fundamentals of electric circuits 3rd edition solutions manual chapter 10

... Trang 1Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: 10482 1024 Trang 21600900 e10q(0) t40sin10eq(t) -30t 30t -Chapter 1, Problem 4 A current of 3.2 ... charge is deposited on the object? Chapter 1, Solution 10 q = it = 8x103x15x10-6 = 120 mC Trang 7Chapter 1, Problem 11 A rechargeable flashlight battery is capable of delivering 85 mA for about ... 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

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... in Fig 3.102, find the values of R, V1, and V2 given that io = 18 mA Figure 3.102 Chapter 3, Solution 57 Assume R is in kilo-ohms V V V V mA x k V2 = 4 Ω 18 = 72 , 1 = 100 − 2 = 100 − 72 = ... circuit of Fig 3.108 Figure 3.108 Trang 88For mesh 2, 20i2 – 10i1 + 4i0 = 0 (1) But at node A, io = i1 – i2 so that (1) becomes i1 = (16/6)i2 (2) For the supermesh, -100 + 50i1 + 10(i1 – i2) ... Chapter 3, Solution 54 Let the mesh currents be in mA For mesh 1, 2 1 2 2 1 3 Using MATLAB, mA 25 10 , mA 5 8 , mA 25 5 25 10 5 8 25 5 3 2 I B A I Trang 75In the circuit of Fig 3.100,

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... Fig 4.100 Figure 4.100 Trang 55Apply Thèvenin's theorem to find V in the circuit of Fig 4.105 oFigure 4.105 Chapter 4, Solution 38 We find Thevenin equivalent at the terminals of the 10-ohm ... terminals a-b of the circuit in Fig 4.106 Trang 583 A+ _ Trang 59Find the Thevenin equivalent at terminals a-b of the circuit in Fig 4.107 + _ 70 V + Vo – 4 Vo + – c c ba Figure 4.107 For Prob ... I1 fc + – Trang 60Chapter 4, Problem 41 Find the Thèvenin and Norton equivalents at terminals a-b of the circuit shown in Fig 4.108 Figure 4.108 Trang 61V A24/)8 8, Trang 62Chapter 4, Problem

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... 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 Let v = output of the first op amp 1 v = input of the second op ... i 0 R ) A 1 ( R v ) A R R ( + + + 4 5 5 4 i 0 i 0 i ) 10 1 ( 100 10 x 10 100 R ) A 1 ( R A R R v v ⋅ + + + = + 9 10 10 1 001 , 100 000 , 100 0.9999990 Trang 5Using the same parameters for the ... amp Trang 6Chapter 5, Solution 6 + - Av d + - v d + i R ) A 1 ( R v + + + − i 0 i 0 R ) A 1 ( R A R R vi 3 5 6 10 x 2 x 10 x 2 1 50 10 10 x 2 x 10 x 50 + + ⋅ + 10 x 2 x 001 , 200 10 x x 000

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... 2 find v(t) for t ≥ 0 Chapter 7, Solution 37 Let v = vh + vp, vp =10 4 /0 tAe v = 10 + −0.25 8 10 2 ) 0 v te v = 10 − 8 −0.25(a) τ = 4 s (b) v ( ∞ ) = 10 V e v = 10 − 8 −0.25 u(t)V Trang ... reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior 4 ) t ( v 25 2 10 x x 10 x 40 RC , e v ) t ( v V 4 ) 24 ( 2 10 2 ) 0 ( v ... 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 )

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... and di 0 / dt 0 find i t for t ! 0 Chapter 8, Solution 9 s2 + 10s + 25 = 0, thus s1,2 = 2 10 10 Therefore, i(t) = [(10 + 50t)e-5t] A Chapter 8, Problem 10 The differential equation that describes ... part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior Chapter 8, Problem 7 A series RLC circuit has R 10 kȍ , L 0 1 mH, and C 10 P F ... reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior The current in an RLC circuit is described by 10 25 0 i d If i 0 10 and di

