ene 104 electric circuit theory -...

ENE 104

Electric Circuit Theory

Lecture 06:

Basic RL and RC Circuits

http://webstaff.kmutt.ac.th/~dejwoot.kha/

Week #6 : Dejwoot KHAWPARISUTH



Objectives : Ch8

Week #6

ENE 104

• time constants for RL and RC circuits

• the natural and forced response

• the total response

• the effect of initial conditions on circuit

response

• RL and RC circuit response


The Source-Free RL Circuit:

Week #6

ENE 104

• A natural response

• A transient response

• A forced response



Week #6

ENE 104

A series RL circuit for which i(t) is

to be determined, subject to the

initial condition that i(0) = I0.

tti

t

tdL

R

i

id

dtL

R

i

di

iL

R

dt

di

dt

diLRivRi

0

)(

L

0

0

LRteItiso

tL

RIi

/

0

0

)(

0 lnln



Week #6

ENE 104

A series RL circuit for which i(t) is

to be determined, subject to the

initial condition that i(0) = I0.


LRteIti /

0 )(

The power: L

Rt

R eRIRip2

2

0

2

The total energy:

0

2

2

0

0

dteRIdtpw L

Rt

RR

2

0

0

2

2

02

1

2LIe

R

LRI L

Rt

Example 8.1:

Week #6

ENE 104 Basic RL and RC Circuits

find the current through the 5-H inductor at t = 200ms

L

Rt

L eIti

0)(

Example :

Week #6


find the current through the 5-H inductor at t = 200ms

L

Rt

L eIti

0)(

4.210

240 I

.8.3244.2.)200( 10 mAemsti t

L

Practice: 8.1

ENE 104 Week #6

Determine the energy remaining in the inductor at t = 2 ns

if it is initially storing 7 µJ


Practice: 8.1

ENE 104 Week #6 Basic RL and RC Circuits

7 µJ stored in a 500 nH inductor

a current magnitude =

6

9

2 7 105.292 A

500 10

9/ 500 10

( ) where 500 ps1000

t

o

Li t I e

R

9

12

2 10(2 ) 5.292exp 96.93 mA

500 10

i ns

2 9 3 21(2 ns) 0.5(500 10 )(96.93 10 )

2w Li

Properties of the Exponential resp:

Week #6


L

Rt

L eIti

0)(

The initial rate of decay:

L

Re

L

R

I

i

dt

d

t

L

Rt

t

000

R

L

3679.0)( 1

0

eI

i 03679.0)( Ii or

Properties of the Exponential resp:

Week #6

ENE 104

A plot of the exponential response versus time.


Practice: 8.2

ENE 104 Week #6

In a source-free series RL circuit, find the numerical value

of the ratio: (a) 𝑖(2𝜏)/𝑖(𝜏), (b) 𝑖(0.5𝜏)/𝑖(0), and (c) 𝑡/𝜏 if

𝑖 𝑡

𝑖 0= 0.2; (d) 𝑡/𝜏 if 𝑖 0 − 𝑖 𝑡 = 𝑖 0 𝑙𝑛2 .


/( ) t

oi t I e (0) oi I

2

1

(2 )

( )

i e

i e

0.5(0.5 )

(0)

i

ei

/( )0.2, so 0.2

(0)

ti t te n

i

The Source-Free RC Circuit:

Week #6

ENE 104

A parallel RC circuit for which

v(t) is to be determined,

subject to the initial condition

that v(0) = V0.


0R

v

dt

dvC

0RC

v

dt

dv

RC

t

RC

t

eVevtv

0)0()(

The Source-Free RC Circuit:

Week #6


RC

Practice: 8.3

ENE 104 Week #6

Find 𝑣 0 and 𝑣(2𝑚𝑠)


Practice: 8.3

ENE 104 Week #6 Basic RL and RC Circuits

With no current flow permitted through the capacitor

(it is assumed any transients have long since died

out), v = 50 V. After the switch is thrown, the only remaining circuit is a simple source-free RC circuit.

