Electric Circuits, 9th Edition P33 ppt

296 Natural and Step Responses of RLC Circuits To find the maximum value of ^ sp , we find the smallest positive value of time where dv &p /dt is zero and then evaluate v sp at this instant The expression for t max is tarn = —tan" 1 ) — ). (8.99) (o d \a J (See Problem 8.67) For the component values in the problem statement, we have R 4 X 10 3 a = —- = —-— = 666.67 rad/s, 2L and f 10 9 <°d = A/TT ~ (666.67) 2 = 28,859.81 rad/s. 1.2 Substituting these values into Eq. 8.99 gives ^max = 53.63 flS. Now use Eq. 8.98 to find the maximum spark plug voltage, v sp (t metx ): VU = -25,975.69 V. b) The voltage across the capacitor at r max is obtained from Eq. 8.97 as ^c(W) = 262.15 V. The dielectric strength of air is approximately 3 x 10 6 V/m, so this result tells us that the switch contacts must be separated by 262.15/3 X 10 6 , or 87.38, fj,m to prevent arcing at the points at ? max . In the design and testing of ignition systems, consideration must be given to nonuniform fuel-air mixtures; the widening of the spark plug gap over time due to the erosion of the plug electrodes; the relationship between available spark plug voltage and engine speed; the time it takes the primary current to build up to its initial value after the switch is closed; and the amount of maintenance required to ensure reliable operation. We can use the preceding analysis of a conventional ignition system to explain why electronic switching has replaced mechanical switching in today's automobiles. First, the current emphasis on fuel economy and exhaust emissions requires a spark plug with a wider gap. This, in turn, requires a higher available spark plug voltage. These higher voltages (up to 40 kV) cannot be achieved with mechanical switching. Electronic switching also permits higher initial currents in the primary winding of the autotransformer. This means the initial stored energy in the system is larger, and hence a wider range of fuel-air mixtures and running condi- tions can be accommodated. Finally, the electronic switching circuit elim- inates the need for the point contacts. This means the deleterious effects of point contact arcing can be removed from the system. NOTE: Assess your understanding of the Practical Perspective by trying Chapter Problems 8.68 and 8.69. Summary Summary 297 The characteristic equation for both the parallel and series RLC circuits has the form s 2 + 2as + O)Q = 0, where a = 1/2RC for the parallel circuit, a = R/2L for the series circuit, and col = ^/LC for both the parallel and series circuits. (See pages 267 and 286.) The roots of the characteristic equation are *1.2 B -a ± Vo 2 " (4- (See page 268.) • The form of the natural and step responses of series and parallel RLC circuits depends on the values of a 2 and col; sucrl responses can be overdamped, underdamped, or critically damped. These terms describe the impact of the dissipative element (R) on the response. The neper frequency, a, reflects the effect of R. (See pages 268 and 269.) • The response of a second-order circuit is overdamped, underdamped, or critically damped as shown in Table 8.2. • In determining the natural response of a second-order circuit, we first determine whether it is over-, under-, or critically damped, and then we solve the appropriate equations as shown in Table 8.3. • In determining the step response of a second-order cir- cuit, we apply the appropriate equations depending on the damping, as shown in Table 8.4. • For each of the three forms of response, the unknown coefficients (i.e., the As, B s, and Ds) are obtained by evaluating the circuit to