fundamentals of electric circuits solutions chapter 3

... of the charging? (b) how much energy is expended? (c) how much does the charging cost? Assume electricity costs 9 cents/kWh 3T33dtidt q 36004 kJ475.2 .) (( = × ×+ 250360040 33600 25010 3 dt3600 ... 40 minutes Burner 3: 15 minutes Burner 4: 45 minutes +++ = = 3.3cents12 Cost kWh3.30.92.4 hr60 30kW1.8hr60 45)1540(20kW21 pt w Trang 20Chapter 1, Problem 30 Reliant Energy (the electric company ... 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 A

Ngày tải lên: 13/09/2018, 13:31

... analysis Chapter 3, Problem 30 Using nodal analysis, find vo and io in the circuit of Fig 3.79 Figure 3.79 Trang 8212 3 1 32 2 3 i i 3 2 1 1 3 0 12 3 1 32 2 12 6 1 32 10 3 6 3 1 10 2 3 Trang 83Calculate ... For loop 3, -i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1 (3) Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A i0 = -i1 = 10.667 A, from fig (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V ... Fig 3.89 Trang 59Use mesh analysis to obtain io in the circuit of Fig 3.90 Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A Trang 60Find current i in the circuit in Fig 3.91

Ngày tải lên: 13/09/2018, 13:31

... 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 ... consider the circuit below 32||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4 i3 = v ’/4 = -1 o For i , consider the circuit below − + + vo’ 2 2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division ... circuit of Fig 4.135 Figure 4.135 Trang 103V6.13636 .13636 .1363 Trang 104Determine the maximum power delivered to the variable resistor R shown in the - + Vx - Trang 1051 1 515 3 5 4 V V

Ngày tải lên: 13/09/2018, 13:31

... at node 2 gives s R R R v 3 1 f R R R R R R R R R = 2 4 3 3 2 4 3 1 = 2 4 3 3 2 4 3 1 R R R R R R R R R R R k f f Trang 27Calculate vo in the op amp circuit of Fig 5.63 Trang 28 0 2 88 4 = k ... 100,000, V1 – V0 = 1000 (V0 + 100,000V1) 0= 1001V0 + 99,999,999[(10VS + V0)/12] 0 = 83,333,332.5 VS + 8,334,334.25 V0 which gives us (V0/ VS) = –10 (for all practical purposes) If VS = 1 mV, then ... amplifier 2 3 v 2 1 1 V 6 ) 4 ( 2 36 R v v Trang 36v v Rv v 2 in11 in but o43 3 R R R v + Combining (1) and (2), 0 v R R v R R v 2 122 1a 22 112 1 R R v R R 1 112 14 R 1 R + 2 112 13 43 R R v R

Ngày tải lên: 13/09/2018, 13:31

... by 1/C 2 C C C 5 30 1 40 1 30 1 30 1 10 1 40 1 10 1 1200 1 300 1 400 1 1200 1 300 1 400 75 23 x 17.39μF in parallel with Ca = 17.39 + 5 = 22.39μF Hence Ceq = 22.39μF Trang 28Chapter 6, Problem ... e 250 v , 2 840 1300 e 1250 J 235 4 ) 15 840 ( 10 12 2 x x x w μF J 3169 0 ) 03 178 ( x 10 x 20 x 2 1 J 6339 0 ) 03 178 ( x 10 x 40 x 2 1 Trang 34Chapter 6, Problem 33.Obtain the Thèvenin ... each of the circuits in Fig 6.51 Figure 6.51 Trang 14Chapter 6, Solution 17 (a) 4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F

Ngày tải lên: 13/09/2018, 13:31

... 10− 3) / 40 = 0 5 ms x R L R Th Trang 143 1 3 1 2 3 13 1 2 R R ) R R ( R R R R R R R + + + = + + = = τ 3 1 3 1 2 3 1 R R ) R R ( R ) R R ( L + + + (b) where 2 1 2 1 L L L + 2 1 2 1 2 1 3 2 ... Ae i ii•h+ h = ⎯ ⎯→ h = − t Let 3 2 ) ( 2 ) ( 3 , 0 ), ) ( 3 2 t u ip = ) ( ) 3 2 t u Ae If i(0) =0, then A + 2/3 = 0, i.e A=-2/3 Thus ) ( ) 1 ( 3 t u e Trang 33PROPRIETARY MATERIAL © 2007 The ... e- 1) = 15 17 30 ) ( v 0 24 - v( 6 1) - -(te ) 30 17 15 ( 30 ) t ( 1) - -(te 83 14 30 ) t ( Thus, = ) t ( 1 t 0 , V e 1 24 -1) (t - t Trang 44t ( v= ) t ( v 10 et 3 V 3 te 10 3 1 - ) 1 0 (

