Alexander_2

8.10 Duality 351Construct the dual of the circuit in Fig. 8.44. Example 8.14Solution: 2Ω t=0 10 mFAs shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two 6 V +− 2Hmeshes and also the ground node 0 for the dual circuit. We draw a linebetween one node and another crossing an element. We replace the line Figure 8.44joining the nodes by the duals of the elements which it crosses. For For Example 8.14.example, a line between nodes 1 and 2 crosses a 2-H inductor, and weplace a 2-F capacitor (an inductor’s dual) on the line. A line betweennodes 1 and 0 crossing the 6-V voltage source will contain a 6-Acurrent source. By drawing lines crossing all the elements, we constructthe dual circuit on the given circuit as in Fig. 8.45(a). The dual circuitis redrawn in Fig. 8.45(b) for clarity. t=0 t=00.5 Ω 2Ω 1 2F 2 6 V +− 2 H 10 mF 1 2F 2 10 mH 6A 0.5 Ω t=0 10 mH 0 6A 0 (a) (b)Figure 8.45(a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn.Draw the dual circuit of the one in Fig. 8.46. Practice Problem 8.14Answer: See Fig. 8.47. 3H 50 mA 3F 4H 50 mV +− 0.1 Ω 4F 10 Ω Figure 8.46 Figure 8.47 For Practice Prob. 8.14. Dual of the circuit in Fig. 8.46.Obtain the dual of the circuit in Fig. 8.48. Example 8.15Solution:The dual circuit is constructed on the original circuit as in Fig. 8.49(a).We first locate nodes 1 to 3 and the reference node 0. Joining nodes1 and 2, we cross the 2-F capacitor, which is replaced by a 2-Hinductor.

352 Chapter 8 Second-Order Circuits 5H 10 V +− i1 2 F i2 20 Ω i3 3A Figure 8.48 For Example 8.15. Joining nodes 2 and 3, we cross the 20-⍀ resistor, which is replaced by a 210-Æ resistor. We keep doing this until all the elements are crossed. The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b). 5F 5H 1 2H 2 1 Ω 3 10 A 2010 V +− 1 2 3 3A 2H 2 F 20 Ω 1 Ω 5F − 3V 20 0 + 0 − 3V +10 A (a) (b)Figure 8.49For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn. To verify the polarity of the voltage source and the direction of the current source, we may apply mesh currents i1, i2, and i3 (all in the clockwise direction) in the original circuit in Fig. 8.48. The 10-V voltage source produces positive mesh current i1, so that its dual is a 10-A current source directed from 0 to 1. Also, i3 ϭ Ϫ3 A in Fig. 8.48 has as its dual v3 ϭ Ϫ3 V in Fig. 8.49(b).Practice Problem 8.15 For the circuit in Fig. 8.50, obtain the dual circuit. Answer: See Fig. 8.51. 5Ω 1 Ω 3 0.2 F 4H 0.2 H 4F2A 3Ω +− 20 V 2 V +− 1 Ω 20 A 5Figure 8.50 Figure 8.51For Practice Prob. 8.15. Dual of the circuit in Fig. 8.50.

8.11 Applications 3538.11 Applications Example 8.16Practical applications of RLC circuits are found in control and com-munications circuits such as ringing circuits, peaking circuits, resonantcircuits, smoothing circuits, and filters. Most of these circuits cannotbe covered until we treat ac sources. For now, we will limit ourselvesto two simple applications: automobile ignition and smoothing circuits.8.11.1 Automobile Ignition SystemIn Section 7.9.4, we considered the automobile ignition system as acharging system. That was only a part of the system. Here, we con-sider another part—the voltage generating system. The system is mod-eled by the circuit shown in Fig. 8.52. The 12-V source is due to thebattery and alternator. The 4-⍀ resistor represents the resistance of thewiring. The ignition coil is modeled by the 8-mH inductor. The 1-mFcapacitor (known as the condenser to automechanics) is in parallel withthe switch (known as the breaking points or electronic ignition). In thefollowing example, we determine how the RLC circuit in Fig. 8.52 isused in generating high voltage. t=0 4 Ω 1 ␮F + vC − i 8 mH 12 V + vL − Spark plug Ignition coil Figure 8.52 Automobile ignition circuit.Assuming that the switch in Fig. 8.52 is closed prior to t ϭ 0Ϫ, findthe inductor voltage vL for t 7 0.Solution:If the switch is closed prior to t ϭ 0Ϫ and the circuit is in steady state,then i(0Ϫ) ϭ 12 ϭ 3 A, vC(0Ϫ) ϭ 0 4At t ϭ 0ϩ, the switch is opened. The continuity conditions require that i(0ϩ) ϭ 3 A, vC(0ϩ) ϭ 0 (8.16.1)We obtain di(0ϩ)͞dt from vL(0ϩ). Applying KVL to the mesh at t ϭ 0ϩyields Ϫ12 ϩ 4i(0ϩ) ϩ vL(0ϩ) ϩ vC(0ϩ) ϭ 0 Ϫ12 ϩ 4 ϫ 3 ϩ vL(0ϩ) ϩ 0 ϭ 0 1 vL(0ϩ) ϭ 0

