Lecture DrHaidi CapacitorAndInductor

8/9/2019 Lecture DrHaidi CapacitorAndInductor

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Dr Haidi bin Ibrahim,Dr Haidi bin Ibrahim,

Room 2.33,Room 2.33,School of Electrical & Electronic Engineering,School of Electrical & Electronic Engineering,Engineering Campus,Engineering Campus,Universiti Sains Malaysia,Universiti Sains Malaysia,13!! "i#ong $e#al, %enang.13!! "i#ong $e#al, %enang.

Email' hai(i)eng.usm.myEmail' hai(i)eng.usm.my

E*t' +22E*t' +22

Semester 2, Session 2!12 2!13

EEU104/3: Electrical $echnology

-ecture 1' Capacitor an( n(uctor



3

Capacitor

In its simplest form, a capacitor is an electrical deviceconstructed of two parallel plates separated by an insulatingmaterial called the dielectric (e.g. paper, mica, air).

Figure taken from:Thomas . Floyd, ! Electronics Fundamentals: Circuits, Devices, and

Applications", #th $dition, %rentice &all, ', page *+.



In the neutral state, both plates have an eual numberof free electrons.

-hen a voltage source is connected to the capacitor,electrons are removed from one plate and an eual

number are deposited on the other plate. o electrons flow through the dielectric (insulator).

The accumulated charge in the plates increase the potential difference (voltage) between the plates.

-hen the potential difference eual to the supplyvoltage, there is no electrons flow.

-hen the supply is removed from the capacitor, the

capacitor retains the stored charge.



+


Applications", #th $dition, %rentice &all, ', page *+/.



/

The amount of charge that a capacitor can store per voltacross the plates is its capacitance (0).

The unit of capacitance is the farad (F).

1ne farad is the amount of capacitance when one coulomb

of charge is stored with one volt across the plates. 2ost capacitors in electronics work have values of

µF (/3# F) or pF (/3/' F).

C =Q

V



EXAMPLE:EXAMPLE: 4 capacitor µF has '56 across its plates. &ow much charge

does it store7

SOLUTION:SOLUTION:

8iven 0 µF, and 6 '56. Thus,



0haracteristic of a capacitor:


Applications", #th $dition, %rentice &all, ', page *+5.


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0apacitance is directly proportional to the physical si;e ofthe plates as determined by the plate area.

0apacitance is inversely proportional to the distance between the plates.

The measure of a material<s ability to establish an electric

field is called the dielectric constant (ε ). 0apacitance is

directly proportional to the dielectric constant.



EXAMPLE:EXAMPLE: 4 capacitor is constructed from two parallel plates with area of

.'m' and a plate separation ./m. The dielectric is mica, which hasa dielectric constant of 5.0alculate its capacitance value.

SOLUTION:SOLUTION:

8iven A .'m', d ./m, ε r = 5. -e know that ε o = @.@5A/3/'F?m.

Thus, the value of capacitance, 0, is:



ENERGY STORED IN A CAPACITOR -e have:

If an alternating voltage is applied to the capacitor:

For a B0 current, we have:

Thus, the current that flow through the capacitor:

C =Q

V ⇒ Q=CV

q=Cv

I =Q

t

i=dq

dt =

d Cv

dt =C

dv

dt



i=C dv

dt v=

/

C ∫ i dt

%ower stored in a capacitor is given by:

In dt second, the energy stored in the capacitor is:

The energy stored in a capacitor when 6 volts is applied acrossit is:

p=vi=vC dvdt

dw= pdt =vC dv ⇒ W =C ∫ v dv

W =C

∫(

V

v dv=C

[ v

'

'

](

V

W =/

'CV

'


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EXAMPLE:EXAMPLE: 0alculate the energy stored by a ./pF capacitor with /6

across its plates7

SOLUTION:SOLUTION:

8iven 0 ./pF, 6 /6, - 7



1/

Csing Dirchoff 6oltage aw (D6)


Applications", #th

$dition, %rentice &all, ', page .

Series capacitors.

vT t =v/t v't v*t ...vnt

/C T ∫ i dt = /

C /∫i dt /

C '∫i dt /

C *∫i dt ... /

C n∫i dt

/

C T

= /

C /

/

C '

/

C *...

/

C n



10

EXAMPLE:EXAMPLE: Find the total capacitance in this circuit.

SOLUTION:SOLUTION:


Applications", #th

$dition, %rentice &all, ', page 5.



1

Csing Dirchoff 0urrent aw (D0)


Applications", #th

$dition, %rentice &all, ', page @.