Ngày tải lên: 13/09/2018, 13:31

... [200e-20t – 300e-30t] or iL = 100C[2e-20t – 3e-30t] But, i is also = 20{[2e-20t – 3e-30t]x10-3} = 100C[2e-20t – 3e-30t] Therefore, C = (0.02/102) = 200 PF L = 1/(200C) = 25 H R = 30L = 750 ohms ... 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  D r D  get we , V 60 ) 0 ( v and , const ... mH, and C 10 P F What type of damping is exhibited by the circuit? Chapter 8, Solution 7 3 6 3 dt t i d Determine: (a) the characteristic equation, (b) the type of damping exhibited by the

Ngày tải lên: 13/09/2018, 13:31

... 9.70, find the value of ZT⋅.450 j 200 z , 333 13 j 30 15 j 450 j 200 z 45 j 20 10 300 j 150 j 200 z 32 − + j 8 Z 821 3 j 70 21 ) 29 j 40 ) 16 j 10 )( 333 13 j 30 ( ) Trang 4630 I 5 j 19 2 j 6 ... 0 ( ) 333 0 j ( ) 1 ( 6 0 j 8 0 1 3 j - 1 1 1 1 o + + + = − + + = ′ 1 . 8 j 0 . 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 ... 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 ) 80 j 60 ( || 20 j 60 Z 60 j 60 ) 80 j 60 )( 20 j ( 60 − + = 63 33 j 23 33 67

Ngày tải lên: 13/09/2018, 13:31

... (3), 32 j ) j 13 ( ) 3 j 9 32 j 20 j 13 ) 3 j 9 ( ) 3 j 9 ( 8 j 19 3 2 I I 69 j 55 24 - 2 356 69 j 167 148 j 324 2 2 Trang 50For mesh 2, 0 12 2 2 ) 3 j 4 ( − I2 − I1− I3+ = 0 12 4 8 j ) 3 ... 103 120 j I I I ) 105 j 240 ( ) 35 j 80 ( ) ) 35 j 80 ( 0 − + -j0.954 - 2.181 - j2.366 0.2641 B -* inv(A) I A 37 96 38 2 366 2 j 2641 0 I A 63 143 38 2 4116 1 j 9167 1 I I A 63 23 38 ... 46where Vo = 2 ( 4 ∠ - 30 ° − I1)Hence, 0 ) 30 - 4 ( 6 30 - 8 ) 4 j 2 ( + I1− ∠ ° + ∠ ° − I1 = 1 ) j 1 ( 30 - 4 ∠ ° = − I ) 30 - 4 )( 2 ( 2 j - 3 2 j - 3 1 o ) 15 2 2 30 - 4 ( 3 j 2 j Trang 47j +

Ngày tải lên: 13/09/2018, 13:31

... 2 5 1 V 533 8 ) 8 ( 15 16 3 t 16 5 1 0 3 2 V P 2 Trang 3515 t 5 t 2 20 )t ( i 15 2 15 5 2 2 20 1 I 5 2 2 5 1 I 3 15 5 3 2 2 3 t 3 t t 10 t 100 5 1 I 332 33 ] 33 83 33 83 [ 5 Trang 36Compute ... 0.352 + j3.038 VThev = 40 – 4I = 40 – 1.408 – j12.152 = 38.59 – j12.152V = 40.46∠–17.479˚V + – Trang 16solve for V1 At node 1, 10 V ) 3333 0 j 2 0 ( V ) 3333 0 j 25 0 ( 0 5 0 V 3 j V V ... 0 j ( V 3333 0 j 0 2 j 40 V 3 j V V 21 21 To calculate the maximum power to the load, |IL|rms = (40.46/(2x0.8233))/1.4141 = 17.376A Pavg = (|IL|rms)20.8233 = 248.58 W Trang 17Chapter 11,