354 Chapter 8 Second-Order CircuitsHence, di(0ϩ) ϭ vL(0ϩ) ϭ 0 (8.16.2) dt LAs t S ϱ, the system reaches steady state, so that the capacitor actslike an open circuit. Then i(ϱ) ϭ 0 (8.16.3)If we apply KVL to the mesh for t 7 0, we obtain Ύdi 1 t 12 ϭ Ri ϩ L ϩ i dt ϩ vC(0) dt C 0Taking the derivative of each term yields d 2i R di i (8.16.4) dt 2 ϩ L dt ϩ LC ϭ 0We obtain the form of the transient response by following the procedurein Section 8.3. Substituting R ϭ 4 ⍀, L ϭ 8 mH, and C ϭ 1 mF, we get a ϭ R ϭ 250, ␻0 ϭ 1 ϭ 1.118 ϫ 104 2L 2LCSince a 6 ␻0, the response is underdamped. The damped naturalfrequency is ␻d ϭ 2␻02 Ϫ a2 Ӎ ␻0 ϭ 1.118 ϫ 104The form of the transient response is it(t) ϭ eϪa(A cos ␻dt ϩ B sin ␻dt) (8.16.5)where A and B are constants. The steady-state response is iss (t) ϭ i(ϱ) ϭ 0 (8.16.6)so that the complete response is (8.16.7)i(t) ϭ it(t) ϩ iss (t) ϭ eϪ250t(A cos 11,180t ϩ B sin 11,180t)We now determine A and B. i(0) ϭ 3 ϭ A ϩ 0 1 A ϭ 3Taking the derivative of Eq. (8.16.7),di ϭ Ϫ250eϪ250t(A cos 11,180t ϩ B sin 11,180t)dt ϩ eϪ250t(Ϫ11,180A sin 11,180t ϩ 11,180B cos 11,180t)Setting t ϭ 0 and incorporating Eq. (8.16.2), 0 ϭ Ϫ250A ϩ 11,180B 1 B ϭ 0.0671Thus, i(t) ϭ eϪ250t(3 cos 11,180t ϩ 0.0671 sin 11,180t) (8.16.8)The voltage across the inductor is then vL(t) ϭ di ϭ Ϫ268eϪ250t sin 11,180t (8.16.9) L dt

