206 Inductance, Capacitance, and Mutual Inductance t = 0 the inductor was switched instantaneously to position b where it remained for 1.6 s before returning instantaneously to position a. The d'Arsonval volt- meter has a full-scale reading of 20 V and a sensitivity of 1000 ft/V. What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movement is negligible? Figure P6.13 3mV +) Voltmeter PSPICE HULTISIH Section 6.2 6.14 The current shown in Fig. P6.14 is applied to a 0.25 /xF capacitor. The initial voltage on the capaci- tor is zero. a) Find the charge on the capacitor at t = 15 /xs. b) Find the voltage on the capacitor at t = 30 /xs. c) How much energy is stored in the capacitor by this current? Figure P6.14 i (mA) t (fis) 6.15 The initial voltage on the 0.5 /xF capacitor shown in PSPICE Fig. P6.15(a) is -20 V. The capacitor current has mnsm the waveform shown in Fig. P6.15(b). a) How much energy, in microjoules, is stored in the capacitor at t = 500 /xs? b) Repeat (a) for t = oo. 6.16 The rectangular-shaped current pulse shown in PSPICE pig. P6.16 is applied to a 0.1 fxF capacitor. The ini- ' tial voltage on the capacitor is a 15 V drop in the reference direction of the current. Derive the expression for the capacitor voltage for the time intervals in (a)-(d). a) 0 < t < 10 /xs; b) 10/AS < t =5 20/xs; c) 20 /xs < t < 40 /xs d) 40 /xs < t < oo c) Sketch v{t) over the interval -10 /xs ^ t ^ 50 ju.s. Figure P6.16 i (mA) 160 PSPICE MULTISIM 100 0 50 10 20 30 40 t(fJLS) 6.17 A 20 fxF capacitor is subjected to a voltage pulse having a duration of 1 s. The pulse is described by the following equations: 30t 2 V, <30(f- 1) 2 V, 0 0 < t < 0.5 s; 0.5 s < t < 1 s; elsewhere. Sketch the current pulse that exists in the capacitor during the 1 s interval. 6.18 The voltage across the terminals of a 0.2 /xF capaci- tor is v = 150 V, t < 0; (Ate**** + A*T saiDt )V, t>0. Figure P6.15 0.5 /xF -20 V v (a) i (n 50 25 0 lA) I 100 50e- 20()() 'mA,/>0 1 1 1 l 200 300 400 500 f(jiS) (b) Problems 207 6.19 PSPICE MULTISIM V = The initial current in the capacitor is 250 mA. Assume the passive sign convention. a) What is the initial energy stored in the capacitor? b) Evaluate the coefficients A { and A 2 . c) What is the expression for the capacitor current? The voltage at the terminals of the capacitor in Fig. 6.10 is known to be -20 V, / < 0; 100 - 40^ 200() '(3 cos 1000? + sin 1000/) V t > 0. Assume C = 4 /xF. a) Find the current in the capacitor for t < 0. b) Find the current in the capacitor for t > 0. c) Is there an instantaneous change in the voltage across the capacitor at t = 0? d) Is there an instantaneous change in the current in the capacitor at t = 0? e) How much energy (in millijoules) is stored in the capacitor at t = oo? Section 6.3 6.20 Assume that the initial energy stored in the induc- PSPICE tors of Fig. P6.20 is zero. Find the equivalent induc- tance with respect to the terminals a,b. Figure P6.20 30FH20H 10 H 8H 6.21 Assume that the initial energy stored in the induc- tors of Fig. P6.21 is zero. Find the equivalent induc- tance with respect to the terminals a,b. Figure P6.21 3H 8H 6.22 Use realistic inductor values from Appendix H to con- struct series and parallel combinations of inductors to yield the equivalent inductances specified below. Try to minimize the number of inductors used. Assume that no initial energy is stored in any of the inductors. a) 3mH b) 250/xH c) 6QfxH 6.23 The three inductors in the circuit in Fig. P6.23 are con- PSPICE nected across the terminals of a black box at t = 0. The resulting voltage for t > 0 is known to be v a = 2000e" 100 ' V. If /,(0) = -6Aand/ 2 (0) = 1 A, find a) U0); b) U0, t > 0; c) ii(f)»* s 0; d) / 2 (/), t > 0: e) the initial energy stored in the three inductors; f) the total energy delivered to the black box; and g) the energy trapped in the ideal inductors. Figure P6.23 6.24 For the circuit shown in Fig. P6.23, how many milli- seconds after the switch is opened is the energy delivered to the black box 80% of the total energy delivered? 