Ngày tải lên: 13/09/2018, 13:31

... 10)512(V3 SI 3 2 2 L At the source, L L L ' )3j1)( 273.31(0240 ' V 819.93j273.271 V 287 04 V Also, at the source, * L ' L ' 3V I S = )273.31)( 819.93j273.271(3 S 078.19273.271 819.93 =θ = cos ... Y-connected load, 3( 4800V 3 SI I p p Chapter 12, Solution 34 3 2203 02.127) 16j10(3 220V Y p a ∠ = 3VLIL 3 220 6.732 -58 2565 58 S Trang 31 Three equal impedances, 60 + j30 Ω each, are delta-connected ... of IAC (b) What is the value of Ib? 1202308 j10 120230 ° ∠ − = 17.96∠–98.66˚ A rms (b) A34.17110.31 684.4j75.30220.11j024.14536.6j729.16 66.3896.1766.15896 .17 8j10 02308 j10 120230 IIII = 31.1∠171.34˚

Ngày tải lên: 13/09/2018, 13:31

... -22I2 – 3I3 = -j10 (9) (8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3 (10) (9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces, I3 = j3.333 + ... and I3 in the circuit of Fig 13.104 For mesh 2, 0= j2I1+(30+ j26)I2 − j12I3 (2) For mesh 3, 0=−j12I2 +(5+ j11)I3 (3) We may use MATLAB to solve (1) to (3) and obtain A 41.214754.15385.03736.1 ... j10 Trang 34 In the circuit of Fig 13.93, (a) find the coupling coefficient, Trang 36 For the network in Fig 13.94, find Zab and Io Figure 13.94 For Prob 13.25 Trang 37Chapter 13, Solution

Ngày tải lên: 13/09/2018, 13:31

... 3 s 2 4 s 1 s 2 3 2 4 s 2s - 2 3 3 ) s 8 ( 4 s s 3 2 s 3 s 3 - + − + − = 2 3 3 2 2 4 s s 8 s 3 4 4 s ) 4 2)(s s 3 (-3 + − + + + + = [ t2cos( 2 t + 30 ° ) ] = 3 2 4 s s 3 s 6 s 3 12 8 + + − − Trang ... - 133 2 551 0 s 133 1 ) s ( G + − + = 449 5 s 133 2 551 0 s 133 1 3 s 3 3 1 3 s s ) s ( F + − + + + ⋅ + + = = ) t e 1778 0 e 0944 0 ) 3 sin( 02778 0 ) 3 cos( 08333 ... 37(c) Let 4 s 3 s ) 4 s )( 3 s ( ) s ( + + + = + + = 13 3 - A = ) 3 s ( C ) s 3 s ( B ) 4 s ( A A = , B = 3 13 , C = 4 13 4 s 4 s 3 3 s 3 - ) s ( G + + + + = ) t 2 sin( 2 ) t 2 cos( 3 e -3 ) t ( g

Ngày tải lên: 13/09/2018, 13:31

... = + = + − LC ' L LC LC L ' C i 3333 1 v 3333 0 ) t ( u i i 666 2 v 3333 1 i 333 5 v 3333 1 i 8 L C' 3333 0 v ; ) t ( u 4 0 i v 3333 1 3333 0 666 2 3333 1 i Trang 68First select the ... = 333 3 ) 2 5 0 /( ) 6 1 ( A = − + − + = , B = ( − 4 + 6 ) /( − 2 + 1 / 2 ) = − 1 3333 2 s 3333 1 2 / 1 s 333 3 Vo + − + 1/s I Trang 32Chapter 16, Problem 20 Find v0(t) in the circuit of ... I s− 1 = 3 s V s 2 V V 1 V 2 s 3 V 2 s 3 3 s 3 3 s 2 2 s 9 s 3 ) 3 s ( s 3 V + + + = s 2 1 2 s 9 s 3 s 9 V 3 s 3 V + + = + s ( H 2 s 9 s 3 s 9 2+ + s 3 2I Trang 576 +Trang 582 I s 2 3 For loop

Ngày tải lên: 13/09/2018, 13:31

... 9 3 2 1 3 n 2 cos n 322 0 5.1 3 4 dt ) t n sin( ) t ( 3 4 = 1 0 2 nt 2 cos n 2 t 3 3 nt 2 sin n 4 9 3 2 3 n 2 sin n 322 odd n 1 n 22 22 3 nt 2 sin 3 n 2 cos n 2 3 n 2 sin n 3 3 nt 2 cos 3 ... ∫3 π 2 4 cos( n t / 3 ) dt ] = 3 2 2 1 2 t n sin n 3 6 16 3 t n sin n 3 3 t n sin n t 3 3 t n cos n 9 6 n cos 3 n 2 cos n 1 24 2 At t = 2, f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3) ... 83 3t n sin n 3 1 3 t n sin n π π 3 1 3 n cos 3 2 n 1 Trang 9+0 t n 2 sin 3 n 4 cos 1 n 3 3 t n 2 cos 3 n sin n 3 1 We can now use MATLAB to check our answer, 1 52 2 5 Trang 102 0Trang 11Chapter

