10 V source -3 A source Absorbed Power (2 V)(-2 A) (8 V)(-2 A) (10 V)[-(-4 A)] (10 V)(-5 A) (10 V)[-(-3 A)] Absorbed Power -4W – 16 W 40 W – 50 W 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. 32 V 8V –16 A 40 V –12 A 20 A 40 V P8V supplied P32V supplied P–16A supplied P40V supplied P–12A supplied Check: = (8)(8) = (32)(8) = (40)(–16) = (40)(20) = = = = 64 W 256 W –640 W 800 W –480 W (source of energy) (source of energy) (source of energy) = (40)( –12) = ? supplied power = 64 + 256 – 640 + 800 – 480 = 0 (ok) 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. © 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 23. 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 v1 = -1 mV Therefore and so vs = v1 = -1 mV 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 24. The voltage across the dependent source = v2 = –2ix = –2(–0. 001) = 2 mV. 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 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. 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. © 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 26. The supply voltage is 110 V, and the maximum dissipated power is 500 W.

The fuses are specified in terms of current, so we need to determine the maximum current that 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 have a maximum 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 maximum power allowed, this fuse will provide adequate protection for the application circuitry. If a fault occurs and the application circuitry attempts to draw too much power, 1000 W for example, the fuse will blow, no current will flow, and the application circuitry will be protected. However, if the application circuitry tries to draw its maximum rated power (500 W), the fuse will also blow. In practice, most equipment will not draw its maximum rated power continuously—although to be safe, we typically assume that it will. 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 27. (a) imax = 5/ 900 = imin = 5/ 1100 = 5. 556 mA 4. 545 mA (b) p = v2 / R so pmin = 25/ 1100 pmax = 25/ 900 = 22. 73 mW = 27. 78 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 28. 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 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 29. (a) Pabs = i2R = [20e-12t] 2 (1200) ? W = [20e-1. 2] 2 (1200) ? W = 43. 54 mW (b) Pabs = v2/R = [40 cos 20t] 2 / 1200 W = [40 cos 2] / 1200 W = 230. 9 mW (c) Pabs = v i = 8t 1. 5 W = 253. 0 mW 2 keep in mind we are using radians 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 30.

It’s probably best to begin this problem by sketching the voltage waveform: v (V) +10 40 20 -10 60 t (ms) (a) vmax = +10 V (b) vavg = [(+10)(20? 10-3) + (-10)(20? 10-3)]/(40? 10-3) = 0 (c) iavg = vavg /R = 0 (d) pabs max = 2 vmax = (10)2 / 50 = 2 W R (e) pabs avg = ? 1 ? (+10) 2 (? 10) 2 ? 20 + ? 20? = 2 W ? R 40 ? R ? 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 1. Since we are informed that the same current must flow through each component, we begin by defining a current I flowing out of the positive reference terminal of the voltage 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 Vs = VR2 R2 (R1 + R2 ) Solving for VR2 we find VR2 = Vs R2 (R1 + R2 ) Q. E. D. 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 32. (a) 6 5 4 3 c urrent (m A ) 2 1 0 -1 -2 -3 -4 -1. 5 -1 -0. 5 0 0. 5 voltage (V ) 1 1. 5 2 2. 5 (b) We see from our answer to part (a) that this device has a reasonably linear characteristic (a not unreasonable degree of experimental 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)] k? = 475 ? Using the last two points instead, we find Reff = 469 ? , so that we can state with some certainty at least that a reasonable estimate of the resistance is approximately 470 ?. (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 voltage (V ) 1 1. 5 2 2. 5 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 0 March 2006 33. Top Left Circuit: Top Right Circuit: I = (5/10) mA = 0. 5 mA, and P10k = V2/10 mW = 2. 5 mW 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 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 0 March 2006 34. The voltage vout is given by vout = -10-3 v? (1000) = – v? Since v? = vs = 0. 01 cos 1000t V, we find that vout = – v? = -0. 01 cos 1000t 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 35. vout = -v? = -vS = -2sin 5t V vout (t = 0) = 0 V vout (t = 0. 324 s) = -2sin (1. 57) = -2 V (use care to employ radian mode on your calculator or convert 1. 7 radians to degrees) 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 36. 18 AWG wire has a resistance of 6. 39 ? / 1000 ft. Thus, we require 1000 (53) / 6. 39 = 8294 ft of wire. (Or 1. 57 miles. Or, 2. 53 km). 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 37. We need to create a 470-? resistor from 28 AWG wire, knowing that the ambient temperature is 108oF, or 42. 22oC. Referring to Table 2. 3, 28 AWG wire is 65. 3 m? /ft at 20oC, 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 m? /ft. Thus, to repair the transmitter we will need (470 ? )/(71. 0 ? 10-3 ? /ft) = 6620 ft (1. 5 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 should 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 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 38. We are given that the conductivity ? f 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/(? 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 = ? r2 = ? (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 m? (b) We assume that the conductivity value specified also holds true at 50oC. The cross-sectional area of the foil is A = (33 ? m)(500 ? )(1 m/106 ? m)( 1 m/106 ? m) = 1. 65? 10-8 m2 So that R = (15 cm)(1 m/100 cm)/[( 5. 8? 107)( 1. 65? 10-8)] = 156. 7 m? 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 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 39. Since R = ? l / A, it follows that ? R A/ l . From Table 2. 4, we see that 28 AWG soft copper wire (cross-sectional area = 0. 0804 mm2) is 65. 3 ? per 1000 ft. Thus, R = 65. 3 ?. l = (1000 ft)(12 in/ft)(2. 54 cm/in)(10 mm/cm) = 304,800 mm. A = 0. 0804 mm2. Thus, ? = (65. 3)(0. 0804)/304800 = 17. 23 ?? / mm or ? = 1. 723 ??. cm which is in fact consistent with the representative data for copper in Table 2. 3. 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 40. (a) From the text, (1) Zener diodes, (2) Fuses, and (3) Incandescent (as opposed to fluorescent) light bulbs This last one requires a few facts to be put together. We have stated that temperature can affect resistance—in other words, if the temperature changes during operation, the resistance will not remain constant and hence nonlinear behavior will be observed. Most discrete resistors are rated for up to a specific power in order to ensure that temperature variation during operation will not significantly change the resistance 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 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 41. The quoted resistivity ? f B33 copper is 1. 7654 ??. cm. A = ? r2 = ? (10–3)2 = 10–6? m2. l = 100 m. (1. 7654 ? 10 Thus, R = ? l / A = ?6 ?cm (1 m/100 cm )(100 m ) 10? 6 ? ) = 0. 5619 ? And P = I2R = (1. 5)2 (0. 5619) = 1. 264 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 42. We know that for any wire of cross-sectional area A and length l , the resistance is given by R = ? / A. If we keep ? 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 short wire determines length of long wire used in circuit. Leads to connect to circuit But this is somewhat impractical, as the leads may turn out to have almost the same resistance unless we have a very long wire, which can also be impractical. One improvement would be to replace the copper wire shown with a coil of insulated copper wire.