6.25 The two parallel inductors in Fig. P6.25 are con- nected across the terminals of a black box at t = 0. The resulting voltage v for t > 0 is known to be 64e -4 ' V. It is also known that /,(0) = -10 A and / 2 (0) = 5 A. a) Replace the original inductors with an equiva- lent inductor and find /(f) for f > 0. b) Find *',(*) for t > 0. c) Find i 2 (t) for t > 0. d) How much energy is delivered to the black box in the time interval 0 < f < oo? e) How much energy was initially stored in the par- allel inductors? f) How much energy is trapped in the ideal inductors? g) Show that your solutions for /, and / 2 agree with the answer obtained in (f). Figure P6.25 208 Inductance, Capacitance, and Mutual Inductance 6.26 Find the equivalent capacitance with respect to the terminals a,b for the circuit shown in Fig. P6.26. 30 V + - OV + 6.27 Find the equivalent capacitance with respect to the terminals a,b for the circuit shown in Fig. P6.27. Figure P6.27 5 V 6.28 Use realistic capacitor values from Appendix H to construct series and parallel combinations of capac- itors to yield the equivalent capacitances specified below. Try to minimize the number of capacitors used. Assume that no initial energy is stored in any of the capacitors. a) 330/AF b) 750 nF c) 150 pF 6.29 The four capacitors in the circuit in Fig. P6.29 are con- nected across the terminals of a black box at t — 0. The resulting current i b for t > 0 is known to be •- 50 ' mA. h -5e If v u (0) = -20 V, v c (0) = -30 V, and v ( ,(0) = 250 V, find the following for / ^ 0: (a) v h (t), (b) vM (<0 »<#). (d) *X0. (e) h(t\ and (f) i 2 (t). Figure P6.29 200 nF: 6.30 For the circuit in Fig. P6.29, calculate a) the initial energy stored in the capacitors; b) the final energy stored in the capacitors; c) the total energy delivered to the black box; d) the percentage of the initial energy stored that is delivered to the black box; and e) the time, in milliseconds, it takes to deliver 7.5 mJ to the black box. 6.31 The two series-connected capacitors in Fig. P6.31 are connected to the terminals of a black box at t = 0. The resulting current i(t) for t > 0 is known to be 800<T 25 ' (xA. a) Replace the original capacitors with an equiva- lent capacitor and find v a (t) for t > 0. b) Find »!(0 for/ > 0. c) Find v 2 (t) for / > 0. d) How much energy is delivered to the black box in the time interval 0 < t < oo? e) How much energy was initially stored in the series capacitors? f) How much energy is trapped in the ideal capacitors? g) Show that the solutions for V\ and v 2 agree with the answer obtained in (f). Figure P6.31 5V- + + 25 V^ - —»- ^2fxF »i + ^8/xF v 2 + > \ / = 0 - *\, 1 Black box 1.25/tF 6.32 Derive the equivalent circuit for a series connection of ideal capacitors. Assume that each capacitor has its own initial voltage. Denote these initial voltages as V\(t[)), ^2(/0)1 and so on. (Hint: Sum the voltages across the string of capacitors, recognizing that the series connection forces the current in each capaci- tor to be the same.) 6.33 Derive the equivalent circuit for a parallel connec- tion of ideal capacitors. Assume that the initial volt- age across the paralleled capacitors is v(t {) ). (Hint: Sum the currents into the string of capacitors, rec- ognizing that the parallel connection forces the voltage across each capacitor to be the same.) Sections 6.1-6.3 6.34 The current in the circuit in Fig. P6.34 is known to be i o = 5<r 2()0,)f (2 cos 4000/ + sin 4000/) A for t > 0 + . Find z; t (0 + ) and v 2 (0 + ). Problems 209 Figure P6.34 40 ft v 2 < 10 mH 6.35 At t = 0, a series-connected capacitor and induc- tor are placed across the terminals of a black box, as shown in Fig. P6.35. For t > 0, it is known that L=1.5 e - ]6 ' 000 '-0.5e- 4000 'A. If v c (0) = -50 V find v () for t Figure P6.35 0. + 25 mH ~625nF Section 6.4 X /-0 >,> + Black box 6.36 a) Show that the differential equations derived in (a) of Example 6.6 can be rearranged as follows: 4-^- + 25/T - 8- - 20/ 2 = 5i g - 8-^; dt dt dt dL dii dig -8—- - 20/, + 16—^ + 80/2 = 16-r- dt dt dt b) Show that the solutions for i h and i 2 given in (b) of Example 6.6 satisfy the differential equations given in part (a) of this problem. 6.37 Let v a represent the voltage across the 16 H inductor in the circuit in Fig. 6.25. Assume v 0 is positive at the dot. As in Example 6.6, i g = 16- 16e _5 'A. a) Can you find v„ without having to differenti- ate the expressions for the currents? Explain. b) Derive the expression for v 0 . c) Check your answer in (b) using the appropri- ate current derivatives and inductances. 