Bài giải mạch P6

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Bài giải mạch P6

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  1. Chapter 6, Solution 1. i=C dv dt ( ) = 5 2e −3t − 6 + e −3 t = 10(1 - 3t)e-3t A p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W Chapter 6, Solution 2. 1 2 1 w1 = Cv1 = (40)(120) 2 2 2 1 2 1 w2 = Cv1 = (40)(80) 2 2 2 ∆w = w 1 − w 2 = 20(120 2 − 80 2 ) = 160 kW Chapter 6, Solution 3. dv 280 − 160 i=C = 40x10 −3 = 480 mA dt 5 Chapter 6, Solution 4. 1 t C ∫o v= idt + v(0) 1 2∫ = 6 sin 4 tdt + 1 = 1 - 0.75 cos 4t Chapter 6, Solution 5. 1 t C ∫o v= idt + v(0) For 0 < t < 1, i = 4t, 1 t − 6 ∫o v= 4t dt + 0 = 100t2 kV 20x10 v(1) = 100 kV
  2. For 1 < t < 2, i = 8 - 4t, 1 t 20x10 −6 ∫1 v= (8 − 4t )dt + v(1) = 100 (4t - t2 - 3) + 100 kV 100t 2 kV, 0 < t
  3. Chapter 6, Solution 8. dv (a) i = C = −100 ACe −100t − 600 BCe −600t (1) dt i (0) = 2 = −100 AC − 600 BC  → 5 = − A − 6B (2) v (0 + ) = v (0 − )  → 50 = A + B (3) Solving (2) and (3) leads to A=61, B=-11 1 2 1 (b) Energy = Cv (0) = x 4 x10 −3 x 2500 = 5 J 2 2 (c ) From (1), i = −100 x61x 4 x10 −3 e −100t − 600 x11x 4 x10 −3 e −600t = − 24.4e −100t − 26.4e −600t A Chapter 6, Solution 9. v(t) = 1 t 12 ( ) ( ∫o 6 1 − e dt + 0 = 12 t + e V −t −t ) v(2) = 12(2 + e-2) = 25.62 V p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t) p(2) = 72(2-e-4) = 142.68 W Chapter 6, Solution 10 dv dv i=C = 2 x10 −3 dt dt  16t , 0 < t < 1µs  v =  16, 1 < t < 3 µs 64 - 16t, 3 < t < 4 µs 
  4.  16 x10 6 , 0 < t < 1µs dv  =  0, 1 < t < 3 µs dt  - 16x10 , 3 < t < 4 µs 6  32 kA, 0 < t < 1µs  i (t ) =  0, 1 < t < 3 µs - 32 kA, 3 < t < 4µs  Chapter 6, Solution 11. 1 t C ∫o v= idt + v(0) For 0 < t < 1, 1 t − 6 ∫o v= 40 x10 −3 dt = 10t kV 4x10 v(1) = 10 kV For 1 < t < 2, 1 t C ∫1 v= vdt + v(1) = 10kV For 2 < t < 3, 1 t ∫ (−40x10 −3 v= )dt + v(2) 4x10 −6 2 = -10t + 30kV Thus 10 t ⋅ kV, 0 < t
  5. Chapter 6, Solution 13. Under dc conditions, the circuit becomes that shown below: i1 10 Ω i2 50 Ω + 20 Ω + 30 Ω v1 v2 − + − 60V − i2 = 0, i1 = 60/(30+10+20) = 1A v1 = 30i2 = 30V, v2 = 60-20i1 = 40V Thus, v1 = 30V, v2 = 40V Chapter 6, Solution 14. (a) Ceq = 4C = 120 mF 1 4 4 (b) = = Ceq = 7.5 mF C eq C 30 Chapter 6, Solution 15. In parallel, as in Fig. (a), v1 = v2 = 100 + v1 − + + C1 + + C1 C2 + C2 100V − v1 v2 100V − v2 − − − (a) (b)
  6. 1 2 1 w20 = Cv = x 20x10 −6 x100 2 = 0.1J 2 2 1 w30 = x30x10 −6 x100 2 = 0.15J 2 (b) When they are connected in series as in Fig. (b): C2 30 v1 = V= x100 = 60, v2 = 40 C1 + C 2 50 1 w20 = x30x10 −6 x 60 2 = 36 mJ 2 1 w30 = x30x10 −6 x 40 2 = 24 mJ 2 Chapter 6, Solution 16 Cx80 C eq = 14 + = 30  → C = 20 µF C + 80 Chapter 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 i.e. Ceq = 3F (b) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6F 1 1 1 1 = + + =1 C eq 2 6 3 Ceq = 1F
