fundamentals of electric circuits 5th edition solutions chapter 7

... Problem 17 Figure 1.28 shows a circuit with five elements If W,30W, 45W, 60W, Trang 15Chapter 1, Problem 19 Find I in the network of Fig 1.30 I 1A + + Trang 17hr 60 4 kW 1.2 Trang 18Chapter ... lamp, (b) the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh Chapter 1, Solution 28 A0.25 2436530pt W b) Chapter 1, Problem 29 An electric stove with ... month, how much will Reliant Energy charge? Chapter 1, Solution 30 Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 968 kWh @ $0.07/kWh= $67.76 Chapter 1, Problem 31 In a household,

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... 16 d d d a V V V V V V V V 7 2 5 150 10 20 4 30 − + = ⎯→ ⎯ − + V V V V d c b a 7 2 0 5 4 0 2 7 0 2 4 1 2 11 736 1 847 7 14 101 V, 736 1 V, 847 7 V, 14 Trang 37Use MATLAB to solve for the ... 4 0 0 4 7 1 1 0 1 4 4 3 2 1 309 2 209 1 7708 01 B A V i.e Trang 39v v 10 v 100 40 v 120 v v 7 6 9 7 o 1 5 54 49 7 6 9 7 = + 8440 7 720 9 280 280 7 − = Δ 5 8440 1 = − = − Δ Δ Io = –5.6 A ... V1 = –7.19V; V2 = –2.78V; V3 = 2.89V Trang 33Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig 3.76 Figure 3.76 Trang 34v v 7 13 41 6 1 4 11 7 3 2 1 , 176 7 13 4

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... Trang 7Chapter 4, Problem 7 Use linearity and the assumption that Vo = 1V to find the actual value of V in Fig 4.75 o + _ 4 Ω 4 V + _ Trang 8Chapter 4, Problem 8 in the circuit of Fig 4.76 ... Th R Trang 92373 337 3 2 1 1007 33 373 337 3123 3127 i2 = Δ/Δ = -120/100 = -1.2 A 2 VTh = 12 + 2i2 = 9.6 V, and IN = V /RTh Th = 8 A Chapter 4, Problem 62. Trang 93Figure 4.128 Chapter 4, Solution ... (c) Trang 76Chapter 4, Problem 53 Find the Norton equivalent at terminals a-b of the circuit in Fig 4.119 Figure 4.119 Trang 77But, (vo/2) = 0.25vo + IN, which leads to IN = 1 A Trang 78Chapter

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... Figure 5.96 Trang 77The output of amplifier A is 9 ) 2 ( 10 30 ) 1 ( 10 30 The output of amplifier B is 14 ) 4 ( 10 20 ) 3 ( 10 20 − + V 2 ) 14 ( Trang 78Chapter 5, Problem 71 Determine v in ... 5.97 o + – Figure 5.97 Trang 79 20k Ω - 40k Ω + 50 1 ( , 8 ) 2 ( 5 20 , v V 10 ) 10 20 ( 80 100 40 Trang 81Find i in the op amp circuit of Fig 5.100 oFigure 5.100 Chapter 5, Solution 74 ... − − = Trang 70o s2 1 Trang 71Find vo in the op amp circuit of Fig 5.92 + – 48 8 Trang 72100 ) 4 ( 2040 20 100 ) 6 ( 25 − = 24 40 20 -4V Trang 73Obtain the output v in the circuit of Fig 5.94

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... circuit in Fig 6.76 Figure 6.76 Chapter 6, Solution 54 ( 10 0 6 12 ) ) 3 9 ( 4 = 4 + 12 ( 0 + 4 ) = 4 + 3 Leq = 7H Trang 49Find Leq in each of the circuits of Fig 6.77 Figure 6.77 Chapter 6, ... to represent the inductive network of Fig 6.79 at the terminals Figure 6.79 Trang 52Chapter 6, Solution 57 Let dt di L 2 2 dt di 4 v di or dt di 5 dt di 7 dt di 5 dt di 5 dt 21 dt Trang 53The ... Figure 6.64 Chapter 6, Solution 32 (a) Ceq = (12x60)/72 = 10 μ F V 1300 e 1250 50 e 1250 ) 0 ( v dt e 30 10 t26 3 V 270 e 250 20 e 250 ) 0 ( v dt e 30 10 t26 3 (b) At t=0.5s, 03 178 270 e 250

