Electric Circuits, 9th Edition P31 potx

276 Natural and Step Responses
of
RLC
Circuits
Therefore, the response is underdamped. Now,
w
(i
= V(4 ~ a
2
= VlO
6
- 4 X 10
4
= 100V96
= 979.80 rad/s,
Sl
= -a + joo
d
= -200 + /979.80 rad/s,
s
2
= -
a
- j(o
ti
= -200 - /979.80 rad/s.
For the underdamped case, we do not ordinarily
solve for S] and s
2
because we do not use them
explicitly. However, this example emphasizes

why
S]
and s
2
are known as complex frequencies.
b) Because v is the voltage across the terminals of a
capacitor, we have
v(0) = v(0
+
) = V
0
= 0.
Because v(0
+
) = 0, the current in the resistive
branch is zero at t = 0
+
. Hence the current in
the capacitor at t = 0
+
is the negative of the
inductor current:
*c(0
+
) = -(-12.25) = 12.25 mA.
Therefore the initial value of the derivative is
dv(0
+
) (12.25)(10""
3

)
,
J
= f- = 98,000 V/s.
dt (0.125)(10
-6
) '
c) From Eqs. 8.30 and 8.31,5, = 0 and
98,000
B
7
100 V.
co
d
Substituting the numerical values of a, co
d
, B\
s
and B
2
into the expression for v(t) gives
v(t) = 100<r
ZUUf
sin 979.80? V, t > 0.
d) Figure 8.9 shows the plot of v(t) versus t for the
first 11 ms after the stored energy is released. It
clearly indicates the damped oscillatory nature
of the underdamped response. The voltage v(t)
approaches its final value, alternating between
values that are greater than and less than the

final value. Furthermore, these swings about the
final value decrease exponentially with time.
v(V)
t(ms)
Figure 8.9 A The voltage response for Example 8.4.
Characteristics of the Underdamped Response
The underdamped response has several important characteristics. First, as
the dissipative losses in the circuit decrease, the persistence of the oscilla-
tions increases, and the frequency of the oscillations approaches w
0
. In
other words, as R
—>
oo, the dissipation in the circuit in Fig. 8.8 approaches
zero because /; = v
2
/R. As R
—>
oo, a
—*
0, which tells us that m
d
—*
w
()
.
When a - 0, the maximum amplitude of the voltage remains constant;
thus the oscillation at a>
()
is sustained. In Example 8.4, if R were increased

to infinity, the solution for v(t) would become
v(t) = 98sinl000rV, t > 0.
Thus,
in this case the oscillation is sustained, the maximum amplitude of
the voltage is 98 V, and the frequency of oscillation is 1000 rad/s.
We may now describe qualitatively the difference between an under-
damped and an overdamped response. In an underdamped system, the
response oscillates, or "bounces," about its final value. This oscillation is
also referred to as ringing. In an overdamped system, the response
approaches its final value without ringing or in what is sometimes
described as a "sluggish" manner. When specifying the desired response of
a second order system, you may want to reach the final value in the short-
est time possible, and you may not be concerned with small oscillations
about that final value. If so, you would design the system components to
achieve an underdamped response. On the other hand, you may be con-
cerned that the response not exceed its final value, perhaps to ensure that
components are not damaged. In such a case, you would design the system
components to achieve an overdamped response, and you would have to
accept a relatively slow rise to the final value.
8.2 The Forms
of
the Natural Response
of a
Parallel
RLC
Circuit
277
^ASSESSMENT PROBLEM
Objective 1—Be able to determine the natural and the step response of parallel
RLC

circuits
8.4 A 10 mH inductor, a 1 /xF capacitor, and a vari-
able resistor are connected in parallel in the
circuit shown. The resistor is adjusted so that
the roots of the characteristic equation are
-8000 ± /6000 rad/s. The initial voltage on the
capacitor is 10 V, and the initial current in the
inductor is 80 mA. Find
a) R;
b) dv(0
+
)/dt;
c) B
x
and B
2
in the solution for v; and
d) idt).
NOTE: Also try Chapter Problems 8.7 and 8.19.
Answer: (a) 62.5 ft;
(b) -240,000 V/s;
(c) B
x
= 10 V, B
2
= -80/3 V;
(d)i
L
(t) = 10e-
80()0f

