fundamentals of electric circuits 3rd edition solutions manual chapter 2

... costs 12 cents per kWh Chapter 1, Solution 28 A0.25 2436530pt W b) Chapter 1, Problem 29 An electric stove with four burners and an oven is used in preparing a meal as follows Burner 1: 20 minutes ... the energy absorbed in 3 s 5 1 e2 3dt3eidt q 4 - 2 0 2t 2 0 2t - 1.4725 C (b) We90 )(t 4 e6dt 0 4t 4 90dt e-90pdt w Trang 11Chapter 1, Problem 16 Figure 1.27 shows the current through and the ... 3T33dtidt q 36004 kJ475.2 .) (( = × ×+ 250360040 33600 25010 3 dt3600 t50103vidt 0 2 0 T 0 t t T cents1.188 475.2 Cost Ws)(J kWs,475.2 W c) Trang 19Chapter 1, Problem 28 A 30-W incandescent

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... V, v2 = 20 V = v3 Trang 21Figure 3.68 Chapter 3, Solution 19 At node 1, 3 2 1 1 2 1 3 4 8 2 2 2 4 2 V V V V V − + 2 3 1 3 7 2 4 36 0 4 2 Using MATLAB, V 267 12 V, 933 4 V, 10 267 12 933 ... 25 0 V 75 0 0 4 V V 2 0 V 2 V 75 0 V 25 0 0 2 2 0 V 4 V V 32 32 2 V 125 1 V 125 0 0 1 0 V 8 V V 4 V 125 1 0 0 125 0 0 75 0 25 0 0 0 25 0 75 0 0 125 0 0 0 125 1 Now we can use MATLAB ... 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3 (1) For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3 (2) For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 +

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... = (2/3)R = 1.2 ohms To get V , we apply mesh analysis to the circuit in Fig (d) 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 ... principle to find v in the circuit of Fig 4.82 oFigure 4.82 Trang 20Chapter 4, Solution 14 Let v = vo o1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively For ... a-b of the circuit in Fig 4.120 + –Figure 4.120 Chapter 4, Solution 54 To find V =V , consider the left loop Th x (1) x o x V =−50 40 =−2000 Combining (1) and (2), mA13000 40001000 22 2, o

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... a, A 2 1 a 2 1 a R R 2 v R R 2 v B 2 1 b 2 1 b R R 2 v R R 2 v R 2 v A B 2 1 1 i 1 2 A R 2 R v v v R v v 2 A = − + − g 2 A R 2 R v 2 / R 0 v R v v 2 / R v g A B B R 2 R v R R v g 2 A R 2 R 1 ... A B R 2 R 1 1 2 v v v Trang 59( B A)12 2 / R R o 2 1 A R 2 R v g 1 i o 2 1 R 2 R 1 1 2 v v R 2 R + ⋅ = Equating (3) and (4), g 1 1 2 i o R 2 R 1 1 R R v v + ⋅ = Trang 602 / Rv v R v v 2 A a ... 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

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... are zero Trang 24C C C v v C = 2 1 1 C C C v + = 2 1 2 C C C v 1C Q C Q = 2 2 1 2 2 2 C C C Q Q C = + or Q2 = 2 1 2C C C + s 2 1 1 C C C Q 1 C C C i + 2 1 2 C C C i + = Trang 25Three capacitors, ... 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 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 ... 2 2 1 2 + = + assuming that the initial conditions are zero Figure 6.81 Trang 55(a) ( )dt di L L 2 1 sL L v2 = 2 , v L L L 2 1 1 2 1 2 L L L v di L v 2 1 1 2 2 1 s i i i = + 2 1 2 1 2 1 2

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... + u(t + 2) 1 Trang 24v1 u t + 1 ) − 2 u ( ) + u t − 1 )(b) v2( t ) = ( 4 − t ) [ u ( t − 2 ) − u ( t − 4 ) ] ) 4 t ( u ) 4 t ( ) 2 t ( u ) 4 t ( ) t ( = ) t ( v2 2 u(t − 2) − r(t − 2) + r(t ... to the following circuit 12 v 6 v 20 v 12 v 5 v 20 A 2 6 v ) 0 ( Since 4 20 || 5 = , 6 1 ) 4 ( 6 4 4 ) ( t 1 6 0 4 e e ) 6 1 2 ( 6 1 ) t ( 20t -e -20) ( 4 0 ( 2 1 dt di L ) t ( = ) t ( ... and L2 flow through the closed switch and become dissipated in the 5 Ω and 20 Ω resistors 1 t 1 1( t ) i ( 0 ) e 2 1 5 5 2 R L1 1 τ = ) t ( i1 2 4 e-2t u ( t ) A 2 t 2 2( t ) i ( 0 ) e 5 1 20