%arallel capacitors.

iT t =i/t i't i* t ...i nt

C T dv t dt =C / dv t dt C ' dv t dt C * dv t dt ...C n dv t dt

C T =C /C 'C *...C n



1

EXAMPLE:EXAMPLE: Find the total capacitance in this circuit.

SOLUTION:SOLUTION:

C T =C /C 'C *C )C 5C #C T =(.(/F(.(''F(.(/F(.()EF(.''F(.(''F

C T =(./**F


Applications", #th

$dition, %rentice &all, ', page @.



o much time is nee(e( to fullycharge a capacitor4


Applications", #th

$dition, %rentice &all, ', page +.

$ransient analysis

Step response



o much time is nee(e( to fullycharge a capacitor4

0onsider the following circuit.

Initially, the voltage across the capacitor is V 0.

R C

+

V S

-

+

V 0

-



Then, we close the switch at time t t 0 s.

The voltage across the capacitor is denoted asv

C , and the current across the capacitor is i

C .

R C

+

V S

-

+

vC

-

i C i

C

i C

+ v R

-



R C

+

V S

-

+

vC

-

i C i

C

i C

+ v R

-

V ! =v "vC

V ! = "icvC



V ! = "i cvC

V ! = "C dvC

dt vC

vc−V ! =− "C dvC

dt

/

vc−V

!

dvC =−/

"C dt

4t t t 0, v

C V

0. 4fter t s we switched on the

switch, the voltage across the capacitor is vc#t$.



∫V

vC t /

vc−V !

dvC =−∫t

t /

"C dt

[ln vc−V ! ]V

vC t =−

t −t

"C

[lnvC t −V !

V −V ! ]=−t −t

"C

vC t =V ! V −V ! e−

t −t

"C , t s

vC t =V ! V −V ! e−

t −t

, t s

-here τ 5 "C 5 time constant




Applications", #th

$dition, %rentice &all, ', page /'.

t =



R C

+

V S

-

+

vC

-

i C i

C

i C

+ v R

-

iC t =i "t

iC t =V ! −vC t

"



iC t =V ! −vC t

"

iC t = / "

V ! −V ! V −V ! e

−t −t

iC t =

/

" V ! −V e

−t −t

, t s

ormally, the capacitor is fully charged after 5 τ



EXAMPLE:Betermine the capacitor voltage %0 µ s after the switch is closed if

the capacitor initially is uncharged. Braw the charging curve.


Applications", #th

$dition, %rentice &all, ', page /*.



SOLUTION:



vC t =5/−e−t /@' s

6


Applications", #th

$dition, %rentice &all, ', page /*.



o much time is nee(e( to fully



o much time is nee(e( to fully(ischarge a capacitor4

0onsider the following circuit.

Initially, the voltage across the capacitor is V 0.

R

C

+

V 0

-



Then, we close the switch at time t t 0 s.

The voltage across the capacitor is denoted as vC , and the

current across the capacitor is iC .

R

C

+

vC

-

i R

i C

- v R

+

The voltage across the resistor is denoted as v ", and the

current across the resistor is i ".



The capacitor discharge and supply voltage tothe circuit.

i "=−iC

vC =v "= "i "

vC =− "iC =− "C dvc

dt

/

vc

dvC =−/

"C dt

∫V

vct /

vc

dvC =∫t

t

−/

"C

dt



∫V

vct /vc

dvC =∫t

t − / "C

dt

lnvct

V

=−t −t

"C

vct =V e−

t −t

"C , t ≥

vct =V e

−t −t

, t ≥

ic t =−i "=−vct

"=−

V

"e−

t −t

, t ≥




Applications", #th

$dition, %rentice &all, ', page /'.

t =



EXAMPLE:Betermine the capacitor voltage &ms after the switch is

closed. Braw the discharging curve.


Applications", #

th

$dition, %rentice &all, ', page /.



SOLUTION:




Applications", #

th

$dition, %rentice &all, ', page /.

R #




Applications", #

th

$dition, %rentice &all, ', page /5.

Remem#er'



Example:Example:

Initially, the capacitor isuncharged and the switch ! is at

position '. 4t time t s, theswitch is switched to position a,

and at t 's, the switch isswitched back to position '.

8iven that E '6, "5Ω andC 'mF. Braw the current andvoltage curves for the followingcircuit from t s to t s.

C

F

v C

3

F v " 3

a

'

! i C



Solutio:Solutio:



i v

t

0as yahcas

i C

vC

(I

(

(

Detika pemindahan

suis dari a ke '

's

charging discharging

5s '5s

'6

m4

3m4

( t



-hen an electric current flow through a conductor, magneticfluA (magnetic field) is created around that conductor.

n(uctor



The magnitude of the fluA is depends on:The magnitude of the current.