8.11 Applications 355This has a maximum value when sine is unity, that is, at 11,180t0 ϭp͞2 or t0 ϭ 140.5 ms. At time ϭ t0, the inductor voltage reaches itspeak, which is vL(t0) ϭ Ϫ268eϪ250t0 ϭ Ϫ259 V (8.16.10)Although this is far less than the voltage range of 6000 to 10,000 Vrequired to fire the spark plug in a typical automobile, a device knownas a transformer (to be discussed in Chapter 13) is used to step up theinductor voltage to the required level.In Fig. 8.52, find the capacitor voltage vC for t 7 0. Practice Problem 8.16Answer: 12 Ϫ 12eϪ250t cos 11,180t ϩ 267.7eϪ250t sin 11,180t V.8.11.2 Smoothing CircuitsIn a typical digital communication system, the signal to be transmitted p(t) D/A vs(t) Smoothing v0(t)is first sampled. Sampling refers to the procedure of selecting samples circuitof a signal for processing, as opposed to processing the entire signal.Each sample is converted into a binary number represented by a series Figure 8.53of pulses. The pulses are transmitted by a transmission line such as a A series of pulses is applied to the digital-coaxial cable, twisted pair, or optical fiber. At the receiving end, the to-analog (D/A) converter, whose outputsignal is applied to a digital-to-analog (D/A) converter whose output is is applied to the smoothing circuit.a “staircase” function, that is, constant at each time interval. In orderto recover the transmitted analog signal, the output is smoothed by let-ting it pass through a “smoothing” circuit, as illustrated in Fig. 8.53.An RLC circuit may be used as the smoothing circuit.The output of a D/A converter is shown in Fig. 8.54(a). If the RLC Example 8.17circuit in Fig. 8.54(b) is used as the smoothing circuit, determine theoutput voltage vo(t).vs10 1 1Ω 1H 34 2+ vs +− 1 F v00−–2 t (s) 00 (a) (b)Figure 8.54For Example 8.17: (a) output of a D/A converter, (b) an RLCsmoothing circuit.Solution:This problem is best solved using PSpice. The schematic is shown inFig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise

356 Chapter 8 Second-Order CircuitsT1=0 V1=0 V V 10 VT2=0.001 V2=4 R1 L1T3=1 V3=4 1 1H 5VT4=1.001 V4=10T5=2 V5=10 + 1 C1 0VT6=2.001 V6=−2 − V1T7=3 V7=−2 0 −5 VT8=3.001 V8=0 0 s 2.0 s 4.0 s 6.0 s V(V1:+) V(C1:1) (a) Time (b)Figure 8.55For Example 8.17: (a) schematic, (b) input and output voltages. linear function. The attributes of V1 are set as T1 ϭ 0, V1 ϭ 0, T2 ϭ 0.001, V2 ϭ 4, T3 ϭ 1, V3 ϭ 4, and so on. To be able to plot both input and output voltages, we insert two voltage markers as shown. We select Analysis/Setup/Transient to open up the Transient Analysis dialog box and set Final Time as 6 s. Once the schematic is saved, we select Analysis/Simulate to run and obtain the plots shown in Fig. 8.55(b).Practice Problem 8.17 Rework Example 8.17 if the output of the D/A converter is as shown in Fig. 8.56. Answer: See Fig. 8.57. vs 8.0 V 8 7 4.0 V 0V 0 1 2 3 4 t (s) −4.0 V –1 0 s 2.0 s 4.0 s 6.0 s –3 V(V1:+) V(C1:1) Figure 8.56 Time For Practice Prob. 8.17. Figure 8.57 Result of Practice Prob. 8.17. 8.12 Summary 1. The determination of the initial values x(0) and dx(0)͞dt and final value x(ϱ) is crucial to analyzing second-order circuits. 2. The RLC circuit is second-order because it is described by a second-order differential equation. Its characteristic equation is