6.38 Let v R represent the voltage across the current source in the circuit in Fig. 6.25. The reference for v g is positive at the upper terminal of the current source. a) Find v g as a function of time when i K = 16 - 16<T 5 ' A. b) What is the initial value of vj c) Find the expression for the power developed by the current source. d) How much power is the current source develop- ing when t is infinite? e) Calculate the power dissipated in each resistor when t is infinite. 6.39 There is no energy stored in the circuit in Fig. P6.39 at the time the switch is opened. a) Derive the differential equation that governs the behavior of / 2 if Lj = 4 H, L 2 = 16 H, M = 2 H, and R 0 = 32 H. b) Show that when i g = 8 - 8e"'A, t > 0, the dif- ferential equation derived in (a) is satisfied when i 2 = e~ l - e~ 2t A, t > 0. c) Find the expression for the voltage V\ across the current source. d) What is the initial value of v{t Does this make sense in terms of known circuit behavior? Figure P6.39 6.40 a) Show that the two coupled coils in Fig. P6.40 can be replaced by a single coil having an inductance of L ab = L\ + L 2 + 2M, (Hint: Express v a ^ as a function of / ab .) b) Show that if the connections to the terminals of the coil labeled L 2 are reversed, L ab = L x + L 2 - 2M. Figure P6.40 6.41 a) Show that the two magnetically coupled coils in Fig. P6.41 (see page 210) can be replaced by a single coil having an inductance of ^ab - L X L 2 M l L { + U 2M (Hint: Let i\ and i 2 be clockwise mesh currents in the left and right "windows" of Fig. P6.41, respec- tively. Sum the voltages around the two meshes. In mesh 1 let v ah be the unspecified applied volt- age. Solve for dijdt as a function of v ab .) b) Show that if the magnetic polarity of coil 2 is reversed, then •^ab — L { L 2 - M 2 L x + L 2 + 2M' 210 Inductance, Capacitance, and Mutual Inductance Figure P6.41 a« M 6.42 The polarity markings on two coils are to be deter- mined experimentally. The experimental setup is shown in Fig. P6.42. Assume that the terminal con- nected to the negative terminal of the battery has been given a polarity mark as shown. When the switch is opened, the dc voltmeter kicks upscale. Where should the polarity mark be placed on the coil connected to the voltmeter? Figure P6.42 / = 0 r-^ >\ VBK- voltmeter 6.43 The physical construction of four pairs of magneti- cally coupled coils is shown in Fig. P6.43. (See page 211.) Assume that the magnetic flux is confined to the core material in each structure. Show two possi- ble locations for the dot markings on each pair of coils. Section 6.5 6.44 The self-inductances of the coils in Fig. 6.30 are L\ = 18 mH and L 2 = 32 mH. If the coefficient of coupling is 0.85, calculate the energy stored in the system in millijoules when (a) i x = 6 A, i 2 = 9 A; (b) i { = -6 A, i 2 = -9 A; (c) i } = -6 A, i 2 = 9 A; and (d) 2, = 6 A, i 2 = -9A. 6.45 The coefficient of coupling in Problem 6.44 is increased to 1.0. a) If /j equals 6 A, what value of i 2 results in zero stored energy? b) Is there any physically realizable value of i 2 that can make the stored energy negative? 6.46 Two magnetically coupled coils have self-inductances of 60 mH and 9.6 mH, respectively.The mutual induc- tance between the coils is 22.8 mH. a) What is the coefficient of coupling? b) For these two coils, what is the largest value that M can have? c) Assume that the physical structure of these cou- pled coils is such that 2?*j = 9> 2 - What is the turns ratio N\/N 2 if N\ is the number of turns on the 60 mH coil? 6.47 The self-inductances of two magnetically coupled coils are 72 mH and 40.5 mH, respectively. The 72 mH coil has 250 turns, and the coefficient of coupling between the coils is %.The coupling medium is non- magnetic. When coil 1 is excited with coil 2 open, the flux linking only coil 1 is 0.2 as large as the flux linking coil 2. a) How many turns does coil 2 have? b) What is the value of <& 2 in nanowebers per ampere? c) What is the value of S^n in nanowebers per ampere? d) What is the ratio (^22/^12)? 6.48 Two magnetically coupled coils are wound on a nonmagnetic core. The self-inductance of coil 1 is 288 mH, the mutual inductance is 90 mH, the coeffi- cient of coupling is 0.75, and the physical structure of the coils is such that SP n = 9^2- a) Find L 2 and the turns ratio Ni/N 2 . b) If N { = 1200, what is the value of S^ and 2P 2 ? 