find the initial value of the response, x(0), and the initial value of the first deriva- tive of the response, dx(Q)/dt. • When two integrating amplifiers with ideal op amps are connected in cascade, the output voltage of the second integrator is related to the input voltage of the first by an ordinary, second-order differential equation. Therefore, the techniques developed in this chapter may be used to analyze the behavior of a cascaded integrator. (See pages 289 and 290.) • We can overcome the limitation of a simple integrating amplifier—the saturation of the op amp due to charge accumulating in the feedback capacitor—by placing a resistor in parallel with the capacitor in the feedback path. (See page 291.) TABLE 8.2 The Response of a Second-Order Circuit is Overdamped, Underdamped, or Critically Damped The Circuit is When Qualitative Nature of the Response Overdamped a 2 > oil Underdamped Critically damped a" < oj{) 2 2 The voltage or current approaches its final value without oscillation The voltage or current oscillates about its final value The voltage or current is on the verge of oscillating about its final value TABLE 8.3 In Determining the Natural Response of a Second-Order Circuit, We First Determine Whether it is Over-, Under-, or Critically Damped, and Then We Solve the Appropriate Equations Damping Natural Response Equations Overdamped x(t) = A^e* 1 ' + A 2 e S2 ' Underdamped x{t) - (B x cos <a d t + B 2 sin o) d t)e' Critically damped x(t) = {Dj. + D 2 )e~°" Coefficient Equations JC(0) = Ai + A 2 ; dx/dt(0) = A { s { + A 2 s 2 x(0) - B i; dx/dt(0) = -aBi + <o d B 2 , where o> d = VOJQ - a 2 -v(0) = D 2 , dx/dt(0) = D i - aD 2 298 Natural and Step Responses of RLC Circuits TABLE 8.4 In Determining the Step Response of a Second-Order Circuit, We Apply the Appropriate Equations Depending on the Damping Damping Step Response Equations 9 Overdamped x(t) = X f + A[ e iV + A 2 e* 2 ' Underdamped x(t) = Xf + (B[ cos <o d t + B' 2 sin a> d t)e~ Critically damped x(t) = X f + D[ te~ al + D' 2 e~ al a where X* is the final value of x(t). Coefficient Equations x(0) = X f + A\ + A 2 ; dx/dt(0) = A\ s, + A 2 s 2 x(0) - X f + B\ ; dx/dt(0) = -aB\ + <o d B' 2 x(0) = X f + D' 2 ; dx/dt(0) = D\ - aD' 2 Problems Sections 8.1-8.2 8.1 The resistance, inductance, and capacitance in a parallel RLC circuit are 2000 ft, 250 mH, and 10 nF, respectively. a) Calculate the roots of the characteristic equation that describe the voltage response of the circuit. b) Will the response be over-, under-, or critically damped? c) What value of R will yield a damped frequency of 12 krad/s? d) What are the roots of the characteristic equation for the value of R found in (c)? e) What value of R will result in a critically damped response? 8.2 The circuit elements in the circuit in Fig. 8.1 are R = 200 ft, C = 200 nF, and L = 50 mH. The ini- tial inductor current is -45 mA, and the initial capacitor voltage is 15 V. a) Calculate the initial current in each branch of the circuit. b) Find v(t) for t > 0. c) Find i L {t) for t > 0. 8.3 The resistance in Problem 8.2 is increased to PSPICE 312.5 ft. Find the expression for v(t) for t > 0. MULTISIM 8.4 The resistance in Problem 8.2 is increased to 250 ft. PSPICE Find the expression for v(t) for t > 0. MULTISIM 8.5 a) Design a parallel RLC circuit (see Fig. 8.1) using component values from Appendix H, with a res- onant radian frequency of 5000 rad/s. Choose a resistor or create a resistor network so that the response is critically damped. Draw your circuit. PSPICE MULTISIM b) Calculate the roots of the characteristic equa- tion for the resistance in part (a). 