1.6 ms RC

3/

3

2 10( ) (0) so (2ms) 50exp

1.6 10

tv t v e v

Example 8.2:

Week #6


determine both i1 and iL in the circuit

Example :

Week #6



.36050

18,0 mAit L

.20090

181 mAi

Example :

Week #6



For t>0, .2.2132

32mHLeq

11050

18090

)12060(90eqR

.20110

102.2 3

sR

L

eq

eq

LRteIti /

0 )(

Example :

Week #6



For t>0,

LRteIti /

0 )(

.360)0()0( mAii LL

.360360360)( 50000mAeeeti t

t

L

Rt

L

Example :

Week #6



For t>0,

LRteIti /

0 )(

.2409060120

60120)0()0(1 mAii L

.240)( 50000

1 mAeti t

Example :

Week #6



)0(.360

)0(.360)(

50000 tmAe

tmAti

tL

)0(.240

)0(.200)(

500001tmAe

tmAti

t

Practice: 8.4

ENE 104 Week #6

at t = 0.15s. find the value of iL, i1, i2


Example :

Week #6


at t = 0.15 s. find the value of iL, i1, i2

For t<0,

.4.0.228

21 AAi

0)0()0(.6.1.228

8IiiAAi LLL

Example :

Week #6


at t = 0.15s. find the value of iL, i1, i2

For t>0, LRteIti /

0 )( 2eqR

.756.06.16.1.)15.0( 4.0

2

Aeesti

t

L

Rt

L

.244.122 Aii L

Example 8.3:

Week #6


find and if )0( v )0(1

i 0)0( Vv

312 // RRRReq

0)0()0( Vvv CC

t

eiti

)0()( 11

CReq

31

3

2

01

31

31

)0(RR

R

R

Vi

RRRR

Practice: 8.5

Week #6


find vC and v0 at t = 0-, 0+, 1.3ms.

Example :

Week #6



For t<0,

)0()0(.100.1202501250

1250

CCC vvVVv

.4.384002100400

2120600

2100400

2)100400()0(

1

Vk

k

k

kvo

Example :

Week #6



For t=0+,

.100)0()0( Vvv CC

.6.254002100400

2

2//500600250

.100)0( V

k

k

k

Vvo

Example :

Week #6



For t=1.3ms.,

625

400850//1250

)400100//(2600250//1250 kReq

0025.0)4()625( FCReq

4001

.100)0()( 400Veevtv t

t

CC

.45.59.)3.1( VmstvC

.22.156.25)0(.)3.1( 400400

00 Veevmstv tt

The Unit-Step Function:

Week #6


0

0

01

0)(

tt

ttttu

01

00)(

t

ttu


Week #6

ENE 104

(a) A voltage-step forcing function is shown as the source driving a

general network. (b) A simple circuit which, although not the exact

equivalent of part (a), may be used as its equivalent in many cases.

(c) An exact equivalent of part (a).



Week #6


The Rectangular Pulse Function:

Week #6

ENE 104

The rectangular pulse

function

tt

ttt

tt

ttv

1

100

0

0

0

V

0

)(


Practice: 8.6

Week #6


Evaluate each of the following at 𝑡 = 0.8:

(a) 3𝑢 𝑡 − 2𝑢 −𝑡 + 0.8𝑢(1 − 𝑡); (b) 4𝑢 𝑡 𝑢(−𝑡);

(c) 2𝑢 𝑡 𝑠𝑖𝑛𝜋𝑡.

(a) 3 - 0 + 0.8 = 3.8

(b) [4] (0) = 0 (c) 2 sin 0.8 ¶ = 1.176

Driven RL Circuits:

Week #6


)()(

)( 0 tuVdt

tdiLtRi

0,0)( tti

for positive time; 0t

0

)()( V

dt

tdiLtRi

yielding; dttRiV

tLdi

)(

)(

0

integrated; ktRiVR

L 0ln

Driven RL Circuits:

Week #6


ktRiVR

L 0ln

setting i = 0 for t = 0, we obtain;

kVR

L 0ln

And, hence,

tVRiVR

L 00 lnln

L

Rt

eV

RiV

0

0

or )()( 00 tueR

V

R

Vti L

Rt

Practice: 8.7

Week #6


The voltage source 60 − 40𝑢(𝑡) V is in series with a 10-

resistor and a 50-mH inductor. Find the magnitudes of the

inductor current and voltage at 𝑡 equal to (a) 0−; (b) 0+;

(c) ∞; (d) 3𝑚𝑠.