[8 cos 6000?
+ (82/3) sin 6000f] mA when t
The Critically Damped Voltage Response
The second-order circuit in Fig. 8.8 is critically damped when
coj,
= a
2
, or
tu
()
= a. When a circuit is critically damped, the response is on the verge of
oscillating. In addition, the two roots of the characteristic equation are
real and equal; that is.
1
Si = $
2
—a =
—
2RC
(8.32)
When this occurs, the solution for the voltage no longer takes the form
of Eq. 8.18. This equation breaks down because if
s%
= .s^ = -a, it pre-
dicts that
{A
x
+
A
2

)e~
at
= A
Q
e
(8.33)
where A
{)
is an arbitrary constant. Equation 8.33 cannot satisfy two inde-
pendent initial conditions
(VQ,
/
0
) with only one arbitrary constant, A
{)
.
Recall that the circuit parameters R and C fix a.
We can trace this dilemma back to the assumption that the solution
takes the form of Eq. 8.18. When the roots of the characteristic equation
are equal, the solution for the differential equation takes a different
form, namely
v(t) = D
{
te~
a
' + D
2
e (8.34)
Thus in the case of a repeated root, the solution involves a simple expo-
nential term plus the product of a linear and an exponential term. The jus-

tification of Eq. 8.34 is left for an introductory course in differential
equations. Finding the solution involves obtaining D
x
and D
2
by following
Ihe same pattern set in the overdamped and underdamped cases: We use
the initial values of the voltage and the derivative of the voltage with
respect to time to write two equations containing D\ and/or D
2
.
< Voltage natural response—critically
damped parallel
RLC
circuit
278 Natural and Step Responses of
RLC
Circuits
From Eq. 8.34, the two simultaneous equations needed to determine
D
x
and D
2
are
,,(0+) = V
()
= D
2
,
dv(Q

+
) /
c
(0
+
)
dt
c
= Di - aDj.
(8.35)
(8.36)
Example 8.5
As we can see, in the case of a critically damped response, both the
equation for v(t) and the simultaneous equations for the constants D\ and
D
2
differ from those for over- and underdamped responses, but the general
approach is the same. You will rarely encounter critically damped systems
in practice, largely because
co
0
must equal a exactly. Both of these quanti-
ties depend on circuit parameters, and in a real circuit it is very difficult to
choose component values that satisfy an exact equality relationship.
Example 8.5 illustrates the approach for finding the critically damped
response of a parallel RLC circuit.
Finding the Critically Damped Natural Response of a Parallel
RLC
Circuit
a) For the circuit in Example 8.4 (Fig. 8.8), find the

value of R that results in a critically damped volt-
age response.
b) Calculate v(t) for t > 0.
c) Plot v(t) versus t for 0 < t =s 7 ms.
Solution
a) From Example 8.4, we know that o>o = 10
6
.
Therefore for critical damping.
or
R
a = 1()-
1
=
10
6
1
2i?C
= 4000 a.
(2000)(0.125)
b) From the solution of Example 8.4, we know that
v(0
+
) = 0 and dv{0
+
)/dt = 98,000 V/s. From
Eqs.
8.35 and 8.36, D
2
= 0 and D

x
= 98,000 V/s.
Substituting these values for a, D
l%
and D
2
into
Eq. 8.34 gives
v(t) = 98,000*e
_1000f
V, t s> 0.
c) Figure 8.10 shows a plot of v(t) versus t in the
interval 0 < r ^ 7 ms.
12 3 4 5 6 7
Figure 8.10 •
The
voltage response for Example 8.5.
L. t (ms)
^ASSESSMENT PROBLEM
Objective 1—Be able to determine the natural and the step response of parallel
RLC
circuits
8.5 The resistor in the circuit in Assessment
Problem 8.4 is adjusted for critical damping.
The inductance and capacitance values are
0.4 H and 10 fiF, respectively. The initial energy
stored in the circuit is 25 mJ and is distributed
equally between the inductor and capacitor.
Find (a) R\ (b) V
0

