Chapter 7
In chapter 6, we noted that an important attribute of inductors and capacitors
is their ability to store energy
In this chapter, we are going to determine the currents and voltages that arise when energy
is either released or acquired by an inductor or capacitor in response to an abrupt change
in a dc voltage or current
Energy acquired by a capacitor
Energy released by a capacitor
+
VS
+
-
R
a
R
vc
-
i
C
VS
+
-
+
i
vc
C
b
-
In this chapter, we will focuses on circuits that consist only of sources, resistors, and
either (but not both) inductors or capacitors
Such configurations are called RL (resistor-inductor) and RC (resistor-capacitor) circuits
Our analysis of the RL and RC circuits will be divided into three phases:
First Phase
we consider the currents and voltages that arise when stored energy
in an inductor or capacitor is suddenly released to a resistive network
This happens when the inductor or capacitor is abruptly disconnected from its dc source and
allowed to discharge through a resistor
R
a
VS
+
-
b
+
i
vc
C
-
Tank of Water
Level of Water
The currents and voltages that arise in this
configuration are referred to as the natural
response of the circuit to emphasize that
the nature of the circuit itself , not external
sources excitation determine its behavior
you can think of this as when a tank of
water is opened suddenly , will the water
in the tank disappear instantaneously in
zero second or will it takes some time no
matter how small to empty the tank
Second Phase we consider the currents and voltages that arise when energy is being acquired
by an inductor or capacitor due to the sudden application to a dc voltage or
current source
R
VS
+
-
+
i
vc
C
This response is referred to as the step response
Level of Water
you can think of this as when a tank of
water is being filled suddenly , will the water
in the tank rise instantaneously in zero
second or will it takes some time no
matter how small for the water to rise in
the tank
Third Phase The process for finding both the natural response ( First Phase ) and
step response ( Second Phase ) is the same thus in the third phase we will
develop a general method that can be used to find the response of RL and RC
circuit to any abrupt change in a dc voltage or current
7.1 The Natural Response of an RL Circuits
Let the circuit shown which contain an inductor is shown
i
switch
+
IS
Ro
L
v
R
Suppose the switch has been in a closed position for a long time
For the time being long time
we will define long time
later
All currents and voltages have reached a constant value
Only constant or DC currents can exist in the circuit just prior to the
switch’s being opened
The inductor appears as a short circuit
v = L
di
dt
0
i
switch
+
IS
Ro
L
Is
v=L
R
v
di
0
dt
-
Because the voltage across the inductive branch is zero
There can be no current in either
All the source current
IS
R0
or
R
appears in the inductive branch
Finding the natural response requires finding the voltages and currents at any branch in the
circuit after the switch has been opened.
t =0
Ro
IS
+
i
L
v
Is
-
If we let t = 0 denote the instant when the switch is opened
For t ≥ 0 the circuit become
i
IS
Ro
L
Is
The problem become one of finding
+
v
i
L
i (0) =I s
v
-
-
v(t)
+
and i(t) ( or i and
v)
for
t ≥ 0
Deriving the Expression for the current
vL
+
i
L
R
+
vR
-
i (0) =I s
KVL around the loop
vL +vR 0
L
di
+ Ri  0
dt
This is a first order differential equation because it contains terms involving the
ordinary derivative of the unknown di/dt.
The highest order derivative appearing in the equation is 1. Hence the first order
We can still describe the equation further.
Since the coefficients in the equation R and L are constant that is not functions of either
the dependent variable
i
or the independent variable
t
Thus the equation can also be described as an ordinary differential equation with constant
coefficients
+
i
i (0) =I s
R
L
v
L
-
di
+ Ri  0
dt
To solve the differential equation we processed as follows:
di
L  - Ri
dt
di
R
dt  - idt
dt
L
di
R
- i
dt
L
di  -
R
idt
L
di
R
 - dt
i
L
Integrating both side to obtain explicit expression for i as a function of

i (t )
i (t 0
dx
R
L
) x
t
 dy
t0
Here t0 = 0

i (t )
i (0)
Based on the definition of the natural logarithm
- R L t
i (t )  i (0)e
dx
R
x
L
t
 dy
e lnx  x
0
t
ln
i (t ) -R

