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

+-

Ccv

+

-

i

a

b

SV

R

Energy released by a capacitor

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

+-

Ccv

+

-

i

SV

R

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

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

Tank of WaterLevel of Water

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

+-

Ccv

+

-

i

a

b

SV

R

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

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

This response is referred to as the step response

+-

Ccv

+

-

i

SV

R

Level of Water

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

Suppose the switch has been in a closed position for a long time we will define long time later

For the time being long time

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 0di

v = Ldt

i

v

+

-

switch

L RoRSI

Because the voltage across the inductive branch is zero

There can be no current in either R0 or R

All the source current IS appears in the inductive branch

0di

v = Ldt

i

v

+

-

switch

L RoRSI

Finding the natural response requires finding the voltages and currents at any branch in the circuit after the switch has been opened.

sI

If we let t = 0 denote the instant when the switch is opened

v

+

-

SI oR

i

L

t = 0

The problem become one of finding v(t) and i(t) ( or i and v ) for t ≥ 0

For t ≥ 0 the circuit become

v

+

-

L

i

(0) = si I

sI

v

+

-

SI oR

i

L

sI

Deriving the Expression for the current

0diL Ridt

+

KVL around the loop

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

L

i

(0) = si I

R Rv-

+Lv

+

-

+ 0L Rv v

0diL Ridt

+ v

+

-

L

i

(0) = si IR

To solve the differential equation we processed as follows:

diL Ridt

- di R idt L

- di Rdt idtdt L

- Rdi idtL

-

di R dti L-

Integrating both side to obtain explicit expression for i as a function of t

0 0( )

( )

t t

t tdx R dyx L

- i

i

Here t0 = 0 0 0( )

( )t tdx R dyx L

- i

i ( )ln

( )0

i

i

tt

R

L

-

Based on the definition of the natural logarithmlnxe x

( ) (0) R L ti t i e -

0diL Ridt

+ v

+

-

L

i

(0) = si IR

( ) (0) R L ti t i e -

Since the current through an indicator can not change abruptly or instanteously  

0( ) ( ) (0)00i i i I+ -

Were I0 is the initial current on the inductor just before the switch opened )or in some cases closed(

In the circuit above I0 = Is

Therefore

0 ( ) R L ti t I e - t 0≥

Which shows that the current start from initial value I0 and decreases exponentially toward zero as t increases

0diL Ridt

+ v

+

-

L

i

(0) = si IR

Note the voltage v(t) here is defined for t > 0 because the voltage across the resistor R is

zero for t < 0 ( all the current Is was going through L and zero current through R)

We derive the voltage across the resistor form direct application of Ohm’s law

0 ( ) R L ti t I e - t 0≥

( ) ( )v t Ri t 0 R L tRI e - +t 0≥

v

+

-

SI oR

i

L

t = 0

R

SI

(0 ) 0v -

0(0 ) Rv I+

Note the voltage v(t) across the resistor and across the indicator can change instanousley or have a jump

0diL Ridt

+ v

+

-

L

i

(0) = si IR

0 ( ) R L ti t I e - t 0≥

We derive the power dissipated in the resistor

p vi2p Ri

2

pR

vor

Whichever form is used, the resulting expression can be reduce to

22

0

R L tI Re

- +t 0≥ 0

2 R L tp R I e -

Note the current i(t) through the resistor is zero for t < 0 0 p t < 0

0diL Ridt

+ v

+

-

L

i

(0) = si IR

0 ( ) R L ti t I e - t 0≥

We derive the energy delivered to the resistor

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

22

0

R L tp I Re

- +t 0≥

0pdxw

t

0

22

0 R xL

dxI Re -t

22

0

11

2 tR L

R LI R e - -

22

0

11

2( ) tR L

LI e - - t 0≥

v

+

-

SI oR

i

L

t = 0

R

SI

Note that just before the switch is opened the current on the indicator is IS = I0

21

2Liw 2

0

1

2LI

v

+

-

L

i

(0) = si IR

22

0

11

2( ) tR L

LI ew - - t 0≥

lim w(t)t→∞

22

0

11

2 lim ( )tR L

LI e -- t→∞

2

0

1

2LI The initial energy stored in the indictor

The significant of the time constant

0diL Ridt

+ v

+

-

L

i

(0) = si IR

0 ( ) R L ti t I e - t 0≥

Example

L

R

V.sAVA

s Seconds

0( ) i I e

-

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

the time constant for the R,L circuit is = L

R

It is convenient to think of the time elapsed after

switching in terms of integral multiple of

10 I e - 00.367I

After one time constant, the maximum of the current I0 has dropped to 37% of its value

0I0.367

te -

t

3.6788 x 10 -

2 1.3534 x 10 -

23 4.0787 x 10 -

35 6.7379 x 10 -

510 4.5400 x 10 -

. .