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... 69.14°) A (c) For ω = 10 , 10 j ) 1 )( 10 ( j L j H 2 j - ) 05 0 )( 10 ( j 1 C j 1 F 05 4 j - 10 j ) 2 j - || 2 10 − + = + 0 4 9 j 1 0 4 V I Hence, io( t ) = 0.4417 cos(10t – 83.66°) A Trang ... 28Calculate v o(t) in the circuit of Fig 9.49 100 j - ) 10 50 )( 200 ( j 1 C j 1 F × = ω ⎯→ ⎯ µ 20 j ) 1 0 )( 200 ( j L j H 1 20 j 40 j2 - 1 j100 - j100 50 ) (50)(-j100 -j100 70 20 j ) 0 60 ( 20 j ... impedance of the circuit in Fig 9.77 Figure 9.77 For Prob 9.70. Trang 53Make a delta-to-wye transformation as shown in the figure below 9 j 7 5 j 15 ) 10 j 15 )( 10 ( 15 j 10 10 j 5 ) 15 j 10 )( 10

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... to find v in the circuit of Fig 10.95 o Figure 10.95 For Prob 10.50 Chapter 10, Solution 50 Trang 7810 cos( 40 j ) 10 4 0 )( 10 ( j L j mH 4 50 j - ) 10 2 0 )( 10 ( j 1 C j 1 F 2 × = ω ... = − 110 3 + j 161 09 o2 1 o 7 276 82 17 40 V V I = − = ∠ − Thus, A ) 17 82 t 200 cos( 276 7 ) t ( Trang 12Chapter 10, Solution 9 10 , 0 10 ) t 10 cos( 10 j L j mH 20 j - ) 10 50 )( 10 ( ... ( 05 31 j 91 Trang 85) 10 j - )(20 j ( 10 ) 10 j - || 20 j 10 Trang 8610 j -30 50 - ) 5 j 8 )( 10 j ( ) 5 j 8 ( || 10 j 32 ) 0 4 ( 5 j 10 j 8 8 I = + = = 5 j 8 320 j 10 58 92 33 Trang 87) 2

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... values of the voltage and of the current (b) the average power dissipated in the load Chapter 11, Solution 45 2 60 20 2 2 = (b) p(t) = v(t)i(t) = 20 + 60cos100t – 10sin100t – 30(sin100t)(cos100t); ... W Trang 9Applying KCL to the right side of the circuit, 0 5 j 10 5 j − + j 10 10 V V j 5 j 10 8 ∠ ° = Vo + j 1 20 80 1 R 2 1 Trang 10In the circuit of Fig 11.40, determine the average power ... Trang 26Find the value of ZL in the circuit of Fig 11.49 for maximum power transfer (80)(-j10) 20 j20 -j10) ( || 80 40 || 40 20 j Z 154 10 j 23 21 Z = = * Th Trang 27Chapter 11, Problem 19

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... = j6 + j8 – j4 = j10 Thus, –j10 + (5 + j10 – j3 + 2)I + 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 ... + (8 + j10 – j5)I 2 + j5I 1 + j5I 1 ++ 10j5j420 12j w = 0.5L 1 i 1 + 0.5L 2 i 2 – Mi 1 i 2 Since ωL1 = 10 and ω = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH w = 0.5(10)(–2.445) 2 + 0.5(10)(–0.8391) ... = 4 – 1 + 7 = 10H or L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 12 L T = 6 + 8 + 10 = 10H Chapter 13, Problem 2 Determine the inductance of the three series-connected inductors of Fig 13.73

