Bài giải phần giải mạch P7

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Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 1 t ∫ i dt + Ri = 0 C -∞ Taking the derivative of each term, i di +R =0 C dt di dt or =− i RC Integrating,  i( t )  - t = ln  I 0  RC i( t ) = I 0 e - t RC v( t ) = Ri( t ) = RI 0 e - t RC or v(t ) = V0e- t RC Chapter 7, Solution 2. τ = R th C where R th is the Thevenin equivalent at the capacitor terminals. R...

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Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 1 t ∫ i dt + Ri = 0 C -∞ Taking the derivative of each term, i di +R =0 C dt di dt or =− i RC Integrating,  i( t )  - t ln =  I 0  RC i( t ) = I 0 e - t RC v( t ) = Ri( t ) = RI 0 e - t RC or v(t ) = V0e- t RC Chapter 7, Solution 2. τ = R th C where R th is the Thevenin equivalent at the capacitor terminals. R th = 120 || 80 + 12 = 60 Ω τ = 60 × 0.5 × 10 -3 = 30 ms Chapter 7, Solution 3. (a) RTh = 10 // 10 = 5kΩ, τ = RTh C = 5 x10 3 x 2 x10 −6 = 10 ms (b) RTh = 20 //(5 + 25) + 8 = 20Ω, τ = RTh C = 20 x0.3 = 6s Chapter 7, Solution 4. τ = R eq C eq C1C 2 R 1R 2 where C eq = , R eq = C1 + C 2 R1 + R 2 R 1 R 2 C1C 2 τ= ( R 1 + R 2 )(C 1 + C 2 )
Chapter 7, Solution 5. v( t ) = v(4) e -(t -4) τ where v(4) = 24 , τ = RC = (20)(0.1) = 2 v( t ) = 24 e -(t - 4) 2 v(10) = 24 e -6 2 = 1.195 V Chapter 7, Solution 6. 2 v o = v ( 0) = (24) = 4 V 10 + 2 2 v( t ) = voe − t / τ , τ = RC = 40 x10−6 x 2 x103 = 25 v( t ) = 4e −12.5t V Chapter 7, Solution 7. v( t ) = v(0) e - t τ , τ = R th C where R th is the Thevenin resistance across the capacitor. To determine R th , we insert a 1-V voltage source in place of the capacitor as shown below. 8Ω i2 i i1 + + + 0.5 V − 10 Ω v=1 1V − − 1 1 − 0.5 1 i1 = = 0.1 , i2 = = 10 8 16 1 13 i = i1 + i 2 = 0.1 + = 16 80 1 80 R th = = i 13 80 8 τ = R th C = × 0.1 = 13 13 v( t ) = 20 e -13t 8 V
Chapter 7, Solution 8. 1 (a) τ = RC = 4 dv -i = C dt - 0.2 e = C (10)(-4) e-4t -4t  → C = 5 mF 1 R= = 50 Ω 4C 1 (b) τ = RC = = 0.25 s 4 1 1 (c) w C (0) = CV02 = (5 × 10 -3 )(100) = 250 mJ 2 2 1 1 1 (d) w R = × CV02 = CV02 (1 − e -2t 0 τ ) 2 2 2 1 0.5 = 1 − e -8t 0  → e -8t 0 = 2 or e =2 8t 0 1 t 0 = ln (2) = 86.6 ms 8 Chapter 7, Solution 9. v( t ) = v(0) e- t τ , τ = R eq C R eq = 2 + 8 || 8 + 6 || 3 = 2 + 4 + 2 = 8 Ω τ = R eq C = (0.25)(8) = 2 v( t ) = 20 e -t 2 V
Chapter 7, Solution 10. io 15 Ω i 10 Ω iT + 10 mF v 4Ω − (10)(3) 15 i o = 10 i  → i o = =2A 15 i.e. if i(0) = 3 A , then i o (0) = 2 A i T (0) = i(0) + i o (0) = 5 A v(0) = 10i(0) + 4i T (0) = 30 + 20 = 50 V across the capacitor terminals. R th = 4 + 10 || 15 = 4 + 6 = 10 Ω τ = R th C = (10)(10 × 10 -3 ) = 0.1 v( t ) = v(0) e - t τ = 50 e -10t dv iC = C = (10 × 10 -3 )(-500 e -10t ) dt i C = - 5 e -10t A By applying the current division principle, 15 i( t ) = ( - i ) = -0.6 i C = 3 e -10t A 10 + 15 C Chapter 7, Solution 11. Applying KCL to the RL circuit, 1 v L ∫ v dt + R = 0 Differentiating both sides, v 1 dv dv R + =0  → + v=0 L R dt dt L v = A e -Rt L
