Use the differential equation approach to find io(t) for t

Use the differential equation approach to find i(t) for t > 0 in the network in Fig. P7.1. t = 0 2 H 12 V 6 6 + i(t) Figure P7.1

Use the differential equation approach to find io(t) for t > 0 in the network in Fig. P7.2. 6 k 6 k 12 V 12 k t = 0 + i o(t) 100 F Figure P7.2

Use the differential equation approach to find o(t) for t > 0 in the network in Fig. P7.3. t = 0 200 F o(t) + 6 k + 3 k 2 k 12 V Figure P7.3

Use the differential equation approach to find io(t) for t > 0 in the network in Fig. P7.4. 4 k 1 k 2 k 4 mA 2 k 300 F i o(t) t = 0 Figure P7.4

Use the differential equation approach to find o(t) for t > 0 in the circuit in Fig. P7.5 and plot the response, including the time interval just prior to switch action. + t = 0 12 k 2 k 6 V 100 F 4 k o(t) + Figure P7.5

Use the differential equation approach to find C(t) for t > 0 in the circuit in Fig. P7.6 and plot the response, including the time interval just prior to closing the switch. + t = 0 2 k 4 k 4 k 6 V C(t) 100 F 2 k + Figure P7.6

Use the differential equation approach to find io(t) for t > 0 in the circuit in Fig. P7.7 and plot the response, including the time interval just prior to closing the switch. + t = 0 12 k 6 k 12 k 6 k 200 F 6 V i o(t) Figure P7.7

Use the differential equation approach to find o(t) for t > 0 in the circuit in Fig. P7.8 and plot the response, including the time interval just prior to opening the switch. 12 V 100 F 6 k 6 k 6 k o(t) + + t = 0 Figure P7.8

Use the differential equation approach to find C(t) for t > 0 in the circuit in Fig. P7.9. + + 3 k 100 F 6 k 12 V C(t) t = 0 Figure P7.9

Use the differential equation approach to find C(t) for t > 0 in the circuit in Fig. P7.10. + 100 F 4 k 9 k 3 k 6 V + C(t) t = 0 Figure P7.10

Use the differential equation approach to find io(t) for t > 0 in the circuit in Fig. P7.11 and plot the response, including the time interval just prior to opening the switch. + 3 k 4 k 12 V 4 k 12 k 12 k 8 k 50 F i o(t) t = 0 Figure P7.11

In the network in Fig. P7.12, find io(t) for t > 0 using the differential equation approach. 2 H 4 6 2 A 12 t = 0 i o(t) Figure P7.12

Use the differential equation approach to find o(t) for t > 0 in the network in Fig. P7.13. o(t) + 4 A 2 t = 0 4 4 4 H Figure P7.13

Use the differential equation approach to find iL(t) for t > 0 in the circuit in Fig. P7.14 and plot the response, including the time interval just prior to opening the switch. + 12 6 6 6 V 2 H 3 i L(t) t = 0 Figure P7.14

Use the differential equation approach to find i(t) for t > 0 in the circuit in Fig. P7.15 and plot the response, including the time interval just prior to opening the switch. t = 0 5 mA 1 k 2 k 1 k 1 mH 5 k 2 k 1 k i(t) Figure P7.15

Using the differential equation approach, find io(t) for t > 0 in the circuit in Fig. P7.16 and plot the response, including the time interval just prior to opening the switch. + + 4 2 H 12 6 V 6 12 V i o(t) t = 0 Figure P7.16

Use the differential equation approach to find o(t) for t > 0 in the circuit in Fig. P7.17 and plot the response, including the time interval just prior to opening the switch. + + o 8 k (t) 12 k 4 k 12 V 6 V 100 F + t = 0 Figure P7.17

Use the step-by-step technique to find io (t) for t > 0 in the network in Fig. P7.18. 12 V 6 k 2 k 4 k 200 F t = 0 i o(t) + Figure P7.18

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.19. 6 V 3 k 6 k 9 k 200 F t = 0 + + 24 V i o(t) Figure P7.19

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.20. 2 k 2 k 1 k 100 F t = 0 + 12 V 4 k i o(t) Figure P7.20

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.21. 6 2 H 24 V i o(t) + 6 t = 0 Figure P7.21

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.22. t = 0 12 k 6 k 12 mA 12 k 200 F i o(t) Figure P7.22

Use the step-by-step technique to find io(t) for t > 0 in the network in Fig. P7.23. + 4 k 4 k 200 F 4 k 4 k 12 V t = 0 i o(t) Figure P7.23

