Alexander_2

Comprehensive Problems 251*6.77 The output vo of the op amp circuit in Fig. 6.92(a) is 6.79 Design an analog computer circuit to solve the shown in Fig. 6.92(b). Let Ri ϭ Rf ϭ 1 M⍀ and following ordinary differential equation. C ϭ 1 mF. Determine the input voltage waveform dy(t) and sketch it. ϩ 4y(t) ϭ f(t) Rf dt C where y(0) ϭ 1 V. − 6.80 Figure 6.93 presents an analog computer designed + to solve a differential equation. Assuming f(t) is known, set up the equation for f (t). Ri + 1 ␮F 1 ␮Fvi +− vo − 1 MΩ 1 MΩ 1 MΩ − 500 kΩ − + − + + (a) vo(t) 100 kΩvo4 100 kΩ 200 kΩ −f (t) − +0 Figure 6.93 1 2 3 4 t (s) For Prob. 6.80. −4 (b)Figure 6.92 6.81 Design an analog computer to simulate the followingFor Prob. 6.77. equation: d 2v6.78 Design an analog computer to simulate dt 2 ϩ 5v ϭ Ϫ2f (t) d 2vo ϩ 2 dvo ϩ vo ϭ 10 sin 2t 6.82 Design an op amp circuit such that dt 2 dt Ύvo ϭ 10vs ϩ 2 vs dtwhere v0(0) ϭ 2 and v0¿(0) ϭ 0. where vs and vo are the input voltage and output voltage, respectively. Comprehensive Problems 6.84 An 8-mH inductor is used in a fusion power6.83 Your laboratory has available a large number of experiment. If the current through the inductor is 10-mF capacitors rated at 300 V. To design a i(t) ϭ 5 sin2 p t mA, t 7 0, find the power being capacitor bank of 40 mF rated at 600 V, how many delivered to the inductor and the energy stored in it 10-mF capacitors are needed and how would you at t ϭ 0.5 s. connect them?

252 Chapter 6 Capacitors and Inductors6.85 A square-wave generator produces the voltage i (A)waveform shown in Fig. 6.94(a). What kind of a 4circuit component is needed to convert the voltagewaveform to the triangular current waveform shownin Fig. 6.94(b)? Calculate the value of thecomponent, assuming that it is initially uncharged. 0 1 2 3 4 t (ms) (b)v (V) Figure 6.94 5 For Prob. 6.85. 0 6.86 An electric motor can be modeled as a series 1 2 3 4 t (ms) combination of a 12-⍀ resistor and 200-mH inductor.−5 If a current i(t) ϭ 2teϪ10t A flows through the series (a) combination, find the voltage across the combination.

First-Order Circuits chapter 7We live in deeds, not years; in thoughts, not breaths; in feelings, not infigures on a dial. We should count time in heart-throbs. He most liveswho thinks most, feels the noblest, acts the best. —P. J. BaileyEnhancing Your CareerCareers in Computer Engineering Computer design of very large scaleElectrical engineering education has gone through drastic changes in integrated (VLSI) circuits.recent decades. Most departments have come to be known as Department Courtesy Brian Fast, Cleveland Stateof Electrical and Computer Engineering, emphasizing the rapid changes Universitydue to computers. Computers occupy a prominent place in modern soci-ety and education. They have become commonplace and are helping tochange the face of research, development, production, business, and enter-tainment. The scientist, engineer, doctor, attorney, teacher, airline pilot,businessperson—almost anyone benefits from a computer’s abilities tostore large amounts of information and to process that information in veryshort periods of time. The internet, a computer communication network,is essential in business, education, and library science. Computer usagecontinues to grow by leaps and bounds. An education in computer engineering should provide breadth in soft-ware, hardware design, and basic modeling techniques. It should includecourses in data structures, digital systems, computer architecture, micro-processors, interfacing, software engineering, and operating systems. Electrical engineers who specialize in computer engineering findjobs in computer industries and in numerous fields where computersare being used. Companies that produce software are growing rapidlyin number and size and providing employment for those who are skilledin programming. An excellent way to advance one’s knowledge ofcomputers is to join the IEEE Computer Society, which sponsorsdiverse magazines, journals, and conferences. 253

254 Chapter 7 First-Order Circuits iC + iR 7.1 Introduction C vR Now that we have considered the three passive elements (resistors, − capacitors, and inductors) and one active element (the op amp) indi- vidually, we are prepared to consider circuits that contain various com-Figure 7.1 binations of two or three of the passive elements. In this chapter, weA source-free RC circuit. shall examine two types of simple circuits: a circuit comprising a resis- tor and capacitor and a circuit comprising a resistor and an inductor. A circuit response is the manner in These are called RC and RL circuits, respectively. As simple as these which the circuit reacts to an circuits are, they find continual applications in electronics, communi- excitation. cations, and control systems, as we shall see. We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws, as we did for resistive circuits. The only difference is that applying Kirchhoff’s laws to purely resistive circuits results in algebraic equations, while applying the laws to RC and RL circuits pro- duces differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyz- ing RC and RL circuits are of the first order. Hence, the circuits are collectively known as first-order circuits. A first-order circuit is characterized by a first-order differential equation. In addition to there being two types of first-order circuits (RC and RL), there are two ways to excite the circuits. The first way is by ini- tial conditions of the storage elements in the circuits. In these so-called source-free circuits, we assume that energy is initially stored in the capacitive or inductive element. The energy causes current to flow in the circuit and is gradually dissipated in the resistors. Although source- free circuits are by definition free of independent sources, they may have dependent sources. The second way of exciting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. (In later chapters, we shall consider sinu- soidal and exponential sources.) The two types of first-order circuits and the two ways of exciting them add up to the four possible situa- tions we will study in this chapter. Finally, we consider four typical applications of RC and RL cir- cuits: delay and relay circuits, a photoflash unit, and an automobile ignition circuit. 7.2 The Source-Free RC Circuit A source-free RC circuit occurs when its dc source is suddenly dis- connected. The energy already stored in the capacitor is released to the resistors. Consider a series combination of a resistor and an initially charged capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combinations of resistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage

7.2 The Source-Free RC Circuit 255v(t) across the capacitor. Since the capacitor is initially charged, wecan assume that at time t ϭ 0, the initial voltage is v(0) ϭ V0 (7.1)with the corresponding value of the energy stored as w(0) ϭ 1 CV 2 (7.2) 2 0Applying KCL at the top node of the circuit in Fig. 7.1 yields iC ϩ iR ϭ 0 (7.3)By definition, iC ϭ C dv͞dt and iR ϭ v͞R. Thus, (7.4a) C dv ϩ v ϭ 0 dt Ror dv ϩ v ϭ 0 (7.4b) dt RCThis is a first-order differential equation, since only the first derivativeof v is involved. To solve it, we rearrange the terms as dv ϭ Ϫ1 dt (7.5) v RCIntegrating both sides, we get ln v ϭ Ϫ t ϩ ln A RCwhere ln A is the integration constant. Thus, ln v ϭ Ϫ t (7.6) A RCTaking powers of e produces v(t) ϭ AeϪt͞RCBut from the initial conditions, v(0) ϭ A ϭ V0. Hence, (7.7) v(t) ϭ V0 eϪt͞RCThis shows that the voltage response of the RC circuit is an exponen-tial decay of the initial voltage. Since the response is due to the initialenergy stored and the physical characteristics of the circuit and not dueto some external voltage or current source, it is called the naturalresponse of the circuit. The natural response of a circuit refers to the behavior (in terms of The natural response depends on the voltages and currents) of the circuit itself, with no external sources of nature of the circuit alone, with no ex- excitation. ternal sources. In fact, the circuit has a response only because of the energyThe natural response is illustrated graphically in Fig. 7.2. Note that at initially stored in the capacitor.t ϭ 0, we have the correct initial condition as in Eq. (7.1). As tincreases, the voltage decreases toward zero. The rapidity with which

256 Chapter 7 First-Order Circuits v V0e−t ⁄␶ the voltage decreases is expressed in terms of the time constant, V0 denoted by t, the lowercase Greek letter tau.0.368V0 The time constant of a circuit is the time required for the response to decay to a factor of 1͞e or 36.8 percent of its initial value.10␶ t This implies that at t ϭ t, Eq. (7.7) becomesFigure 7.2 V0eϪt͞RC ϭ V0eϪ1 ϭ 0.368V0The voltage response of the RC circuit. or t ϭ RC (7.8) In terms of the time constant, Eq. (7.7) can be written as v(t) ϭ V0eϪt͞t (7.9) TABLE 7.1 With a calculator it is easy to show that the value of v(t)͞V0 is as shown in Table 7.1. It is evident from Table 7.1 that the voltage v(t)Values of v(t)͞V0 ϭ eϪt͞␶. is less than 1 percent of V0 after 5t (five time constants). Thus, it ist v(t)͞V0 customary to assume that the capacitor is fully discharged (or charged) after five time constants. In other words, it takes 5t for the circuit to t 0.36788 reach its final state or steady state when no changes take place with2t 0.13534 time. Notice that for every time interval of t, the voltage is reduced3t 0.04979 by 36.8 percent of its previous value, v(t ϩ t) ϭ v(t)͞e ϭ 0.368v(t),4t 0.01832 regardless of the value of t.5t 0.00674 Observe from Eq. (7.8) that the smaller the time constant, the more v rapidly the voltage decreases, that is, the faster the response. This is V0 illustrated in Fig. 7.4. A circuit with a small time constant gives a fast 1.0 response in that it reaches the steady state (or final state) quickly due to quick dissipation of energy stored, whereas a circuit with a large time constant gives a slow response because it takes longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants. With the voltage v(t) in Eq. (7.9), we can find the current iR(t), iR(t) ϭ v(t) ϭ V0 eϪt͞t (7.10) R R0.75 1 The time constant may be viewed from another perspective. Evaluating the derivative of v(t) in Eq. (7.7) at t ϭ 0, we obtain0.50 Tangent at t = 00.37 d a v b 2 ϭ Ϫ1 eϪt͞t 2 ϭ Ϫ10.25 dt V0 tϭ0 t tϭ0 t 0 ␶ 2␶ 3␶ 4␶ 5␶ Thus, the time constant is the initial rate of decay, or the time taken for v͞V0 to decay from unity to zero, assuming a constant rate of decay. This initial slope interpretation ofFigure 7.3Graphical determination of the time t (s) the time constant is often used in the laboratory to find t graphically from the responseconstant t from the response curve. curve displayed on an oscilloscope. To find t from the response curve, draw the tangent to the curve at t ϭ 0, as shown in Fig. 7.3. The tangent intercepts with the time axis at t ϭ t.

7.2 The Source-Free RC Circuit 257 ␶=2 v = e−t ⁄␶V0 1 ␶=1 ␶ = 0.5 01234 5tFigure 7.4Plot of v͞V0 ϭ eϪt͞t for various values of the time constant.The power dissipated in the resistor is p(t) ϭ viR ϭ V 2 eϪ2t͞t (7.11) 0 RThe energy absorbed by the resistor up to time t is t t V 2 0 p(l)dl ϭΎ ΎwR(t) ϭ R eϪ2l͞tdl 00 (7.12) tV 2 t 1 0 2ϭ Ϫ eϪ2l͞t 2 ϭ CV 2 (1 Ϫ eϪ2t͞t), t ϭ RC 2R 0 0Notice that as t S ϱ, wR(ϱ) S 21CV 20, which is the same as wC (0),the energy initially stored in the capacitor. The energy that was initiallystored in the capacitor is eventually dissipated in the resistor.In summary: The Key to Working with a Source-Free RC Circuit The time constant is the same regard- Is Finding: less of what the output is defined to be. 1. The initial voltage v(0) ϭ V0 across the capacitor. 2. The time constant t. When a circuit contains a single capacitor and several resistors andWith these two items, we obtain the response as the capacitor voltage dependent sources, the TheveninvC(t) ϭ v(t) ϭ v(0)eϪt͞t. Once the capacitor voltage is first obtained, equivalent can be found at theother variables (capacitor current iC, resistor voltage vR, and resistor cur- terminals of the capacitor to form arent iR) can be determined. In finding the time constant t ϭ RC, R is simple RC circuit. Also, one can useoften the Thevenin equivalent resistance at the terminals of the capacitor; Thevenin’s theorem when severalthat is, we take out the capacitor C and find R ϭ RTh at its terminals. capacitors can be combined to form a single equivalent capacitor.In Fig. 7.5, let vC (0) ϭ 15 V. Find vC, vx, and ix for t 7 0. Example 7.1Solution:We first need to make the circuit in Fig. 7.5 conform with the standardRC circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin

258 Chapter 7 First-Order Circuits 8Ω resistance at the capacitor terminals. Our objective is always to first obtain capacitor voltage vC. From this, we can determine vx and ix. ix The 8-⍀ and 12-⍀ resistors in series can be combined to give a ++ 20-⍀ resistor. This 20-⍀ resistor in parallel with the 5-⍀ resistor can5Ω 0.1 F vC 12 Ω vx be combined so that the equivalent resistance is − −Figure 7.5 Req ϭ 20 ϫ 5 ϭ 4 ⍀For Example 7.1. 20 ϩ 5 Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous to Fig. 7.1. The time constant is + t ϭ ReqC ϭ 4(0.1) ϭ 0.4 s Req v 0.1 F Thus, − v ϭ v(0)eϪt͞t ϭ 15eϪt͞0.4 V, vC ϭ v ϭ 15eϪ2.5t V From Fig. 7.5, we can use voltage division to get vx; soFigure 7.6 vx ϭ 12 v ϭ 0.6(15eϪ2.5t ) ϭ 9eϪ2.5t VEquivalent circuit for the circuit in 12 ϩ 8Fig. 7.5. Finally, ix ϭ vx ϭ 0.75eϪ2.5t A 12Practice Problem 7.1 Refer to the circuit in Fig. 7.7. Let vC(0) ϭ 60 V. Determine vC, vx, and io for t Ն 0. io 8 Ω Answer: 60eϪ0.25t V, 20eϪ0.25t V, Ϫ5eϪ0.25t A. +12 Ω + 1 F 3 vC 6 Ω vx − −Figure 7.7For Practice Prob. 7.1.Example 7.2 The switch in the circuit in Fig. 7.8 has been closed for a long time, and it is opened at t ϭ 0. Find v(t) for t Ն 0. Calculate the initial 3Ω t=0 1Ω energy stored in the capacitor.20 V +− 9Ω + 20 mF Solution: v For t 6 0, the switch is closed; the capacitor is an open circuit to dc, − as represented in Fig. 7.9(a). Using voltage divisionFigure 7.8 vC (t) ϭ 9 9 (20) ϭ 15 V, t60For Example 7.2. ϩ 3 Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at t ϭ 0Ϫ is the same at t ϭ 0, or vC (0) ϭ V0 ϭ 15 V

7.3 The Source-Free RL Circuit 259 For t 7 0, the switch is opened, and we have the RC circuit 20 V +− 3Ω 1Ωshown in Fig. 7.9(b). [Notice that the RC circuit in Fig. 7.9(b) is 9Ωsource free; the independent source in Fig. 7.8 is needed to provide (a) +V0 or the initial energy in the capacitor.] The 1-⍀ and 9-⍀ resistors vC (0)in series give − Req ϭ 1 ϩ 9 ϭ 10 ⍀ 1ΩThe time constant is + t ϭ ReqC ϭ 10 ϫ 20 ϫ 10Ϫ3 ϭ 0.2 s 9 Ω Vo = 15 V 20 mFThus, the voltage across the capacitor for t Ն 0 is − v(t) ϭ vC (0)eϪt͞t ϭ 15eϪt͞0.2 Vor (b) v(t) ϭ 15eϪ5t V Figure 7.9 For Example 7.2: (a) t 6 0, (b) t 7 0.The initial energy stored in the capacitor iswC (0) ϭ 1 CvC2 (0) ϭ 1 ϫ 20 ϫ 10Ϫ3 ϫ 152 ϭ 2.25 J 2 2If the switch in Fig. 7.10 opens at t ϭ 0, find v(t) for t Ն 0 and wC (0). Practice Problem 7.2Answer: 8eϪ2t V, 5.333 J. t=0 6Ω + 24 V +− 1 F v 12 Ω 4 Ω 6 −7.3 The Source-Free RL Circuit Figure 7.10 For Practice Prob. 7.2.Consider the series connection of a resistor and an inductor, as shown iin Fig. 7.11. Our goal is to determine the circuit response, which wewill assume to be the current i(t) through the inductor. We select the + +inductor current as the response in order to take advantage of the idea −that the inductor current cannot change instantaneously. At t ϭ 0, we R vRassume that the inductor has an initial current I0, or − i(0) ϭ I0 (7.13) L vLwith the corresponding energy stored in the inductor as w(0) ϭ 1 I 2 (7.14) Figure 7.11 L 0 A source-free RL circuit. 2Applying KVL around the loop in Fig. 7.11, vL ϩ vR ϭ 0 (7.15)But vL ϭ L di͞dt and vR ϭ iR. Thus, di ϩ Ri ϭ 0 L dt

260 Chapter 7 First-Order Circuits or di ϩ R ϭ 0 (7.16) i dt L Rearranging terms and integrating gives Ύ Ύi(t) di tR ϭ Ϫ dt iL I0 0 i(t) Rt t Rt ln i 2 ϭ Ϫ 2 1 ln i(t) Ϫ ln I0 ϭ Ϫ L ϩ 0 L I0 0 or ln i(t) ϭ Ϫ Rt (7.17) I0 L Taking the powers of e, we have i(t) ϭ I0eϪRt͞L (7.18)i(t) This shows that the natural response of the RL circuit is an exponen- I0 tial decay of the initial current. The current response is shown in Fig. 7.12. It is evident from Eq. (7.18) that the time constant for the RL circuit is0.368I0 Tangent at t = 0 L (7.19) I0 e −t ⁄ ␶ tϭ R with t again having the unit of seconds. Thus, Eq. (7.18) may be written as 0␶ tFigure 7.12 i(t) ϭ I0eϪt͞t (7.20)The current response of the RL circuit. With the current in Eq. (7.20), we can find the voltage across the resistor asThe smaller the time constant t of a vR(t) ϭ i R ϭ I0 ReϪt͞t (7.21)circuit, the faster the rate of decay ofthe response. The larger the time con- The power dissipated in the resistor isstant, the slower the rate of decay ofthe response. At any rate, the response p ϭ vR i ϭ I 2 ReϪ2t͞t (7.22)decays to less than 1 percent of its 0initial value (i.e., reaches steady state)after 5t. The energy absorbed by the resistor isFigure 7.12 shows an initial slope inter- t t t t tϭ Lpretation may be given to t. Ϫ R p(l)dl ϭ 2 0 Ύ ΎwR(t) ϭ I 2 eϪ2l͞t dl ϭ I 2 ReϪ2l͞t 2 , 0 0 0 0 or wR (t) ϭ 1 I 2 (1 Ϫ eϪ2t͞t) (7.23) L 0 2 Note that as t S ϱ, wR(ϱ) S 1 L I 20, which is the same as wL(0), 2 the initial energy stored in the inductor as in Eq. (7.14). Again, the energy initially stored in the inductor is eventually dissipated in the resistor.