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... 5 7 ) t ( ( 1 e ) mA 5 7 ) t ( Thus, mA e 5 7 5 7 30 ) t ( = ) t ( Trang 48Chapter 7, Problem 51 Rather than applying the short-cut technique used in Section 7.6, use KVL to obtain Eq (7.60) ... Calculate the capacitor voltage for t < 0 and t > 0 for each of the circuits in Fig 7.106 Figure 7.106 For Prob 7.39 Trang 34Chapter 7, Solution 39 (a) Before t = 0, = + = ( 20 ) 1 4 1 ) t ( ... 6Chapter 7, Problem 7 Assuming that the switch in Fig 7.87 has been in position A for a long time and is moved to position B at t =0, find v0(t) for t ≥ 0 Trang 7PROPRIETARY MATERIAL © 2007

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... i o o 00 00 Z ! D 21 21 t32.372t679.21 2o2 e e 928 6 ) t ( i A 928 6 A to leads This 240 A 32 37 A 679 2 dt ) 0 ( di , A A 0 ) 0 ( i e A e A ) t ( i 32 37 , 679 2 3 10 20 300 20  r  Z ... prior In the circuit of Fig 8.71, the switch instantaneously moves from position A to B at t 0 Find v t for all t t 0 Figure 8.71 For Prob 8.17 Chapter 8, Solution 17 is which , 20 4 1 2 10 ... diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45 diR(0+)/dt = -6.1778 A/s Also, iR = iC + iL diR(0+)/dt = diC(0+)/dt + diL(0+)/dt -6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s (c) As t approaches infinity,

Ngày tải lên: 13/09/2018, 13:31

... 933 2 . 028 27 . 41 1 o Y 2271 0 j 4378 0 41 27 - 4932 0 Y 773 4 j 4378 0 5 j Y 97 22 773 4 j 4378 0 5 0 773 4 j 4378 0 1 2 1 1eq − + = + + = Y 2078 0 j 5191 0 1eq 2078 0 j 5191 ... 4 )( 8 4 j 2 25 7 j 5 1 4 6 j 6 9 j 4 46 Z 688 6 j 574 3 8 4 j 4 6 9 j 4 46 6 9 j 4 46 12 j 574 3 ) 88 61 583 7 )( 90 6 ( j7.25) -j4)(1.5 20 j 727 1 ) 07 79 11 9 )( 90 12 ... j)(150) -12 ( j 12 150 - ab + = + = ) 76 97 25 4 )( 24 175 457 12 ( Trang 49At ω = 103 rad/s find the input admittance of each of the circuits in Fig 9.74 10 ( j 1 C j 1 F 5 × = ω ⎯→ ⎯ µ )

Ngày tải lên: 13/09/2018, 13:31

... 14 57 4 08 25 08 14 j 1 2 j 25 V 25 V 08 0 j 2 j V ) 14 Now to solve for io, mA 8784 0 j 367 12 2000 7567 1 j 2666 0 25 2000 V 25 io = 12.398cos(4x103t + 4.06˚) mA + _ Trang 7V 3 ... + 0 ) 5529 1 j 7955 5 ( V V ) 2 j 1 ( 3 025 0 ) 01 0 j 025 − Using MATLAB, V = inv(A)*B leads to V1 = − 70 63 − j 127 23 , V2 = − 110 3 + j 161 09 o2 1 o 7 276 82 17 40 V V I = − = ... frequency-domain version of the circuit is shown below, where o2 75 3 j 2 ( 21 o I 75 3 j 2 ( I 25 0 j 1 ( 30 − + − + I 75 3 j 2 25 0 j 1 25 0 j 1 75 3 j 2 10 j 32 17 10 Solving this leads