Review Questions 357 s2 ϩ 2a s ϩ ␻02 ϭ 0, where a is the damping factor and ␻0 is the undamped natural frequency. For a series circuit, a ϭ R͞2L, for a parallel circuit a ϭ 1͞2RC, and for both cases ␻0 ϭ 1͞0 1LC.3. If there are no independent sources in the circuit after switching (or sudden change), we regard the circuit as source-free. The com- plete solution is the natural response.4. The natural response of an RLC circuit is overdamped, under- damped, or critically damped, depending on the roots of the char- acteristic equation. The response is critically damped when the roots are equal (s1 ϭ s2 or a ϭ ␻0), overdamped when the roots are real and unequal (s1 s2 or a 7 ␻0), or underdamped when the roots are complex conjugate (s1 ϭ s*2 or a 6 ␻0).5. If independent sources are present in the circuit after switching, the complete response is the sum of the transient response and the steady-state response.6. PSpice is used to analyze RLC circuits in the same way as for RC or RL circuits.7. Two circuits are dual if the mesh equations that describe one circuit have the same form as the nodal equations that describe the other. The analysis of one circuit gives the analysis of its dual circuit.8. The automobile ignition circuit and the smoothing circuit are typ- ical applications of the material covered in this chapter. Review Questions 8.4 If the roots of the characteristic equation of an RLC circuit are Ϫ2 and Ϫ3, the response is:8.1 For the circuit in Fig. 8.58, the capacitor voltage at (a) (A cos 2t ϩ B sin 2t)eϪ3t t ϭ 0Ϫ ( just before the switch is closed) is: (b) (A ϩ 2Bt)eϪ3t (a) 0 V (b) 4 V (c) 8 V (d) 12 V (c) AeϪ2t ϩ BteϪ3t (d) AeϪ2t ϩ BeϪ3t t=0 where A and B are constants. 2Ω 4Ω 8.5 In a series RLC circuit, setting R ϭ 0 will produce:12 V +− 1H 2F (a) an overdamped responseFigure 8.58 (b) a critically damped responseFor Review Questions 8.1 and 8.2. (c) an underdamped response (d) an undamped response 8.2 For the circuit in Fig. 8.58, the initial inductor (e) none of the above current (at t ϭ 0) is: 8.6 A parallel RLC circuit has L ϭ 2 H and C ϭ 0.25 F. (a) 0 A (b) 2 A (c) 6 A (d) 12 A The value of R that will produce unity damping factor is: 8.3 When a step input is applied to a second-order (a) 0.5 ⍀ (b) 1 ⍀ (c) 2 ⍀ (d) 4 ⍀ circuit, the final values of the circuit variables are found by: 8.7 Refer to the series RLC circuit in Fig. 8.59. What kind of response will it produce? (a) Replacing capacitors with closed circuits and inductors with open circuits. (a) overdamped (b) underdamped (b) Replacing capacitors with open circuits and (c) critically damped inductors with closed circuits. (d) none of the above (c) Doing neither of the above.

358 Chapter 8 Second-Order Circuits 1Ω 1H R L 1F vs +− C L R is CFigure 8.59 (a) (b)For Review Question 8.7. R C18.8 Consider the parallel RLC circuit in Fig. 8.60. What R1 R2 vs +− L C2 type of response will it produce? C2 is (a) overdamped C1 (b) underdamped (c) critically damped (c) (d) (d) none of the above R1 R2 R1 C vs +− C L is 1Ω 1H 1F L R2Figure 8.60 (f)For Review Question 8.8. (e) Figure 8.61 For Review Question 8.9. 8.10 In an electric circuit, the dual of resistance is:8.9 Match the circuits in Fig. 8.61 with the following (a) conductance (b) inductance items: (c) capacitance (d) open circuit (e) short circuit (i) first-order circuit (ii) second-order series circuit Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b, (iii) second-order parallel circuit 8.9 (i)-c, (ii)-b, e, (iii)-a, (iv)-d, f, 8.10a. (iv) none of the aboveProblemsSection 8.2 Finding Initial and Final Values 8.2 Using Fig. 8.63, design a problem to help other students better understand finding initial and final 8.1 For the circuit in Fig. 8.62, find: values. (a) i(0ϩ) and v(0ϩ), (b) di(0ϩ)͞dt and dv(0ϩ)͞dt, iR R1 R2 (c) i(ϱ) and v(ϱ). t=0 6Ω 4Ω iC iL C L 12 V +− i R3 2 H 0.4 FFigure 8.62 + v +−For Prob. 8.1. v t=0 − Figure 8.63 For Prob. 8.2.