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... response 2 4 82 2 o Z i(t) = (Acos2t + Bsin2t)e-2t i(0) = 6 = A di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-Dtdi(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0 Thus, ... described by 24 8 r   r  v(t) = Vs + (A1cos2t + A2sin2t)e-2t8Vs = 24 means that Vs = 3 v(0) = 0 = 3 + A1 leads to A1 = -3 dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t 0 = dv(0)/dt ... (d) s2 + 2s +5 = 0, s1,2 = -1 + j2, -1 – j2 i(t) = [Is + (Acos2t + Bsin2t)e-t], where 5Is = 10 or Is = 2 i(0) = 4 = 2 + A or A = 2 di/dt = [-(Acos2t + Bsin2t)e-t] + [(-2Asin2t + 2Bcos2t)e-t]

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... transformation j 1 2 j 2 j 2 2 j 4 j 2 j 2 Z j 1 2 2 j 2 Z j2 -2 j - 2 j 2 Z 8 0 j 6 1 3 j 1 ) j 1 )( 4 j ( ) j 1 ( || 4 j ) j 1 )( 1 ( ) j 1 ( = Z 6 0 j 2 2 1 2 j 2 - 1 2 j - 1 1 Trang 56Calculate ... - 2 j 8 12 j20 3 j26 25 261 31 26 2 1 j 2 3 j26 - 4 + Vo − Trang 41At ω = 377 rad/s, find the input impedance of the circuit shown in Fig 9.63 2 ⎯ ⎯→ j ω L = j j C 2 1 j 6 2 j 2 2 j ) j 2 ... shift ) 3 j 1 ( 4 2 j 1 20 j 20 - 40 j 20 ) 20 j 20 )( 20 j ( ) 20 j 20 ( + = + = Z ) j 1 ( 3 1 3 j 6 3 j 1 ) 0 1 ( 12 j 24 12 j 4 + = + Z Z V = + 3 j ) j 1 ( 3 1 j 1 j 20 j 20 20 j Trang 61Design

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... 17At node 1, 10 20 2 20 o V1 V1 V2 + + 10 j 2 o V I = 10 20 10 j 2 20 V2 V1 V1− V2 + + = 2 1 ( 2 j 4 ) 3 400 = V − + V (1) At node 2, 10 j 20 j - 10 10 2 1 - 3 j 2 ) 2 2 ( 2 j 4 ) ( 1 j 0 ... - 2 6 j 8 2 2 x 2 1 − + + = + + V V V 2 1 ( 12 j 41 ) ) 3 j 104 2 1 3 j 104 41 j 12 V V − + = (2) Substituting (2) into (1), 2 2 ( 4 j 3 ) 3 j 104 ) 41 j 12 ( ) 3 j 29 ( 45 45 200 2 V 2 2 2 ... 2 V 2 2 2 o 25 8 j 6 - 3 j 4 2 j - 2 j 5 j 4 2 j - V V V = + = − + 45 200 25 13 233 10 Trang 294 j - 43 1 1 2 2 + + = − 3 2 ) 2 j 2 ( At node 3, 4 2 j 2 0 ) 2 j 1 - ) 2 j 8 0 j 2 1 0 = V1+

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... to each of the passive elements 6828 1 P 38 25 6828 1 2 j 2 6 j ) 2 j 2 ( 6 j 1 40 8 I 21 − + P3H = P0.25F = 0 W 097 5 2 2 Trang 83 320 mH ⎯⎯ → j L ω = j 10 20 10 x x − = j 20 25 j 10 x ... Vo 20Ix + – 0 V 20 j 10 I 20 V − − + + − + − But 25 j 50 1 ) 57 26 9 55 )( 43 63 36 22 ( 20 43 63 36 . 22 1 6 V 25 j 50 1 ) 25 j 50 ( 1 ) 20 j 10 ( 20 20 j 10 1 o oo = + + + − + (0.0232 ... a 2- Ω resistor when the voltage is applied across the resistor 0 2 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

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... + j2 Ω Solve for the line currents and neutral current Chapter 12, Solution 9 =+ ° ∠ =+ = 15j20 0120 Y L an a Z Z Trang 722010 j27 02202 A an a Z V I 22 120-2202 B bn b Z V I ° ∠ =+ 62.2213 ... and the load impedance ZY Chapter 12, Solution 37 206.0 12pf P kVA16j1220 1020I 3 p 2 p L p (3)(55.51) 10)16j12(I Trang 340110 ° ∠ =+++ )110(2 12 1 2 2 2 Y 2 a )110(2 33 2 = S S = S 551 86+j 735 ... 3 2 1 5 2.525 .1-120- Adding (1) and (6), 3 1 (10.5 j6)) 6j75.16(65.21j5 15j38-18j4465 .12j5.87 200 I I 25.26j5 192 − = ∆ , ∆1 =900.25− j935.2, ∆3 =110.3−j1327.6 144.4j242.538.33-682.67.76-28.194

