Elec 3202 Chap 6

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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

Chapter (6)

First Order Transient Circuits

This chapter deals with the transient response of first order circuit .

first order circuit are those circuits whose response can beexpressed by a first order differential equations.

Example of such circuits include the RL and RC circuits

RL circuit : circuit that contains an indicator and a resistor.RC circuit : circuit that contains a capacitor and resistor

For example: RC circuitConsider the following circuit .

+

- iVS

C

R

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S

S

Vi(t)R(t)v

0i(t)R(t)vV

c

C

=+

=++

Sc

c

c

Vdt

(t)dvCR(t)v

dt

(t)dvCi(t)since

=+

=

Scc V

CR1(t)v

CR1

dt

(t)dv

=+

b(t)va

dt

(t)dv

VCR

1

b,CR

1

alet

cc

S

=+

==This is the general form of a linear

first order differential equation

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This is a first order linear ordinary non-homogenous differential

equation describing the response of the capacitor voltage.

It is first order because the highest degree of derivative is one.

it is called linear because the differential equation is linear

function of (dvc /dt) and vc(t)

Example of non-linear differential equation :

b(t)vdt

(t)dv(t)v2

cc

c =+

it is called ordinary because it deals only with ordinary

derivatives ( not partial derivative )

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It is called non-homogenous because b 0.

Example of first order linear ordinary homogenous differential

equation is

0(t)va

dt

(t)dvc

c =+

Example :

R L circuit

+

- iL(t)

R

VSL

KVL around the loop

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L

V(t)i

L

R

dt

(t)di

0dt

(t)diL(t)iRV

SL

L

LLS

=+

=++

b(t)iadt

(t)di

LVb,

LRalet

LL

S

=+

==

This is a first order linear ordinary non-homogenous differential

equation

To find the response of the Vc(t) in RC circuit or iL(t) in the RL

circuit we need to solve these differentials

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Solution of the first order differential equation :

Consider the first order linear ordinary non-homogenousdifferential equation :

We want to find X(t) that satisfies (*)

(*)bx(t)adt

dx(t)

LL=+

Theorem : ( in differential equation)

If x (t) = xP (t) is any solution of equation (*) and x (t) = xc (t) is

any solution of the homogenous differential equation

(**)0(t)xadt

(t)dxc

cLL=+

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then

(t)x(t)xx(t) cP +=

Where :

xP(t) = particular solution ( forced solution)

xC(t) = complementary solution ( natural solution)

Hence we need to solve 2 differential equations

(*)b(t)xadt

(t)dx PP

LL=+

What is the function xP(t) that if its differential is summed to a*xP(t)

will give a constant (b) The solution xp(t) must be constant

xP(t) = k1

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Use xP(t) in the non-homogenous differential equation

a

bk(t)x

a

bk

bka)(kdt

d

1P

1

11

==

=

=+

Particular (forced) response

Consider the homogenous differential equation:

adt

(t)dx

(t)x

1

(*)0(t)xa

dt

(t)dx

c

c

cc

=

=+ LL

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( )[ ]

( )[ ] a(t)xlndt

d

dt

(t)dx

(t)x

1(t)xln

dt

dsince

c

c

c

c

=

=

[ ]

[ ]Cat

c

c

c

e(t)x

Cta(t)xln

dta(t)xln

+=+=

= Take the integral of both sides

at

2c

c

2

Catc

ek(t)x

eklet

ee(t)x

=

==

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t/

21 ekkx(t)+=

Hence : let = 1/a time constant

Time constant : a parameter that determines the rate of decrease of

x(t)

Lets find the solution of RC & RL circuits :

at21

cP

ekkx(t)

(t)x(t)xx(t)

+=

+=

Hence, a general form of the solution is:

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Example :

+

- ic(t)RVS

C

S1

CRt

21

at

21c

S

c

c

VCR

1

CRVs

a

bk

ekkekk(t)v

CR

V(t)v

CR

1

dt

(t)dv

===

+=+=

=+

Assume vc(0) = v0

To find k2 , we need the initial condition of vc(t)