  7. Chapter 6, Solution 18. For the capacitors in parallel C1 = 15 + 5 + 40 = 60 µF eq 1 1 1 1 1 Hence = + + = C eq 20 30 60 10 Ceq = 10 µF Chapter 6, Solution 19. We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60 µ F. The 60 - µ F capacitor in series with another 60- µ F capacitor gives 30 µ F. 30 + 50 = 80 µ F, 80 + 40 = 120 µ F The circuit is reduced to that shown below. 12 120 12 80 120- µ F capacitor in series with 80 µ F gives (80x120)/200 = 48 48 + 12 = 60 60- µ F capacitor in series with 12 µ F gives (60x12)/72 = 10 µ F Chapter 6, Solution 20. 3 in series with 6 = 6x3/(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below: 20 1 6 2 8
  8. 6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus Ceq = 4 mF Chapter 6, Solution 21. 4µF in series with 12µF = (4x12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF Hence Ceq = 2.5µF Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below: a b 40 µF 60 µF 30 µF 20 µF Combining the capacitors in series gives C1 , where eq 1 1 1 1 1 1 = + + = C1 = 10µF eq C eq 60 20 30 10 Thus Ceq = 10 + 40 = 50 µF
  9. Chapter 6, Solution 23. (a) 3µF is in series with 6µF 3x6/(9) = 2µF v4µF = 1/2 x 120 = 60V v2µF = 60V 3 v6µF = (60) = 20V 6+3 v3µF = 60 - 20 = 40V (b) Hence w = 1/2 Cv2 w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ Chapter 6, Solution 24. 20µF is series with 80µF = 20x80/(100) = 16µF 14µF is parallel with 16µF = 30µF (a) v30µF = 90V v60µF = 30V v14µF = 60V 80 v20µF = x 60 = 48V 20 + 80 v80µF = 60 - 48 = 12V 1 2 (b) Since w = Cv 2 w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w80µF = 1/2 x 80 x 10-6 x 144 = 5.76mJ Chapter 6, Solution 25. (a) For the capacitors in series, v1 C 2 Q1 = Q2 C1v1 = C2v2 = v 2 C1
  10. C2 C + C2 C1 vs = v1 + v2 = v2 + v2 = 1 v2 v2 = vs C1 C1 C1 + C 2 C2 Similarly, v1 = vs C1 + C 2 (b) For capacitors in parallel Q1 Q 2 v1 = v2 = = C1 C 2 C C + C2 Qs = Q1 + Q2 = 1 Q 2 + Q 2 = 1 Q2 C2 C2 or C2 Q2 = C1 + C 2 C1 Q1 = Qs C1 + C 2 dQ C1 C2 i= i1 = is , i2 = is dt C1 + C 2 C1 + C 2 Chapter 6, Solution 26. (a) Ceq = C1 + C2 + C3 = 35µF (b) Q1 = C1v = 5 x 150µC = 0.75mC Q2 = C2v = 10 x 150µC = 1.5mC Q3 = C3v = 20 x 150 = 3mC 1 1 (c) w= C eq v 2 = x35x150 2 µJ = 393.8mJ 2 2
  11. Chapter 6, Solution 27. 1 1 1 1 1 1 1 7 (a) = + + = + + = C eq C1 C 2 C 3 5 10 20 20 20 Ceq = µF = 2.857µF 7 (b) Since the capacitors are in series, 20 Q1 = Q2 = Q3 = Q = Ceqv = x 200µV = 0.5714mV 7 1 1 20 (c) w = C eq v 2 = x x 200 2 µJ = 57.143mJ 2 2 7 Chapter 6, Solution 28. We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C. Cb 50 µF Ca Cc 20 µF  1  1   1  1   1  1     +    +    1 10 40 10 30 30 40 =          Ca 1 30 3 1 1 2 = + + = 40 10 40 10 Ca = 5µF 1 1 1 + + 1 2 = 400 300 1200 = C6 1 30 10 Cb = 15µF
  12. 1 1 1 + + 1 4 = 400 300 1200 = Cc 1 15 40 Cc = 3.75µF Cb in parallel with 50µF = 50 + 15 = 65µF Cc in series with 20µF = 23.75µF 65x 23.75 65µF in series with 23.75µF = = 17.39µF 88.75 17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF Hence Ceq = 22.39µF Chapter 6, Solution 29. (a) C in series with C = C/(2) C/2 in parallel with C = 3C/2 3C Cx 3C 2 = 3C in series with C = 2 C 5 5 2 C C 3 in parallel with C = C + 3 = 1.6 C 5 5 (b) 2C Ceq 2C 1 1 1 1 = + = C eq 2C 2C C Ceq = C