7.3 The Source-Free RL Circuit 261In summary: The Key to Working with a Source-Free RL Circuit When a circuit has a single inductor Is to Find: and several resistors and dependent sources, the Thevenin equivalent can 1. The initial current i(0) ϭ I0 through the inductor. be found at the terminals of the induc- 2. The time constant t of the circuit. tor to form a simple RL circuit. Also, one can use Thevenin’s theorem whenWith the two items, we obtain the response as the inductor current several inductors can be combined toiL(t) ϭ i(t) ϭ i(0)eϪt͞t. Once we determine the inductor current iL, form a single equivalent inductor.other variables (inductor voltage vL, resistor voltage vR, and resistorcurrent iR) can be obtained. Note that in general, R in Eq. (7.19) is theThevenin resistance at the terminals of the inductor.Assuming that i(0) ϭ 10 A, calculate i(t) and ix (t) in the circuit of Example 7.3Fig. 7.13.Solution: i 4Ω + 3iThere are two ways we can solve this problem. One way is to obtain 0.5 H ix −the equivalent resistance at the inductor terminals and then use 2ΩEq. (7.20). The other way is to start from scratch by using Kirchhoff’svoltage law. Whichever approach is taken, it is always better to first Figure 7.13obtain the inductor current. For Example 7.3.■ METHOD 1 The equivalent resistance is the same as theThevenin resistance at the inductor terminals. Because of the depend-ent source, we insert a voltage source with vo ϭ 1 V at the inductorterminals a-b, as in Fig. 7.14(a). (We could also insert a 1-A currentsource at the terminals.) Applying KVL to the two loops results in 1 (7.3.1)2(i1 Ϫ i2) ϩ 1 ϭ 0 1 i1 Ϫ i2 ϭ Ϫ 2 5 (7.3.2)6i2 Ϫ 2i1 Ϫ 3i1 ϭ 0 1 i2 ϭ 6 i1Substituting Eq. (7.3.2) into Eq. (7.3.1) gives i1 ϭ Ϫ3 A, io ϭ Ϫi1 ϭ 3 A io a 4Ω 4Ωvo = 1 V +− i1 2 Ω i2 + 3i1 + − − 0.5 H i1 2 Ω i2 3i b (b) (a)Figure 7.14Solving the circuit in Fig. 7.13.

262 Chapter 7 First-Order CircuitsHence, Req ϭ RTh ϭ vo ϭ 1 ⍀ io 3The time constant is t ϭ L ϭ 1 ϭ 3 s Req 2 2 1 3Thus, the current through the inductor is i(t) ϭ i(0)eϪt͞t ϭ 10eϪ(2͞3)t A, t70■ METHOD 2 We may directly apply KVL to the circuit as inFig. 7.14(b). For loop 1, 1 di1 ϩ 2(i1 Ϫ i2) ϭ 0 2 dtor di1 ϩ 4i1 Ϫ 4i2 ϭ 0 (7.3.3) dtFor loop 2, 5 (7.3.4) 6i2 Ϫ 2i1 Ϫ 3i1 ϭ 0 1 i2 ϭ 6 i1Substituting Eq. (7.3.4) into Eq. (7.3.3) gives di1 ϩ 2 ϭ 0 dt 3 i1Rearranging terms, di1 ϭ Ϫ 2 dt i1 3Since i1 ϭ i, we may replace i1 with i and integrate: ln i(t) ϭ Ϫ2t 2t i2 i(0) 3 0or i(t) ϭ Ϫ2t ln i(0) 3Taking the powers of e, we finally obtain t70 i(t) ϭ i(0)eϪ(2͞3)t ϭ 10eϪ(2͞3)t A,which is the same as by Method 1. The voltage across the inductor is v ϭ L di ϭ 0.5(10) aϪ 2 b eϪ(2͞3)t ϭ Ϫ 10 eϪ(2͞3)t V dt 3 3

7.3 The Source-Free RL Circuit 263Since the inductor and the 2-⍀ resistor are in parallel, ix (t) ϭ v ϭ Ϫ1.6667eϪ(2͞3)t A, t70 2Find i and vx in the circuit of Fig. 7.15. Let i(0) ϭ 12 A. Practice Problem 7.3Answer: 12eϪ2t A, Ϫ12eϪ2t V, t Ͼ 0. 1Ω i + vx − 2Ω 2H 6Ω + 2vx − Figure 7.15 For Practice Prob. 7.3.The switch in the circuit of Fig. 7.16 has been closed for a long time. Example 7.4At t ϭ 0, the switch is opened. Calculate i(t) for t 7 0. t=0 4Ω 2ΩSolution: +− 40 V 12 Ω 16 Ω i(t) 2HWhen t 6 0, the switch is closed, and the inductor acts as a shortcircuit to dc. The 16-⍀ resistor is short-circuited; the resulting circuitis shown in Fig. 7.17(a). To get i1 in Fig. 7.17(a), we combine the 4-⍀and 12-⍀ resistors in parallel to get 4 ϫ 12 ϭ 3 ⍀ Figure 7.16 4 ϩ 12 For Example 7.4.Hence, i1 ϭ 40 ϭ 8 A i1 2 Ω 4Ω ϩ 12 Ω 2 3 i(t)We obtain i(t) from i1 in Fig. 7.17(a) using current division, by 40 V +−writing i(t) ϭ 12 4 i1 ϭ 6 A, t60 (a) 12 ϩ 4ΩSince the current through an inductor cannot change instantaneously, i(t) i(0) ϭ i(0Ϫ) ϭ 6 A 2H When t 7 0, the switch is open and the voltage source is 12 Ω 16 Ωdisconnected. We now have the source-free RL circuit in Fig. 7.17(b).Combining the resistors, we have (b) Req ϭ (12 ϩ 4) ʈ 16 ϭ 8 ⍀ Figure 7.17The time constant is Solving the circuit of Fig. 7.16: (a) for t 6 0, (b) for t 7 0.Thus, L 21 tϭ ϭ ϭ s Req 8 4 i(t) ϭ i(0)eϪt͞t ϭ 6eϪ4t A