6.49 The self-inductances of two magnetically coupled coils are L { = 180/xH and L 2 = 500 /xH. The coupling medium is nonmagnetic. If coil 1 has 300 turns and coil 2 has 500 turns, find <?fi n and 5P 2 i (in nanowebers per ampere) when the coefficient of coupling is 0.6. 6.50 a) Starting with Eq. 6.59, show that the coefficient of coupling can also be expressed as ¢1 ¢12 02 b) On the basis of the fractions 4> 2 \f4>\ and <f>\ 2 /(f> 2 , explain why k is less than 1.0. Sections 6.1-6.5 6.51 Rework the Practical Perspective example, except PERSSE tnat this time, put the button on the bottom of the divider circuit, as shown in Fig. P6.51. Calculate the output voltage v(t) when a finger is present. Figure P6.51 >M&) Fixed capacitor' 25 pF Button + (A)»(0 6.52 Some lamps are made to turn on or off when the PERSPECTIVEb ase * s touched. These use a one-terminal variation of the capacitive switch circuit discussed in the Practical Perspective. Figure P6.52 shows a circuit model of such a lamp. Calculate the change in the voltage v(t) when a person touches the lamp. Assume all capacitors are initially discharged. Problems 211 Figure P6.43 Figure P6.52 Figure P6.53 10 pF Lamp Person 10 pF «*>© 10 pF! t • + tt loo pF; 6.53 In the Practical Perspective example, we calculated PRACTICAL the output voltage when the elevator button is the PERSPECTIVE r ° upper capacitor in a voltage divider. In Problem 6.51, we calculated the voltage when the button is the bottom capacitor in the divider, and we got the same result! You may wonder if this will be true for all such voltage dividers. Calculate the volt- age difference (finger versus no finger) for the cir- cuits in Figs. P6.53 (a) and (b), which use two identical voltage sources. — ' Fixed Button 25 pF capacitor 25 pF + "•* v{t) No finger (a) 25 pF 25 pF Button 25 pF Fixcd -" capacitor 25 pF + "* v(t) Finger (b) 7.1 The Natural Response of an RL Circuit p. 214 7.2 The Natural Response of an RCCircuit p. 220 7.3 The Step Response of RL and RC Circuits p. 224 7.4 A General Solution for Step and Natural Responses p. 231 7.5 Sequential Switching p. 235 7.6 Unbounded Response p. 240 7.7 The Integrating Amplifier p. 242 1 Be able to determine the natural response of both RL and RC circuits. 2 Be able to determine the step response of both RL and RC circuits. 3 Know how to analyze circuits with sequential switching. 4 Be able to analyze op amp circuits containing resistors and a single capacitor. 212 Response of First-Order RL and RC Circuits In Chapter 6, we noted that an important attribute of inductors and capacitors is their ability to store energy. We are now in a position to determine the currents and voltages that arise when energy is either released or acquired by an inductor or capacitor in response to an abrupt change in a dc voltage or current source. In this chapter, we will focus on circuits that consist only of sources, resistors, and either (but not both) inductors or capaci- tors. For brevity, such configurations are called RL (resistor- inductor) and RC (resistor-capacitor) circuits. Our analysis of RL and RC circuits will be divided into three phases. In the first phase, we consider the currents and voltages that arise when stored energy in an inductor or capacitor is sud- denly released to a resistive network. This happens when the inductor or capacitor is abruptly disconnected from its dc source. Thus we can reduce the circuit to one of the two equivalent forms shown in Fig. 7.1 on page 214. The currents and voltages that arise in this configuration are referred to as the natural response of the circuit, to emphasize that the nature of the circuit itself, not exter- nal sources of excitation, determines its behavior. In the second phase of our analysis, we consider the currents and voltages that arise when energy is being acquired by an induc- tor or capacitor due to the sudden application of a dc voltage or current source. This response is referred to as the step response. The process for finding both the natural and step responses is the same; thus, in the third phase of our analysis, we develop a general method that can be used to find the response of RL and RC cir- cuits to any abrupt change in a dc voltage or current source. Figure 7.2 on page 214 