t
i (0) L
+
i
i (0) =I s
R
L
v
L
di
+ Ri  0
dt
i (t )  i (0)e - R L t
-
Since the current through an indicator can not change abruptly or instanteously
i (0-)  i (0+ )  i (0)  I 0
Were I0 is the initial current on the inductor just before the switch opened
( or in some cases closed)
In the circuit above
Therefore
I0 = Is
i (t )  I 0 e - R L t
t ≥0
Which shows that the current start from initial value I0
and decreases exponentially toward zero as t increases
+
i
i (0) =I s
L
R
L
v
-
di
+ Ri  0
dt
v(t)
t =0
Ro
Is
was going through L and zero current through R)
i
L
IS
Note the voltage
have a jump
t ≥ 0+
here is defined for t > 0 because the voltage across the resistor R is
zero for t < 0 ( all the current
IS
t ≥0
We derive the voltage across the resistor form direct
application of Ohm’s law
v (t )  Ri (t )  RI 0 e - R L t
Note the voltage
i (t )  I 0 e - R L t
R
+
v (0- )  0
v
v (0+ )  I 0 R
-
v(t) across the resistor and across the indicator can change instanousley or
We derive the power dissipated in the resistor
+
i
i (0) =I s
L
R
v
di
L
+ Ri  0
dt
-
p  Ri
p  vi
i (t )  I 0 e - R L t
2
or
p
t ≥0
v2
R
Whichever form is used, the resulting expression can be reduce to
p  R I 0 e
- R L t

2
 I Re
2
-2 R L t
0
Note the current i(t) through the resistor is zero for t < 0
t ≥ 0+
p 0
t<0
We derive the energy delivered to the resistor
+
i
L
i (0) =I s
R
v
di
L
+ Ri  0
dt
-
p  I Re
2
0
-2 R L t
i (t )  I 0 e - R L t
t ≥0
t ≥0
+
The energy deliver to the resistor is after the switch is opened because before that there was
no current passing through the resistor and the voltage across it was zero
w

 pdx   I 0
t
0

t
2
Re
-2 R L x
dx
0
1 2
-2 R L t
L I (1 - e
)
2 0
t≥0

1
-2 R L t
I 2 R 1 - e   
2R L  0
t =0
Ro
IS
+
i
L
R
v
IS
-
Note that just before the switch is opened the current on the indicator is IS = I0
w

1 2
Li
2
R
L
1 2
LI
2 0
+
i
i (0) =I s

v
w

1 2
-2 R L t
L I (1 - e
)
2 0
t≥0
-
→
→
1 2
1 2
-2 R L t

lim
L
I
(
1
e
)

LI
lim
w(t) t ∞
t ∞
2
0
2
0
The initial energy stored in the indictor
The significant of the time constant
+
i
i (0) =I s
R
L
v
-
di
L
+ Ri  0
dt
i (t )  I 0 e - R L t
t ≥0
The coefficient at the exponential term namely R/L determine
the rate at witch the exponential term in the current approaches
zero
The reciprocal of this ratio (L/R ) is the time constant which has the units of seconds
L
R
V.s
 A
V
A
s
Seconds
the time constant for the R,L circuit is
It is convenient to think of the time elapsed after
switching in terms of integral multiple of 
Example
-
i ( )  I 0 e     I 0 e -1  0.367I 0
 0.367I 0

After one time constant, the maximum of the current I0 has dropped to 37% of its value
 =
L
R
Table showing the value of
exponential
multiples of
e-t/

for integral
Note that when the time elapsed after switching
exceed five time constants 5 
t


e

3.6788 x 10-
2
3
4
1.3534 x 10-
4.0787 x 10-2
1.8316 x 10-2
5
6
6.7379 x 10-3
2.4788 x 10-3
- t
.
.
10
The current is less than 1% of its initial value
.
.
4.5400 x 10-5
0.067 I 0
< 1%
5 
Thus sometime we say that five time constant
after switching has occurred, the currents and
voltages have for most practical purposes
reached their final values
Thus with 1% accuracy along time
Five or more time consent
I0
+
i
i (0) =I s
L
R
i (t )  I 0 e - R L t
t ≥0
v
-
The existence of the current in the RL circuit is momentary event and is referred to as
transient response of the circuit
The response that exist a long time after the switching has taken place is called
steady-state response
The steady-state response in the RL circuit is zero , that were the i(t) will go to
as t goes to infinity
Determining the time constant