Practice: 8.7

Week #6


แหล่งจ่ายแรงดันขนาด V ต่ออนุกรมอยู่กับตัว

ต้านทานขนาด 10- และตัว inductor ขนาด 50-mH ให้หา

ขนาดของกระแสในตัว inductor และค่าแรงดันท่ี t เท่ากับ (a)

; (b) ; (c) ; (d) 3 ms

)(4060 tu

00

Natural and Forced Response:

Week #6


The solution of any linear differential equation

may be expressed as the sum of two parts:

- the complementary solution (natural response) - the particular solution (forced response)

let consider; QPi

dt

di

dtQdtPidi

PtPtPt AedtQeei

;Q = a forcing function


Week #6


PtPtPt AedtQeei

For a source free circuit, Q must be zero, and the solution is the natural response:

Pt

n Aei

The first term is also called the steady state

response, the particular solution, or the particular

integral. When Q is a constant,

P

Qi f


Week #6


PtPtPt AedtQeei

or the complete response;

Pt

nf AeP

Qiiti )(

Determination of the complete res.:

Week #6


nf iiti )(

L

Rt

n Aeti

)(

R

Vi f

0

R

VAeititi L

Rt

fn0)()(

apply initial condition to evaluate A, i = 0 when t = 0;

R

VA 00 so

R

Ve

R

Vti L

Rt

00)(

Determination of the complete res.:

Week #6


R

Ve

R

Vti L

Rt

00)(

Example 8.4:

Week #6

ENE 104

Circuit for which a

complete response i(t) is

desired.

The desired current

response as a function of

time.


Example :

Week #6


0,)(

tAeAetit

L

tR

n

eq

5.1

62

626//2eqR

nf iiti )(

.502

100Ai f

0,50)( 3

5.1

tAetit

To find A, @t=0-; .252

50)0( Ai

A 5025 25; A

0,2550)( 5.0 teti t or .)(12525)( 5.0 Atueti t

Example :

Week #6


.)(12525)( 5.0 Atueti t

Practice: 8.8

Week #6


A voltage source, 𝑣𝑠 = 20𝑒−100𝑡𝑢(𝑡) V, is in series with a

200-Ω resistor and a 4-H inductor. Find the magnitude of

the inductor current at 𝑡 equal to (a) 0−; (b) 0+; (c) 8 ms;

(d) 15 ms.

(0 ) 0 so (0 ) S Lv i

(0 ) (0 ) so (0 ) L L Li i i

Practice: 8.8

Week #6


Example 8.5:

Week #6


find the current response

Example :

Week #6



)()()( 21 tititi

0,1)( 01

teR

Vti L

Rt

0

)(

02 ,1)(

0

tteR

Vti L

ttR

00 0,1)( tte

R

Vti L

Rt

and 0

)(

00 ,11)(0

tteR

Ve

R

Vti L

ttR

L

Rt

Example :

Week #6



00 0,1)( tte

R

Vti L

Rt

0

)(

00 ,11)(0

tteR

Ve

R

Vti L

ttR

L

Rt

Summary:

Week #6


Practice: 8.9

Week #6


The circuit shown has been in the form shown for a very

long time. The switch opens at 𝑡 = 0. find 𝑖𝑅 at 𝑡 equal to

(a) 0−; (b) 0+; (c) ∞; (d) 1.5 ms.

Practice: 8.9

Week #6


Driven RC Circuits: Example

Week #6

ENE 104

Find the capacitor voltage vC(t) and the current i(t) in the 200-

resistor of the circuit shown in (a).

(a) Original circuit; (b) circuit valid for t 0; (c) circuit for t 0.



Week #6

ENE 104




)0(1001201050

50)0(

CC vv

CnCfC vvtv )(

.3.192260

50)0( mAi


Week #6

ENE 104




50//200//60eqR

241

501

2001

601

2.1)(t

CR

t

Cn AeAetv eq

.205060 )20050(

20050

)20050(20050

VvCf

CnCfC vvtv )(


Week #6

ENE 104




2.120)(

t

C Aetv

To find A, @t=0;

Avv CC 20100)0()0(

80A

2.18020)(;

t

C etv


Week #6

ENE 104




To find A, @t=0;

.5.0200

100

200

)0()0( A

vi C

.1.020050

50

60

50)(

)20050(20050

Atii f

2.11.0)()(t

nf Aetiiti

Ai 1.05.0)0(

4.0A 0,4.01.0)( 2.1

tetit


Week #6

ENE 104




2.18020)(t

C etv

2.14.01.0)(t

eti

Practice: 8.10

Week #6


For the circuit, find 𝑣𝐶(𝑡) at 𝑡 equal to (a) 0−; (b) 0+; (c) ∞;

(d) 0.08 s.

Practice: 8.10

Week #6


Summary :

Week #6


Reference:

ENE 104

W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.

Copyright ©2002 McGraw-Hill. All rights reserved.

Circuit Analysis and Electrical Engineering

ene 104 electric circuit theory -...

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