fundamentals of electric circuits 3rd edition solutions manual 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

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... 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

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... 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

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... 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

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... 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

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... 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 (

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... [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 ... -2A1 +2A2 or A2 = A1 = -3 v(t) = [3 – 3(cos2t + sin2t)e-2t] volts Trang 31PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed,

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

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... (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 +

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... 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,

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... 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˚

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... -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

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... Q of 80, and has a resistive branch of 40 k Ω Determine the values of L and C in the other two branches Chapter 14, Solution 34 10 x 60 x 10 x 1 LC 1 63 1 RC 1 (c) Q = ωoRC = 1 443 x 103x ... Bode magnitude plot of H(ω ) is shown in Fig 14.75 Find H( ω ) Figure 14.75 For Prob 14.23. Chapter 14, Solution 23 A zero of slope + 20 dB / dec at the origin ⎯ ⎯→ j ω A pole of slope - 20 dB ... 2 10 j ) 1 ( H = 3 9 j 2 7 4 743 - 34.7 j 2 6 j 1 3 ) 1 ( H = = 20 log 4 743 HdB 10 13.521, φ = –34.7˚ Trang 8A ladder network has a voltage gain of H ( ω ) = ) 10 )( j 1 ( 1 ) ( ω +

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

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... = + = + − 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

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... 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

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... F 2 3 x 4 ) ( G , t 3 5 f 10 t 3 2 f 4 t g 5 3 j 6 1 e 2 3 ⋅ = j 4 ( e j 3 / 2 1 ) j 10 ( e j 3 / 5− 1 ) ω + − ω ω − ω − Trang 301 (Trang 311 +⎯→ ⎯ − − + 1 ( u 2 ) 1 ( − + ω = ω Trang 32Chapter ... 2 2 I 2 3 4 j 4 j 2 2 5 1 4 j 4 j 2 I ω − ω + ω = ( )2 2 o j 5 1 4 j 4 j 2 I V ω + ω + 8 j 3 8 j 3 4 V ω + ω + ω = 2 2 2 2 3 8 j 3 4 3 16 3 8 j 3 4 j 3 4 4 3 8 sin e 657 5 ) ( u t 3 8 cos ... 2) + 3 δ (t) – 2u(t – 2) Chapter 18, Solution 14 Trang 18(a) ) cos( 3 t + π ) = cos 3 t cos π − sin 3 t sin π = cos 3 t ( − 1 ) − sin 3 t ( 0 ) = − cos( 3 t) t ( u t 3 cos e ) t ( = − − tF(ω)

Ngày tải lên: 13/09/2018, 13:31

... h I V 3 2 3 2 3 6 6 2 1 12 2 V V = ] 4 3 2 2 8 The equivalent circuit of the given circuit is shown in Fig (c) 10 3 2 1 1 1 30 29 45 3 2 4 9 || 5 3 2 I I Trang 7010 32 30 29 ) 8 = = 252 300 2 ... (d) 091 9 100 || 091 9 300 50 091 091 309 091 9 V V 091 309 34 34 091 309 2 2 o I V ) 091 309 )( 1 1 ( 34.34 - 110 100 o 1 I I Thus, = ] [ g ⎢⎣ ⎡ - 5.96 34.34 Ω ⎥⎦ ⎤ 0.101 - S 02929 ... 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 j55) -15 ( 13 6 j 3 j2) - (3 - ) 18 j 24 ( j - - 1 1 2 V B Ω + = -6.923 j25.385 B = ] Ω + + j0.6923

Ngày tải lên: 13/09/2018, 13:32

... Δ Chapter 2, Solution 48 10 100100100R RRRRRR 3 1 3 3 2 2 50x2050x3020x30 Ra ,15520 RR c b a c a R1 = R2 = R3 = 4 Ω ++ 103060 30x60 R3 R 1 = 18 Ω, R 2 = 6 Ω, R 3 = 3 Ω Trang 35What value of ... 35What value of R in the circuit of Fig 2.114 would cause the current source to deliver 800 4 3R4 RxR )R4/( 3)R4/( )RxR3(R RR2 3R3 R2 3Rx3R2 3R3R4 3R4 3R3 =+ Trang 36Obtain the equivalent resistance ... and i3 in Fig 2.73 Trang 4In the circuit in Fig 2.67 decrease in R3 leads to a decrease of: (a) current through R3 3A -2A 2 3 At node 1, 4 + 3 = i1 i1 = 7A At node 3, 3 + i2 = -2 i 2 = -5A Chapter

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... consent of McGraw-Hill Education.Solution 1.26 (a) Clearly 10.78 watt-hours = (voltage)(current)(time) = 3.85I(3) or I = 10.78/[(3.85)(3)] = 933.3 mA (b) p = energy/time = 10.78/3 = 3.593 W (c) ... v(t)i(t); v(1) = 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 ... written consent of McGraw-Hill Education.3T33dtidt q 36004 4h TLet (a) T 0 kJ475.2 .) (( = × ×+ 250360040 33600 250103 dt3600 t50103vidt pdtW b) 3600 4 0 2 0 T 0 t t T cents1.188 475.2Cost Ws)(J

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