shows the four possibilities for the gen- eral configuration of RL and RC circuits. Note that when there are no independent sources in the circuit, the Thevenin voltage or Norton current is zero, and the circuit reduces to one of those shown in Fig. 7.1; that is, we have a natural-response problem. RL and RC circuits are also known as first-order circuits, because their voltages and currents are described by first-order differential equations. No matter how complex a circuit may Practical Perspective A Flashing Light Circuit You can probably think of many different applications that require a flashing light. A still camera used to take pictures in low light conditions employs a bright flash of light to illumi- nate the scene for just long enough to record the image on film. Generally, the camera cannot take another picture until the circuit that creates the flash of light has "re-charged." Other applications use flashing lights as warning for haz- ards, such as tall antenna towers, construction sites, and secure areas. In designing circuits to produce a flash of light the engineer must know the requirements of the application. For example, the design engineer has to know whether the flash is controlled manually by operating a switch (as in the case of a camera) or if the flash is to repeat itself automati- cally at a predetermined rate. The engineer also has to know if the flashing light is a permanent fixture (as on an antenna) or + V s 1 i K C*-^ k^i , c 1 < + ^^ ""^p v =s % 5H rt% p\ Lamp a temporary installation (as at a construction site). Another question that has to be answered is whether a power source is readily available. Many of the circuits that are used today to control flashing lights are based on electronic circuits that are beyond the scope of this text. Nevertheless we can get a feel for the thought process involved in designing a flashing light circuit by analyzing a circuit consisting of a dc voltage source, a resis- tor, a capacitor, and a lamp that is designed to discharge a flash of light at a critical voltage. Such a circuit is shown in the figure. We shall discuss this circuit at the end of the chapter. 213 214 Response of First-Order RL and RC Circuits R eq C, eq. (a) (b) Figure 7.1 • The two forms of the circuits for natural response, (a) RL circuit, (b) RC circuit. R rii L i v (b) (d) Figure 7.2 A Four possible first-order circuits. (a) An inductor connected to a Thevem'n equivalent. (b) An inductor connected to a Norton equivalent. (c) A capacitor connected to a Thevem'n equivalent. (d) A capacitor connected to a Norton equivalent. ^V r = () Ro Rkv Figure 7.3 • An RL circuit. /(0) = I \ L RZv appear, if it can be reduced to a Thevenin or Norton equivalent connected to the terminals of an equivalent inductor or capacitor, it is a first-order circuit. (Note that if multiple inductors or capacitors exist in the original circuit, they must be interconnected so that they can be replaced by a sin- gle equivalent element.) After introducing the techniques for analyzing the natural and step responses of first-order circuits, we discuss some special cases of interest. The first is that of sequential switching, involving circuits in which switching can take place at two or more instants in time. Next is the unbounded response. Finally, we analyze a useful circuit called the integrating amplifier. 7.1 The Natural Response of an RL Circuit The natural response of an RL circuit can best be described in terms of the circuit shown in Fig. 7.3. We assume that the independent current source generates a constant current of I s A, and that the switch has been in a closed position for a long time. We define the phrase a long time more accurately later in this section. For now it means that all currents and volt- ages have reached a constant value. Thus only constant, or dc, currents can exist in the circuit just prior to the switch's being opened, and therefore the inductor appears as a short circuit (Ldi/dt = 0) prior to the release of the stored energy. Because the inductor appears as a short circuit, the voltage across the inductive branch is zero, and there can be no current in either R () or R. Therefore, all the source current / s appears in the inductive branch. Finding the natural response requires finding the voltage and current at the