Ngày tải lên: 13/09/2018, 13:31

... 200 Contents xi Chapter 2 Basic Laws 27 Chapter 3 Methods of Analysis 75 PART 1 DC CIRCUITS 1 Chapter 1 Basic Concepts 3 Chapter 4 Circuit Theorems 119 Chapter 5 Operational Amplifiers 165 f51-cont.qxd 3/ 16/00 ... 528 13. 2 Mutual Inductance 528 13. 3 Energy in a Coupled Circuit 535 13. 4 Linear Transformers 539 13. 5 Ideal Transformers 545 13. 6 Ideal Autotransformers 552 † 13. 7 Three-Phase Transformers 556 13. 8 PSpice ... 694 xiii CONTENTS PART 3 ADVANCED CIRCUIT ANALYSIS 6 43 Chapter 15 The Laplace Transform 645 Chapter 12 Three-Phase Circuits 477 Chapter 13 Magnetically Coupled Circuits 527 Chapter 14 Frequency Response 5 83 | |...

Ngày tải lên: 08/05/2014, 15:37

... Manage a Web of Partners,” California Management Review, 37 (3) , 146-1 63. Moore, J. (1996), The Death of Competition, Harper Business, New York. Narver, J. and S. Slater (1994), “The Effect of a Market ... run into the hundreds of billions! The estimates of the size of business-to-business transactions over the Internet are ten times those of the consumer marketplace. The lack of research notwithstanding, ... “The Effect of a Market Orientation on Business Profitability,” Journal of Marketing, 585 (4), 20 -35 . Peter Drucker (1954) spoke of the importance of marketing to business thinking and its pervasive...

Ngày tải lên: 06/07/2014, 07:20

... INVESTMENT OWNERSHIP YEAR 3- YEAR 5-YEAR 7-YEAR 10-YEAR 1 33 % 20% 14% 10% 24 532 2518 31 5191714 4 7 12 13 12 51199 6697 797 847 9 7 10 6 11 3 100% 100% 100% 100% NOTES: a. We developed these recovery ... projects: PROJECT SIZE IRR A $ 750,000 14.0% B 1,250,000 13. 5 C 1,250,000 13. 2 D 1,250,000 13. 0 E 750,000 12.7 F 750,000 12 .3 G 750,000 12.2 Assume that each of these projects is independent and that each ... coefficient of variation? PROBABILITY UNIT SALES SALES NPV SCENARIO OF OUTCOME VOLUME PRICE (IN 000’S) Worst case 0 .30 10,800 $6,480 Ϫ$10,800 Base case 0.50 18,000 7,560 ϩ 23, 400 Best case 0.20 23, 400...

Ngày tải lên: 20/01/2014, 18:20

... in- formation: (1) name of the proposed corporation, (2) types of activities it will pursue, (3) amount of capital stock, (4) number of directors, and (5) names and addresses of directors. The charter ... value of money and its effects on asset values and rates of return. Part III covers the valuation of stocks and bonds. Chapter 8 focuses on bonds, and Chapter 9 considers stocks. Both chapters ... MANAGEMENT 4 The purpose of this chapter is to give you an idea of what financial management is all about. After you finish the chapter, you should have a reasonably good idea of what finance majors...

Ngày tải lên: 25/01/2014, 23:20

... return of 7 percent, a standard deviation of expected returns of 35 percent, a correlation coefficient with the market of Ϫ0 .3, and a beta coefficient of Ϫ0.5. Security B has an expected return of ... ؍ (6) Strong 0 .3 100% 30 % 20% 6% Normal 0.4 15 6 15 6 Weak 0 .3 (70) (21)1 10 13% 1.0 k ˆ ϭ 15% k ˆ ϭ 15% Calculation of Expected Rates of Return: Payoff Matrix 4 The second form of the equation ... expected return of 10 percent, a beta coefficient of 0.9, and a standard deviation of expected returns of 35 percent. Stock Y has an expected return of 12.5 per- cent, a beta coefficient of 1.2, and...

Ngày tải lên: 25/01/2014, 23:20