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 Equationsa

Sz1Overdarnped x(t) = X f + A; eS,1 + Aze

Underdarnped

Critically damped x(t) = Xf + D; te-at + Dze-o:l

Problems

Sections 8.1-8.2

8.1 The resistance, inductance, and capacitance in a parallel RLC circuit are 2000 fl, 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 PSPICE R = 200 fl, C = 200 nF, and L = 50 mHo The ini­

/-lUlTISIM tial inductor current is -45 rnA, and the initial capacitor voltage is 15 V.

a) Calculate the initial current in each branch of the circuit.

b) Find vet) for t :2: O.

c) Find iL(t) for t :2: O.

8.3 The resistance in Problem 8.2 is increased to PSPICE 312.5 fl. Find tbe expression for vet) for t :2: O.

/·,UlTIsm

8.4 The resistance in Problem 8.2 is increased to 250 fl. PSPICE Find the expression for vet) for t :2: O.

/-\ULTISlf.I

8.5 a) Design a parallel RLC circuit (see Fig. 8.1) using component values from Appendix H, with a res­onant radian frequency of 5000 rad/s. Choose a resistor or create a resistor network so that the response is critically damped. Draw your circuit.

Coefficient Equations

xeD) = XI + A; + A z; dx/dt(D) = A; S1 + Azs2

xeD) = X f + B; ;

dx/dl(D) = -aB; + WdBz

xeD) = X f + Dz; dx/dl(D) = D; - aDz

-------. ._._------­

b) Calculate the roots of the characteristic equa­tion 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 over­damped. Continue to use components from Appendix H. Calculate the roots of the character­istic equation for this new resistance.

8.7 The natural voltage response of the circuit in Fig. 8.1 is

vet) = 75e-80001(cos 6000t - 4 sin 6000t)V, t;:>: 0,

when the inductor is 400 mB. Find (a) C; (b) R; (c) Vo; (d) f o; and (e) iL(l).

8.8 Suppose the capacitor in the circuit shown in Fig. 8.1 has a value of 0.1 J.LF and an initial voltage of 24 V. The initial current in the inductor is zero. The resulting voltage response for l 2: 0 is

_8e-ZSOI + 32e-lOOOt V.vet) =

a) Determine the numerical values of R, L, 0',

and Wo.

b) Calculate in(t), iLCl), and icCt) for t 2: 0+.

8.9 The voltage response for the circuit in Fig. 8.1 is known to be

vet) = Djle-5001 + Dze-S001, t:2: O.

(

8 PSf

:·.Ul1

8 PSP:

MULf

Problems 299

The initial current in the inductor (10) is -10 rnA, and the initial voltage on the capacitor (Va) is 8 V. The inductor has an inductance of 4 H.

a) Find the values of R, C, D j , and D2.

b) Find ic(t) for t 2: 0+.

8.10 The natural response for the circuit shown in Fig. 8.1 is known to be

vet) = _l1e- IOOI -I- 20e-400I Y, t 2: O.

If C = 2 p,F and L = 12.5 H, find iL(O+) in milli­amperes.

8.11 The initial value of the voltage v in the circuit in Fig. 8.1 is zero, and the initial value of the capacitor current, ic(O+), is 45 rnA. The expression for the capacitor current is known to be

ic(t) = Ale-200t -I- A2e-SOOt, t 2: 0+,

when R is 250 D. Find

a) the values of Cl', Wo, L, C, At> and A 2

diL(O+) _ diR(O+) = -v(O) _ l idO+») dt dt L R C

b) the expression for vet), t 2: 0,

c) the expression for iR(t) 2: 0,

d) the expression for iL(t) 2: O.

8.12 Assume the underdamped voltage response of the circuit in Fig. 8.1 is written as

vet) = (AI + A2)e-ot cos wdt + j(AI - A 2)e-at sin wdt

The initial value of the inductor current is 10, and the initial value of the capacitor voltage is Va. Show that A2 is the conjugate of A 1. (Hint: Use the same process as outlined in the text to find AI and A2 .)