A small 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 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 43. x 10 16 14 12 10 c urrent (A) 8 6 4 2 0 -2 -0. 7 (a) We need to plot the negative and positive voltage ranges separately, as the positive voltage range is, after all, exponential! 6 10 -4 10 -5 c urrent (A ) 10 -6 10 -7 10 -0. 6 -0. 5 -0. 4 -0. 3 -0. 2 voltage (V ) -0. 1 0 0. 1 -8 0 0. 01 0. 02 0. 03 0. 04 voltage (V) 0. 05 0. 06 0. 07 (b) To determine the resistance of the device at V = 550 mV, we compute the corresponding current: I = 10-9 [e39(0. 55) – 1] = 2. 068 A Thus, R(0. 55 V) = 0. 55/2. 068 = 266 m? (c) R = 1 ? 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 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. Engineering Circuit Analysis, 7th Edition Chapter Two Solutions 10 March 2006 44. We require a 10-? resistor, and are told it is for a portable application, implying that size, weight or both would be important to consider when selecting a wire gauge. We have 10,000 ft of each of the gauges listed in Table 2. 3 with which to work. Quick inspection of the values listed eliminates 2, 4 and 6 AWG wire as their respective resistances are too low for only 10,000 ft of wire.

Using 12-AWG wire would require (10 ? ) / (1. 59 m? /ft) = 6290 ft. Using 28-AWG wire, the narrowest available, would require (10 ? ) / (65. 3 m? /ft) = 153 ft. Would the 28-AWG wire weight less? Again referring to Table 2. 3, we see that the cross-sectional area of 28-AWG wire is 0. 0804 mm2, and that of 12-AWG wire is 3. 31 mm2. The volume of 12-AWG wire required is therefore 6345900 mm3, 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 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 45. Our target is a 100-? resistor. We see from the plot that at ND = 1015 cm-3, ? n ~ 2×103 cm2/V-s, yielding a resistivity of 3. 121 ? -cm. At ND = 1018 cm-3, ? n ~ 230 cm2/ V-s, yielding a resistivity of 0. 02714 ? -cm. Thus, we see that the lower doping level clearly provides material with higher resistivity, requiring less of the available area on the silicon wafer. Since R = ?