Use the step-by-step method to find o(t) for t > 0 in the network in Fig. P7.24. 12 mA 6 k 6 k 100 F + o(t) t = 0 Figure P7.24

Use the step-by-step method to find o(t) for t > 0 in the network in Fig. P7.25. 2 2 4 12 V 12 V o(t) + + + H1 3 t = 0 Figure P7.25

Find C (t) for t > 0 in the network in Fig. P7.26 using the step-by-step method. + C 2 k (t) 100 F 2 k 2 k 12 V + t = 0 Figure P7.26

Find o(t) for t > 0 in the network in Fig. P7.27 using the step-by-step method. + + 2 k 2 k 4 k 4 k 50 F 24 V 12 V + o(t) t = 0 Figure P7.27

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.28. t = 0 24 V 4 k 2 k 4 k 4 k 200 F + i o(t) Figure P7.28

Find o(t) for t > 0 in the network in Fig. P7.29 using the step-by-step technique. t = 0 24 V 4 k 2 k 4 k 4 k 200 F + o(t) + Figure P7.29

Find o(t) for t > 0 in the circuit in Fig. P7.30 using the stepby-step method. + 4 k 4 k 12 V 8 k 50 F + o(t) t = 0 Figure P7.30

Find io(t) for t > 0 in the network in Fig. P7.31 using the step-by-step method. 2 k 2 k 2 k 2 k 4 k t = 0 12 V 200 F + i o(t) Figure P7.31

Use the step-by-step technique to find o(t) for t > 0 in the circuit in Fig. P7.32. + 12 V 6 k 6 k 6 k 6 k t = 0 50 F + o(t) Figure P7.32

Find C(t) for t > 0 in the circuit in Fig. P7.33 using the step-by-step method. t = 0 10 10 5 5 6 A 20 0.5 F 5 C + (t) Figure P7.33

Use the step-by-step method to find io(t) for t > 0 in the network in Fig. P7.34. 12 V 6 V 2 k t = 0 1 k 2 k i o(t) + + 100 F Figure P7.34

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.35. o(t) + 6 V t = 0 1 k 1 k 1 k + 100 F Figure P7.35

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.36. o(t) + 9 k t = 0 3 k 6 k 2 k 4 k 12 mA 60 F Figure P7.36

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.37. 36 V 12 V 4 k t = 0 3 k 3 k 3 k 2 k i o(t) + + 150 F Figure P7.37

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.38. 200 F 3 k 6 k 18 V 3 k 6 V t = 0 + o(t) + + Figure P7.38

Use the step-by-step method to find io(t) for t > 0 in the network in Fig. P7.39. 2 k 1 k 24 V 3 k + t = 0 i o(t) 100 F 4 mA Figure P7.39

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.40. 1 k 12 V t = 0 2 k 3 k + i o(t) 100 F 12 mA Figure P7.40

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.41. 12 V 6 V 2 k t = 0 1 k 2 k i o(t) + + 100 F Figure P7.41

Use the step-by-step method to find o(t) for t > 0 in the network in Fig. P7.42. + 12 V 8 V 2 F 2 2 1 2 + o(t) + t = 0 Figure P7.42

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.43. 12 k 4 k 2 k 12 V 6 k o(t) + t = 0 100 F + Figure P7.43

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.44. 12 k 6 k 3 k 6 k 6 k 12 V o(t) + t = 0 100 F + Figure P7.44

Use the step-by-step method to find io(t) for t > 0 in the network in Fig. P7.45.

The switch in the circuit in Fig. P7.46 is opened at t = 0. Find o(t) for t > 0 using the step-by-step technique. 16 k 12 k 16 k o(t) 10 F 4 k + 12 V 24 V + + t = 0 Figure P7.46

Find io(t) for t > 0 in the circuit in Fig. P7.47. 2 k 4 k 6 mA 150 F 10 k i o(t) t = 0 Figure P7.47

Use the step-by-step method to find io(t) for t > 0 in the network in Fig. P7.48. + 6 k 6 k 12 V 6 k 50 F i o(t) t = 0 Figure P7.48

Find o(t) for t > 0 in the network in Fig. P7.49. 9 k 8 k 6 k 4 k 4 V 200 F o(t) + + t = 0 Figure P7.49

Find o(t) for t > 0 in the network in Fig. P7.50. 6 k 12 k 3 k 3 k 6 k 12 V 160 F o(t) + t = 0 + Figure P7.50

Find o(t) for t > 0 in the circuit in Fig. P7.51. 2 k 3 k 4 k 1 k 12 V 4 k 6 V 100 F o(t) + + + t = 0 Figure P7.51