8.13 Show that the results obtained from Problem 8.12­that is, the expressions for Al 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 kD, L = 8 H, e = 125 nF, Va = 30 V, and 10 = 6 rnA.

~U1.7!5::·

a) Find v(t) for t 2: O.

b) Find the first three values of t for which dvldt is zero. Let these values of t be denoted t 1, t2>

and 13'

c) Show that t3 - tl = Td·

d) Show that l2 - t1 = T d12. e) Calculate vet,), V(t2), and v(t3).

f) Sketch vet) versus tfor 0 :5 t :5 t2'

8.15 a) Find vet) for t 2: 0 in the circuit in Problem 8.14 P5PICE if the 5 kD resistor is removed from the circuit.

':(Jm5Ir·;

b) Calculate the frequency of vet) in hertz.

c) Calculate the maximum amplitude of vet) in volts.

8.16 In the circuit shown in Fig. 8.1, a 2.5 H inductor is FSP[CE shunted by a 100 nF capacitor, the resistor R is

;·iUlTISli·! adjusted for critical damping, Va = -15 Y, and 10 = -SmA.

a) Calculate the numerical value of R.

b) Calculate vet) for t 2: O.

c) Find vet) when icCt) = O.

d) What percentage of the initially stored energy remains stored in the circuit at the instant idt) is O?

8.17 The resistor in the circuit in Example 8.4 is changed PSPICE to 3200 D.

a) Find the numerical expression for vet) when t 2: O.

b) Plot vet) versus t for the time interval o :5 t :5 7 ms. Compare this response with the one in Example 8.4 (R = 20 kD) and Example 8.5 (R = 4 kD). In paJ"ticular, compare peak values of vet) and the times when these peak values occur.

8.18 The two switches in the circuit seen in Fig. P8.18 PSFICE 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 vo(t) for t 2: O.

Figure P8.l8

500 n

8.19 The resistor in the circuit of Fig. P8.18 is increased rsPICE from 100 n to 200 n. Find vo(t) for I 2: O.

l·iULTISJI·\

8.20 The resistor in the circuit of Fig. P8.18 is increased FSPICE from 100 D to 125 D. Find vo(t) for t 2: O.

";ULTISlI·l

8.21 The switch in the circuit of Fig. P8.21 has been in F5PICE position a for a long time. At t = 0 the switch

;·;ULTISm moves instantaneously to position b. Find vo(t) for t 2: O.

Figure P8.2l

{ --II 16 X 103 i,p

1 kn - -f

a / b 24kflI',7.5V 4kD 40kD

4 nF l5.625 H I,', •

5kD240V

25 p.F

300 Natural and Step Responses of RLC Circuits r 8.22 The inductor in the circuit of Fig. P8.21 is decreased 8.30 The resistance in the circuit in Fig. P8.29 is changed

to 10 H. Find vo(t) for t 2: O. PSPICE to 312.5 n. Find iL(t) for t 2: O. /·IULTlSI/·,

8.23 The inductor in the circuit of Fig. P8.21 is decreased 8.31 The resistance in the circuit in Fig. P8.29 is changed to 6.4 H. Find v()(t) for t 2: O. PSPICE to 250 n. Find iL(t) for t 2: O.

r·1ULTISl/o1

8.32 The switch in the circuit in Fig. P8.32 has been Section 8.3 PSPICE open a long time before closing at t = O. Find iL(t)

/·1 ULTlSJI·' for t 2: O. 8.24 For the circuit In Example 8.6, find, for t 2: 0,

PSPICE (a) vet); (b) iR(t); and (c) ie(t). Figure P8.32I~ULTlSlt-1

8.25 For the circuit in Example 8.7, find, for t 2: 0, PSPICE (a) vet) and (b) idt).

'·IULTlml

8.26 For the circuit in Example 8.8, find vet) for t 2: O. PSPICE 15 V 9rnA

/·IULTISI/·1

8.27 The switch in the circuit in Fig. P8.27 has been open PSPICE

a long time before closing at t = O. Find ~'ULTlSlI-'

8.33 Switches 1 and 2 in the circuit in Fig. P8.33 are syn­a) vo(t) for t 2: 0+, PSPICE chronized. When switch 1 is opened, switch 2 closes

b) iL(t) fort 2: O. /·1ULTlSIJ.I and vice versa. Switch 1 has been open a long time before closing at t = O. Find iL(t) for f 2: O.