8.6 a) Change the resistance for the circuit you designed in Problem 8.5(a) so that the response is underdamped. Continue to use components from Appendix H. Calculate the roots of the characteristic equation for this new resistance. b) Change the resistance for the circuit you designed in Problem 8.5(a) so that the response is over- damped. Continue to use components from Appendix H. Calculate the roots of the character- istic equation for this new resistance. 8.7 The natural voltage response of the circuit in Fig. 8.1 is v(t) = 75<r 800,) '(cos 6000/ - 4 sin 60000V, t > 0, when the inductor is 400 mH. Find (a) C; (b) R; (c)V 0 ;(d)/ 0 ;and(e)/ L (/). 8.8 Suppose the capacitor in the circuit shown in Fig. 8.1 has a value of 0.1 juF and an initial voltage of 24 V. The initial current in the inductor is zero. The resulting voltage response for / s 0 is v(t) = -8e- 250t + 32^ 1000 ' V. a) Determine the numerical values of R, L, a, and <w 0 . b) Calculate i R (t), i L (t), and i c (t) for t > 0 + . 8.9 The voltage response for the circuit in Fig. 8.1 is known to be 500* v(f) = Dite'™ + D 2 e -500/ t >0. Problems 299 The initial current in the inductor (/ () ) is -10 mA, and the initial voltage on the capacitor (VQ) is 8 V. The inductor has an inductance of 4 H. a) Find the values of R, C, D h and D 2 . b) Find i c (t) for t > 0 + . 8.10 The natural response for the circuit shown in Fig. 8.1 is known to be v(t) = -l\e- im + 20e- 400 ' V, t > 0. If C = 2 /xF and L = 12.5 H, find i L (i) + ) in milli- amperes. 8.11 Tlie initial value of the voltage v in the circuit in Fig. 8.1 is zero, and the initial value of the capacitor current, / c (0 + ), is 45 mA. The expression for the capacitor current is known to be i c (t) = A x e- m)t + A 2 e~ m \ t > 0 + , when R is 250 ft. Find a) the values of a, co {) , L, C, A h and A 2 Hint: di c (0 + ) di L (0 + ) di R (Q + ) -v(0) 1 / c (0 + ) dt ell dt L R C b) the expression for v(t), t 2: 0, c) the expression for i R (t) > 0, d) the expression for i L {t) £: 0. 8.12 Assume the underdamped voltage response of the circuit in Fig. 8.1 is written as v(t) = (A x + A 2 )e~ al cos (o ( ,t + y'(^i _ A 2 )e~ at sin a> d t The initial value of the inductor current is / () , and the initial value of the capacitor voltage is V {) . Show that A 2 is the conjugate of A]_. (Hint: Use the same process as outlined in the text to find A\ and A 2 .) 8.13 Show that the results obtained from Problem 8.12— that is, the expressions for A x and A 2 —are consistent with Eqs. 8.30 and 8.31 in the text. 8.14 In the circuit in Fig. 8.1, R = 5 kft, L = 8 H, PSPICE c = 125 n p, v 0 = 30 V, and / 0 = 6 mA. MULTISIM a) Find v{t) for t > 0. b) Find the first three values of t for which dv/dt is zero. Let these values of t be denoted *j, t 2 , and r 3 . c) Show that t 3 — t\ — T (l . d) Show that t 2 - t x = T d /2. e) Calculate v(t { ), v(t 2 ), and v(t 3 ). f) Sketch v(t) versus t for 0 < t < t 2 . 8.15 a) Find v(t) for t > 0 in the circuit in Problem 8.14 if the 5 kfi resistor is removed from the circuit. MULTISIM b) Calculate the frequency of v(t) in hertz. c) Calculate the maximum amplitude of v(t) in volts. 8.16 In the circuit shown in Fig. 8.1, a 2.5 H inductor is PSPICE shunted by a 100 nF capacitor, the resistor R is MULTISIM adjusted for critical damping, V 0 = -15 V, and /() = —5 mA. a) Calculate the numerical value of R. b) Calculate v(t) for t > 0. c) Find v{t) when i c (t) = 0. d) What percentage of the initially stored energy remains stored in the circuit at the instant / c (r) isO? 8.17 The resistor in the circuit in Example 8.4 is changed «PICE to 3200 a. MULTISIM a) Find the numerical expression for v{t) when t > 0. b) Plot v(t) versus t for the time interval 0 s f < 7 ras. Compare this response with the one in Example 8.4 (R = 20kft) and Example 8.5 (R = 4 kH). In particular, compare peak values of v(t) and the times when these peak values occur. 