If the initial current is I 0 , then v(0) = I 0 R = A L v = I 0 R e -t τ , τ= R 1 t i= L -∞ ∫ v(t ) dt - τ I 0 R -t τ t i= e -∞ L i = - I 0 R e -t τ i( t ) = I 0 e - t τ Chapter 7, Solution 12. When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω resistor is short-circuited so that the resulting circuit is as shown in Fig. (a). 3Ω 12 V + i(0-) 4Ω 2H − (a) (b) 12 i (0 − ) = =4A 3 Since the current through an inductor cannot change abruptly, i(0) = i(0 − ) = i(0 + ) = 4 A When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b). L 2 τ = = = 0.5 R 4 Hence, i( t ) = i(0) e - t τ = 4 e -2t A
Chapter 7, Solution 13. L τ= R th where R th is the Thevenin resistance at the terminals of the inductor. R th = 70 || 30 + 80 || 20 = 21 + 16 = 37 Ω 2 × 10 -3 τ= = 81.08 µs 37 Chapter 7, Solution 14 Converting the wye-subnetwork to delta gives 16 Ω R2 80mH R1 R3 30 Ω 10 x 20 + 20 x50 + 50 x10 1700 1700 R1 = = 1700 / 20 = 85Ω, R2 = = 34Ω , R3 = = 170Ω 20 50 10 30//170 = (30x170)/200 = 25.5 Ω , 34//16=(34x16)/50 =10.88 Ω 85 x36.38 L 80 x10 −3 RTh = 85 //( 25.5 + 10.88) = = 25.476Ω, τ= = = 3.14 ms 121.38 RTh 25.476
Chapter 7, Solution 15 L (a) RTh = 12 + 10 // 40 = 20Ω, τ= = 5 / 20 = 0.25s RTh L (b) RTh = 40 // 160 + 8 = 40Ω, τ= = (20 x10 −3 ) / 40 = 0.5 ms RTh Chapter 7, Solution 16. L eq τ= R eq R 1R 3 R 2 (R 1 + R 3 ) + R 1 R 3 (a) L eq = L and R eq = R 2 + = R1 + R 3 R1 + R 3 L( R 1 + R 3 ) τ= R 2 (R 1 + R 3 ) + R 1 R 3 L1 L 2 R 1R 2 R 3 (R 1 + R 2 ) + R 1 R 2 (b) where L eq = and R eq = R 3 + = L1 + L 2 R1 + R 2 R1 + R 2 L1L 2 (R 1 + R 2 ) τ= (L 1 + L 2 ) ( R 3 ( R 1 + R 2 ) + R 1 R 2 ) Chapter 7, Solution 17. L 14 1 i( t ) = i(0) e - t τ , τ= = = R eq 4 16 i( t ) = 2 e -16t di v o ( t ) = 3i + L = 6 e-16t + (1 4)(-16) 2 e-16t dt v o ( t ) = - 2 e -16t V
Chapter 7, Solution 18. If v( t ) = 0 , the circuit can be redrawn as shown below. + 0.4 H Req vo(t) i(t) − 6 L 2 5 1 R eq = 2 || 3 = , τ= = × = 5 R 5 6 3 i( t ) = i(0) e = e -t τ -3t di - 2 v o ( t ) = -L = (-3) e -3t = 1.2 e -3t V dt 5 Chapter 7, Solution 19. 1V i i1 i2 − + 10 Ω i1 i/2 i2 40 Ω To find R th we replace the inductor by a 1-V voltage source as shown above. 10 i1 − 1 + 40 i 2 = 0 But i = i2 + i 2 and i = i1 i.e. i1 = 2 i 2 = 2 i 1 10 i − 1 + 20 i = 0  → i = 30 1 R th = = 30 Ω i L 6 τ= = = 0.2 s R th 30 i( t ) = 2 e -5t A