264 Chapter 7 First-Order CircuitsPractice Problem 7.4 For the circuit in Fig. 7.18, find i(t) for t 7 0. Answer: 5eϪ2t A, t 7 0. t=0 12 Ω 8Ω24 Ω 15 Ai(t) 5Ω 2HFigure 7.18For Practice Prob. 7.4.Example 7.5 In the circuit shown in Fig. 7.19, find io, vo, and i for all time, assum- ing that the switch was open for a long time. 2Ω 3Ω io10 V +− + vo − 6Ω i Solution: t=0 2 H It is better to first find the inductor current i and then obtain otherFigure 7.19 quantities from it.For Example 7.5. For t 6 0, the switch is open. Since the inductor acts like a short circuit to dc, the 6-⍀ resistor is short-circuited, so that we have the circuit shown in Fig. 7.20(a). Hence, io ϭ 0, and i(t) ϭ 2 10 3 ϭ 2 A, t60 ϩ t60 2Ω 3Ω10 V +− + vo − io i vo (t) ϭ 3i(t) ϭ 6 V, 6Ω Thus, i(0) ϭ 2. For t 7 0, the switch is closed, so that the voltage source is short- (a) circuited. We now have a source-free RL circuit as shown in 3Ω Fig. 7.20(b). At the inductor terminals, + vo − RTh ϭ 3 ʈ 6 ϭ 2 ⍀ 6Ω io i so that the time constant is 2H + L vL tϭ ϭ1s − RTh (b) Hence, i(t) ϭ i(0)eϪt͞t ϭ 2eϪt A, t 7 0Figure 7.20The circuit in Fig. 7.19 for: (a) t 6 0, Since the inductor is in parallel with the 6-⍀ and 3-⍀ resistors,(b) t 7 0. vo(t) ϭ ϪvL ϭ ϪL di ϭ Ϫ2(Ϫ2eϪt) ϭ 4eϪt V, t70 dt and io(t) ϭ vL ϭ Ϫ 2eϪt A, t70 6 3

7.4 Singularity Functions 265 tThus, for all time, 0 A, t60 6 V, t60 2io(t) ϭ c Ϫ 2eϪt A, , vo(t) ϭ b 4eϪt V, t70 i(t) 3 t70 2 A, t60 i(t) ϭ b 2eϪt A, tՆ0We notice that the inductor current is continuous at t ϭ 0, while the − 2 io(t)current through the 6-⍀ resistor drops from 0 to Ϫ2͞3 at t ϭ 0, and 3the voltage across the 3-⍀ resistor drops from 6 to 4 at t ϭ 0. We alsonotice that the time constant is the same regardless of what the output Figure 7.21 A plot of i and io.is defined to be. Figure 7.21 plots i and io.Determine i, io, and vo for all t in the circuit shown in Fig. 7.22. Practice Problem 7.5Assume that the switch was closed for a long time. It should be notedthat opening a switch in series with an ideal current source creates an 3Ωinfinite voltage at the current source terminals. Clearly this is impossi-ble. For the purposes of problem solving, we can place a shunt resis- 24 A t=0 i 1Htor in parallel with the source (which now makes it a voltage source 4Ωin series with a resistor). In more practical circuits, devices that act like iocurrent sources are, for the most part, electronic circuits. These circuits +will allow the source to act like an ideal current source over its oper-ating range but voltage-limit it when the load resistor becomes too large 2 Ω vo(as in an open circuit). −Answer: Figure 7.22 For Practice Prob. 7.5. 16 A, t60 8 A, t60i ϭ b 16eϪ2t A, t Ն 0, io ϭ b Ϫ5.333eϪ2t A, , t70 vo ϭ b 32 V, V, t60 10.667eϪ2t t707.4 Singularity FunctionsBefore going on with the second half of this chapter, we need to digressand consider some mathematical concepts that will aid our under-standing of transient analysis. A basic understanding of singularityfunctions will help us make sense of the response of first-order circuitsto a sudden application of an independent dc voltage or current source. Singularity functions (also called switching functions) are very use-ful in circuit analysis. They serve as good approximations to theswitching signals that arise in circuits with switching operations. Theyare helpful in the neat, compact description of some circuit phenom-ena, especially the step response of RC or RL circuits to be discussedin the next sections. By definition, Singularity functions are functions that either are discontinuous or have discontinuous derivatives.

266 Chapter 7 First-Order Circuits u(t) The three most widely used singularity functions in circuit analy- 1 sis are the unit step, the unit impulse, and the unit ramp functions. The unit step function u(t) is 0 for negative values of t and 1 for pos- itive values of t. In mathematical terms, 0 tFigure 7.23 u(t) ϭ b 0, t60 (7.24)The unit step function. 1, t70u(t − t0) The unit step function is undefined at t ϭ 0, where it changes abruptly 1 from 0 to 1. It is dimensionless, like other mathematical functions such as sine and cosine. Figure 7.23 depicts the unit step function. If the abrupt change occurs at t ϭ t0 (where t0 7 0) instead of t ϭ 0, the unit step function becomes 0 t0 t u (t Ϫ t0) ϭ b 0, t 6 t0 (7.25) (a) 1, t 7 t0u(t + t0) which is the same as saying that u (t) is delayed by t0 seconds, as shown in Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace 1 every t by t Ϫ t0. If the change is at t ϭ Ϫt0, the unit step function becomes 0, t 6 Ϫt0 (7.26) u (t ϩ t0) ϭ b 1, t 7 Ϫt0−t0 0 t meaning that u (t) is advanced by t0 seconds, as shown in Fig. 7.24(b). We use the step function to represent an abrupt change in voltage (b) or current, like the changes that occur in the circuits of control systemsFigure 7.24 and digital computers. For example, the voltage(a) The unit step function delayed by t0, 0, t 6 t0(b) the unit step advanced by t0. v(t) ϭ b t 7 t0 V0, (7.27) may be expressed in terms of the unit step function as v(t) ϭ V0u (t Ϫ t0) (7.28)Alternatively, we may derive If we let t0 ϭ 0, then v(t) is simply the step voltage V0 u (t). A voltageEqs. (7.25) and (7.26) from Eq. (7.24) source of V0 u (t) is shown in Fig. 7.25(a); its equivalent circuit is shownby writing u [f (t )] ϭ 1, f (t) 7 0, in Fig. 7.25(b). It is evident in Fig. 7.25(b) that terminals a-b are short-where f (t ) may be t Ϫ t0 or t ϩ t0. circuited (v ϭ 0) for t 6 0 and that v ϭ V0 appears at the terminals t=0 V0u(t) +− a a b = V0 +− b (a) (b) Figure 7.25 (a) Voltage source of V0u(t), (b) its equivalent circuit.