8.18 The two switches in the circuit seen in Fig. P8.18 PSPICE operate synchronously. When switch 1 is in position a, switch 2 is in position d. When switch 1 moves to position b, switch 2 moves to position c. Switch 1 has been in position a for a long time. At t = 0, the switches move to their alternate positions. Find v 0 (t) for t > 0. Figure P8.18 in 8.19 The resistor in the circuit of Fig. P8.18 is increased PSPICE f r om 100 H to 200 O. Find v a (t) for t > 0. MULTISIM 8.20 The resistor in the circuit of Fig. P8.18 is increased "sn« from 100 ft to 125 ft. Find vjt) for t s 0. MULTISIM 8.21 The switch in the circuit of Fig. P8.21 has been in PSPICE position a for a long time. At J = 0 the switch moves instantaneously to position b. Find v 0 (t) for /s0. t = () a\ / b 16 X lO 3 /* 7.5 V %4kft 24 kft !4nF ? 15.625 H A 300 Natural and Step Responses of RLC Circuits 8.22 The inductor in the circuit of Fig. P8.21 is decreased to 10 H. Find v a {t) for t > 0. 8.23 The inductor in the circuit of Fig. P8.21 is decreased to 6.4 H. Find v 0 (t) for t > 0. Section 8.3 8.24 For the circuit in Example 8.6, find, for t > 0, PSPICE (a) v{t) . (b) iR{t) . and (c) /c(0 . MULTISIM 8.25 For the circuit in Example 8.7, find, for t 3: 0, «"« (a) v(t) and (b) i c (t\ MULTISIM v \ / «-\ / 8.26 For the circuit in Example 8.8, find v(t) for t > 0. 8.27 The switch in the circuit in Fig. P8.27 has been open MULTISIM a l° n § tnne ^ e f° re closing at t = 0. Find a) v 0 (t) for t > 0 + , b) i L {t) for t > 0. Figure P8.27 156.25 a 25 V v 0 < 312.5 mH PSPICE MULTISIM 8.28 Use the circuit in Fig. P8.27 a) Find the total energy delivered to the inductor. b) Find the total energy delivered to the equivalent resistor. c) Find the total energy delivered to the capacitor. d) Find the total energy delivered by the equiva- lent current source. e) Check the results of parts (a) through (d) against the conservation of energy principle. Assume that at the instant the 60 mA dc current source is applied to the circuit in Fig. P8.29, the ini- tial current in the 50 mH inductor is -45 mA, and the initial voltage on the capacitor is 15 V (positive at the upper terminal). Find the expression for i L {t) for/ > 0 if # equals 200 H. Figure P8.29 8.29 PSPICE MULTISIM 60 mA '/.(0M5( 8.30 The resistance in the circuit in Fig. P8.29 is changed PSPICE to 312.5 a. Find i L (t) for t > 0. MULTISIM 8.31 The resistance in the circuit in Fig. P8.29 is changed PSPICE to 250 ft. Find i L {t) for t > 0. MULTISIM 8.32 The switch in the circuit in Fig. P8.32 has been PSPICE open a long time before closing at t = 0. Find i r (t) MULTISIM for fe a Figure P8.32 3 kfl 15 V 8.33 Switches 1 and 2 in the circuit in Fig. P8.33 are syn- PSPKE chronized. When switch 1 is opened, switch 2 closes and vice versa. Switch 1 has been open a long time before closing at / = 0. Find i L (t) for t 2: (). 8.34 The switch in the circuit in Fig. P8.34 has been open PSPICE for a long time before closing at t = 0. Find v 0 (t) mnsiM for/ > 0> Figure P8.34 12 V 400 a ^ / = 0 1.25/xF + Ml .251 8.35 a) For the circuit in Fig. P8.34, find i 0 for t > 0. PSPICE b) Show that your solution for L is consistent with MULTISIM ' ' v the solution for v 0 in Problem 8.34. 