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... 5 ) 10 1 )( 10 40 ( = ω C 4 L 4 j R ) 4 ( 0 0 0 = ω ) 10 1 )( 10 5 ( 4 10 40 4 10 5 j 2000 ) 4 3 0 Z ) 5 4000 50 ( j 2000 ) = ω C 2 L 2 j R ) 2 ( 0 0 0 = ω ) 10 1 )( 10 5 ( 2 ) 10 40 ( 2 ) 10 ... 200+j(100-2000/5) = ω 2 ) ( 0 = ω C 2 1 L 2 j R ) 2 ( 0 0 = ω ) 10 1 )( 10 5 )( 2 ( 1 ) 10 40 )( 10 5 )( 2 ( j 2000 ) = ω C 4 1 L 4 j R ) 4 ( 0 0 = ω ) 10 1 )( 10 5 )( 4 ( 1 ) 10 40 )( 10 5 )( ... parallel resonant RLC circuit with ωo= 10rad/s and Q = 20 Calculate the bandwidth of the circuit Let R = 10 Ω L L X rad/s 10 x 796 8 10 x 40 10 x 5 10 x 10 x 2 X Trang 32A parallel RLC circuit

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... ) ( (c) 10 s 6 s D Cs 1 s B s A ) s ( + + + + + + = 10 1 A = , B = - 1 5 ) s s ( D ) s s ( C ) s 10 s 6 s ( B ) 10 s 16 s 7 s ( A 1 A = , B = -2A = - 1 5 , C = A = 1 10 , 10 4 A 4 10 s 6 s ... transform of the function in Fig 15.27 ( ) ( ) se e 1 5 s32 ss Trang 15Obtain the Laplace transform of f(t) in Fig 15.28 = ) s ( e 5 e 3 e 3 5 s 1 − + − Trang 16Find the Laplace transform of f(t) ... 32Determine the inverse Laplace transform of each of the following functions: ( 2 )( 4 ) 3 1 4 2 + + + − s s s s (c) ( 3 ) ( 4 5 ) 12 2+ + + + s s s s Chapter 15, Solution 32 (a) 4 s C 2 s B

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... Io 2 Trang 8s s+ + 20 10 + _ 10 Ω v 20 s I Trang 9Chapter 16, Problem 7 Find v0(t), for all t > 0, in the circuit of Fig 16.41 Figure 16.41 For Prob 16.7 Trang 10Chapter 16, Solution 7 ... 22 11 + + + = + + 10 4 s 25 2 s 25 0 5 2 I 22 22 + + = + + 10 s Trang 18Chapter 16, Problem 12 Find vo(t) in the circuit of Fig 16.46 Figure 16.46 For Prob 16.12 Trang 19Chapter 16, Solution ... Solution 12 We apply nodal analysis to the s-domain form of the circuit below o o o sV 2 4 V s 3 s V 1 s 10 + = + − + 1 s 15 s 15 10 15 1 s 10 V ) s s 25 = + + = + + 1 s 25 0 s C Bs 1 s A ) 1 s

Ngày tải lên: 13/09/2018, 13:31

... −0.3183 (b) ωn = nωo = 10 or n = 10 a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π) Trang 29Thus the magnitude is A10 = 210 a10 + = 1/(5π) = 0.06366 and the phase is φ10 = tan−1(bn/an) = −90° ... Trang 1710 2 ) ( n n nt n ) 4 / n sin( 1 n 10 ) nt 2 cos( ) 4 / n cos( 1 n 10 2 + = 10 sin 1 10 cos 1 4 10 ) ( n nt n nt n t f (a) in a cosine and angle form (b) in a sine and angle form Chapter ... 102 0Trang 11Chapter 17, Problem 9 Determine the Fourier coefficients an and bn of the first three harmonic terms of the rectified cosine wave in Fig 17.52 2 , T 183 3 10 4 / sin ) 4 ( 4 10