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... resistor 3 1 Trang 13Chapter 11, Problem 11 For the network in Fig 11.43, assume that the port impedance is RC C 10 )( 377 ( ) 754 0 ( 1 k 10 Z 2ab mA ) 68 t 377 cos( 2 ) 22 t 377 sin( 2 ) t ( ° ... a-b of the circuits in Fig 11.52 so that the maximum power is transferred to the load? || 100 10 j - ) 30 j 40 )( 100 ( ) 30 j 40 ( 634 4 j 707 81 ) 634 4 j 707 31 )( 50 ( ) 634 4 j 707 ... 2 1 S kVA 937 1 j 4 )) 9 0 ( sin(cos 9 0 4 j 1 I V S = 104 j 74 22 90 100 10 ) 137 1 j 2 5 )( 2 ( = = V S I 104 j 74 22 I Similarly, sin(cos ( 0 707 )) 2 ( 1 j ) kVA 707 0 2 j 1 I V

Ngày tải lên: 13/09/2018, 13:31

... voltages Trang 751.667 L L Z Z = = = 1.6669 Ω Ω Trang 76 Find the power absorbed by the 10-Ω resistor in the ideal transformer circuit of Fig 13.121 Figure 13.121 For Prob 13.56 Trang 77Chapter 13, ... + 8 = 0 I = (– 8 + j10)/ (7 + j7) But, –j10 + (5 + j6)I – j2I + VTh = 0 VTh = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7) VTh = 5.349∠34.11° + – Trang 17To obtain Z Th , we set all ... -j0.5 (7) Multiplying [(6)+(5)] by 12 yields (-22 + j20)I2 + 7I3 = 0 (8) Multiplying [(5)+(7)] by 20 yields -22I2 – 3I3 = -j10 (9) (8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3

Ngày tải lên: 13/09/2018, 13:31

... 1 s 25 0 s ( A 25 4 3 2 1 s 2 3 3 2 7 155 4 3 2 1 s 2 1 s 7 40 1 s 1 7 40 4 3 2 1 s 7 135 s 7 40 - 1 s + − + = + + + 7 ( ) 2 )( 155 ( t 2 3 cos e 7 40 e 7 40 ) + − 10/(s + 1) Vo Trang 20V ... 1 10 : − + + − 7071 0 ) 1 ( 7071 0 414 1 7071 0 ) 1 ( 1 1 3 20 5 1 2 2 1 3 20 s s s s s s s s Trang 36Chapter 16, Problem 22 Find the node voltages v1 and v2 in the circuit of Fig 16.56 using ... E D 4 7 18 : 0 F 4, E , 8 D D 8 3 3 or F D 8 9 9 : 64 23 ) 8 7 s ( 2 / 7 64 23 ) 8 7 s ( ) 8 / 7 s ( 4 ) 1 s ( 8 ) 8 9 s 4 7 s ( s 4 ) 1 s ( 8 V 22 2 1 + + − + + + + + = + + + + Trang 39Chapter

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... the circuit of Fig 17.71 is the same as function f2(t) in Fig 17.56(b), determine the dc component and the first three nonzero harmonics of vo(t) Figure 17.71 For Prob 17.35 Chapter 17, Solution ... Chapter 17, Problem 18 Trang 21Determine the fundamental frequency and specify the type of symmetry present in the functions in Fig 17.56 Figure 17.56 For Probs 17.18 and 17.63 Chapter 17, ... sin n 2 1 3 n 2 cos n 3 Chapter 17, Problem 26 Trang 32Find the Fourier series representation of the signal shown in Fig 17.64 Figure 17.64 For Prob 17.26 Chapter 17, Solution 26 T = 4, ωo

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... transforms of these functions: ω − → ω π → − 2 t 4 ω π − 4 (b) e−at 2 2 a a 2 ω + e 4 Chapter 18, Problem 17 Trang 22Find the Fourier transforms of: (a) cos 2tu(t) (b) sin 10tu(t) Chapter 18, ... transform of each term (jω)2G(jω) = 10jω – 5jωe–jω – 5e–jω + 5e–j2ω which leads to G(jω) = (–10jω + 5jωe–jω + 5e–jω – 5e–j2ω )/ω2 Chapter 18, Problem 7 Trang 10Find the Fourier transforms of the ... Trang 32Chapter 18, Problem 27 Find the inverse Fourier transforms of the following functions: (a) F(ω ) = ) 10 ( 100 + 2 ( 10 + + ω j j j (c) H(ω ) = 1300 40 1 ( ) ( + ω ω δ j j Chapter 18,