8.36 The switch in the circuit in Fig. P8.36 has been PSPICE open a long time before closing at t = 0. At the time the switch closes, the capacitor has no stored energy. Find v a for t > 0. Figure P8.36 7.5 V 250 a ^ / = 0 !4H + v 0 , 25 fiF Figure P8.33 5 kfl Switch 1 | )60mA Problems 301 8.37 There is no energy stored in the circuit in Fig. P8.37 PSPICE when the switch is closed at t = 0. Find v () (t) ™ forr>0. Figure P8.37 12V 400 n ^ t = 0 1.25/nF: + 1 vAl.25H 8.38 a) For the circuit in Fig. P8.37, find i a for t > 0. b) Show that your solution for i a is consistent with the solution for v„ in Problem 8.37. PSPICE MULTISIM Section 8.4 8.39 The initial energy stored in the 31.25 nF capacitor in the circuit in Fig. P8.39 is 9 /xJ. The initial energy stored in the inductor is zero. The roots of the char- acteristic equation that describes the natural behav- ior of the current i are -4000 s _1 and -16,000 s _1 a) Find the numerical values of R and L. b) Find the numerical values of /(0) and di(0)/dt immediately after the switch has been closed. c) Find i(t) for d) How many microseconds after the switch closes does the current reach its maximum value? e) What is the maximum value of/ in milliamperes? f) Find v L {t) for t > 0. Figure P8.39 31.25 nF 8.40 a) Design a series RLC circuit (see Fig. 8.3) using component values from Appendix H, with a res- onant radian frequency of 20 krad/s. Choose a resistor or create a resistor network so that the response is critically damped. Draw your circuit. b) Calculate the roots of the characteristic equa- tion for the resistance in part (a). 8.41 a) Change the resistance for the circuit you designed in Problem 8.40(a) so that the response is underdamped. Continue to use components from Appendix H. Calculate the roots of the characteristic equation for this new resistance. b) Change the resistance for the circuit you designed in Problem 8.40(a) so that the response is overdamped. Continue to use components from Appendix Ff. Calculate the roots of the characteristic equation for this new resistance. 8.42 The current in the circuit in Fig. 8.3 is known to be i = 5 1 <r 2(,,),) ' cos 1500f + B 2 e~ 2mt sin 1500/, t > 0. The capacitor has a value of 80 nF; the initial value of the current is 7.5 mA; and the initial voltage on the capacitor is -30 V. Find the values of R, L, B h and B 2 . 8.43 Find the voltage across the 80 nF capacitor for the circuit described in Problem 8.42. Assume the refer- ence polarity for the capacitor voltage is positive at the upper terminal. 8.44 In the circuit in Fig. P8.44, the resistor is adjusted PSPICE for critical damping. The initial capacitor voltage is iiimsm 15 y anc j tne j n { t i a l inductor current is 6 mA. a) Find the numerical value of R. b) Find the numerical values of i and di/dt immedi- ately after the switch is closed. c) Find v c (t) for t a 0. Figure P8.44 + TA !320nF R -vw- 125 mH 8.45 The switch in the circuit shown in Fig. P8.45 has PSPICE been in position a for a long time. At t = 0, the switch is moved instantaneously to position b. Find /(0 for t > 0. Figure P8.45 8012 AM- 10 H 8.46 The switch in the circuit in Fig. P8.46 on the next page has been in position a for a long time. At / = 0, the switch moves instantaneously to position b. a) What is the initial value of t> fl ? b) What is the initial value of dvjdtl c) What is the numerical expression for vjf) for t > 0? PSPICE MULTISIM 302 Natural and Step Responses of RLC Circuits Figure P8.46 Figure P8.50 75 V 1 kO AAA- o.i MF: / = 0 2kO -AW- 3 kO -AAA— 400 mH 8.47 The switch in the circuit shown in Fig. P8.47 has PSPICE been closed for a long time. The switch opens at *™ r = 0.Find a) i () (t) for t > 0, b) v a {t) for t > 0. Figure P8.47 f = 0 300 a >v 80 V 6 500 n| f /,,(0 + 2.5 mH • :4() nF 8.48 The switch in the circuit shown in Fig. P8.48 has been closed for a long time. The switch opens at t = 0. Find v t) (t) for t > 0. Figure P8.48 8.49 The circuit shown in Fig. P8.49 has been in operation PSPICE for a long time. At t = 0, the source voltage suddenly MULTI5IM jumps to 250 V. Find v () {t) for t > 0. Figure P8.49 8kH ^AV- 160mH 50V! 10 nF! + v„[t) 8.50 The initial energy stored in the circuit in Fig. P8.50 PSPICE is zero. Find v () {t) for t > 0. 