. .

Table showing the value of

exponential e-t for integral

multiples of

24 1.8316 x 10 -

36 2.4788 x 10 -

0I<0 67

1%.0

Note that when the time elapsed after switching exceed five time constants

The current is less than 1% of its initial value I0

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

5

v

+

-

L

i

(0) = si IR

0 ( ) R L ti t I e - t 0≥

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

(1) If the RL circuit can be put asL R

Were L is an equivalent inductor and R is an equivalent resistorSeen by the equivalent indictor

L1

1R

L2

2R3R eq 1 2 +L L L eq 1 2 3( || ) +R R R R

eq

eq

L

R

Example

(2) If we know the differential equation

0diL Ridt

+ v

+

-

L

i

(0) = si IR

0di R idt L

+

is the inverse of this constant

Example

Suppose the differential equation is given as

3 6 0di idt

+ 3

6 0di idt

+ 3

0.56

=

(2) If a trace or graph of i(t) is given

Differentiating i(t) we have

0 d ( )

dtt ti t I e

--

0 d (0)

dti I

-

v

+

-

L

i

(0) = si IR

0diL Ridt

+ 0 ( ) R L ti t I e - t 0≥

i

v

+

-

switch

L RoRSI

sI

( ) ( )v t Ri t 0 R L tRI e - +t 0≥

22

0

R L tI Re

- +t 0≥ 0

2 R L tp R I e -

0

22

0

11

2( ) tR L

pdx LI ew - - t 0t

≥

(0 ) 0v - 0(0 ) Rv I+

2

0

1

2lim w(t) LIt→∞

We see the circuit at 0

( just befor you move the switch)

t -

0t -

The inductor is short

30(0 ) 6

5Li A- - -

We see the circuit at t

( ) 0Li

We see the circuit at 0t

0t eq

eq

L

R

10 (5)(10)5

6 5 10eqL H + +

eqR

Teq

T

VR

I

2x

T x

iI i -

2xi

100T

x

Vi

1002

T

T

V

I 200

TV T

eqT

VR

I 200

eq

eq

L

R 35

25 X 10 s200

-

( ) (0) 0t

L Li t i e t-

3 4025X10( ) 6 = 6

tt

Li t e e--

-- -

1

( )( ) L

eq

di tv t L

dt- 1

(5)(10) 50

5 10 15eqL H +

4050( ) 6

15td

v t edt

- - -

40506( 40)

15te- - - -

40800 Vte--

We need to Find v (t)

0

01 ( ) ( )

t

t

i v d i tL

+Since the current through an inductor is given as

1 2Therefor to find ,i i

The circuit reduce to

Finding v (t)

( ) 8 ( )v t i t 28 12 te -

( ) 10

v t

3( ) 15

10 10 15v t

i + 2 +5.7 A t 0te -

( )

4 15 10v t

+

2 +3 5.7 A t 0ti e -

2 ( )( )

8v t

p t =

The voltage across the capacitor =

For t >= 0

KCL on the upper junction

Solving for t

36 V

a b

12

6 0 st

3

6

1 F16

+-

2 xixi

( )v t+

-

For the circuit shown above, the switch was in position a for a long time ,then at time t=0 the switch move to position b .

Find the voltage v(t) for t > 0?