  13. Chapter 6, Solution 30. 1 t C ∫o vo = idt + i(0) For 0 < t < 1, i = 60t mA, 10 −3 t 3x10 −6 ∫o vo = 60tdt + 0 = 10 t 2 kV vo(1) = 10kV For 1< t < 2, i = 120 - 60t mA, 10 −3 t 3x10t −6 ∫1 vo = (120 − 60t )dt + v o (1) = [40t – 10t2 ] 1 + 10kV = 40t – 10t2 - 20 10t 2 kV, 0 < t
  14. = t 2 − 5 + 3 +5kV = t 2 − 5t + 11kV t  t 2 kV, 0 < t
  15. Chapter 6, Solution 33 Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. 3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors. Therefore, we get: VTh = 7.5 V, CTh = 2.5 F Chapter 6, Solution 34. i = 6e-t/2 di 1 v = L = 10 x10 −3 (6) e − t / 2 dt 2 -t/2 = -30e mV v(3) = -300e-3/2 mV = -0.9487 mV p = vi = -180e-t mW p(3) = -180e-3 mW = -0.8 mW Chapter 6, Solution 35. di V 60 x10 −3 v=L L= = = 200 mH dt ∆i / ∆t 0.6 /(2)
  16. Chapter 6, Solution 36. di 1 v=L = x10 −3 (12)(2)(− sin 2 t )V dt 4 = - 6 sin 2t mV p = vi = -72 sin 2t cos 2t mW But 2 sin A cos A = sin 2A p = -36 sin 4t mW Chapter 6, Solution 37. di v=L = 12 x10 −3 x 4(100) cos100t dt = 4.8 cos 100t V p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t t 11 / 200 w= ∫ pdt = ∫ o o 9.6 sin 200 t 9.6 =− cos 200t 11 / 200 J o 200 = −48(cos π − 1)mJ = 96 mJ Chapter 6, Solution 38. = 40x10 −3 (e − 2 t − 2te − 2 t )dt di v=L dt = 40(1 − 2t )e −2 t mV, t > 0
  17. Chapter 6, Solution 39 di 1 t v=L  i = ∫ 0 idt + i(0) → dt L 1 i= t (3t 2 + 2t + 4)dt + 1 −3 ∫ 0 200x10 t = 5( t 3 + t 2 + 4t ) +1 0 i(t) = 5t3 + 5t2 + 20t + 1 A Chapter 6, Solution 40 di di v=L = 20 x10 −3 dt dt  10t , 0 < t < 1 ms  i =  20 - 10t, 1 < t < 3 ms - 40 + 10t, 3 < t < 4 ms   10 x10 3 , 0 < t < 1 ms di  = - 10x10 3 , 1 < t < 3 ms dt  10x10 , 3 3 < t < 4 ms  200 V, 0 < t < 1 ms  v = - 200 V, 1 < t < 3 ms  200 V, 3 < t < 4 ms  which is sketched below. v(t) V 200 0 1 2 3 4 t(ms) -200
  18. Chapter 6, Solution 41. i= 1 t L 1 t ( ) ∫0 vdt + i(0) =  2 ∫o 20 1 − 2 dt + 0.3 −2 t    1  = 10 t + e − 2t  to +0. 3 = 10t + 5e − 2t − 4. 7 A  2  At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A 1 2 w= L i = 35.72J 2 Chapter 6, Solution 42. 1 t 1 t L ∫o i= vdt + i(0) = ∫ v( t )dt − 1 5 o 10 t For 0 < t < 1, i = ∫ dt − 1 = 2t − 1 A 5 0 For 1 < t < 2, i = 0 + i(1) = 1A 1 ∫ 10dt + i(2) = 2t t +1 2 For 2 < t < 3, i = 5 = 2t - 3 A For 3 < t < 4, i = 0 + i(3) = 3 A 1 t 5 ∫4 For 4 < t < 5, i = 10dt + i(4) = 2 t 4 +3 t = 2t - 5 A  2t − 1A, 0 < t
  19. Chapter 6, Solution 43. 2 t 1 1 w = L ∫ idt = Li( t ) − Li (−∞) −∞ 2 2 1 ( = x80 x10 −3 x 60x10 −3 − 0 2 ) = 144 µJ Chapter 6, Solution 44. 1 t 1 t i= L ∫to vdt + i(t o ) = 5 ∫o (4 + 10 cos 2t )dt − 1 = 0.8t + sin 2t -1 Chapter 6, Solution 45. 1 t L ∫o i(t) = v( t ) + i(0) For 0 < t < 1, v = 5t 1 t i= 10x10 −3 ∫ 5t dt + 0 o = 0.25t2 kA For 1 < t < 2, v = -10 + 5t 1 t i= 10x10 −3 ∫ (−10 + 5t )dt + i(1) 1 t = ∫ (0.5t − 1)dt + 0.25kA 1 = 1 - t + 0.25t2 kA 0.25t 2 kA, 0 < t
  20. Chapter 6, Solution 46. Under dc conditions, the circuit is as shown below: 2Ω iL + 4Ω vC 3A − By current division, 4 iL = (3) = 2A, vc = 0V 4+2 1 2 11 2 wL = L i L =  (2) = 1J 2 22 1 1 wc = C v c = (2)( v) = 0J 2 2 2 Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below: R + vC − iL 2Ω 5A 2 10 10R iL = (5) = , v c = Ri L = R+2 R+2 R+2
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