For example , if we know vc(0) = V0

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S02

2S0

CRt

2Sc

VVk(1)kVV

ekV(0)v

=+=

+=

CRt

S0S

CR

t

21c

e)V(VV

ekk(t)v

+=

+=

As a special case , lets consider the natural response

Natural response :

Circuit response when no source is affecting the response

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Case 1 : Natural RC response

ic(t) R

C

0(t)vCR

1

dt

(t)dv

0Rdt

(t)dv

C(t)v

0R(t)i(t)v

cc

c

c

cc

=+

=+

=+Assume vc(0)=V0

First order linear ordinary homogenous differential equation :

CRt

2c ek(t)v

=

We can find k2 from initial conditions

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CRt

0c

02

2CR

t

20c

0c

eV(t)v

Vk

kekV(0)v

thenV(0)vAssume

=

====

=vc(t)

t

V0

RCt

21 ekkx(t)+=

Note : the general solution of the forced response is

RCt

0c

0S02

S1S

eV(t)V

V)V(Vkand

0Vk0Vsince

=

==

===

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Case 2 : Natural response of RL circuit

iL(t)

R

L

0(t)iL

R

dt

(t)di

0dt

(t)diL(t)iR

L

L

LL

=+

=+

Assume :iL(0) = i0

KVL :

This is a first order linear ordinary homogenous differential equation

tL

R

2L ek(t)i

=

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To find k2, we need initial condition iL(0)

tL

R

0L

02

2(0)L

R

20L

ei(t)i

ik

keki(0)i

=

==== iL(t)

t

i0

Note :

In the forced response , we have

RVk

ekk(t)i

S1

tL

R

21L

=

+=

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Since VS here is 0 then k1 = 0

tL

R

0L

S0S

02

ei(t)i

0)V(sinceiR

V

ik

=

==

=

Analysis Techniques :

1. The differential equation approach

Here , a differential equation that describe the behavior of the

circuit is used.This first order differential equation is expressed in tems of the

voltage across the capacitor or current through the inductor.

Then the solution of this differential equation is obtained

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Example :

Find iL(t) , t 0 ?

First, find the initial condition iL(0- )

At t = 0-

, the inductor behaves as a short circuit .

2 H20 A 1.0

40iL (t)

2

10

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Hence , we assume that the current through the inductor doesnt

change instantaneously

iL(0- ) = iL(0) = iL(0

+) = 20 A

20 A 1.040

iL (0-)

= 20 A

2

10

iL(0-

) = 20 A

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At t = 0 the switch is open

Where Req = (40 // 10) + 2 = 400/5 + 2 =10

L= 2 H iL (t) 10

Req

=+

-

2 H 40iL (0-)= 20 A

10

2

OR

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KVL around the loop :

0(t)i5dt

(t)di

0(t)i10dt

(t)di2

0(t)iRdt

(t)diL

0v(t)v

LL

LL

LeqL

eqL

=+

=+

=+

=+

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We know that

5t

2L

t2

at2L

ek(t)i5

1

a

15awhere

ekek(t)i

=

===

==

We can find k2 from the initial condition

iL(0 ) = 20 = k2 e0 = k2

iL(t) = 20 e-5t A

iL(t)

t

20

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Example :

Find vc(t) , t 0 ?

The switch has been closed for long time .

The capacitor behave as open circuit .

7.5 m A

k80k50

k20

F0.4

t = 0

+

-

vc (t)

7.5 m Ak80 k50

k20

vc (0-)

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At t = 0 , the switch is open


0(t)v50dt

(t)dv

F0.4C,0

dt

(t)dvCk50(t)v

0(t)ik50(t)v

cc

cc

cc

=+

==

+

=+

( )

( ) V200k150

k80Am7.5k50

k70k80

k80Am7.5k50)(0v(0)v)(0v ccc

=

=

+=== +

k50+

-

vc(t)F0.4

Vc(t) = 200 e-50t V.

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Example :

Find vc(t) , t 0 ?

For t < 0 , the capacitor behave as open circuit .