Solution

( ) (0) t

eqv t v e R C -

To be find at t < 0

To be find at t < 0

36 V

a b

12

6 0 st

3

6

1 F16

+-

2 xixi

( )v t+

-

( ) (0) t

eqv t v e R C -

To be find at t < 0 To be find at t < 0

For t < 0 capacitor is open

36 V

6

3

6

+- ( )v t

+

-Capacitor r is open

3

3 Voltage division (t) = (t)= (36) = 27 V

3+6

(0 ) = 27 V

v v

v

-

For t > 0

36 V

a b

12

6 0 st

3

6

1 F16

+-

2 xixi

( )v t+

-

( ) (0) t

eqv t v e R C -

To be find at t < 0 To be find at t < 0

b

12 1 F16

2 xixi

( )v t+

-

1 = F . To find we apply a test voltage

16eq eqR C C R

12 2 xixi

+-T

V

TI

12 2 xixi

+-T

V

TI

4

12 < 0 ( )

12 > 0t

tv t

e t-

T x x T x2 + 3 I i i I i KCL T

x 12

Vi

T T

T

T

4 12

V V3 eqRI

I

1 1 4 s

16 4eqR C

4(1/4) ( ) (0) 12 12tt

tv t v e e e-- -

We need to Find i(t)

0

01( ) ( ) ( )

t

t

v t i d v tC

+

Similar to the previous example 7.2 , the voltage across a capacitor is given as

1 2Therefor to find ( ), ( )v t v t

Question: how to find i(t) ?

We need to Find i(t)

0

01( ) ( ) ( )

t

t

v t i d v tC

+1 2Therefor to find ( ), ( )v t v t

Question: how to find i(t) ?

Do we find i(t) through the capacitor voltage 1 21 2

( ) ( )( ) or ( ) dv t dv ti t C i t Cdt dt

- -

The answer is no of course because we do not know 1 2( ), ( )t tv v

However we can find i(t) through the resistor 250 k as 3

( )( )(250X10 )

v ti t

we need to find v(t) as will be shown next

The circuit reduce to

(5)(20)(5 20)+

-

/0( ) 0tv t V e t-

The solution of the voltage v(t) is given as

RC 3 6(250 10 )(4 10 ) 1 sX X -

0 20 VV

were

( ) 20 0tv t e t-

The circuit reduce to

( ) 20 0tv t e t-

Note the direction of the current i )t( and the initial voltagesand the polarity of 1 2( ), ( )t tv v

The circuit reduce to ( ) 20 0tv t e t-

( ) 20 0tv t e t-

( ) 20 0tv t e t-

d) Show that the total energy delivered to the 250 kW resistor is the difference between the results in (b) and (c)

)a(

)b(

)c(

Comparing the results obtained in (b) and in (c)

v

+

-

L

i

(0) = si IR

0diL Ridt

+ KVL

0 ( ) R L ti t I e - t 0≥

0 ( ) ( ) R L tv t Ri t RI e - +t 0≥

22 2

0( ) ( )

R L tp t Ri t RI e

- +t 0≥

0pdxw

t

2

0

1

2lim w(t) LIt→∞

0dv vCdt R

+ KCL

0 ( ) V RC

t

v t e-

t 0≥

0 0 ( ) V V( )

RCRC

ttv t ei t e

R RR

--

+t 0≥

2

0

1

2lim w(t) CVt→∞

0di R idt L

+ 1 0dv v

dt RC+

1/ L 1/ C

2220

V( )RC

tv tp eRR

- +t 0≥

0pdxw

t

22

0

11

2( ) tR L

LI e - - t 0≥2

2

0

11

2( ) RC

t

CV e-

- 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 circuitswhen 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

+-

L)t(v

+

-

SV

R= 0t

After the switch has being closed, we have

KVL s

diV Ri L

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

sRi Vdi

dt L

- + sVR

iL R

- -

sVRdi i dt

L R

-

-

i+-

L)t(v

+

-

SV

R

s

diV Ri L

dt +

sRi Vdi

dt L

- + sVR

iL R

- -

sVRdi i dt

L R

-

-

s

di Rdt

i V R L

-

-

0

( )