250 n 60 V 8.51 The capacitor in the circuit shown in Fig. P8.50 is changed to 4 ju.F. The initial energy stored is still zero. Find v () (t) for t SE 0. 8.52 The capacitor in the circuit shown in Fig. P8.50 is changed to 2.56 ^tF. The initial energy stored is still zero. Find v a (t) for t > 0. 8.53 The switch in the circuit of Fig. P8.53 has been in PSPICE position a for a long time. At t = 0 the switch moves instantaneously to position b. Find a) v o (0 + ) b) dv () (Q + )/dt c) v„(t) for t > 0. Figure P8.53 8.54 The switch in the circuit shown in Fig. P8.54 has been closed for a long time before it is opened at t = 0. Assume that the circuit parameters are such that the response is underdamped. a) Derive the expression for vjt) as a function of Vg, a, co d , C, and R for t > 0. b) Derive the expression for the value of t when the magnitude of v 0 is maximum. Figure P8.54 t = 0 \i- R -AMr- c + Liv a {t) 8.55 The circuit parameters in the circuit of Fig. P8.54 PSPICE are R = 4800 ft, L = 64 mH, C = 4 nF, and «iunsm ^ = _ 72V a) Express v a (t) numerically for t & 0. b) How many microseconds after the switch opens is the inductor voltage maximum? Problems 303 c) What is the maximum value of the inductor voltage? d) Repeat (a)-(c) with R reduced to 480 ft. 8.56 The two switches in the circuit seen in Fig. P8.56 PSPICE operate synchronously. When switch 1 is in position a, switch 2 is closed. When switch 1 is in position b, switch 2 is open. Switch 1 has been in position a for a long time. At ( = 0, it moves instan- taneously to position b. Find v c (t) for t > 0. Figure P8.56 w 4 .(') £1811 8.57 Assume that the capacitor voltage in the circuit of Fig. 8.15 is underdamped. Also assume that no energy is stored in the circuit elements when the switch is closed. a) Show that dvc/dt = (a)Q/a) (i )Ve~ lxl s'mw d t. b) Show that dv c /dt = 0 when t = mrfo) th where n = 0,1,2 c) Let t n = mr/aj, and show that v c (t„) = y _ y/_|\n e -t«Hr/«Brf d) Show that 1 v c { h ) - V (X = 777" m :—: —, T d v c (h) - V where T d - t 3 - t { . 8.58 The voltage across a 100 nF capacitor in the circuit of Fig. 8.15 is described as follows: After the switch has been closed for several seconds, the voltage is constant at 100 V. The first time the voltage exceeds 100 V, it reaches a peak of 163.84 V. This occurs 7r/7 ms after the switch has been closed. The second time the voltage exceeds 100 V, it reaches a peak of 126.02 V. This second peak occurs 3-77-/7 after the switch has been closed. At the time when the switch is closed, there is no energy stored in cither the capacitor or the inductor. Find the numerical values of R and L. (Hint: Work Problem 8.57 first.) Section 8.5 8.59 Show that, if no energy is stored in the circuit shown in Fig. 8.19 at the instant v s jumps in value, then dvjdt equals zero at t = 0. 8.60 a) Find the equation for v ( ,(t) for 0 < t < r sat in the circuit shown in Fig. 8.19 if u„i(0) = 5 V and vJQ) = 8 V. b) How long does the circuit take to reach saturation? 