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... I 1 = (3 – j2)i 2 – 2 (2) Substituting (2) into (1), 8∠30° + (2 + j4)2 = (14 + j7)I 2 I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12° Vx = 2I2 = 2.074∠21.12° 2 Ω + – Trang 12 Use mesh analysis ... 5.1812 1.08.02.1 ++ − = X j X j j I Z Th 1.08.02.1 5.18121 −+ = − = 62472 75.108 .0)1.02.1( )5.18(1220 | 2 2 2 2 −+ = ⎯→ ⎯+ − −+ Trang 45127.38j30j100Trang 46(a) Find the input impedance of the ... jωLa + ω2Ma2/(jωLa + Zin) = (–ω2La2 + ω2Ma2 + jωLaZin)/( jωLa + Zin) (2) Substituting (1) into (2) gives, = b 2 b 2 2 b 2 b a b 2 b 2 2 b 2 b a 2 a 2 2 a 2 LjR ML RLjLj LjR )ML RLj(LjML ω+ ω+ω −ω+ω

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... 14.22. Chapter 14, Solution 22 10 k k log 20 20 = 10 ⎯ ⎯→ = A zero of slope + 20 dB / dec at ω = 2 ⎯ ⎯→ 1 + j ω 2 A pole of slope - 20 dB / dec at 20 j 1 1 20 ω + ⎯→ ⎯ = ω A pole of slope - 20 ... j 2000 ) = ω C 2 L 2 j R ) 2 ( 0 0 0 = ω ) 10 1 )( 10 5 ( 2 ) 10 40 ( 2 ) 10 5 ( j 2000 ) 2 3 0 Z Z(ω0/2) = 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 1 2 1 2 ω + ω = Z 2 2 1 2 2 2 2 1 2 1 2 1 2 2 1 2 1 1 2 1 2 )] C R R L ( j LCR LCR R )[ L R j LC R R - + ω + ω − ω − + ω − ω − ω − ω + ω Trang 43PROPRIETARY MATERIAL © 2007 The McGraw-Hill

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... 30 t 2 cos( 2 2 d ds d 2 3 s 2 4 s 2 3 ds d ( ) 2 2 2 2 2 2 1 s 2 3 ) s 8 ( 4 s 2 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 ... 0 t 2 ) t ( h1 ) 2 t ( u ) 2 t ( 2 ) 1 t ( u t 2 ) 1 t ( u ) 1 t ( u t 2 ) t ( u t 2 ) t ( ) 2 t ( u ) 2 t ( 2 ) 1 t ( u ) 1 t ( 4 ) t ( u t 2 ) t ( 2 s - 2 2 -2s s 2 2 s 2 s e 2 e s 4 s 2 ) ... 2 3 20 36 6 2 + + + + + = s s s s s − 54 108 20 17 ( 2)( 1) Trang 27Find the inverse Laplace transform of: 26 2 2 + + + = s s 2 22 22 3 ) 2 s ( 3 3 2 3 ) 2 s ( ) 2 s ( 2 s ; 3 ) 2 s ( B As s

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... 2 j )( 3229 1 j 5 . 1 ( ) 3229 1 j 5 0 ( 2 s 1 ) 3229 1 j 5 0 s )( 3229 1 j 5 0 s )( 2 s ( s 2 Vs I ) 3229 1 j 5 0 s )( 3229 1 j 5 0 s )( 2 s ( s 2 2 s s s 2 2 s 1 2 s 2 1 s 1 1 2 ... 2 1 s 1 2 2 1 s 3 2 s 2 + + = ) 2 s ( s 5 s 4 2 - ∆ 3 s 2 s 2 C Bs 2 s A ) 3 s 2 s 2 )( 2 s ( 13 2s - I 2 2 2 + + + = + + + 0 5 1 s s 714 2 s 7145 1 2 s 7143 0 3 s 2 s 2 429 5 s 429 3 2 ... V 2 2 V ) 2 s ( 4 ( + + = + − + o s 1 V s 1 2 1 2 2 s 2 + + = + But I Vo = V1+ 2 and s 1 V = 2 s 2 V s s ) 2 s ( s 2 V V s ) 1 V ( 2 V + − = + 2 V 2 s s s 2 2 s s 1 2 2 + oV s 2 1 s 1 s 1 2 2