The time constant can be determined as follows
L
(1) If the RL circuit can be put as
R
Were L is an equivalent inductor and R is an equivalent resistor
Seen by the equivalent indictor
Example
R1
L1
R2
R3
Leq  L1 + L2
L2
 
L eq
R eq
Req  R1 + (R2 || R3 )
(2) If we know the differential equation
+
i
L
i (0) =I s
R
v
-
L
di
+ Ri  0
dt
di
R
+
i 0
dt
L

is the inverse of this constant
Example
Suppose the differential equation is given as
di
3
+ 6i  0
dt
di
6
+ i 0
dt
3
3
6
 =  0.5
(2) If a trace or graph of i(t) is given
Differentiating i(t) we have
di (t )
I
 - 0 e -t  t

dt
di (0)
I
- 0

dt
i
switch
+
IS
L
v
Is
L
p  R I 0 e
w


t
0
pdx 

R
-
v (t )  Ri (t )  RI 0 e - R L t
2
L
 I Re
2
t ≥ 0+
-2 R L t
0
1 2
-2 R L t
L I (1 - e
)
2 0
v
-
i (t )  I 0 e - R L t
di
+ Ri  0
dt
- R L t
R
i (0) =I s
t≥0
t ≥0
v (0- )  0 v (0+ )  I 0 R
t ≥0
+
1 2
lim
w(t)

LI
t ∞
→
Ro
+
i
2
0
Since the current through an inductor is given as
t

i  1 v ( )d + i (t 0 )
L
t0
Therefor to find i1,i 2
We need to Find
v (t)
2A
4H
6W
Finding
v (t)
The circuit reduce to

v (t )  8i (t )  8 12e -2t

v (t )
 v (t )
10
4+15 10 
 v (t )  15
-2t
i3  
10+15  5.7e A
10


t  0+
i 3  5.7e -2t A
t  0+
v 2 (t )
p (t ) 
=
8
The voltage across the capacitor =
For t >= 0
KCL on the upper junction
Solving for t
Similar to the previous example 7.2 , the voltage across a capacitor is given as
t
v (t )  1
C

i ( )d +v (t 0 )
t0
Question: how to find i(t) ?
Therefor to find v 1(t ),v 2 (t )
We need to Find
i(t)
t
v (t )  1
C

i ( )d +v (t 0 )
t0
Therefor to find v 1(t ),v 2 (t )
We need to Find
i(t)
Question: how to find i(t) ?
Do we find i(t) through the capacitor voltage
i (t )  - C1
The answer is no of course because we do not know
However we can find i(t) through the resistor 250 kW
as
we need to find v(t) as will be shown next
dv 1(t )
dv (t )
or i (t )  - C 2 2
dt
dt
v 1(t ), v 2 (t )
i (t ) 
v (t )
(250X103 )
The circuit reduce to
-

(5)(20)
(5 + 20)
The solution of the voltage v(t) is given as
v (t ) V 0e
-t / 
t 0
were
  RC
 (250X 103 )(4X 10-6 )  1 s
V 0  20 V
v (t )  20e -t
t 0
The circuit reduce to
v (t )  20e -t
t 0
Note the direction of the current i (t) and the initial voltages
and the polarity of v 1(t ), v 2 (t )
The circuit reduce to
v (t )  20e -t
t 0
v (t )  20e -t
t 0
v (t )  20e -t
t 0
(a)
(b)
(c)
d) Show that the total energy delivered to the 250 kW resistor is the difference between
the results in (b) and (c)
Comparing the results obtained in (b)
and in (c)
+
i
i (0) =I s
R
L
v
-
KVL
L
di
+ Ri  0
dt
i (t )  I 0 e
- R L t

t ≥0
di
R
+
i 0
dt
L
1 / L
KCL
C
dv
v
+
0
dt
R
v (t )  V0 e
-
v (t )  Ri (t )  RI 0 e
- R L t
p (t )  Ri 2 (t )  RI 2 e
-2 R L t
0
w


v (t ) V0e
i (t ) 

R
R
t ≥ 0+
t ≥ 0+
w
pdx
V0 - RCt
 e
R

t ≥ 0+
t ≥ 0+
 pdx
t
0
0
t≥0
1
 CV 2 (1 - e
2
0
-2t
RC
1
lim
w(t)

CV 2
t ∞
→
1 2
-2 R L t
L I (1 - e
)
2 0
1 2
lim
w(t)