V40)(0v(0)v)(0v ccc ===+

At t = 0 , the switch is moved

k60

k601

k20

F0.25

+

-

vc (t)

+

-+

-

40 V

75 V

k8 k40t = 0

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k601

+

-

vc (t) +-

75 V

k40k8

k601+

-

vc (t)

k40k8

1.875 m A

Source transformation

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+

-vc (t) +

-

60 VF0.25

32 k

8 k

+

-

vc (t)

k23

k8

1.875 m A

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6000-b,100a

6000(t)v100dt

(t)dv

0(t)v

dt

(t)dvCk4060

0(t)v(t)ik4060

cc

cc

cc

==

=+

=++

=++

601006000

abk

ekk(t)v

1

at

21c

===

+=

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Ve10060(t)v

100k40k60

40kk(0)v

t100

c

22

21c

+=

==+

=+=

vc(t)

t

30

- 60

To find k2 , we use initial condition:

40

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Case 2 : Step by step approach

1. Assume the solution is x(t) = k1 + k2 e-t/2. Assume that the circuit is in steady state before the switch moves

replace a capacitor by open circuit

replace a inductor by short circuitThen find vc(0

-) or iL(0

-)

3. The switch is now in the new location :

Replace the capacitor by a voltage source = vc(0

-

) Replace the inductor by a current source = iL(0

-)

And solve for x(0)

4. Assume t = , find x (t = )replace capacitor by open circuit and inductor by short circuit

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5. Find the time constant :

HOW ??

- Find the Thevenin equivalent resistance w.r.t the terminals of

the capacitor or inductor.

= RTH C or = L/RTH

6. Find the constants :

k1 = x()

k1+k2 = x(0) k2 = x(0)-x()

x(t) = x()+[x(0) - x()] e-t/

x(t) = final value + [ initial value final value ] et/

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Example :

[ ] t

000

t

210

e)(V(0)V)(V

ekk(t)v(1)

+=

+=

Find V0 (t) ?

F2

+

-

v0 (t)+-

12 V 2

+

-

8 V

2

2

1

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(1)20i2-i5

08)i-(i2i312-

21

211

LL=

=++

(2) Assume steady state , replace capacitor by open circuit .

Mesh

V812412(1)i)(0v

A0i,A4i

(2)8i4i2

0)i(i28i2

1c

21

21

122

=+=+=

===+

=++

LL

v0 (0-)12 V

2

+

-8 V

2

2

1

+

-

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3. The switch is moved now t = 0 ,

replace the capacitor by a voltage source = vc(0-) and solve

for V0(0)

V4

2

18

22

2)(0v(0)V c0 =

=

+

=

vc (0-)

= 10 V

12 V

2

21

+

- +

-

v0 (t)

+

-

=8V

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4. At t = replace capacitor by open circuit

V5

24

5

212

122

212)(V0 =

=

++=

12 V

2

21

+

-

v0 (t)

+

-)(vc

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5. Find the time constant

find the Thevenin equivalent resistance w.r.t x ,y

RTH = 1 // (2+2) = 1 // 4 = 4/5

12 V2

21

+

- x

y

( )5

8F2

5

4CR TH =

==

t

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[ ]

t8

5

0

t8

5

t

0000

e5

4

5

24

(t)v

e5244

524

e)(v(0)v)(v(t)V6.

=

+=

+=

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Example :

[ ] t

000

t

210

e)(v)(0v)(v

ekk(t)v:Step(1)+

+=

+=

+

-

v0

(t)

24 V

Vx4

4

F2+

-

2 Vx

t = 0

3 A

+-

Step (2) : assume steady-state , replace capacitor by open circuit

Find v0

(0-) ?

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V60)(0v

V60243624(12)3

24v3v24v2)(0v

V12(3)4v

0

xxx0

x

==+=+=

+=++=

==

+

-

v0 (0-)

24 V

Vx4

4

+

-

2 Vx

3 A

+-

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Step (3) : now switch is moved .Find v0(0)

(t)vx(t)hence

V60)(0v(0)v)(0v

c

000

==== +

24 V

Vx4

4

+

-

2 Vx

+-

( )0v

Step (4) : assume t = Find V0() ?