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... 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 n 2 sin n 2 1 3 n 2 cos n 3 Chapter 17, Problem 26 Trang 32Find ... + π 2 ( j n 802 [ n 200 j 2 2π − + π π − = 2 2 2 2 2 2 1 ) 1200 n ( ) 802 ( n )} n 802 /( ) 1200 n {( tan 90 200 − π + π π − π + 1 k n n sin( n t ) I 200 20 ° = n 802 1200 n 2 tan 90 2 2 1 ... of Fig 17.70 if Trang 42tan ) 2 / ( ) 4 / n [(4 n n 20 ) n / ) 2 n (( tan ) 2 n ( n n ) 2 / ) 4 / n (( 20 2 1 2 2 2 1 2 2 2 2 − − π − π ∠ + = − ∠ − + π − π 2 1 2 2 n tan 2 4 n nt cos 4 n n 20

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... x 20 − j V RC − = j 2 j 2 4 0 1 − = = j 2 j 2 125 0 mA 20 V o ( ) ω πδ − ω + + ω − j 2 125 0 j 125 0 π − + − 2 125 0 t u e 125 0 ) t sgn( 125 0 ) io(t) = 0 625 − 0 25 u ( ) + 0 125 ... πδ − ω πδ + + ω + + ω πδ = ω 2 j 1 2 2 j 1 2 2 1 H − ω + + ω − − ω δ + + ω δ π = 2 2 2 2 2 j 2 2 2 4 j 2 2 ω − − ω δ + + ω δ π (b) G(ω) = F [ 0 ] [ F ( 0) ( F 0) ] 2 j ) t ( t where F(ω) = F [ ... + ω δ + ω = 2 4 j j d 1 e 2 j 2 1 2 4 j j d 1 e 2 j 2 1 2 t j 2 t j 2 4 j 1 e 2 j 2 1 2 4 j 1 e 2 j 2 + + − + + + 1 e 17 4 j 1 j 2 2 1 ) t ( v jt jt o − − + Trang 61Chapter 18, Problem 50 Determine

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... (4), 22221 12 22121 y 1 V V y V y V + − = ) y 5 0 )( y 1 ( y 1 y s / 2 V s 2 V y V y ) y 5 0 )( y 1 ( V 2211 21122 212221 2211 s ) 1 s ( 8 0 s 1 s s 2 1 2 s 1 1 s 1 ) 1 s ( s / 2 V 22 2 + + ... that V 12 21 22 11 21 2 = where ∆ is the determinant of [g] matrix g Trang 77Chapter 19, Solution 41 For the g parameters 2 12 1 11 2 22 1 21 But V1 = Vs − I1Zs and 2 22 1 21 L 2 2 L 22 1 ) ( ... 121 x 20 ) 121 40 ( 8 57 V ) 121 40 ( 8 57 V 8 57 V 12 ) 160 V 13 ( 60 V 1 2211 11 xx x2 22222 60 V 12 V 150 V I , V 3 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

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... circuit of Fig 2.121 Chapter 2, Solution 57 12 21612 6x81212x6 4x8x2x Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω Trang 4428028 43 7x36736Ω = 30 3x273 + 567.26 7.2x18867.57.218 7.2x18 868.518 7.2x868.5 ... Ω 10Ω 20 Ω Figure 2.98 For Prob 2.34 Chapter 2, Solution 34 40//(10 + 20 + 10)= 20 Ω, 40//(8+12 + 20) = 20 Ω Trang 22Calculate V o and I o in the circuit of Fig 2.99 100 30x70 Trang 23Find ... 60 Ω + 20x1010x4040x20R RRRRR R 3 1 3 3 2 2 Trang 42Determine V in the circuit of Fig 1.120 15x1212x1010 Req = 19.688||(12 + 16.667) = 11.672Ω By voltage division, 16672.11 672.11+ = 42.18 V

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... consent of McGraw-Hill Education.e2 006.0dt0.006eidt q 4 - 2 0 2t 2 0 2t - 2.945 mC (b) 0.012e-2t(10) 0.12e-2t dt di10 v= =− =− V this leads to p(t) = v(t)i(t) = (-0.12e-2t)(0.006e-2t) = –720e ... written consent of McGraw-Hill Education.Solution 1.20 p30 volt source = 30x(–6) = –180 W p12 volt element = 12x6 = 72 W p28 volt e.ement with 2 amps flowing through it = 28x2 = 56 W p28 volt element ... consent of McGraw-Hill Education.Solution 1.30 Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 2,436–250 kWh = 2,186 kWh @ $0.07/kWh= $153.02 Total = $164.02 Trang 32Copyright © 2017

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