Chapter 7, Solution 20. L 1 (a). τ= =  → R = 50L R 50 di -v= L dt - 150 e = L(30)(-50) e -50t  -50t → L = 0.1 H R = 50L = 5 Ω L 1 (b). τ= = = 20 ms R 50 1 1 (c). w = L i 2 (0) = (0.1)(30) 2 = 45 J 2 2 (d). Let p be the fraction 1 1 L I 0 ⋅ p = L I 0 ( 1 − e -2t 0 τ ) 2 2 p = 1− e -(2)(10) 50 = 1 − e -0.4 = 0.3296 i.e. p = 33% Chapter 7, Solution 21. The circuit can be replaced by its Thevenin equivalent shown below. Rth Vth + 2H − 80 Vth = (60) = 40 V 80 + 40 80 R th = 40 || 80 + R = +R 3 Vth 40 I = i(0) = i(∞) = = R th 80 3 + R 2 1 1  40  w = L I 2 = (2)  =1 2 2  R + 80 3  40 40 =1 → R = R + 80 3 3 R = 13.33 Ω
Chapter 7, Solution 22. L i( t ) = i(0) e - t τ , τ= R eq 2 R eq = 5 || 20 + 1 = 5 Ω , τ= 5 i( t ) = 10 e -2.5t A Using current division, the current through the 20 ohm resistor is 5 -i io = (-i) = = -2 e -2.5t 5 + 20 5 v( t ) = 20 i o = - 40 e -2.5t V Chapter 7, Solution 23. Since the 2 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel, they always have the same voltage. 2 2 -i = + = 1.5 → i(0) = -1.5 2 3 +1 The Thevenin resistance R th at the inductor’s terminals is 4 L 13 1 R th = 2 || (3 + 1) = , τ= = = 3 R th 4 3 4 i( t ) = i(0) e - t τ = -1.5 e -4t , t > 0 di v L = v o = L = -1.5(-4)(1/3) e -4t dt v o = 2 e V, t > 0 -4t 1 vx = v = 0.5 e -4t V , t > 0 3 +1 L Chapter 7, Solution 24. (a) v( t ) = - 5 u(t) (b) i( t ) = -10 [ u ( t ) − u ( t − 3)] + 10[ u ( t − 3) − u ( t − 5)] = - 10 u(t ) + 20 u(t − 3) − 10 u(t − 5)
(c) x ( t ) = ( t − 1) [ u ( t − 1) − u ( t − 2)] + [ u ( t − 2) − u ( t − 3)] + (4 − t ) [ u ( t − 3) − u ( t − 4)] = ( t − 1) u ( t − 1) − ( t − 2) u ( t − 2) − ( t − 3) u ( t − 3) + ( t − 4) u ( t − 4) = r(t − 1) − r(t − 2) − r(t − 3) + r(t − 4) (d) y( t ) = 2 u (-t ) − 5 [ u ( t ) − u ( t − 1)] = 2 u(-t ) − 5 u(t ) + 5 u(t − 1) Chapter 7, Solution 25. v(t) = [u(t) + r(t – 1) – r(t – 2) – 2u(t – 2)] V Chapter 7, Solution 26. (a) v1 ( t ) = u ( t + 1) − u ( t ) + [ u ( t − 1) − u ( t )] v1 ( t ) = u(t + 1) − 2 u(t ) + u(t − 1) (b) v 2 ( t ) = ( 4 − t ) [ u ( t − 2) − u ( t − 4) ] v 2 ( t ) = -( t − 4) u ( t − 2) + ( t − 4) u ( t − 4) v 2 ( t ) = 2 u(t − 2) − r(t − 2) + r(t − 4) (c) v 3 ( t ) = 2 [ u(t − 2) − u(t − 4)] + 4 [ u(t − 4) − u(t − 6)] v 3 ( t ) = 2 u(t − 2) + 2 u(t − 4) − 4 u(t − 6) (d) v 4 ( t ) = -t [ u ( t − 1) − u ( t − 2)] = -t u(t − 1) + t u ( t − 2) v 4 ( t ) = (-t + 1 − 1) u ( t − 1) + ( t − 2 + 2) u ( t − 2) v 4 ( t ) = - r(t − 1) − u(t − 1) + r(t − 2) + 2 u(t − 2) Chapter 7, Solution 27. v(t) is sketched below. v(t) 2 1 0 1 2 3 4 t -1
Chapter 7, Solution 28. i(t) is sketched below. i(t) 1 0 1 2 3 4 t -1 Chapter 7, Solution 29 x(t) (a) 3.679 0 1 t (b) y(t) 27.18 0 t