LI
t ∞
→
t ≥0
2
t
2
t
RC
2t
v 2 (t ) V0 - RC
p
 e
R
R
0

t
RC
-
dv
1
+
v 0
dt
RC
1 / C
2
0
)
t≥0
7.3 Step Response of RL and RC Circuits
We are going to find the currents and voltages in 1st order RL and RC circuits
when a DC voltage or current is suddenly applied.
The Step Response of an RL Circuit
The switch is closed at t = 0 , the task is to find the expressions for the current in the circuit
and for the voltage across the inductor after the switch has been closed
R
t =0
+
VS
+
-
L
v (t)
-
After the switch has being closed, we have
R
+
VS
+
-
i
L
v (t)
-
KVL
V s  Ri + L
di
dt
We can solve the differential equation similar to what we did previously with the natural response by
separating the variables i and t and integrating as follows
di - Ri + V s

dt
L

-R
L
-R
 Vs 
i

di



L
R 

 Vs 
 i -  dt
R 

R
V s  Ri + L
+
VS
+
-
i
L
di - Ri + V s
-R


dt
L
L
v (t)
-
We know separate the variables
di
dt
i
and
-R
 Vs 
i

di



L
R 

di
-R

dt
i - V s R  L
t
Integrating both side to obtain explicit expression for i as a function of

i (t )
I0
dx
-R

x - V s R  L
 Vs 
 i -  dt
R 

t
t

dy
Were I0 is the current at t = 0 and i(t) is
the current at any t > 0
0
Performing the Integrating and the substitution of the limits
i (t ) - V s R  - R
i (t ) - V s R 
- R
 ln

t
e
I 0 - V s R 
L
I 0 - V s R 
When no initial current on the inductor I0 = 0
L t
i (t ) 
i (t ) 
Vs 
V
+I0 - s
R 
R
Vs
V - R
- se
R
R
L t
 - R
e

L t
V s  Ri + L
t =0
R
+
VS
+
-
i
L
v (t)
i (t ) 
Vs
V - R
- se
R
R
i (t ) 
Vs 
V
+I0 - s
R 
R
-
di
dt
When no initial current
on the inductor I0 = 0
L t
 - R
e

L t
When initial current on
the inductor I0 ≠ 0
The equation for the no initial current (I0 = 0) indicate that the after the switch is closed the
current will increase exponentially to its final value of Vs / R
i (t ) 
Vs
V - R
- se
R
R
L t
When no initial current
on the inductor I0 = 0
At one time constant t = L / R the current will be
i ( ) 
Vs
V
V
- s e -1  0.6321 s
R
R
R
The current will reached 63% of Its final value
i (t ) 
Vs
V - R
- se
R
R
L t
The rate of change of i(t) or di(t)/dt current will be
V - R
di (t )
 - se
dt
R
L t
 R
- 
 L

V s - R
e
L
L t
The rate of change of i(t) or di(t)/dt at t = 0 ( The tangent at t=0 ) will
be
di (0) V s

dt
L
Which when drawn on the plot of i(t) will be as
Here  = L/R therefore the slop of the tangent at
t =0 is
Vs
Vs
R  R V s
L
L

R
di
dt
V s  Ri + L
t =0
R
+
VS
+
-
i
L
i (t ) 
v (t)
-
Vs 
V
+I0 - s
R 
R
 - R
e

L t
When initial current on
the inductor I0 ≠ 0
The voltage across the inductor will be
v (t )  L
di (t )
V

 L I 0 - s
dt
R

 - R
e

L t
 R

 L
 V s - I 0 R  e
- R L t
Before the switch close the voltage across the inductor is zero
t0
+
-
v (0 )  0
Just after the switch close the voltage across the inductor will jump to
v (0 )  V s - I 0 R e
+
- R L (0 )
+
 V s - I 0 R 
because initial current on the inductor I0.This will make the
voltage drop across the resistor after the switch was closed at t =
0+ to be RI0 → voltage drop across the inductor is (Vs-RI0)
If the initial current on the inductor I0 = 0, the current will be v (t ) V s e - R
and the voltage across the inductor will jump to
L t
v (0+ ) V s
Note the inductor voltage can jump however the current is not allowed to jump
t  0+
V s  Ri + L
t =0
R
+
VS
+
-
L
i
v (t)
-
i (t ) 
Vs
V - R
- se
R
R
 0
v (t )  
- R
V
I
R
e