terminals of the resistor after the switch has been opened, that is, after the source has been disconnected and the inductor begins releasing energy. If we let t = 0 denote the instant when the switch is opened, the problem becomes one of finding v(t) and i(t) for f>0. For t S 0, the cir- cuit shown in Fig. 7.3 reduces to the one shown in Fig. 7.4. Deriving the Expression for the Current To find i(i), we use Kirchhoff s voltage law to obtain an expression involv- ing i, R, and L. Summing the voltages around the closed loop gives r dl L— + Ri dt 0, (7.1) where we use the passive sign convention. Equation 7.1 is known as a first- order ordinary differential equation, because it contains terms involving the ordinary derivative of the unknown, that is, di/dt. The highest order derivative appearing in the equation is 1; hence the term first-order. We can go one step further in describing this equation. The coeffi- cients in the equation, R and L, are constants; that is, they are not func- tions of either the dependent variable i or the independent variable f.Thus the equation can also be described as an ordinary differential equation with constant coefficients. To solve Eq. 7.1, we divide by L, transpose the term involving i to the right-hand side, and then multiply both sides by a differential time dt. The result is Figure 7.4 A The circuit shown in Fig. 7.3, for t s 0. —dt= —-idt. dt L (7.2) 7.1 The Natural Response of an RL Circuit 215 Next, we recognize the left-hand side of Eq. 7.2 as a differential change in the current /, that is, di. We now divide through by i, getting di R J — = —rdt. i L (7.3) We obtain an explicit expression for i as a function of f by integrating both sides of Eq. 7.3. Using x and y as variables of integration yields *(0 dx R i(/„) X U < (7.4) in which /(f 0 ) is the current corresponding to time fg, and /(f) is the current corresponding to time f. Here, f () = 0. Therefore, carrying out the indi- cated integration gives /(f) In /(0) R = ~1 L Based on the definition of the natural logarithm, /(f) = i(0)e~ ls/L K (7.5) (7.6) Recall from Chapter 6 that an instantaneous change of current cannot occur in an inductor. Therefore, in the first instant after the switch has been opened, the current in the inductor remains unchanged. If we use 0~ to denote the time just prior to switching, and 0 + for the time immediately following switching, then ,(0-) = /(0 + ) = / 0 , -4 Initial inductor current where, as in Fig. 7.1, / () denotes the initial current in the inductor. The initial current in the inductor is oriented in the same direction as the reference direction of /. Hence Eq. 7.6 becomes /(f) = / 0 <T WL) ', f ^ 0, (7.7) -4 Natural response of an RL circuit which shows that the current starts from an initial value I () and decreases exponentially toward zero as f increases. Figure 7.5 shows this response. We derive the voltage across the resistor in Fig. 7.4 from a direct appli- cation of Ohm's law: v = iR = I 0 Re- WL)t , t > 0 + . (7.8) Note that in contrast to the expression for the current shown in Eq. 7.7, the voltage is defined only for t > 0, not at f = 0. The reason is that a step change occurs in the voltage at zero. Note that for t < 0, the derivative of the current is zero, so the voltage is also zero. (This result follows from v = Ldi/dt = 0.) Thus ?;(CT) = 0, v(0 + ) = I 0 R, (7.9) (7.10) where v(0 + ) is obtained from Eq. 7.8 with f = 0" 1 ". 1 With this step change at an instant in time, the value of the voltage at f = 0 is unknown. Thus we use f > 0 + in defining the region of validity for these solutions. Figure 7.5 A The current response for the circuit shown in Fig. 7.4. 1 We can define the expressions ()"" and u + more formally. The expression x(0~) refers to the limit of the variable x as / —»0 from the left, or from negative time. The expression .v(O') refers to the limit of the variable x as / —*• 0 from the right, or from positive time. . that is, we have a natural-response problem. RL and RC circuits are also known as first-order circuits, because their voltages and currents are described by first-order differential equations After introducing the techniques for analyzing the natural and step responses of first-order circuits, we discuss some special cases of interest. The first is that of sequential switching,