L/A, where we know R = 100 ? and ? = 3. 121 ? -cm for a phosphorus concentration of 1015 cm-3, we need only define the resistor geometry to complete the design. We choose a geometry as shown in the figure; our contact area is arbitrarily chosen as 100 ? m by 250 ? m, so that only the length L remains to be specified. Solving, L= R ? A= (100 ? )(100 ? m)(250 ? m) = 80. 1 ? m ( 3. 121 ? cm ) 104 ? m/cm ( ) Design summary (one possibility): ND = 1015 cm-3 L = 80. 1 ? m Contact width = 100 ? m (Note: this is somewhat atypical; in the semiconductor industry contacts are typically made to the top and/or bottom surface of a wafer.

So, there’s more than one solution based on geometry as well as doping level. ) 100 ? m Wafer surface 250 ? m contact 80. 1 ? m contact 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 Three Solutions 10 March 2006 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. Engineering Circuit Analysis, 7th Edition Chapter Three Solutions 10 March 2006 2. (a) six nodes; (b) nine branches. 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 Three Solutions 10 March 2006 3. (a) Four nodes; (b) five branches; (c) path, yes – loop, no. 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 Three Solutions 10 March 2006 4. (a) Five nodes; (b) seven branches; (c) path, yes – loop, no. 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 Three Solutions 10 March 2006 5. (a) The number of nodes remains the same – four (4). (b) The number of nodes is increased by one – to five (5). (c) i) ii) iii) iv) v) YES NO YES NO NO – does not return to starting point – does not return to starting point – point B is crossed twice 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 Three Solutions 10 March 2006 . (a) By KCL at the bottom node: So iZ = 9 A. 2 – 3 + iZ – 5 – 3 = 0 (b) If the left-most resistor has a value of 1 ? , 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 R= 3 = 600 m? . ? (? 5) 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 Three Solutions 10 March 2006 7. a) 3 A; (b) –3 A; (c) 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 Three Solutions 10 March 2006 8. 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 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 Three Solutions 10 March 2006 9. 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 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 Three Solutions 10 March 2006 10. 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 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 Three Solutions 10 March 2006 1. 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 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 Three Solutions 10 March 2006 12. 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 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 Three Solutions 10 March 2006 13. (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 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 Three Solutions 10 March 2006 14. (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, (c) no value – this is impossible. iy = 300 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. Engineering Circuit Analysis, 7th Edition Chapter Three Solutions 10 March 2006 15. 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 ?. (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 o 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 I x = G · VG or G = 90 mS By Ohm’s law, So G = 13. 5/ 150 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 Three Solutions 0 March 2006 16. (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 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 Three Solutions 10 March 2006 17. Circuit I. 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.

Circuit II. 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 ? resistor) Solving, we find that i = 9/8 A = 1. 125 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. Engineering Circuit Analysis, 7th Edition Chapter Three Solutions 10 March 2006 18. Begin by defining a clockwise current i. vS + v1 + v2 = 0 and hence i = vS . R1 + R2 R1 R2 vS and v2 = R2i = vS . R1 + R2 R1 + R2 so vS = v1 + v2 = i(R1 + R2) Thus, v1 = R1i = Q. E. D. 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 Three Solutions 10 March 2006 19. 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 ? resistor, and i2 flowing right through the 470 ? resistor. -5 + 100i1 + Vd = 0 -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. Thus, Eq. [2] reduces to -5 + 570(5/100) + Vout = 0 or Vout = -23. 5 V (hence, the circuit is acting as a voltage amplifier. ) [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. Engineering Circuit Analysis, 7th Edition Chapter Three Solutions 10 March 2006 20. (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. 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 Three Solutions 10 March 2006 21. (a) Working from left to right, v1 = 60 V v2 = 60 V i2 = 60/20 = 3 A i4 = v1/4 = 60/4 = 15 A v3 = 5i2 = 15 V By KVL, -60 + v3 + v5 = 0 v5 = 60 – 15 = 45 V v4 = v5 = 45 i5 = v5/5 = 45/5 = 9 A i3 = i4 + i5 = 15 + 9 = 24 A i1 = i2 + i3 = 3 + 24 = 27 v1 = 60 V v2 = 60 V v3 = 15 V v4 = 45 V v5 = 45 V i1 = 27 A i2 = 3 A i3 = 24 A i4 = 15 A i5 = 9 A (b) It is now a simple matter to compute the power absorbed by each element: p1 p2 p3 p4 p5 = -v1i1 = v2i2 = v3i3 = v4i4 = v5i5 = -(60)(27) = (60)(3) =

Solution Manual-Engineering Circuit Analysis

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