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... Chapter 18, Solution 27 + + = + 10 s B s A 10 s s 100 s F 10 10 100 B , 10 10 10 j 10 F + ω − ω = ω f(t) = 5 sgn ( ) t − 10 e−10 tu ( ) t − 3 s B s 2 A s 3 s 2 s 10 s G 6 5 30 B , 4 5 20 ( ) 3 ... − 10 ) ⎥ ⎦ ⎤ 1 10 10 j 1 10 2 j G j 2 j 10 10 10 2 j 2 − ω − − ω δ − + ω δ π Chapter 18, Problem 18 Trang 23Given that F( ω ) = F[f(t)], prove the following results, using the definition of ... 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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... 1 V 50 100 V z 2 2 704 3 I V z I 704 3 I 3 11 11 V 3 1 V V 2 1122 22 704 3 88 29 ] z [ Trang 14) 4 j 10 ( IV z I 4 j I 10 V 1 2211 V z I 8 j 10 5 ( V 2 2222 ) 4 j 10 ( I V z I 4 j 10 ( V ... Fig (c) 10 j 20 j 10 j 15 j -j10) )( 15 j ( -) -j10)(-j20 ( Z 20 j j15 - -j20) )( 15 j ( Trang 846923 0 j 5385 0 2 j 3j 3 - 1 ) 3 40 j 20 )( 20 j ( 10 − + = ) 18 j 24 ( j ] 7 j 9 ( 2 10 j V ... circuit in Fig.(d) Thus, = ] [T ⎢⎣ ⎡ Y 1 ⎥⎦ ⎤ 0 1 Trang 8210 j 20 j15 j10)(-j5) ( o 3 2 5 j 10 j - 10 j - I I Trang 83I 3 40 j 20 ) 3 10 j 20 To find B and D , consider the circuit in Fig (b)

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... a-b of each circuit in Fig 2.109 Trang 33Find I in the circuit of Fig 2.110 25 20x 6 3 = = 2Ω 9 36 Trang 34Convert the circuits in Fig 2.112 from Y to Δ Chapter 2, Solution 48 10 100100100R ... resistance R ab at terminals a-b for each of the circuits in 20x401020 120 160 Rab = 80(10+10) = + = 10020 Trang 29For each of the circuits in Fig 2.108, obtain the equivalent resistance at terminals ... from 1 A to 10 A Calculate the values of R and R x to achieve this Chapter 2, Solution 64 When Rx = 0, ix =10A R = = 11Ω 10 110 When Rx is maximum, ix = 1A + = =110Ω 1 110R full-Chapter 2,

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... distribution without the prior written consent of McGraw-Hill Education.Solution 1.9 (a) =∫ =∫1 =10C 010dtidt q (b) C5.2255.715 152 1510110idt 0 =++ = ×+ Trang 10Copyright © 2017 McGraw-Hill Education ... 20sin(4) = 20sin(229.18°) = –15.135 volts; and i(1) = 10(1+e-2)(10–3) = 10(1.1353)(10–3) = 11.353 m-amps p(1) = (–15.125)(11.353)(10–3) = –171.71 mW Trang 16Copyright © 2017 McGraw-Hill ... reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education.q = it = 10x103x15x10-6 = 150 mC Trang 11Copyright © 2017 McGraw-Hill Education All rights reserved

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... Trang 102 - Consolidation Of Financial Information Fundamentals of Advanced Accounting 6th Edition Solutions Manual Test Bank by Hoyle Schaefer Doupnik Complete download complete SOLUTIONS MANUAL ... The fair value of any noncontrolling interest also adds to the valuation of the acquired firm and is covered beginning in Chapter 4 of the text 3 Any excess of the fair value of the consideration ... complete SOLUTIONS MANUAL for Fundamentals of Advanced Accounting 6th Edition by Joe Ben Hoyle, Thomas Schaefer, Timothy Doupnik : Click HERE Chapter 2 Consolidation of Financial Information Accounting

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... The fair value of any noncontrolling interest also adds to the valuation of the acquired firm and is covered beginning in Chapter 4 of the text 3 Any excess of the fair value of the consideration ... fair value as of the date of acquisition 2 Any portion of the payment made in excess of the fair value of these assets and Trang 43 If the price paid was below the fair value of the assets ... Financial information from the members of a business combination must be Fundamentals of Advanced Accounting 6th Edition Solutions Manual Test Bank by Hoyle Schaefer Doupnik Trang 21 If the acquired

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