0

ti t

I s

dx Rdy

x V R L

-

- Were I0 is the current at t = 0 and i(t) is the current at any t > 0

Performing the Integrating and the substitution of the limits

0

( ) ln s

s

i t V R Rt

I V R L

- -

-

0

( ) R L ts

s

i t V Re

I V R-

-

- 0( ) R L ts sV V

i t I eR R

-

+

-

When no initial current on the inductor I0 = 0 ( ) R L ts sV V

i t eR R

- -

i+-

L)t(v

+

-

SV

R

Integrating both side to obtain explicit expression for i as a function of t

We know separate the variables i and t

( ) R L ts sV Vi t e

R R- -

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

+-

L)t(v

+

-

SV

R= 0t

i

s

diV Ri L

dt +

When no initial current on the inductor I0 = 0

0( ) R L ts sV V

i t I eR R

-

+

- When initial current on the inductor I0 ≠ 0

( ) R L ts sV Vi t e

R R- -

At one time constant t = L / R the current will be

1( ) 0.6321 s s sV V Vi e

R R R - -

The current will reached 63% of Its final value

When no initial current on the inductor I0 = 0

( ) R L ts sV Vi t e

R R- -

The rate of change of i(t) or di(t)/dt current will be

( ) R L tsVdi t R

edt R L

- - -

The rate of change of i(t) or di(t)/dt at t = 0 ( The tangent at t=0 ) will be

(0) sVdi

dt L

R L tsVe

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

sV

R

s

R

V

RL

sV

L

The voltage across the inductor will be

( )( )

di tv t L

dt +

0 t 0R L t

sV I R e - -

Before the switch close the voltage across the inductor is zero

0 L R L tsV R

I eR L

- - -

(0 ) 0v -

Just after the switch close the voltage across the inductor will jump to

+-

L)t(v

+

-

SV

R= 0t

i

s

diV Ri L

dt +

When initial current on the inductor I0 ≠ 0

(0 )

0(0 ) R L

sv V I R e+-+ - 0sV I R -

If the initial current on the inductor I0 = 0, the current will be

(0 ) sv V+

+( ) t 0R L t

sv t V e -

and the voltage across the inductor will jump to

Note the inductor voltage can jump however the current is not allowed to jump

0( ) R L ts sV V

i t I eR 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(

( ) R L ts sV Vi t e

R R- -

+-

L)t(v

+

-

SV

R= 0t

i

s

diV Ri L

dt +

When no initial current on the inductor I0 = 0

0( ) R L ts sV V

i t I eR R

-

+

- When initial current on the inductor I0 ≠ 0

+0

0 t < 0( )

t 0R L t

s

v tV I R e -

-

When initial current on the inductor I0 ≠ 0

+

0 t < 0( )

t 0R L t

s

v tV e -

When no initial current on the inductor I0 = 0

( ) R L ts sV Vi t e

R R- -

+-

L)t(v

+

-

SV

R= 0t

i

s

diV Ri L

dt +

When no initial current on the inductor I0 = 0

+

0 t < 0( )

t 0R L t

s

v tV e -

+-

L)t(v

+

-

SV

R= 0t

i

s

diV Ri L

dt +

0( ) R L ts sV V

i t I eR R

-

+

-

Solving the differential equation when the initial current on the inductor I0 ≠ 0 , we have

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)

0( ) ts sV Vi t I e

R R-

+ - R L

The constant part which is thesteady state value of the current

or the value of the current at t = ∞ ( i(∞) )

initial current on the inductor i(0)

reciprocal of the time constant

+-

L)t(v

+

-

SV

R= 0t

i

s

diV Ri L

dt +

0( ) R L ts sV V

i t I eR R

-

+

-

(( ) (( ) ) ) ti t e - + -ii i

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

EQ

L

R

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.

(( ) (( ) ) ) ti t e - + -ii i

iInductor is short= ( ) 8 A-i

For t < 0.

(( ) (( ) ) ) ti t e - + -ii i

i Inductor is short24

12 A2

( ) i

For t

We need ( the time constant) We look at the circuit t > 0

(( ) (( ) ) ) ti t e - + -ii i

= ( ) 8 A-i 2412 A

2( ) i

We deactivate independent sources ( leave dependent)

We now find the equivalent resistant seen by the indictor

eq 2R

The time constant

L 200 X 10 = 100 msR 2

-

(( ) (( ) ) ) ti t e - + -ii i

= ( ) 8 A-i 2412 A

2( ) i

cv

+

-i

SI R C

= 0t

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

After the switch has being closed, we have

KC L C Cs

v dvI C

R dt +

cv

+

-C

C

dvi = C

dt

SI R C

CR

vi =

R

We can solve the differential equation similar to what we did previously with the step

response for RL circuit by separating the variables v and t and integrating we obtain

0( ) t RCC s sv t RI V RI e - + - t 0 Were V0 is the initial voltage at the capacitor

KC L C Cs

v dvI C

R dt +

cv

+

-C

C

dvi = C

dt

SI R C

CR

vi =

R

0( ) t RC

C s sv t RI V RI e - + - t 0

Let us look at the the capacitor voltage current v(t)