 s
0
 0
v (t )   - R
V s e
t<0
t  0+
t<0
L t
Vs 
V
+I0 - s
R 
R
 - R
e

When no initial current
on the inductor I0 = 0
L t
L t
i (t ) 
t0
+
di
dt
When initial current on
the inductor I0 ≠ 0
When no initial current
on the inductor I0 = 0
L t
When initial current on
the inductor I0 ≠ 0
V s  Ri + L
t =0
R
When no initial current
on the inductor I0 = 0
di
dt
+
VS
+
-
i
L
v (t)
-
i (t ) 
Vs
V - R
- se
R
R
 0
v (t )   - R
V s e
L t
t<0
L t
t0
+
t =0
R
+
VS
+
-
i
L
v (t)
-
V s  Ri + L
di
dt
Solving the differential equation when the initial current
on the inductor I0 ≠ 0 , we have
i (t ) 
Vs 
V
+I0 - s
R 
R
 - R
e

L t
What we want to do next is to find the indictor current i(t) without finding the differential
equation or solving it
Let us look at the the indictor current i(t)
i (t ) 
The constant part which is the
steady state value of the current
or the value of the current at
t = ∞ ( i(∞) )
Vs

+  I0
R

-
initial current on
the inductor i(0)
Vs
R
 e

R
L
t
reciprocal of the
time constant 
t =0
R
+
VS
+
-
L
i
v (t)
-
di
V s  Ri + L
dt
Vs 
Vs
i (t ) 
+I0 R 
R
 - R
e

L t
i (t )  i () +  i () - i ()  e -t 
You find it after you move the
switch and the inductor in the
DC state or is short
You find it before you move
the switch and the inductor in
the DC state or is short

The time constant 
You find it after you
move the switch and it
is
L
R EQ
The resistant seen by the inductor
Example 7.5
The switch in the circuit shown has been in position a for a long time
At t = 0, the switch moves from position a to position b.
(a) Find the expression for i(t) for t >0.
i (t )  i () +  i () - i ()  e -t 
For t < 0.
Inductor is short
i
i ( ) = - 8 A
i (t )  i () +  i () - i ()  e -t 
For t  
i
Inductor is short
24
i ( ) 
 12 A
2
i (t )  i () +  i () - i ()  e -t 
i ( ) = - 8 A
We need  ( the time constant)
24
i ( ) 
 12 A
2
We look at the circuit t > 0
We deactivate independent sources ( leave dependent)
We now find the equivalent resistant seen by the indictor R eq  2W
i (t )  i () +  i () - i ()  e -t 
i ( ) = - 8 A
24
i ( ) 
 12 A
2
The time constant
L 200 X 10
 = 
R
2
-
100 ms
The Step Response of an RC Circuit
The switch is closed at t = 0 , the task is to find the expressions for the voltage across
And the current through the capacitor after the switch has been closed
t =0
+
IS
R
C
i
vc
-
After the switch has being closed, we have
+
C
R
IS
iR =
vC
R
iC = C
dv C
dt
KCL
Is 
vC
dv
+ C C
R
dt
vc
-
We can solve the differential equation similar to what we did previously with the step
response for RL circuit by separating the variables
v C (t )  RI s + V 0 - RI s  e -t
RC
t0
v and t
and integrating we obtain
Were V0 is the initial voltage at the capacitor
+
KCL
C
R
IS
iR =
vC
R
iC = C
Is 
vC
dv
+ C C
R
dt
vc
dv C
dt
-
Let us look at the the capacitor voltage current v(t)
v C (t )  RI s +
The steady state
value of the voltage
vC(∞)
 V0
- RI s
 e -t
Initial voltage on
the capacitor v(0)
RC
t0
The time constant 
t =0
+
IS
R
C
i
Is 
KCL
vC
dv
+ C C
R
dt
vc
-
v C (t )  RI s + V 0 - RI s  e -t
RC
t0
v C (t )  v C () + v C () -v C ()  e -t 
t0
t =0
+
R
IS
C
i
KCL
vc
Is 
vC
dv
+ C C
R
dt
v C (t )  RI s + V 0 - RI s  e -t
-
RC
t0
When initial voltage on
the capacitor V0 ≠ 0
The current through the capacitor will be
dv (t )
i C (t )  C C
dt
 C V 0 - RI s  e -t
RC
1 