(c) z (t ) = cos 4tδ (t − 1) = cos 4δ (t − 1) = −0.6536δ (t − 1) , which is sketched below. z(t) 0 1 t -0.653 δ (t ) Chapter 7, Solution 30. ∫ 10 (a) 0 4 t 2 δ( t − 1) dt = 4 t 2 t =1 =4 ∞ (b) ∫ -∞ cos(2πt ) δ( t − 0.5) dt = cos(2πt ) t = 0.5 = cos π = - 1 Chapter 7, Solution 31. ∫ [ e δ(t − 2)] dt = e ∞ (a) -∞ - 4t 2 = e = 112 × 10 - 4t 2 t=2 -16 -9 ∫ [ 5 δ(t ) + e δ(t ) + cos 2πt δ(t )] dt = ( 5 + e + cos(2πt )) ∞ (b) -∞ -t -t t =0 = 5 +1+1 = 7 Chapter 7, Solution 32. t t t (a) ∫ u (λ )dλ = ∫ 1dλ = λ 1 1 1 = t −1 4 1 4 t2 ∫ r (t − 1)dt = ∫ 0dt + ∫ (t − 1)dt = − t 1 = 4 .5 4 (b) 0 0 1 2 5 ∫ (t − 6) δ (t − 2)dt = (t − 6) 2 = 16 2 (c ) t =2 1
Chapter 7, Solution 33. 1 t i( t ) = ∫ v(t ) dt + i(0) L 0 10 -3 ∫ 20 δ(t − 2) dt + 0 t i( t ) = 10 × 10 -3 0 i ( t ) = 2 u( t − 2 ) A Chapter 7, Solution 34. d [u ( t − 1) u ( t + 1)] = δ( t − 1)u ( t + 1) + (a) dt u ( t − 1)δ( t + 1) = δ( t − 1) • 1 + 0 • δ( t + 1) = δ( t − 1) d [r ( t − 6) u ( t − 2)] = u ( t − 6)u ( t − 2) + (b) dt r ( t − 6)δ( t − 2) = u ( t − 6) • 1 + 0 • δ( t − 2) = u ( t − 6) d [sin 4t u (t − 3)] = 4 cos 4t u ( t − 3) + sin 4tδ( t − 3) dt (c) = 4 cos 4t u ( t − 3) + sin 4x3δ( t − 3) = 4 cos 4t u ( t − 3) − 0.5366δ( t − 3) Chapter 7, Solution 35. (a) v( t ) = A e -5t 3 , v(0) = A = -2 v( t ) = - 2 e -5t 3 V (b) v( t ) = A e 2t 3 , v(0) = A = 5 v( t ) = 5 e 2t 3 V
Chapter 7, Solution 36. (a) v( t ) = A + B e-t , t > 0 A = 1, v(0) = 0 = 1 + B or B = -1 v( t ) = 1 − e -t V , t > 0 (b) v( t ) = A + B e t 2 , t > 0 A = -3 , v(0) = -6 = -3 + B or B = -3 v( t ) = - 3 ( 1 + e t 2 ) V , t > 0 Chapter 7, Solution 37. Let v = vh + vp, vp =10. • 1 vh + 4 v h =0  → v h = Ae −t / 4 v = 10 + Ae −0.25t v(0) = 2 = 10 + A  → A = −8 v = 10 − 8e −0.25t (a) τ = 4 s (b) v(∞) = 10 V (c ) v = 10 − 8e −0.25t Chapter 7, Solution 38 Let i = ip +ih • i h + 3ih = 0  → ih = Ae −3t u (t ) • 2 Let i p = ku (t ), ip = 0, 3ku (t ) = 2u (t )  → k= 3
2 ip = u (t ) 3 2 i = ( Ae −3t + )u (t ) 3 If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus 2 i= (1 − e −3t )u (t ) 3 Chapter 7, Solution 39. (a) Before t = 0, 1 v( t ) = (20) = 4 V 4 +1 After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ τ = RC = (4)(2) = 8 , v(0) = 4 , v(∞) = 20 v( t ) = 20 + (8 − 20) e - t 8 v( t ) = 20 − 12 e -t 8 V (b) Before t = 0, v = v1 + v 2 , where v1 is due to the 12-V source and v 2 is due to the 2-A source. v1 = 12 V To get v 2 , transform the current source as shown in Fig. (a). v 2 = -8 V Thus, v = 12 − 8 = 4 V After t = 0, the circuit becomes that shown in Fig. (b). 2F 4Ω 2F + − v2 + + 8V 12 V − − 3Ω 3Ω (a) (b)