Use the step-by-step technique to find io(t) for t > 0 in the network in Fig. P7.52. 4 k 2 k 2 k 2 k 12 k 2 k 100 F 4 mA 4 V i o(t) + t = 0 Figure P7.52

Find io(t) for t > 0 in the network in Fig. P7.53. 1 k 4 k 2 k 3 k 150 F 6 V i o(t) + t = 0 Figure P7.53

Use the step-by-step technique to find io(t) for t > 0 in the network in Fig. P7.54. 200 F 300 F 600 F 4 k 2 k 2 k 1 mA 12 V t = 0 io(t) + Figure P7.54

Find io(t) for t > 0 in the circuit in Fig. P7.55 using the step-by-step technique. 50 F 150 F 6 k 200 F 2 k 3 k 1 k 1 k 12 V 6 V i o(t) + + t = 0 Figure P7.55

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.56. 6 k 2 k 2 H 12 V + i o(t) t = 0 Figure P7.56

Use the step-by-step method to find o(t) for t > 0 in the network in Fig. P7.57. 30 V 3 3 3 H 3 + 3 o(t) + t = 0 Figure P7.57

The switch in the circuit in Fig. P7.58 is opened at t = 0. Find i(t) for t > 0. + + i(t) 2 9 V 3 4 6 V 0.5 H t = 0 Figure P7.58

The switch in the circuit in Fig. P7.59 is moved at t = 0. Find iL(t) for t > 0 using the step-by-step technique. 4 k 12 k 12 mH 2 k 6 k + 30 V t = 0 i L(t) + 12 V Figure P7.59

The switch in the circuit in Fig. P7.60 is moved at t = 0. Find iR(t) for t > 0 using the step-by-step technique. 5 k 6 k 5 mH 3 k 3 k + 20 V 12 V i R + (t) t = 0 Figure P7.60

The switch in the circuit in Fig. P7.61 is moved at t = 0. Find iL(t) for t > 0 using the step-by-step technique. 2 k 6 k 12 k 6 k 20 mH 6 k + 24 V 16 V t = 0 i L(t) + Figure P7.61

Find o(t) for t > 0 in the circuit in Fig. P7.62 using the step-by-step method. t = 0 6 3 6 4 2 H 6 A o(t) + Figure P7.62

The switch in the circuit in Fig. P7.63 has been closed for a long time and is opened at t = 0. Find i(t) for t > 0. t = 0 0.3 H 2 k 2 k 30 V 10 V 1 k 1 k + + i(t) Figure P7.63

The switch in the circuit in Fig. P7.64 has been closed for a long time and is moved at t = 0. Find io(t) for t > 0. 5 mA 2 k 2/3 H 5 k t = 0 5 k 5 k + 10 V i o(t) Figure P7.64

Find o(t) for t > 0 in the network in Fig. P7.65 using the step-by-step technique. 2 2 3 6 12 V 2 H o(t) + + t = 0 Figure P7.65

Find i(t) for t > 0 in the circuit in Fig. P7.66 using the step-by-step method. t = 0 0.5 H 6 30 V 2 2 30 V 5 2 i(t) + + t = 0 Figure P7.66

Find i(t) for t > 0 in the circuit in Fig. P7.67 using the step-by-step method. i(t) 6 6 105 V 12 10 15 1 H + t = 0 Figure P7.67

Use the step-by-step technique to find o(t) for t > 0 in the circuit in Fig. P7.68. 2 4 12 6 12 V 2 H o(t) + + t = 0 Figure P7.68

Find o(t) for t > 0 in the circuit in Fig. P7.69 using the step-by-step method. + 6 2 3 3 1 24 V + + 12 V t = 0 5 H o(t) Figure P7.69

Use the step-by-step method to find io(t) for t > 0 in the circuit in Fig. P7.70. 4 mH 12 mH 10 mA 10 k 10 k i o(t) t = 0 Figure P7.70

Find io(t) for t > 0 in the network in Fig. P7.71. t = 0 12 V 12 6 V 6 3 3 0.2 H i o(t) + + Figure P7.71

Find io(t) for t > 0 in the circuit in Fig. P7.72. 6 V 2 H + t = 0 i o(t) 4 2 6 Figure P7.72

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.73. o(t) + 12 V 4 A t = 0 + 6 3 2 H 3 Figure P7.73

Use the step-by-step method to find io(t) for t > 0 in the network in Fig. P7.74. 1 H t = 0 6 4 3 2 + 12 V i o(t) 4 A Figure P7.74

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.75. 4 H 6 V 4 A 6 6 6 o(t) + t = 0 + Figure P7.75

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.76. 4 H 2 A 6 6 6 o(t) + t = 0 + 24 V Figure P7.76