; (c) /
()
; (d) D
x
and D
2
in the
solution for v; and (e) i
R
, t S 0
+
.
Answer: (a) 100 O;
(b)50V;
(c) 250 mA;
(d) -50,000 V/s, 50 V;
(e) i
R
(t) = (-500te~
5m
+ 0.50e
-500
') A,
t £= 0
+
.
NOTE: Also try Chapter Problems 8.9 and 8.20.
8.2
The
Forms

of
the Natural Response
of
a Parallel RLC Circuit
279
A Summary of the Results
We conclude our discussion of the parallel RLC circuit's natural response
with a brief summary of the results. The first step in finding the natural
response is to calculate the roots of the characteristic equation. You then
know immediately whether the response is overdamped, underdamped, or
critically damped.
If the roots are real and distinct (c_
2
, < a
2
), the response is over-
damped and the voltage is
v(t) =
A
{
e
Sit
+
A
2
e
s
*,
where
S\ = —a + voT

s
2
= —a — Va
2
1
2RC
2_
1
"° " LC-
-
w
{
2
,,
-
<4h
The values of A
{
and A
2
are determined by solving the following simulta-
neous equations:
v(0
+
) = Ai + A
2
,
dv{0
+
) /

C
(0
+
)
If the roots are complex
OJ
2
}
> a" the response is underdamped and
the voltage is
v(t) = B
{
e~
n
' cos o>/ + B
2
e~
a!
sin ctjt,
where
to
(l
= vo){)
—
or.
The values of B\ and B
2
are found by solving the following simultaneous
equations:
v(Q

+
) - % = B
h
dv(0
+
) ic(0
+
) _ , _
_____ = _______ _
aB] +
^
If the roots of the characteristic equation are real and equal
(OJQ
= a"),
the voltage response is
v(t) =
D^e-o"
+
D
2
e~
a
\
where a is as in the other solution forms. To determine values for the con-
stants D) and D
2
, solve the following simultaneous equations:
v(0
+
) = V

0
= D
2
,
dv(0
+
)
i
c
(0
+
)
_
-_-_ — __
A
- aD,
: 8,3 The Step Response of a Parallel
RLC
Circuit
• Finding the step response of a parallel RLC circuit involves finding the
voltage across the parallel branches or the current in the individual
branches as a result of the sudden application of a dc current source.
There may or may not be energy stored in the circuit when the current
source is applied. The task is represented by the circuit shown in Fig. 8.11.
To develop a general approach to finding the step response of a second-
order circuit, we focus on finding the current in the inductive branch (//).
This current is of particular interest because it does not approach zero as
t increases. Rather, after the switch has been open for a long time, the
inductor current equals the dc source current I. Because we want to focus
on the technique for finding the step response, we assume that the initial

energy stored in the circuit is
zero.
This assumption simplifies the calcula-
tions and doesn't alter the basic process involved. In Example 8.10
we will see how the presence of initially stored energy enters into the
general procedure.
To find the inductor current i
L
, we must solve a second-order differ-
ential equation equated to the forcing function /, which we derive as fol-
lows.
From Kirchhoffs current law, we have
k + hi + k = U
or
Because
v _ dv
lL +
-
+
C- = I. (8.37)
dir
v = L—-, (8.38)
we get
dt '
dv
T
d
2
i
L

Substituting Eqs. 8.38 and 8.39 into Eq. 8.37 gives
L dir afit
R dt dt
For convenience, we divide through by LC and rearrange terms:
d
2
i, 1 di
L
i
L
I ,
—£ + + —^ = . (8.41)
dt
1
RC dt LC LC
Comparing Eq. 8.41 with Eq. 8.3 reveals that the presence of a nonzero
term on the right-hand side of the equation alters the task. Before show-
ing how to solve Eq. 8.41 directly, we obtain the solution indirectly.
When we know the solution of Eq. 8.41, explaining the direct approach
will be easier.
The Indirect Approach
We can solve for i
L
indirectly by first finding the voltage v. We do this with
the techniques introduced in Section 8.2, because the differential equation
that v must satisfy is identical to Eq. 8.3. To see this, we simply return to
Eq. 8.37 and express i
L
as a function of v; thus
If, v „dv

li
vdT
+
J
+
c
T
t
=
'•
(842)
Differentiating Eq. 8.42 once with respect to t reduces the right-hand side
to zero because I is a constant. Thus
v
1 dv d
2
v
— + — — + C—r = 0,
L R dt dt
2
or
d
2
v 1 dv v
^
+
7^
+
^
= 0