Figure P8.27 8.34 The switch in the circuit in Fig. P8.34 has been open

156.25 D PSPICE for a long time before closing at t = O. Find vo(t)

!·!UlTlSI/·/ for f 2: O.

25V 625 D 312.5 mH Figure P8.34

/" .. 8.28 Use the circuit in Fig. P8.27

PSPICE

/·IUlTISIJ.I a) Find the total energy delivered to the inductor. 12V /" 1.25 H b) Find the total energy delivered to the equivalent

resistor.

c) Find the total energy delivered to the capacitor. 8.35 a) For the circuit in Fig. P8.34, find io for f 2: O.

d) Find the total energy delivered by the equiva­ PSPICE b) S hi' f . .'.IULTlSlI.1 how t at your so utlOn 'or io IS consistent withlent current source.

the solution for Vo in Problem 8.34. e) Check the results of parts (a) through (d)

8.36 The switch in the circuit in Fig. P8.36 has beenagainst the conservation of energy principle. PSPICE open a long time before closing at t = O. At the

8.29 Assume that at the instant the 60 rnA dc current /·\UlTISlI-l time the switch closes, the capacitor has no storedPSPICE source is applied to the circuit in Fig. P8.29, the ini­MULTISI/·, energy. Find Vo for f 2: O.

tial current in the 50 mH inductor is -45 rnA, and the initial voltage on the capacitor is 15 V (positive at the upper terminal). Find the expression for iL(t) Figure P8.36

for t 2: 0 if R equals 200 n. 250£2

Figure P8.29 7.5 V 4H

60 rnA 200 nF R

L- ~---~e___--------lFigure P8.33

5kD

1kD

:d

n t)

n )

8.37 There is no energy stored in the circuit in Fig. P8.37 PSPICE when the switch is closed at t = O. Find vo(t)

MUlTISIH for t :::::: O.

Figure P8.37

400 D.

+ 12V v(/ 1.25 H

8.38 a) For the circuit in Fig. P8.37, find io for t :::::: O. PSPICE ) Show that your solution for io is consistent with MUlTISIM b

the solution for va in Problem 8.37.

Section 8.4

8.39 The initial energy stored in the 31.25 nF capacitor in the circuit in Fig. P8.39 is 9 f.LJ. The initial energy stored in the inductor is zero. The roots of the char­acteristic equation that describes the natural behav­ior of the current i are -4000 S-1 and -16,000 S-1

a) Find the numerical values of Rand L.

b) Find the numerical values of i(O) and di(O)jdt 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 i in milliamperes?

f) Find VL(t) for t :::::: O.

Figure P8.39

R

(= 0

31.25 nF

8.40 a) Design a series RLC circuit (see Fig. 8.3) using component values from Appendix H, with a res­onant radian frequency of 20 kradjs. 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 equa­tion 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.

Problems 301

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 H. 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 = B1e-zooor cos 1500t + Bze-zoool sin 1500t, t:::::: O.

The capacitor has a value of 80 nF; the initial value of the current is 7.5 rnA; and the initial voltage on the capacitor is -30 V. Find the values of R, L, B], and Bz.

8.43 Find the voltage across the 80 nF capacitor for the circuit described in Problem 8.42. Assume the refer­ence 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

r~ULTIslt~ 15 V, and the initial inductor current is 6 rnA.

a) Find the numerical value of R.

b) Find the numerical values of i and dijdt immedi­ately after the switch is closed.

c) Find vc(t) for t :::::: O.