V24)(V

24

v24v2)(V

0Vx

0

xx0

=

=

++=

=

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Step (5) : Find the time constantfind the RTH w.r.t the terminals of capacitor

Voc = 24 V

24 V

Vx4

+

-

2 Vx+

-

+

-

voc4

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A4

2

8

2

VI

V8V

24V3

0V24V2

xsc

x

x

xx

===

=

=

=++

24 V

Vx4

+

-

2 Vx+-

4 I SC

24 V

Vx2

+

-

2 Vx+-

I SC

Now, find Isc

sec.12F)(2)(6CR

64

24

I

VR

TH

SC

OCTH

===

===

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[ ]

[ ]

Ve3624(t)V

e246024(t)V

e)(V(0)V)(V(t)V:Step(6)

12

t

0

12

t

0

t

0000

+=

+=

+=

t

210 ekk(t)i

+=

Example :

Find i0(t) , t > 0 ?

Step (1) :

12 V

k4

k2

F200

t = 0

+-

i0(t)

k2

k2

k2

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Step 2 : assume steady state ( for t < 0 ) replace capacitor by

open circuit .

V4)(0v(0)v)(0v

V4k6

2k)12()(0v

ccc

c

===

==

+

12 V

k2

+-

i0(t)

k2

k2

k2

vC(0-) -+

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Step 3 : now switch is moved , replace capacitor by voltage

source = vc(0) ,

Now find i0(0)

12 V

k2

+

-

i0(0+)

k2

k2

k2

+-

4 V

k4

i 2

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Am2.66

k3

k1Am8(0)i0 =

=

i0

k2k1

8 m A

i0(0+)

k2k2

6 m A2 m A

k2

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Step 4 : assume t = , find i0() . Steady state

Replace capacitor by open circuit

12 V

k2

+-

k2

k2

k2

)(i0

k4

12 V

+-

k2k2

)(i0

12

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k2k2

k2

k2k4

RTH

Am3k4

12)(i0 ==

Step 5 : find time constant .First find RTH at terminals of the capacitor

( )

( ) ( ) sec0.6F200k3CRk3

k2k2//k4//k4R

TH

TH

===

=

+=

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Step 6 : find the solution i0(t)

[ ]

Ame0.333(t)i

Ame3)(2.663

e)(i(0)i)(i(t)i

0.6t

0

0.6t

t

0000

=

+=

+=

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Example:

k21

F50

k4 k8k3

k21

V12k4

+-t = 0

i0 (t)

Find io(t) , using Step by step approach.

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Step 1 : assume i0

(t) = k1

+ k2

e t/

Step 2 : assume t < 0 ( steady state)

Replace capacitor by open circuit and find voc(0-)

As we have done before , vc(0

-

) = vc(0)= vc(0+

) = - 4 V.

Step 3 : now the switch is moved

replace the capacitor by voltage source of value -4 and find i0(0)

k4 k8k21

k4

i0 (0+)

i

k3

4-+-+-

k21

244

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( )

Am2

1

16

12

3

2

k4k12

k12

3

2(0)i

Am3

2

k6

4

k12//k4k3

4i(0)

0

=

=

+

=

=

=

+

=

Step 4 : assume t = , steady statereplace capacitor by open circuit and find i0()

i0(

) = 0

k12k3 k4

)(i0

St 5 fi d

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Step 5 : find [ ]

( ) ( ) sec0.350k6CR

k6k3k4//k12R

TH

TH

===

=+=

[ ]

0tAme

2

1(t)i

Ame2

1

0

e)(i(0)i)(i(t)i

0.3t

0

0.3t

t

0000

=

=

+=

Step 6 :k12k3 k4

RTH

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Example :

Step 1 : assume i0(t) = k1 + k2 e -t/

Step 2 : assume t < 0 ( steady state )

Replace inductor by short circuit and find

iL(0-) = i0(0

-)

k4 k2

t = 0

i0 (t)

Am10

k4

k2k5

Hm10

find i0(t) using step by step approach ?

k5

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Am5k2k2

k2Am10)(0i0 =

+=

k4 k2i0 (0)

Am10

k4

k2

Am10

k2

k2k5

Step 3 : no the s itch is mo ed t 0 replace ind ctor b a

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Step 3 : now the switch is moved t = 0 replace inductor by a

current source of value = 5 mA.