Use the step-by-step method to find o(t) for t > 0 in the circuit in Fig. P7.77. 6 3 2 2 12 V 2 2 H 2 A o(t) i o(t) + + t = 0 t = 0 Figure P7.77

Use the step-by-step method to find io(t) for t > 0 in the network in Fig. P7.78. 3 H 3 A 2 H 12 V 4 4 6 12 t = 0 i o(t) + Figure P7.78

Use the step-by-step method to find o(t) for t > 0 in the network in Fig. P7.79. 6 k 2 k 4 k 6 F 6 F 6 F 12 V t = 0 + o(t) + Figure P7.79

Find iL(t) for t > 0 in the circuit in Fig. P7.80 using the step-by-step method. 10 V 4 A 2 2 H 12 t = 0 51 (t) 1 (t) i L(t) + + + Figure P7.80

Find io(t) for t > 0 in the network in Fig. P7.81 using the step-by-step method. 10 mA 2 k 2 k 2 k 2 k 2 mH 2i A t = 0 i o(t) i A Figure P7.81

Use the step-by-step technique to find o(t) for t > 0 in the network in Fig. P7.82. 3 k 3 k 6 k 2 k 2000i A i A 24 V 6 V 200 F t = 0 o(t) + + + + Figure P7.82

The current source in the network in Fig. P7.83a is defined in Fig. P7.83b. The initial voltage across the capacitor must be zero. (Why?) Determine the current io(t) for t > 0. 2 2 F 2 (a) (b) 2 i o(t) i(t) t (s) 6 0 4.5 i(t) (A) Figure P7.83

Determine the equation for the voltage o(t) for t > 0 in Fig. P7.84a when subjected to the input pulse shown in Fig. P7.84b. 2 k 6 k 3 k 200 F (a) (b) t (s) 12 10 + o (t) (t) + (t) (V) Figure P7.84

The voltage (t) shown in Fig. P7.85a is given by the graph shown in Fig. P7.85b. If iL(0) = 0, answer the following questions: (a) How much energy is stored in the inductor at t = 3 s? (b) How much power is supplied by the source at t = 4 s? (c) What is i(t = 6 s)? (d) How much power is absorbed by the inductor at t = 3 s? (t) (t) 2 i(t) i L(t) + 2 H t (s) 10 V 10 V 2 5 (a) (b) Figure P7.85

Given that C1(0) = 10 V and C2(0) = 20 V in the circuit in Fig. P7.86, find i(0+). 4 2 F 4 F t = 0 i(t) C1 (t) + C2(t) + Figure P7.86

In the circuit in Fig. P7.87, R(t) = 100e400t V for t < 0. Find R(t) for t > 0. 25 100 R(t) L + t = 0 Figure P7.87

Find the output voltage o(t) in the network in Fig. P7.88 if the input voltage is i (t) = 5(u(t) u(t 0.05)) V. 100 k 1 F + o(t) + i (t) Figure P7.88

In the network in Fig. P7.89, find i(t) for t > 0. If C1(0) = 10 V, calculate C2(0). 0.6 F 0.3 F i(t) C1(t) C2 10e (t) 5t u(t) V + + + Figure P7.89

The switch in the circuit in Fig. P7.90 is closed at t = 0. If i1(0) = 2 A, determine i2(0+), R(0+), and i1(t = ). 2 H 4 3 H + i 1(t) i 2(t) R(t) t = 0 Figure P7.90

Given that i(t) = 2.5 + 1.5e4t A for t > 0 in the circuit in Fig. P7.91, find R1, R2, and L. i(t) L R1 + R2 20 V t = 0 Figure P7.91

The switch in the circuit in Fig. P7.92 has been closed for a long time and is opened at t = 0. If C(t) = 20 8e0.05t V, find R1, R2, and C. C(t) R2 2 A R1 C + t = 0 Figure P7.92

Given that i(t) = 13.33et 8.33e0.5t A for t > 0 in the network in Fig. P7.93, find the following: (a) c(0). (b) c(t = 1 s). (c) the capacitance C. C(t) i(t) C 2 4 + + 40e tu(t) V Figure P7.93

For the circuit in Fig. P7.94, choose R1, R2, and L such that o(t) = 20 e 106 t V and the current i1 never exceeds 1 A. L R2 10 V R1 t = 0 i1 + + o(t) Figure P7.94

For the network in Fig. P7.95, choose C so the time constant will be 120 s for t > 0. 7 k 4 k 2 k 18 k C S 23 k o(t) + t = 0 + Figure P7.95

The differential equation that describes the current io(t) in a network is d2 io (t) dt2 + 6 dio (t) dt + 4io(t) = 0 Find (a) the characteristic equation of the network. (b) the networks natural frequencies. (c) the expression for io(t).