7.4 Singularity Functions 267for t 7 0. Similarly, a current source of I0 u (t) is shown in Fig. 7.26(a),while its equivalent circuit is in Fig. 7.26(b). Notice that for t 6 0,there is an open circuit (i ϭ 0), and that i ϭ I0 flows for t 7 0. t=0 i aa b=I0u(t) I0 b (a) (b)Figure 7.26(a) Current source of I0u (t), (b) its equivalent circuit. The derivative of the unit step function u(t) is the unit impulsefunction d(t), which we write as 0, t60 (7.29) ␦(t) (∞)d(t) ϭ d u(t) ϭ c Undefined, tϭ0 t70 dt 0,The unit impulse function—also known as the delta function—is 0 tshown in Fig. 7.27. Figure 7.27 The unit impulse function d(t) is zero everywhere except at t ϭ 0, The unit impulse function. where it is undefined.Impulsive currents and voltages occur in electric circuits as a result ofswitching operations or impulsive sources. Although the unit impulsefunction is not physically realizable (just like ideal sources, idealresistors, etc.), it is a very useful mathematical tool. The unit impulse may be regarded as an applied or resulting shock.It may be visualized as a very short duration pulse of unit area. Thismay be expressed mathematically as 0ϩ (7.30)Ύ d(t) dt ϭ 1 0Ϫwhere t ϭ 0Ϫ denotes the time just before t ϭ 0 and t ϭ 0ϩ is the time 10␦(t)just after t ϭ 0. For this reason, it is customary to write 1 (denotingunit area) beside the arrow that is used to symbolize the unit impulse 5␦(t + 2)function, as in Fig. 7.27. The unit area is known as the strength of theimpulse function. When an impulse function has a strength other than −2 −1 0 1 2 3 tunity, the area of the impulse is equal to its strength. For example, animpulse function 10d(t) has an area of 10. Figure 7.28 shows the −4␦(t − 3)impulse functions 5d(t ϩ 2), 10d(t), and Ϫ4d(t Ϫ 3). Figure 7.28 To illustrate how the impulse function affects other functions, let Three impulse functions.us evaluate the integral b (7.31)Ύ f(t)d(t Ϫ t0)dt a

268 Chapter 7 First-Order Circuits where a 6 t0 6 b. Since d(t Ϫ t0) ϭ 0 except at t ϭ t0, the integrand is zero except at t0. Thus, bb Ύ Ύf (t)d(t Ϫ t0)dt ϭ f (t0)d(t Ϫ t0)dt aa b Ύϭ f(t0) d(t Ϫ t0)dt ϭ f(t0) a or b (7.32) Ύ f(t)d(t Ϫ t0)dt ϭ f(t0) a This shows that when a function is integrated with the impulse func- tion, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function known as the sampling or sifting property. The special case of Eq. (7.31) is r(t) for t0 ϭ 0. Then Eq. (7.32) becomes (7.33) 1 0ϩ Ύ f(t)d(t)dt ϭ f(0) 0Ϫ Integrating the unit step function u(t) results in the unit ramp func- tion r(t); we write 01 t t (7.34)Figure 7.29 Ύr(t) ϭ u(l)dl ϭ tu(t)The unit ramp function. Ϫϱ or r (t − t0) r(t) ϭ b 0, tՅ0 (7.35) 1 t, tՆ0 The unit ramp function is zero for negative values of t and has a unit slope for positive values of t. 0 t0 t0 + 1 t Figure 7.29 shows the unit ramp function. In general, a ramp is a func- (a) tion that changes at a constant rate. r(t + t0) The unit ramp function may be delayed or advanced as shown in Fig. 7.30. For the delayed unit ramp function, 0, t Յ t0 (7.36) r (t Ϫ t0) ϭ b t Ϫ t0, t Ն t0 1 and for the advanced unit ramp function, 0, t Յ Ϫt0 (7.37) r (t ϩ t0) ϭ b t ϩ t0, t Ն Ϫt0 −t0 −t0 + 1 0 t We should keep in mind that the three singularity functions (impulse, step, and ramp) are related by differentiation as (b) du (t) dr (t) (7.38)Figure 7.30 d(t) ϭ , u(t) ϭThe unit ramp function: (a) delayed by t0,(b) advanced by t0. dt dt

















7.5 Step Response of an RC Circuit 277the response using Eq. (7.53). This technique equally applies to RL cir- Example 7.10cuits, as we shall see in the next section. Note that if the switch changes position at time t ϭ t0 instead ofat t ϭ 0, there is a time delay in the response so that Eq. (7.53)becomes v(t) ϭ v(ϱ) ϩ [v(t0) Ϫ v(ϱ)]eϪ(tϪt0)͞t (7.54)where v(t0) is the initial value at t ϭ t0ϩ. Keep in mind that Eq. (7.53)or (7.54) applies only to step responses, that is, when the input exci-tation is constant.The switch in Fig. 7.43 has been in position A for a long time. At t ϭ 0,the switch moves to B. Determine v(t) for t 7 0 and calculate its valueat t ϭ 1 s and 4 s. 3 kΩ A B 4 kΩ 0.5 mF 24 V +− 5 kΩ t=0 +− 30 V + v − Figure 7.43 For Example 7.10.Solution:For t 6 0, the switch is at position A. The capacitor acts like an opencircuit to dc, but v is the same as the voltage across the 5-k⍀ resistor.Hence, the voltage across the capacitor just before t ϭ 0 is obtainedby voltage division as v(0Ϫ) ϭ 5 (24) ϭ 15 V 5ϩ3Using the fact that the capacitor voltage cannot change instantaneously, v(0) ϭ v(0Ϫ) ϭ v(0ϩ) ϭ 15 VFor t 7 0, the switch is in position B. The Thevenin resistanceconnected to the capacitor is RTh ϭ 4 k⍀, and the time constant is t ϭ RThC ϭ 4 ϫ 103 ϫ 0.5 ϫ 10Ϫ3 ϭ 2 sSince the capacitor acts like an open circuit to dc at steady state,v(ϱ) ϭ 30 V. Thus, v(t) ϭ v(ϱ) ϩ [v(0) Ϫ v(ϱ)]eϪt͞t ϭ 30 ϩ (15 Ϫ 30)eϪt͞2 ϭ (30 Ϫ 15eϪ0.5t ) VAt t ϭ 1, v(1) ϭ 30 Ϫ 15eϪ0.5 ϭ 20.9 VAt t ϭ 4, v(4) ϭ 30 Ϫ 15eϪ2 ϭ 27.97 V