8.61 a) Rework Example 8.14 with feedback resistors Ri and R 2 removed. b) Rework Example 8.14 with v ol (0) = -2Vand v o (0) = 4 V. 8.62 a) Derive the differential equation that relates the output voltage to the input voltage for the circuit shown in Fig. P8.62. b) Compare the result with Eq. 8.75 when Rtd - R 2 C 2 = RC in Fig. 8.18. c) What is the advantage of the circuit shown in Fig. P8.62? Figure P8.62 8.63 The voltage signal of Fig. P8.63(a) is applied to PSPICE the cascaded integrating amplifiers shown in MULT1SIM Fig. P8.63(b). There is no energy stored in the capacitors at the instant the signal is applied. a) Derive the numerical expressions for v a (t) and v a i(t) for the time intervals 0 < t < 0.5 s and 0.5 s < t < t m . b) Compute the value of t S(lt . Figure P8.63 MrrAO 80 0 -40 . 0.5 1 1 f(s) (a) 304 Natural and Step Responses of RLC Circuits 8.64 The circuit in Fig. P8.63(b) is modified by adding a PSPICE i Mil resistor in parallel with the 500 nF capacitor mTISIM and a 5 MH resistor in parallel with the 200 nF capacitor. As in Problem 8.63, there is no energy stored in the capacitors at the time the signal is applied. Derive the numerical expressions for v n (t) and v o[ (t) for the time intervals 0 < t < 0.5 s and t > 0.5 s. 8.65 We now wish to illustrate how several op amp cir- cuits can be interconnected to solve a differential equation. a) Derive the differential equation for the spring- mass system shown in Fig. P8.65(a). Assume that the force exerted by the spring is directly proportional to the spring displacement, that the mass is constant, and that the frictional force is directly proportional to the velocity of the moving mass. b) Rewrite the differential equation derived in (a) so that the highest order derivative is expressed as a function of all the other terms in the equa- tion. Now assume that a voltage equal to d 2 x/dt 2 is available and by successive integrations gen- erates dx/dt and x. We can synthesize the coeffi- cients in the equations by scaling amplifiers, and we can combine the terms required to generate d 2 x/dt 2 by using a summing amplifier. With these ideas in mind, analyze the interconnection shown in Fig. P8.65(b). In particular, describe the purpose of each shaded area in the circuit and describe the signal at the points labeled B, Figure P8.65 K M -*(0 — (a) Ri (b) Problems 305 C, D, E, and F, assuming the signal at A repre- sents d 2 x/dt 2 . Also discuss the parameters R; R], Ci; R 2 , C2; R$, R4', R5, Re', and R 7 , i? 8 in terms of the coefficients in the differential equation. Sections 8.1-8.5 8.66 a) Derive Eq. 8.92. mSSmb) Derive Eq. 8.93. c) Derive Eq. 8.97. 8.67 Derive Eq. 8.99. PRACTICAL 8.68 a) Using the same numerical values used in the Practical Perspective example in the text, find the instant of time when the voltage across the capacitor is maximum. PRACTICAL PERSPECTIVE b) Find the maximum value of v c . c) Compare the values obtained in (a) and (b) with and v c (r max ). PRACTICAL pjcr PERSPECTIVE 6 V 8.69 The values of the parameters in the circuit in 8.21 are R = 3 O; L = 5 mH; C = 0.25 juF; i c — 12 V; and a = 50. Assume the switch opens when the primary winding current is 4 A. a) How much energy is stored in the circuit at t = 0 + ? b) Assume the spark plug does not fire. What is the maximum voltage available at the spark plug? c) What is the voltage across the capacitor when the voltage across the spark plug is at its maxi- mum value? 8.70 Repeat Problem 8.68 using the values given in Problem 8.69. . voltage across the capacitor at r max is obtained from Eq. 8.97 as ^c(W) = 262.15 V. The dielectric strength of air is approximately 3 x 10 6 V/m, so this result tells us that the switch