Figure P8.44

R

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

~IULTISH~ switch is moved instantaneously to position b. Find i(t) for t :::::: O.

Figure P8.45

SOD. 20D.

lOB240 V

8.46 The switch in the circuit in Fig. P8.46 on the next PSPICE page has been in position a for a long time. At t = 0,

HULTISH·' the switch moves instantaneously to position b.

a) What is the initial value of va?

b) What is the initial value of dval dt?

c) What is the numerical expression for va(t) for t :::::: O?

F 302 Natural and Step Responses of RLC Circuits

Figure PB.46 Figure PB.5D

62.5 mH 250 D 2kD 3kD

I' I', (il75V 400mH 60V 6.25 f-LF'f

8.51 The capacitor in the circuit shown in Fig. P8.50 is 8.47 The switch in the circuit shown in Fig. P8.47 has changed to 4 p.,F. The initial energy stored is still

PSPICE been closed for a long time. The switch opens at zero. Find vo(t) for t :2: O. /·IULTISf1.\

t = O. Find 8.52 The capacitor in the circuit shown in Fig. P8.50 is

a) io(t) for t :2: 0, changed to 2.56 p.,F. The initial energy stored is still b) vo(t) for t 2: O. zero. Find vu(t) for t :2: O.

8.53 The switch in the circuit of Fig. P8.53 has been inFigure P8.47 P5PICE position a for a long time. At t = 0 the switch

{ = (J 1·\ULTl5H·1 moves instantaneously to position b. Find300D

a) voCO+) 500 D ,i,,(I) b) dvu(O+)/ dt

40nF c) vo(t) fort :2: O.SOV

Figure P8.53

b 3 kD

8.48 The switch in the circuit shown in Fig. P8.48 has {= (J

4kD a 6kD 500mHbeen closed for a long time. The switch opens at t = O. Find vo(!) for t :2: O.

Figure P8.48

IOD

5 mF

1= (J

20D70D

lID

lH 8.54 The switch in the circuit shown in Fig. P8.54 has

been closed for a long time before it is opened at I = O. Assume that the circuit parameters are such that the response is underdamped.

I'.. a) Derive the expression for vo(t) as a function of Yg, a, Wd, C, and R for t :2: O.

b) Derive the expression for the value of t when the magnitude of va is maximum.

Figure P8.54

40 V 781.25 nF

.J

8.49 PsprCE

1·'ULTISIl-I

The circuit shown in Fig. P8.49 has been in operation for a long time. At t = 0, the source voltage suddenly jumps to 250 V. Find vo(t) for I :2: O.

r = (J

R

Figure P8.49

8.50 PSPlCE

1,IUlTISlI-l

111e initial energy stored in the circuit in Fig. P8.50 is zero. Find vo(!) for t :2: O.

8.55 P5PICE

!.lULTtSIf.i

The circuit parameters in the circuit of Fig. P8.54 are R = 4800 D, L = 64 mH, C = 4 nF, and vg = -72 V.

a) Express vo(t) numerically for t :2: O.

b) How many microseconds after the switch opens is the inductor voltage maximum?

SD

Problems 303

b) How long does the circuit take to reach saturation?

8.61 a) Rework Example 8.14 with feedback resistors R j and R2 removed.

b) Rework Example 8.14 with Vol(O) = -2 V and vo(O) = 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 RtC J = R2C2 = RC in Fig. 8.18.

c) What is the advantage of the circuit shown in Fig. P8.62?

8.63 The voltage signal of Fig. P8.63(a) is applied to PSPIEE the cascaded integrating amplifiers shown in

1,IULTISlf.1 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 Vo(f) and Vol (f) for the time intervals 0 :S f :S 0.5 sand 0.5 s :S f ':S f sllt

b) Compute the value of '.<0/'

Figure P8.63

vg (mY)

SO 1--------,

-,0,n------:-------1..------ I (s)-40 0.51'- --' _

(a)

500 nF 200nF

400 kn

l'"l

(b)

c) What is the maximum value of the inductor voltage?

d) Repeat (a)-(c) with R reduced to 480.0..