278 Chapter 7 First-Order CircuitsPractice Problem 7.10 Find v(t) for t 7 0 in the circuit of Fig. 7.44. Assume the switch has been open for a long time and is closed at t ϭ 0. Calculate v(t) at t=0 t ϭ 0.5. 2Ω 6Ω + −+ Answer: (9.375 ϩ 5.625eϪ2t) V for all t 7 0, 7.63 V.15 V +− v 1 F 7.5 V − 3Figure 7.44For Practice Prob. 7.10.Example 7.11 In Fig. 7.45, the switch has been closed for a long time and is opened at t ϭ 0. Find i and v for all time. 10 Ω i t=0 + 30u(t) V +− 20 Ω v 1 F +− 10 V − 4 Figure 7.45 For Example 7.11. Solution: The resistor current i can be discontinuous at t ϭ 0, while the capacitor voltage v cannot. Hence, it is always better to find v and then obtain i from v. By definition of the unit step function, 10 Ω i 30u(t) ϭ b 0, t 6 0 30, t 7 0 + v +− 10 V For t 6 0, the switch is closed and 30u(t) ϭ 0, so that the 30u(t) − voltage source is replaced by a short circuit and should be regarded as contributing nothing to v. Since the switch has been closed for a long time, the capacitor voltage has reached steady state and the capacitor acts like an open circuit. Hence, the circuit becomes that shown in Fig. 7.46(a) for t 6 0. From this circuit we obtain 20 Ω v ϭ 10 V, v i ϭ Ϫ ϭ Ϫ1 A 10 (a) Since the capacitor voltage cannot change instantaneously, 10 Ω i v(0) ϭ v(0Ϫ) ϭ 10 V30 V +− + 1 For t 7 0, the switch is opened and the 10-V voltage source is 4 disconnected from the circuit. The 30u(t) voltage source is now operative, 20 Ω v F so the circuit becomes that shown in Fig. 7.46(b). After a long time, the − circuit reaches steady state and the capacitor acts like an open circuit again. We obtain v(ϱ) by using voltage division, writing (b) v(ϱ) ϭ 20 20 (30) ϭ 20 VFigure 7.46 ϩ 10Solution of Example 7.11: (a) for t 6 0,(b) for t 7 0.

7.5 Step Response of an RC Circuit 279The Thevenin resistance at the capacitor terminals is RTh ϭ 10 ʈ 20 ϭ 10 ϫ 20 ϭ 20 ⍀ 30 3and the time constant is t ϭ RTh C ϭ 20 1 5 ؒϭs 3 43Thus, v(t) ϭ v(ϱ) ϩ [v(0) Ϫ v(ϱ)]eϪt͞t ϭ 20 ϩ (10 Ϫ 20)eϪ(3͞5)t ϭ (20 Ϫ 10eϪ0.6t) VTo obtain i, we notice from Fig. 7.46(b) that i is the sum of the currentsthrough the 20-⍀ resistor and the capacitor; that is, v dviϭ ϩC 20 dt ϭ 1 Ϫ 0.5eϪ0.6t ϩ 0.25(Ϫ0.6)(Ϫ10)eϪ0.6t ϭ (1 ϩ eϪ0.6t) ANotice from Fig. 7.46(b) that v ϩ 10i ϭ 30 is satisfied, as expected.Hence, ϭ 10 V, t60 v b tՆ0 (20 Ϫ 10eϪ0.6t) V, t60 i ϭ b(Ϫ11ϩAe, Ϫ0.6t) A, t70Notice that the capacitor voltage is continuous while the resistor currentis not.The switch in Fig. 7.47 is closed at t ϭ 0. Find i(t) and v(t) for all time. Practice Problem 7.11Note that u(Ϫt) ϭ 1 for t 6 0 and 0 for t 7 0. Also, u(Ϫt) ϭ 1 Ϫ u(t). 5Ω i t=0 0.2 F 10 Ω 20u(−t) V +− + 3A v − Figure 7.47 For Practice Prob. 7.11.Answer: i (t) ϭ b0Ϫ,2(1 ϩ eϪ1.5t) A, t60 t 7 0,v ϭ b1200(V1 ,ϩ eϪ1.5t) V, t60 t70

280 Chapter 7 First-Order Circuits R 7.6 Step Response of an RL Circuit Vs +− t=0 i Consider the RL circuit in Fig. 7.48(a), which may be replaced by the L circuit in Fig. 7.48(b). Again, our goal is to find the inductor current i + as the circuit response. Rather than apply Kirchhoff’s laws, we will use v (t) the simple technique in Eqs. (7.50) through (7.53). Let the response be − the sum of the transient response and the steady-state response, (a) i ϭ it ϩ iss (7.55) Vsu(t) +− R We know that the transient response is always a decaying exponential, that is, i + it ϭ AeϪt͞t, tϭ L L v (t) R − (7.56) (b) where A is a constant to be determined. The steady-state response is the value of the current a long time afterFigure 7.48An RL circuit with a step input voltage. the switch in Fig. 7.48(a) is closed. We know that the transient response essentially dies out after five time constants. At that time, the inductor becomes a short circuit, and the voltage across it is zero. The entire source voltage Vs appears across R. Thus, the steady-state response is iss ϭ Vs (7.57) R Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives i ϭ AeϪt͞t ϩ Vs (7.58) R We now determine the constant A from the initial value of i. Let I0 be the initial current through the inductor, which may come from a source other than Vs. Since the current through the inductor cannot change instantaneously, i(0ϩ) ϭ i(0Ϫ) ϭ I0 (7.59) Thus, at t ϭ 0, Eq. (7.58) becomes I0 ϭ A ϩ Vs R From this, we obtain A as i(t) A ϭ I0 Ϫ Vs I0 R Vs Substituting for A in Eq. (7.58), we get R i(t) ϭ Vs ϩ aI0 Ϫ Vs b eϪt͞t (7.60) 0t R RFigure 7.49 This is the complete response of the RL circuit. It is illustrated inTotal response of the RL circuit with Fig. 7.49. The response in Eq. (7.60) may be written asinitial inductor current I0. i(t) ϭ i(ϱ) ϩ [i(0) Ϫ i(ϱ)]eϪt͞t (7.61)