-
(8
'
43)
As discussed in Section 8.2, the solution for v depends on the roots of the
characteristic equation. Thus the three possible solutions are
v =
A
x
e
Slt
+
A
2
e
&2
\
(8.44)
v = B
x
e~
a
' cos o)
d
t + B
2
e~
al
sin a)/, (8.45)
v = D

x
te~
a
' +
D
2
e~
at
.
(8.46)
A word of caution: Because there is a source in the circuit for t > 0, you
must take into account the value of the source current at t = 0
+
when you
evaluate the coefficients in Eqs. 8.44-8.46.
To find the three possible solutions for /
L
, we substitute Eqs. 8.44-8.46
into Eq. 8.37. You should be able to verify, when this has been done, that
the three solutions for i
L
will be
i
L
= I +
A\e
Sli
+
A'
2

e
s
-
1
,
(8.47)
//, = I + B\e~
at
cos o)
d
t + B'
2
e~
at
sin wj, (8.48)
i
L
= I + D[te-
at
+
D'
2
e~
at
,
(8.49)
where A\, A
2
, B[, B
2

, D\, and D'
2
, are arbitrary constants.
In each case, the primed constants can be found indirectly in terms of
the arbitrary constants associated with the voltage solution. However, this
approach is cumbersome.
The Direct Approach
It is much easier to find the primed constants directly in terms of the ini-
tial values of the response function. For the circuit being discussed, we
would find the primed constants from /
L
(0) and dii(0)/dt.
The solution for a second-order differential equation with a constant
forcing function equals the forced response plus a response function
282 Natural
and
Step Responses of
RLC
Circuits
identical in form to the natural response. Thus we can always write the
solution for the step response in the form
._ [function of the same form!
' \ as the natural response J'
or
v = V
f
+
{
function of the same form
1

as the natural response J'
(8.50)
(8.51)
where If and Vf represent the final value of the response function. The
final value may be zero, as was, for example, the case with the voltage v in
the circuit in Fig. 8.8.
Examples 8.6-8.10 illustrate the technique of finding the step
response of a parallel RLC circuit using the direct approach.
Example 8.6
Finding the Overdamped Step Response of a Parallel
RLC
Circuit
The initial energy stored in the circuit in Fig. 8.12 is
zero.
At t = 0, a dc current source of 24 mA is
applied to the circuit. The value of the resistor is
400 ft.
a) What is the initial value of i
L
l
b) What is the initial value of dijdtl
c) What are the roots of the characteristic equation?
d) What is the numerical expression for //,(/
1
) when
t > 0?
'(p'X
Figure 8.12 •
The
circuit for Example 8.6.

Solution
a) No energy is stored in the circuit prior to the
application of the dc current source, so the initial
current in the inductor is zero. The inductor pro-
hibits an instantaneous change in inductor cur-
rent; therefore i
L
(0) = 0 immediately after the
switch has been opened.
b) The initial voltage on the capacitor is zero
before the switch has been opened; therefore it
will be zero immediately after. Now, because
v = Ldifjdt,
di
dt
-(0
+
) = 0.
c) From the circuit elements, we obtain
, 1 10
12
(-0() =
a =
LC (25)(25)
1 10
y
2RC (2)(400)(25)
16 x 10*,
= 5 x 10
4

rad/s,
or
a
2
= 25 X 10
8
.
Because a»o < a
2
, the roots of the characteristic
equation are real and distinct. Thus
Sl
= -5 X 10
4
+ 3 X 10
4
= -20,000 rad/s,
s
2
= -5 X 10
4
- 3 X 10
4
= -80,000 rad/s.
d) Because the roots of the characteristic equation
are real and distinct, the inductor current response
will be overdamped. Thus i
L
(t) takes the form of
Eq. 8.47, namely,

i
L
= I
f
+ A[e
Sit
+
A!
2
e
s
-'.
• Inductor current in overdamped parallel
RLC
circuit step response
Hence, from this solution, the two simultaneous
equations that determine A\ and A^are
*
t
(0) = If + A\ + A'
2
= 0,
di
L
di
(0) = s
x
A\ +
s
2

A
,
2
= 0.
Solving for A\ and A'
2
gives
A[ = -32 mA and A'
2
= 8 mA.
The numerical solution for i
L
(t) is
i
L
{t) = (24 - 32e-
20
-
000
' + 8e-"
)mk
) mA, t > 0.
8.3 The Step Response
of a
Parallel
RLC
Circuit
283
Finding the Underdamped Step Response of a Parallel
RLC