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... Trang 4428028 43 7x36736Ω = 30 3x273 + 567.26 7.2x18867.57.218 7.2x18 868.518 7.2x868.5 Rcn )145964.0()368.7977.3(829.14 =5.829+11.34614.5964= 12.21 Ω i = 20/(Req) = 1.64 A Trang 45The lightbulb ... circuit of Fig 1.120 15x1212x1010 Req = 19.688||(12 + 16.667) = 11.672Ω By voltage division, 16672.11 672.11+ = 42.18 V Trang 43Find Req and I in the circuit of Fig 2.121 Chapter 2, Solution 57 ... circuit of Fig 2.71 Figure 2.71 For Prob 2.7 Trang 3Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig 2.72 Find i1, i ,2 and i3 in Fig 2.73 Trang 4In the circuit in Fig 2.67 decrease

Ngày tải lên: 26/03/2019, 11:40

... this leads to p(t) = v(t)i(t) = (-0.12e-2t)(0.006e-2t) = –720e –4t µW (c) 3 0 6 4t - 3 0 4t - 10e4 720dt e-0.72pdt Trang 17Copyright © 2017 McGraw-Hill Education All rights reserved No reproduction ... (–15.125)(11.353)(10–3) = –171.71 mW Trang 16Copyright © 2017 McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education.e2 ... 2017 McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education.(a) At t = 1ms, = = = 2 30dt dq Trang 7Copyright © 2017

Ngày tải lên: 20/08/2020, 12:02

... Equations 57 Slope 62 Writing Equations of Lines 73 Applications and Uses of Graphs 79 Chapter Summary 86 Chapter Review Problems 87 Chapter Test 88 Suggested Laboratory Exercises 88 chapter three ... by Graphing 270 Solving Systems Algebraically 276 Applications of Linear Systems 286 Contents 6-4 Systems of Nonlinear Functions 292 Chapter Summary 296 Chapter Review Problems 297 Chapter Test ... Personal Income 208 Chapter Summary 217 Chapter Review Problems 218 Chapter Test 220 Suggested Laboratory Exercises 221 chapter five ADDITIONAL APPLICATIONS OF ALGEBRAIC MODELING 227 5-1 5-2 5-3 Models

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... distribution Trang 21 17 A sample of the ages of 10 employees of a company is shown below 18 The following data set shows the number of hours of sick leave that some of the employees of Bastien's, Inc ... showing the relative frequency b a graphical form of representing data c a tabular summary of a set of data showing the frequency of items in each of several nonoverlapping classes d a graphical ... frequency 6 The sum of frequencies for all classes will always equal a 1 b the number of elements in a data set c the number of classes d a value between 0 and 1 Trang 27 Fifteen percent of the students

Ngày tải lên: 27/10/2017, 09:24

... A) devalued; $20.67/oz to $35.00/oz of gold B) devalued; $35.00/oz to $20.67/oz of gold C) revalued; $20.67/oz to $35.00/oz of gold D) revalued; $35.00/oz to $20.67/oz of gold Answer: A Diff: ... the gold standard of currency exchange that existed from 1879 to 1914, an ounce of gold cost $20.67 in U.S dollars and £4.2474 in British pounds Therefore, the exchange rate of pounds per dollar ... exchanges has the implicit effect of A) making currencies float relative to the price of gold B) limiting the growth of a country's money supply subject to the ability of the official authorities to obtain

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... charge of $1,000 per month What is the amount of total assets of Kirk Corporation at the end of the month? Beginning total assets of $850,000 + Equipment purchased of $6,000 + Supplies purchased of ... normal debit balance Beginning debit balance of $68,900 + Debit of $18,300 – Credit of $16,000 = Ending debit balance of $71,200 [QUESTION] 73 Which of the following statements about normal account ... the best interpretation of the word “credit” is: A) left side of an account B) increase side of an account C) right side of an account D) decrease side of an account 67 The Accounts Payable

Ngày tải lên: 27/10/2017, 09:25