7.6 Step Response of an RL Circuit 281where i(0) and i(ϱ) are the initial and final values of i, respectively.Thus, to find the step response of an RL circuit requires three things: 1. The initial inductor current i(0) at t ϭ 0. 2. The final inductor current i(ϱ). 3. The time constant t.We obtain item 1 from the given circuit for t 6 0 and items 2 and 3from the circuit for t 7 0. Once these items are determined, we obtainthe response using Eq. (7.61). Keep in mind that this technique appliesonly for step responses. Again, if the switching takes place at time t ϭ t0 instead of t ϭ 0,Eq. (7.61) becomes i(t) ϭ i(ϱ) ϩ [i(t0) Ϫ i(ϱ)]eϪ(tϪt0)͞t (7.62)If I0 ϭ 0, then 0, t60 (7.63a) i(t) ϭ c Vs (1 Ϫ eϪt͞t), t70 Ror i(t) ϭ Vs (1 Ϫ eϪt͞t)u(t) (7.63b) RThis is the step response of the RL circuit with no initial inductor cur-rent. The voltage across the inductor is obtained from Eq. (7.63) usingv ϭ L di͞dt. We get v(t) ϭ di ϭ Vs L eϪt͞t, t ϭ L t70 L tR , R dtor v(t) ϭ VseϪt͞tu(t) (7.64)Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64). i(t) v(t) Vs Vs R 0 t0 t (a) (b) Figure 7.50 Step responses of an RL circuit with no initial inductor current: (a) current response, (b) voltage response.

282 Chapter 7 First-Order CircuitsExample 7.12 Find i(t) in the circuit of Fig. 7.51 for t 7 0. Assume that the switch has been closed for a long time. t=0 Solution: 2Ω 3Ω When t 6 0, the 3-⍀ resistor is short-circuited, and the inductor acts i like a short circuit. The current through the inductor at t ϭ 0Ϫ (i.e., just 10 V +− 1 H before t ϭ 0) isFigure 7.51 3For Example 7.12. i(0Ϫ) ϭ 10 ϭ 5 A 2 Since the inductor current cannot change instantaneously, i(0) ϭ i(0ϩ) ϭ i(0Ϫ) ϭ 5 A When t 7 0, the switch is open. The 2-⍀ and 3-⍀ resistors are in series, so that i(ϱ) ϭ 2 10 3 ϭ 2 A ϩ The Thevenin resistance across the inductor terminals is RTh ϭ 2 ϩ 3 ϭ 5 ⍀ For the time constant, tϭ L ϭ 1 ϭ 1 s 3 RTh 5 15 Thus, i(t) ϭ i(ϱ) ϩ [i(0) Ϫ i(ϱ)]eϪt͞t t70 ϭ 2 ϩ (5 Ϫ 2)eϪ15t ϭ 2 ϩ 3eϪ15t A, Check: In Fig. 7.51, for t 7 0, KVL must be satisfied; that is, di 10 ϭ 5i ϩ L dt 5i ϩ di ϭ [10 ϩ 15eϪ15t] ϩ c 1 (3)(Ϫ15)eϪ15t d ϭ 10 L dt 3 This confirms the result.Practice Problem 7.12 The switch in Fig. 7.52 has been closed for a long time. It opens at t ϭ 0. Find i(t) for t 7 0. i 1.5 H Answer: (4 ϩ 2eϪ10t) A for all t 7 0.5Ω t=0 10 Ω 6AFigure 7.52For Practice Prob. 7.12.

7.6 Step Response of an RL Circuit 283At t ϭ 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later. Example 7.13Find i(t) for t 7 0. Calculate i for t ϭ 2 s and t ϭ 5 s. 4 Ω S1 t = 0 P 6Ω S2 i t=4 40 V +− 2Ω 5H 10 V +− Figure 7.53 For Example 7.13.Solution:We need to consider the three time intervals t Յ 0, 0 Յ t Յ 4, andt Ն 4 separately. For t 6 0, switches S1 and S2 are open so that i ϭ 0.Since the inductor current cannot change instantly, i(0Ϫ) ϭ i(0) ϭ i(0ϩ) ϭ 0 For 0 Յ t Յ 4, S1 is closed so that the 4-⍀ and 6-⍀ resistors arein series. (Remember, at this time, S2 is still open.) Hence, assumingfor now that S1 is closed forever, i(ϱ) ϭ 4 40 6 ϭ 4 A, RTh ϭ 4 ϩ 6 ϭ 10 ⍀ ϩ tϭ L ϭ 5 ϭ1s RTh 10 2Thus, i(t) ϭ i(ϱ) ϩ [i(0) Ϫ i(ϱ)]eϪt͞t 0ՅtՅ4 ϭ 4 ϩ (0 Ϫ 4)eϪ2t ϭ 4(1 Ϫ eϪ2t) A, For t Ն 4, S2 is closed; the 10-V voltage source is connected, andthe circuit changes. This sudden change does not affect the inductorcurrent because the current cannot change abruptly. Thus, the initialcurrent is i(4) ϭ i(4Ϫ) ϭ 4(1 Ϫ eϪ8) Ӎ 4 ATo find i(ϱ), let v be the voltage at node P in Fig. 7.53. Using KCL, 40 Ϫ v ϩ 10 Ϫ v ϭ v 1 v ϭ 180 V 4 26 11 i(ϱ) ϭ v ϭ 30 ϭ 2.727 A 6 11The Thevenin resistance at the inductor terminals is RTh ϭ 4 ʈ 2 ϩ 6 ϭ 4 ϫ 2 ϩ 6 ϭ 22 ⍀ 6 3and L 5 15 RTh 22 t ϭ ϭ 22 ϭ s 3