The terminal current in a network is described by the equation d2 io (t) dt2 + 8 dio (t) dt + 16io(t) = 0 Find (a) the characteristic equation of the network, (b) the networks natural frequencies, and (c) the equation for io(t).

The voltage 1(t) in a network is defined by the equation d2 1 (t) dt2 + 2 [ d1 (t) dt ] + 51(t) = 0 Find (a) the characteristic equation of the network. (b) the circuits natural frequencies. (c) the expression for 1(t).

The output voltage of a circuit is described by the differential equation d2 o (t) dt2 + 8 do (t) dt + 10o(t) = 0 Find (a) the characteristic equation of the circuit, (b) the networks natural frequencies, and (c) the equation for o(t).

A parallel RLC circuit contains a resistor R = 1 and an inductor L = 2 H. Select the value of the capacitor so that the circuit is critically damped.

A series RLC circuit contains a resistor R = 2 and a capacitor C = 12 F. Select the value of the inductor so that the circuit is critically damped.

The parameters for a parallel RLC circuit are R = 1 , L = 12 H, and C = 12 F. Determine the type of damping exhibited by the circuit.

In the critically damped circuit shown in Fig. P7.103, the initial conditions on the storage elements are iL(0) = 2 A and C(0) = 5 V. Determine the voltage (t). 0.01 F (t) 10 4 H + C(0) + i L(0) Figure P7.103

Find C(t) for t > 0 in the circuit in Fig. P7.104. 1 A 8 0.04 F 1 H + C(t) i L(t) t = 0 Figure P7.104

Find o(t) for t > 0 in the circuit in Fig. P7.105 and plot the response, including the time interval just prior to closing the switch. 2 12 V 1 H 10 1 400 + o(t) + F t = 0 Figure P7.105

The switch in the circuit in Fig P7.106 has been closed for a long time and is opened at t = 0. Find i(t) for t > 0. 6 1.5 0.1 F 1.25 H 12 V 6 V i(t) + + t = 0 Figure P7.106

In the circuit shown in Fig. P7.107, find (t) > 0. 8 V t = 0 + (t) + 12 5 H 1 12 2 F Figure P7.107

Find o(t) for t > 0 in the network in Fig. P7.108 and plot the response, including the time interval just prior to moving the switch. 10 k 100 mH 10 k 5 k 100 V 2 k t = 0 10 F + o(t) + Figure P7.108

Find o(t) for t > 0 in the circuit in Fig. P7.109 and plot the response, including the time interval just prior to moving the switch. 6.25 F 1 k 2 k t = 0 1 mH o + (t) + 12 V Figure P7.109

For the underdamped circuit shown in Fig. P7.110, determine the voltage (t) if the initial conditions on the storage elements are iL(0) = 1 A and C (0) = 10 V. 2 H 5 + + 1 40 F C(0) i L(0) (t) Figure P7.110

Find C(t) for t > 0 in the circuit in Fig. P7.111 if C(0) = 0. 1 k 100 mH + 12 V 1 F C(t) + t = 0 Figure P7.111

The switch in the circuit in Fig. P7.112 has been closed for a long time and is opened at t = 0. Find i(t) for t > 0. 5 5 5 1 H 20 V 0.04 F 10 V i(t) + + t = 0 Figure P7.112

Find o(t) for t > 0 in the circuit in Fig. P7.113 and plot the response, including the time interval just prior to closing the switch.

Given the network in Fig. P7.114, plot o(t) over a 10-s interval starting at t = 0, using a 100-ms step size. 2 4 6 2 A 2 H 12 V 2 1 F + o(t) + t = 0 Figure P7.114

Given the network in Fig. P7.115, plot o(t) over a 10-s interval starting at t = 0, using a 100-ms step size. 5 k 20 k 10 k 10 mH 6 V 12 V + o(t) + + 50 F t = 0 Figure P7.115

Given the network in Fig. P7.116a and the input voltage shown in Fig. P7.116b, plot the voltage o(t) over the interval 0 t 4 s, using a 20-ms step size. 1 M 2 mH 100 k 0.1 F (a) (b) t(s) 10 10 + o (t) in(t) in(t) V + Figure P7.116

Design a parallel RLC circuit with R 1 k that has the characteristic equation s 2 + 4 107 s + 3 1014 = 0

Design a parallel RLC circuit with R 1 k that has the characteristic equation s 2 + 4 107 s + 4 1014 = 0