 RC 
V 

  I s - 0  e - t RC

R 
Before the switch close the current through the capacitor is
zero ( open circuit )
t  0+
i C (0- )  0
Just after the switch close the current through the capacitor will jump to
V 0  -0+ RC 
V 

i C (0 )   I s -  e
 Is - 0 

R 

R 
+
because initial voltage on the capacitor V0.This will make the
current through the resistor after the switch was closed at t = 0+
to be V0/R → current through the capacitor is (Is-V0/R)
Note the capacitor current can jump however the voltage is not allowed to jump
Summary of Step Response of RL and RC Circuits
t =0
R
t =0
+
VS
+
-
L
i
+
C
R
IS
v (t)
i
-
KVL V s  Ri + L
i (t ) 
Vs 
V
+I0 - s
R 
R
di
dt
 - R
e


0
v (t )  
- R
V
I
R
e



s
0

L t
KCL
t0
t < 0
L t
t  0+
because initial current on the inductor I0.
This will make the voltage drop across the resistor
after the switch was closed at t = 0+ to be
RI0 → voltage drop across the inductor is
(Vs-RI0)
Is 
vc
-
vC
dv
+ C C
R
dt
v C (t )  RI s + V 0 - RI s  e -t
0

i C (t )  
V 0  - t RC
I
 s R  e

RC
t0
t <0
t  0+
because initial voltage on the capacitor V0.
This will make the current through the
resistor after the switch was closed at t = 0+ to be
V0/R → current through the capacitor is
(Is-V0/R)
Example 7.11
For t ≤ 0
For t < 0 both switches are closed causing the
150 mH inductor to short the 18 W resistor
iL(0-) = 6 A
Using source transformations, we find that iL(0-) = 6 A
For 0 ≤ t ≤ 35 ms
Switch 1 is open ( switch 2 is closed)
The 60 V voltage source and 4 W and 12 W
are disconnected from the circuit
For t ≤ 0
For 0 ≤ t ≤ 35 ms
Switch 1 is open ( switch 2 is closed)
The indictor is no longer behaving as a short
circuit because the DC source is no longer in
The circuit
The 60 V voltage source and 4 W and 12 W
are disconnected from the circuit
The 18 W resistor is no longer
short-circuited
6W
R EQ  (3W+ 6W) 18W = 6 Ω
i L (t )  i L (0)e - R
EQ
L t
 6e -6 0.15t
 6e -40t A
+
iL
vL
150 mH
-
iL(0-) = 6 A
For 0 ≤ t ≤ 35 ms
For t ≤ 0
iL(0-) = 6 A
For 0 ≤ t ≤ 35 ms
6W
+
iL
vL
150 mH
-
i L (t )  6e -40t A
R EQ  (3W+ 6W) 18W = 6 Ω
t ≥ 35 ms
9W
i L (t )  i L (0.035)e -9 0.15(t -0.035)
+
iL
vL
150 mH
-
i L (0.035)
We find iL(0.035-) from the circuit before
 i L (0.035)e -60(t -0.035)
i L (0.035)  i L (0.035-)
For 0 ≤ t ≤ 35 ms
For 0 ≤ t ≤ 35 ms
6W
+
iL
vL
150 mH
-
R EQ  (3W+ 6W) 18W = 6 Ω
i L (t )  6e -40t A
t ≥ 35 ms
9W
+
vL
150 mH
-
i L (0.035)  i L (0.035-)
i L (t )  i L (0.035)e -9 0.15(t -0.035)
iL
i L (0.035)  i L (0.035-)
We find iL(0.035-) from the circuit before
For 0 ≤ t ≤ 35 ms
 6e -40(0.035)  6e -1.4 1.48 A
i L (t ) 1.48e -60(t -0.035) A
For 0 ≤ t ≤ 35 ms
iL(0-) = 6 A
For 0 ≤ t ≤ 35 ms
6W
+
iL
vL
150 mH
-
i L (t )  6e -40t A
t ≥ 35 ms
9W
+
iL
vL
150 mH
-
i L (t ) 1.48e -60(t -0.035) A
6
 -40t
i L (t )  6e
1.48e -60(t -0.035)

t<0
0  t  35 ms
t  35ms
Unbounded Response
Example 7.11
The Integrating Amplifier