The steady state value of the voltage

vC (∞)

Initial voltage on the capacitor v(0)

The time constant

cv

+

-i

SI R C

= 0t

KC L C Cs

v dvI C

R dt +

0( ) t RCC s sv t RI V RI e - + - t 0

( ) (( ( ))) C CC Ctv vvv t e - + - t 0

The current through the capacitor will be

t +0 t 0RCs

VI e

R-

- 0

1 C t RC

sV RI eRC

- - -

cv

+

-i

SI R C

= 0t

KC L C Cs

v dvI C

R dt +

0( ) t RCC s sv t RI V RI e - + - t 0

( )( ) C

C

dv ti t C

dt

When initial voltage on the capacitor V0 ≠ 0

Before the switch close the current through the capacitor is zero ( open circuit )

(0 ) 0Ci-

Just after the switch close the current through the capacitor will jump to

0s

VI

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(

+00(0 ) RCC s

Vi I e

R+ -

-

Note the capacitor current can jump however the voltage is not allowed to jump

Summary of Step Response of RL and RC Circuits

+-

L)t(v

+

-

SV

R= 0t

i

KV L s

diV Ri L

dt +

0( ) R L ts sV V

i t I eR R

-

+ -

t 0

+0

0 t < 0( )

t 0R L t

s

v tV I R e -

-

cv

+

-i

SI R C

= 0t

KC L C Cs

v dvI C

R dt +

0( ) t RCC s sv t RI V RI e - + - t 0

t +0

0 t 0

( ) t 0

C RCs

i t VI e

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)

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 both switches are closed causing the 150 mH inductor to short the 18 resistor

Using source transformations, we find that iL)0-( = 6 A

For 0 ≤ t ≤ 35 ms

For t ≤ 0

Switch 1 is open ) switch 2 is closed( The 60 V voltage source and 4 and 12 are disconnected from the circuit

iL)0-( = 6 A

For 0 ≤ t ≤ 35 ms

Switch 1 is open ) switch 2 is closed( The 60 V voltage source and 4 and 12 are disconnected from the circuit

The indictor is no longer behaving as a short circuit because the DC source is no longer in The circuit

The 18 resistor is no longer short-circuited

For t ≤ 0

Li

6 150 mH

+

-

Lv

(3 6 ) 18EQR + = 6 Ω

( ) (0) EQ LR tL Li t i e - 0.1566 te - 40 6 Ate - For 0 ≤ t ≤ 35 ms

iL)0-( = 6 A

For 0 ≤ t ≤ 35 ms

Li

6 150 mH

+

-

Lv

40 ( ) 6 AtLi t e -

t ≥ 35 ms

iL)0-( = 6 A

For t ≤ 0

(3 6 ) 18EQR + = 6 Ω

Li

9 150 mH

+

-

Lv

9 0. 5 (1 )( ) ( ) tL Li t i e - - 0.0350.035

We find iL(0.035-) from the circuit before

( ) ( )L Li i -0.035 0.035

For 0 ≤ t ≤ 35 ms

(60 )( ) tLi e - - 0.0350.035

( )Li 0.035

For 0 ≤ t ≤ 35 ms

Li

6 150 mH

+

-

Lv40 ( ) 6 At

Li t e -

t ≥ 35 ms

(3 6 ) 18EQR + = 6 Ω

Li

9 150 mH

+

-

Lv

9 0. 5 (1 )( ) ( ) tL Li t i e - - 0.0350.035

We find iL(0.035-) from the circuit before

( ) ( )L Li i -0.035 0.035

For 0 ≤ t ≤ 35 ms

( ) ( )L Li i -0.035 0.035 40( )6e - 0.035 1.46e - 1.48 A(60 ) ( ) 1.48 At

Li t e - - 0.035

For 0 ≤ t ≤ 35 ms

iL)0-( = 6 A

For 0 ≤ t ≤ 35 ms

Li

6 150 mH

+

-

Lv40 ( ) 6 At

Li t e -t ≥ 35 ms

Li

9 150 mH

+

-

Lv

(60 ) ( ) 1.48 AtLi t e - - 0.035

(

4

)60

0 0 t 35

t

t 3

<

5

0

s

ms

m

1.48

66( )L

t

ti te

e-

-

-

0.035

Example 7.11

Unbounded Response

The Integrating Amplifier