Problems 3598.3 Refer to the circuit shown in Fig. 8.64. Calculate: Rs R (a) iL(0ϩ), vC(0ϩ), and vR(0ϩ), Vsu(t) +− C + vR − + L (b) diL(0ϩ)͞dt, dvC(0ϩ)͞dt, and dvR(0ϩ)͞dt, vL (c) iL(ϱ), vC(ϱ), and vR(ϱ). − 40 Ω Figure 8.67 For Prob. 8.6. + IL + vC 1 F Section 8.3 Source-Free Series RLC Circuit vR 10 Ω 4 − − 2u(t) A 1 HFigure 8.64 8For Prob. 8.3. +− 10 V 8.7 A series RLC circuit has R ϭ 20 k⍀, L ϭ 0.2 mH, and C ϭ 5 mF. What type of damping is exhibited8.4 In the circuit of Fig. 8.65, find: by the circuit? (a) v(0ϩ) and i(0ϩ), (b) dv(0ϩ)͞dt and di(0ϩ)͞dt, 8.8 Design a problem to help other students better (c) v(ϱ) and i(ϱ). understand source-free RLC circuits. 8.9 The current in an RLC circuit is described by d 2i di ϭ ϩ 10 ϩ 25i 0 dt 2 dt If i(0) ϭ 10 A and di(0)͞dt ϭ 0, find i(t) for t 7 0. 3 Ω 0.25 H 8.10 The differential equation that describes the voltage in an RLC network is d 2v dv ϩ ϭ i ϩ 5 4v 0 0.1 F + dt 2 dt4u(–t) V +− v 5Ω 4u(t) A − Given that v(0) ϭ 0, dv(0)͞dt ϭ 10 V/s, obtain v(t).Figure 8.65For Prob. 8.4. 8.11 The natural response of an RLC circuit is described by the differential equation d 2v dv ϩ 2 ϩ v ϭ 0 dt 2 dt8.5 Refer to the circuit in Fig. 8.66. Determine: for which the initial conditions are v(0) ϭ 10 V and (a) i(0ϩ) and v(0ϩ), dv(0)͞dt ϭ 0. Solve for v(t). (b) di(0ϩ)͞dt and dv(0ϩ)͞dt, (c) i(ϱ) and v(ϱ). 8.12 If R ϭ 50 ⍀, L ϭ 1.5 H, what value of C will make an RLC series circuit: 1H (a) overdamped,4u(t) A i 1 F + (b) critically damped, 4Ω 4 (c) underdamped? 6Ω v − 8.13 For the circuit in Fig. 8.68, calculate the value of R needed to have a critically damped response.Figure 8.66 60 Ω 0.01 F 4 HFor Prob. 8.5. R 8.6 In the circuit of Fig. 8.67, find: Figure 8.68 (a) vR(0ϩ) and vL(0ϩ), For Prob. 8.13. (b) dvR(0ϩ)͞dt and dvL(0ϩ)͞dt, (c) vR(ϱ) and vL(ϱ).

360 Chapter 8 Second-Order Circuits8.14 The switch in Fig. 8.69 moves from position A to 100 V + 5Ω 1F position B at t ϭ 0 (please note that the switch must − t=0 connect to point B before it breaks the connection at A, a make-before-break switch). Let v(0) ϭ 0, find 1 Ω 0.25 H v(t) for t 7 0. 30 Ω A t=0 4H Figure 8.72 For Prob. 8.18.80 V −+ B + 0.25 F v (t) 10 Ω − 8.19 Obtain v(t) for t 7 0 in the circuit of Fig. 8.73.Figure 8.69 10 Ω +For Prob. 8.14. t=0 v 1F 8.15 The responses of a series RLC circuit are 120 V +− − vC(t) ϭ 30 Ϫ 10eϪ20t ϩ 30eϪ10t V iL(t) ϭ 40eϪ20t Ϫ 60eϪ10t mA 4H where vC and iL are the capacitor voltage and Figure 8.73 inductor current, respectively. Determine the values For Prob. 8.19. of R, L, and C. 8.20 The switch in the circuit of Fig. 8.74 has been closed 8.16 Find i(t) for t 7 0 in the circuit of Fig. 8.70. for a long time but is opened at t ϭ 0. Determine i(t) for t 7 0. t=0 10 Ω 60 Ω 40 Ω i(t) i(t) 1 H 2Ω 1 mF 230 V +− 2.5 H 30 V t=0 +−Figure 8.70 1 FFor Prob. 8.16. 4 8.17 In the circuit of Fig. 8.71, the switch instantaneously Figure 8.74 moves from position A to B at t ϭ 0. Find v(t) for all For Prob. 8.20. t Ն 0. *8.21 Calculate v(t) for t 7 0 in the circuit of Fig. 8.75. A t = 0 0.25 H 15 Ω B 12 Ω 6Ω t=05A 10 Ω 0.04 F + 60 Ω + 25 Ω 4Ω v (t) v – − 24 V +− 3HFigure 8.71 1 FFor Prob. 8.17. 27 8.18 Find the voltage across the capacitor as a function of Figure 8.75 time for t 7 0 for the circuit in Fig. 8.72. Assume For Prob. 8.21. steady-state conditions exist at t ϭ 0Ϫ. * An asterisk indicates a challenging problem.