Circuit
The resistor in the circuit in Example 8.6 (Fig. 8.12)
is increased to 625 ft. Find i
L
(t) for t 2s 0.
Solution
Because L and C remain fixed, col has the same
value as in Example 8.6; that is, col = 16 X 10
8
.
Increasing R to 625 ft decreases a to
3.2 X 10
4
rad/s. With col > oc\ the roots of the
characteristic equation are complex. Hence
s
x
= -3.2 X 10
4
+ /2.4 X 10
4
rad/s,
s
2
= -3.2 X 10
4
- /2.4 X 10
4
rad/s.
The current response is now underdamped and

given by Eq. 8.48:
kit) = I/ +
B[e~°"
cos
co
d
t
+ B'
2
e~
at
sin
to
d
t.
• Inductor current in underdamped parallel
RLC
circuit step response
Here, a is 32,000 rad/s, co
d
is 24,000 rad/s, and
If is 24 mA.
As in Example 8.6, B[ and B'
2
are determined
from the initial conditions. Thus the two simultane-
ous equations are
I'ZXO) = I
f
+ B[ = 0,

^(0)=co
(l
B
2
-aB[ = Q.
Then,
and
B\
= -24 mA
54 = -32 mA.
The numerical solution for i
L
(t) is
i
L
(t) = (24 - 24e-
3Z000
'cos24,000r
- 32e"
32
'
000
' sin 24,000?) mA, t > 0.
Example 8.8
Finding the Critically Damped Step Response of a Parallel
RLC
Circuit
The resistor in the circuit in Example 8.6 (Fig. 8.12)
is set at 500 ft. Find i
L

for t > 0.
Solution
We know that
col
remains at 16 X 10
8
. With R set at
500 ft, a becomes 4 X 10
4
s~\ which corresponds
to critical damping. Therefore the solution for i
L
(t)
takes the form of Eq. 8.49:
kit) = I
f
+ D\te~
at
+ D'
2
e~°".
• Inductor current in critically damped parallel
RLC
circuit step response
Again, D\ and D
2
are computed from initial
conditions, or
i
L

(0) = If + D'
2
= 0,
~(0) = D\ - aD'o = 0.
at
Thus
D\
= -960,000 mA/s and D'
2
= -24 mA.
The numerical expression for i
L
{i) is
i
L
(t) = (24 - 960,000re-
4(X()0{)
' - 24e
_40
'
000
') mA, t > 0.
284 Natural and Step Responses of
RLC
Circuits
Example
8.9
Comparing
the
Three-Step Response Forms

a) Plot
on a
single graph, over
a
range from
0 to
220 jus,
the
overdamped, underdamped,
and
critically damped responses derived
in
Examples 8.6-8.8.
b)
Use the
plots
of (a) to
find
the
time required
for
i
L
to
reach 90%
of
its final value.
c)
On the
basis

of the
results obtained
in
(b), which
response would
you
specify
in a
design that puts
a premium
on
reaching 90%
of the
final value
of
the output
in the
shortest time?
d) Which response would
you
specify
in a
design
that must ensure that
the
final value
of the
cur-
rent
is

never exceeded?
i
L
(mA)
Underdamped
{R = 625
il)
jOverdamped
(R
=
400 SI)
I Critically damped
{R = 500
Q)
tQa)
0
20 60 100 140 180
Figure 8.13
A
The
current plots for
Example
8.9.
Solution
a) See Fig. 8.13.
b)
The
final value
of i
L

is 24
mA,
so we can
read
the
times
off
the plots corresponding to
i
L
=
21.6
mA.
Thus
t
od
=
130 ^is,
t
cd
=
97 /xs,
and t
ud
= 74 fis.
c)
The
underdamped response reaches 90%
of the
final value

in the
fastest time,
so it is the
desired
response type when speed
is the
most important
design specification.
d) From
the
plot,
you can see
that
the
under-
damped response overshoots
the
final value
of
current, whereas neither
the
critically damped
nor
the
overdamped response produces currents
in excess
of
24 mA. Although specifying either
of
the latter

two
responses would meet
the
design
specification,
it is
best
to use the
overdamped
response.
It
would
be
impractical
to
require
a
design
to
achieve
the
exact component values
that ensure
a
critically damped response.
Example
8.10
Finding Step Response
of a
Parallel