8.56 The two switches in the circuit seen in Fig. P8.56 PlPIEE operate synchronously. When switch 1 is in

HUlTISIII 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 f = 0, it moves instan­taneously to position b. Find ve(!) for ( 2: O.

Figure P8.56

4D [=(1

100 mH 2

" 1',.(1 )2mF 1S n

Asswne 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) Showthatdvcldf = (w5Iwd)Ve-CXl sin Wdl.

b) Show that dveldt = 0 when f = 17.7;1Wd, where n = 0,1, 2, ....

c) Let til = n7llwd, and show that VC(tIl) = V - V(-l)"e-crmrj,"".

d) Show that

8.57

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. TI1e first time the voltage exceeds 100 V, it reaches a peak of 163.84 V. This occurs 71/7 ms after the switch has been closed. The second time the voltage exceeds 100 V, it reaches a peak of 126.02 V. TI1is second peak occurs 371/7 after the switch has been closed. At the time when the switch is closed, there is no energy stored in either the capacitor or the inductor. Find the numerical values of Rand 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 vI! jumps in value, then dVol df equals zero at t = O.

8.60 a) Find the equation for vo(t) for 0 :S ( :S I Sal in the circuit shown in Fig. 8.19 if Vol(O) = 5 V and vo(O) = 8 V.

Figure P8.62

R R

c c

~ Vee f!{/i

1'"

304 Natural and Step Responses of RLC Circuits

8.64 The circuit in Fig. P8.63(b) is modified by adding a PSPICE 1 Mil resistor in parallel with the 500 nF capacitor

MUmSlM and a 5 Mil 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 vo(t) and vol (t) for the time intervals a s:; t s:; 0.5 sand t ;::: 0.5 s.

8.65 We now wish to illustrate how several op amp cir­cuits can be interconnected to solve a differential equation.

a) Derive the differential equation for the spring­mass system shown in Fig. P8.65(a). Assume that the force exerted by the spring is directly proportional to the spring displacement, that

Figure P8.65

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 equa­tion. Now assume that a voltage equal to d2x/dt2

is available and by successive integrations gen­erates dx/dt and x. We can synthesize the coeffi­cients in the equations by scaling amplifiers, and we can combine the terms required to generate d2x/dP 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,

II-X(t)-.­M -~f(t)

'--- /D

(a)

1

5

4f( t) _./I,fI/\.--+--~

6 +e+-------------'

(b)

Problems 305

C, D, E, and F, assuming the signal at A repre­ b) Find the maximum value of vc. sents d2x/dt2

. Also discuss the parameters R; R] , c) Compare the values obtained in (a) and (b) with C]; R2 , C2 ; R3 , R4 ; Rj , R6 ; and R7 , Rs in terms

t max and vc(tmaJ.of the coefficients in the differential equation.

8.69 The values of the parameters in the circuit in P~~~~~~E Fig. 8.21 are R = 3 fl; L = 5 mH; C = 0.25 }J-F;

Sections 8.1-8.5 Vdc = 12 V; and a = 50. Assume the switch opens when the primary winding current is 4 A.

8.66 a) Derive Eg. 8.92. a) How much energy is stored in the circuit at t = o+?F:~~~;~~E b) Derive Eg. 8.93.

c) Derive Eg. 8.97. b) Assume the spark plug does not fire. What is the maximum voltage available at. the spark plug?

8.67 Derive Eq. 8.99. PRACTICAL c) What is the voltage across the capacitor when

PERSPECTIVE the voltage across the spark plug is at its maxi­8.68 a) Using the same numerical values used in the mum value?

PRACTICAL Practical Perspective example in the text, find PERSPECTIVE

the instant of time when the voltage across the 8.70 Repeat Problem 8.68 using the values given in capacitor is maximum. Problem 8.69.