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.