Since it is inductor i0(0-) = i0(0)

Step 4 : assume t = ( steady state )

Replace inductor by short circuit and find i0()

k4

k2i0 (t)

Am10

k4

k2k5

k2

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Am5k2k2

k2Am10)(i0 =

+=

Step 5 : find where = L / RTHLets find RTH at the terminals of the inductor

RTH = (4 k // 4 k) + 2 k

= 4 k = L / RTH = 10m/4k = 2.5 sec

i0 (t)

Am10

k4

k2k5

k4

k2k5

k4 RTH

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[ ]

[ ]Am5(t)i

Aem5m5m5

e)(i(0)i)(i(t)i

0

s2.5t

t

0000

=

+=

+=

Step 6 : find i0(t)

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Example :

Find i0(t) using step by step approach ?

Step 1 : assume i0(t) = k1 + k2 et /

Step 2 : assume t < 0 ( steady state )

Replace inductor by short circuit and find iL (0-)

4

86

6

21

4

+

-

t = 0

24 V

1 H

i L(t)

i 0(t)

46 4

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iL(0-) = 24 / 6 = 4 A

i0(0-) = 0 A

Step 3 : switch is movedReplace inductor by current source of value (4 A) and find i0(0)

4

86

6

21

4

+

-

24 V

i L(0-)

i 0(0-)

4

4

6

+

-

24 V

i L(0-)

i 0(0-)

44

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4

86i L(0

-)

i 0(0+)

21 4 A

4

86

4

i 0(0+)

21

+

-

48 V

4

16

i 0(0+)

48 V+

-i0(0) = - 48 / 20i0(0) = -2.4 A

St 4 t ( t d t t )

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Step 4 : assume t = ( steady state)

Replace inductor by short circuit and find i0()

i0() = 0 A

Step 5 : find , = L / RTHSo find RTH across the terminal of the inductor

4

86

4

21

RTH

( )[ ]{ } 4 812//46//84R =++=

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Step 6 : find i0(t)

[ ]

( )

0tfor,0(t)i

Ae2.4(t)i

0t,Ae02.40

Ae)(i(0)i)(i(t)i

0

t4.8

0

t4.8

t

0000

=

=

+=

+=

i0(t)

t

- 2.4

( )[ ]{ }

sec4.8

1

R

L

4.812//46//84R

TH

TH

==

=++=

Example :

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Example :

Solve previous example using differential equation approach

For t < 0 , steady stateReplace the inductor by short circuit and find iL(0

-) and i0(0

-)

As before iL(0-) = 4 A = iL(0

+

)i0(0

-) = 0 A

For t > 0 ,

4

86

4

i 0(t)

i L(t)

L = 1 H21

4

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(t)i5

3(t)i

(t)i

20

12

k8k12

k12(t)i(t)i

L0

LL0

=

=

+=

4

4

i 0(t)

i L(t)

21

So we need to find iL(t) first

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KVL :

0(t)i4.8dt

(t)di

0(t)i4.8dt

(t)diL

0(4.8)(t)i(t)V

LL

LL

LL

=+

=+

=+

218

i 0(t)

i L(t)4.8i L(t)

L

0k( )i t4 8

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( )

0t,0(t)i

0t,e2.4(t)i

e

5

12(t)i

e453(t)i

53(t)i

0

t4.8

0

t4.8

0

t4.8L0

=

=

=

==

i0(t)

t

- 2.4

0t,e4(t)i

k4(0)i

0t,ek(t)i

t4.8

L

L

t4.8

L

=

==

=

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Elec 3202 Chap 6