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v(∞) = 12 , v(0) = 4 , τ = RC = (2)(3) = 6 v( t ) = 12 + (4 − 12) e -t 6 v( t ) = 12 − 8 e -t 6 V Chapter 7, Solution 40. (a) Before t = 0, v = 12 V . After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v(∞) = 4 , v(0) = 12 , τ = RC = (2)(3) = 6 v( t ) = 4 + (12 − 4) e -t 6 v( t ) = 4 + 8 e - t 6 V (b) Before t = 0, v = 12 V . After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ After transforming the current source, the circuit is shown below. t=0 2Ω 4Ω 12 V + 5F − v(0) = 12 , v(∞) = 12 , τ = RC = (2)(5) = 10 v = 12 V Chapter 7, Solution 41. 30 v(0) = 0 , v(∞) = (12) = 10 16 (6)(30) R eq C = (6 || 30)(1) = =5 36 v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 10 + (0 − 10) e - t 5 v( t ) = 10 ( 1 − e -0.2t ) V
Chapter 7, Solution 42. (a) v o ( t ) = v o (∞) + [ v o (0) − v o (∞)] e - t τ 4 v o (0) = 0 , v o (∞) = (12) = 8 4+2 4 τ = R eq C eq , R eq = 2 || 4 = 3 4 τ = (3) = 4 3 v o ( t ) = 8 − 8 e -t 4 v o ( t ) = 8 ( 1 − e -0.25t ) V (b) For this case, v o (∞) = 0 so that v o ( t ) = v o (0) e -t τ 4 v o (0) = (12) = 8 , τ = RC = (4)(3) = 12 4+2 v o ( t ) = 8 e -t 12 V Chapter 7, Solution 43. Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source. vo vo 0.5i = 2 − , i= 40 80 1 vo vo 320 Hence, = 2− → v o = = 64 2 80 40 5 vo i= = 0.8 A 80 After t = 0, the circuit is as shown in Fig. (b). v C ( t ) = v C (0) e - t τ , τ = R th C To find R th , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). 0.5i vC i + 1V 0.5i 80 Ω − (c)
vC 1 0.5 i= = , i o = 0.5 i = 80 80 80 1 80 R th = = = 160 Ω , τ = R th C = 480 i o 0.5 v C (0) = 64 V v C ( t ) = 64 e - t 480 dv C  1  0.5 i = -i C = -C = -3   64 e - t 480 dt  480  i( t ) = 0.8 e -t 480 A Chapter 7, Solution 44. R eq = 6 || 3 = 2 Ω , τ = RC = 4 v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ Using voltage division, 3 3 v(0) = (30) = 10 V , v(∞) = (12) = 4 V 3+ 6 3+ 6 Thus, v( t ) = 4 + (10 − 4) e - t 4 = 4 + 6 e - t 4 dv  - 1 i( t ) = C = (2)(6)   e - t 4 = - 3 e -0.25t A dt 4 Chapter 7, Solution 45. For t < 0, v s = 5 u ( t ) = 0  → v(0) = 0 4 5 For t > 0, v s = 5 , v(∞) = (5) = 4 + 12 4 R eq = 7 + 4 || 12 = 10 , τ = R eq C = (10)(1 2) = 5 v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 1.25 ( 1 − e -t 5 ) V dv  1  - 5  - 1  - t 5 i( t ) = C =     e dt  2  4  5  i( t ) = 0.125 e -t 5 A
Chapter 7, Solution 46. τ = RTh C = (2 + 6) x0.25 = 2s, v(0) = 0, v(∞) = 6i s = 6 x5 = 30 v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ = 30(1 − e − t / 2 ) V Chapter 7, Solution 47. For t < 0, u ( t ) = 0 , u ( t − 1) = 0 , v(0) = 0 For 0 < t < 1, τ = RC = (2 + 8)(0.1) = 1 v(0) = 0 , v(∞) = (8)(3) = 24 v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 24( 1 − e - t ) For t > 1, v(1) = 24( 1 − e -1 ) = 15.17 - 6 + v(∞) - 24 = 0  → v(∞) = 30 v( t ) = 30 + (15.17 − 30) e -(t-1) v( t ) = 30 − 14.83 e -(t-1) Thus, (  24 1 − e - t V , v( t ) =  ) 0 0, u (-t) = 0 , v(∞) = 0 R th = 20 + 10 = 30 , τ = R th C = (30)(0.1) = 3 v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 10 e -t 3 V dv  - 1 i( t ) = C = (0.1)  10 e - t 3 dt 3 - 1 -t 3 i( t ) = e A 3