RLC
Circuit with Initial Stored Energy
Energy
is
stored
in the
circuit
in
Example
8.8
(Fig. 8.12, with
R = 500 n) at the
instant
the dc
cur-
rent source
is
applied.
The
initial current
in the
inductor
is 29
mA,
and the
initial voltage across
the
capacitor
is 50 V.
Find

(a)
/
L
(0);
(b)
di
L
{0)/dt;
(c)
i
L
{t) for t > 0; (d) v(t) for t > 0.
Solution
a) There cannot
be an
instantaneous change
of
cur-
rent
in an
inductor, so the initial value
of i
L
in the
first instant after
the dc
current source
has
been
applied must

be 29 mA.
b)
The
capacitor holds
the
initial voltage across
the
inductor
to
50
V.
Therefore
c) From
the
solution
of
Example 8.8, we know that
the current response
is
critically damped. Thus
idO
= if +
D\te~
at
+
D
2
e~
at
,

where
2RC
= 40,000 rad/s
and I
f
= 24 mA.
Notice that
the
effect
of the
nonzero initial
stored energy
is on the
calculations
for the
con-
stants
D[ and
D'
2
, which
we
obtain from
the
ini-
tial conditions. First
we use the
initial value
of
the inductor current:

Lf«n
=
50,
^(0+)
= § X 10
3
=
2000
A/s.
at
25
i
L
(0)
= I
f
+ D'
2
=
29
mA,
from which
we get
D'
2
= 29 - 24 = 5 mA.
8.4 The Natural and Step Response of
a
Series
RLC

Circuit 285
The solution for D[ is
~(0
+
) = D{ - aD'
2
= 2000,
or
D\
= 2000 + aD'
2
= 2000 + (40,000)(5 X 10~
3
)
= 2200 A/s = 2.2 X 10
6
m
A/s.
Thus the numerical expression for i
L
(t) is
i
L
(t) = (24 + 2.2 X lO'Ve"
40
-
000
'
+ 5e~
4{UMH

) mA, t > 0.
d) We can get the expression for v(t), t > 0 by
using the relationship between the voltage and
current in an inductor:
dij
v(t) = L^
= (25 X 10~
3
)[(2.2 X 10
6
)(-40,000)te-
4ao,1()
'
+ 2.2 X ioV
40
-
000
'
+ (5)(-40,000)<r
4(M)00
'] X 10~
3
= -2.2 x
lO
6
^
40
-
000
'

+ 50e^
wu
* V, / s> 0.
To check this result, let's verify that the initial
voltage across the inductor is 50 V:
v(0) = -2.2 X 10
6
(0)(1) + 50(1) = 50 V.
^ASSESSMENT PROBLEM
Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits
8.6 In the circuit shown, R = 500 O, L = 0.64 H,
C =
1
fxF,
and I =
—1
A. The initial voltage
drop across the capacitor is
40
V and the initial
inductor current is
0.5
A. Find (a) //?(0
+
);
(b) i
c
(0
+
); (c) di

L
(0
+
)/dt; (d) s
t
,
s
2
;
(e) i
L
(t) for
t > 0; and (f) v(t) for t > 0
+
.
Answer: (a) 80 mA;
(b)-1.58 A;
(c) 62.5 A/s;
(d) (-1000 + ;750) rad/s,
(-1000 - /750) rad/s;
(e)[-l + <r
1000
'[1.5cos750f
+ 2.0833 sin 750f] A, for t > 0;
(f) £r
1000
'(40cos750; - 2053.33 sin
750r)
V,
for t > 0

+
.
NOTE: Also try
Chapter Problems
8.29-8.31.
8.4 The Natural and Step Response
of a Series
RLC
Circuit
The procedures for finding the natural or step responses of a series RLC
circuit are the same as those used to find the natural or step responses of a
parallel RLC circuit, because both circuits are described by differential
equations that have the same form. We begin by summing the voltages
around the closed path in the circuit shown in
Fig.
8.14.
Thus
/* 1 p
Ri + L~ + - idr + V
0
= 0.
dt CJ
()
We now differentiate Eq. 8.52 once with respect to t to get
n
di , dri i
R—
+ L— +
—
= 0,

dt dt
2
C
(8.52)
(8.53)
R
•"•"N,
iy
L
X
) d
+
^V
a
Figure 8.14 A A circuit used to illustrate the natural
response of
a
series
RLC
circuit.