%roperties of the core

%hysical properties of the coil (length and area)

The 1hmGs law for fluA:

= F m

ℜ I =

V

"

=fluA F m=magnetomotiveforce

ℜ=reluctance



= F mℜ

= ) I A

l

= ) I

ℜ

= ) I r ( A

l

F m= ) I

ℜ= l

A= permeability=r

"= l

A =

l

A

=×/−E

-b /4t.m



v= )

'

ℜ

d i

d t

v= )

d

) iℜ

d t

v= *d i

d t

*= ) '

ℜ

*= ) '

r A

l


Applications", #th $dition, %rentice &all, ', page 5+.



Inductance is a measure of a coilGs ability toestablish an induced voltage as result of achange in its current.

The inductance is / henry (/&) when currentthrough the coil changing at the rate of /4?s,induces one volt across the coil.




Applications", #th $dition, %rentice &all, ', page 5/*.



Example:Example:

4n inductor has /@ turns, with its core is made from aniron with relative permeability of /5. The core has the

length of *mm and cross sectional area of [email protected]'.Betermine the inductance.

Solutio:Solutio:

E*ercise'



E*ercise'

0urrent shown in the figure below flowing through '5m&inductor. Braw the graph of the voltage across thatinductor.

t (ms)

i (m4)

'

@

!



++

v= * didt

i= / *∫ v dt

%ower stored in an inductor is given by:

In dt second, the energy stored in the inductor is:

The energy stored in an inductor when I amperes is appliedthrough it is:

p=vi= *

di

dt i

dw= pdt = *i di ⇒ W = *∫ i di

W = *∫(

I

i di= *

[

i'

'

](

I

W =/

' *I

'



+/

Csing Dirchoff 6oltage aw (D6)


Applications", #th $dition, %rentice &all, ', page 5/.

Series in(uctors.

vT t =v/t v't v*t ...vn t

*T

di

dt = */

di

dt *'

di

dt **

di

dt ... *n

di

dt

*T = */ *' **... *n



+0

Csing Dirchoff 0urrent aw (D0)


Applications", #th $dition, %rentice &all, ', page 5/#.

%arallel in(uctors.

iT t =i/t i't i*t ...i nt

/

*T ∫ v dt = /

*/∫v dt /

*'∫ v dt /

**∫ v dt ... /

*n∫v dt

*T = /

*/

/

*'

/

**

... /

*n−/

E*ercise'



+

E*ercise'

Find the total inductance between terminal 4and H.

* '

* /

* *

# m&

'5 m&

/5 m&

4

H

E*ercise'



+

Betermine the total energy stored by inductors in thefollowing circuit. 8iven that inductor *

+ stores '@m

energy.

* '

* /

* *

/

'

*5( m&

/5( m&

/(( m&

: #(( m4

/

'

E*ercise'



R

L

+

V S

-

Then, we close the switch at time t 0s.

The voltage across the inductor is denoted as v *,

and the current across the capacitor is i *.

+

v L

-

i L i

L

i L

+ v R

-



R

L

+

V S

-

+

v L

-

i L i

L

i L

+ v R

-

V ! =v "v *

V ! = "i *v *



V ! = "i *v *

V ! =i * " *di *

dt

V s−i * "= * di *

dt

"

* dt =

di *

V s

" −i *

4t t t

0, i

* I

0. 4fter t s we switched on the switch,

the current flow across the inductor is i *#t$.



∫t

t "

* dt =∫ I

i * t di *

V s

" −i *

[−ln V !

"−i *]

I

i *t

=t −t

*/ "

−ln V !

" −i *t

V !

" − I = t −t

* / "

V !

" −i *t

V !

" − I

=e−

t −t

*/ "



V !

" −i *t

V !

" − I

=e−

t −t

*/ "

i *t =V !

"−

V !

"− I e

− t −t

*/ " , t s

i *t =V !

"

I −

V !

"

e

−t −t

*/ " , t s

i * t =V !

" I −

V !

" e−

t −t

, t s

-here τ *" time constant. ormally, fully energi;ed after 5τ.



i * t =V !

" I −

V !

" e− t −t

, t s

$he voltage across the in(uctor'

v *t = * di *t dt

v *t = * d

dt V !

" I −

V !

" e−

t −t

=− *

I −V !

" e−

t −t

v *t =V ! − I "e−

t −t




Applications", #th $dition, %rentice &all, ', page 5/@.

t =



Then, open switch J/ and close switch J' at time t t 0

s.

The voltage across the capacitor is denoted as v *, and the

current across the capacitor is i *.

+

vC

-

The voltage across the resistor is denoted as v ", and the

current across the resistor is i ".

R

L

+

V S

-

S1

S2 i R i

L

- v R

+



The inductor deenergi;e and supply current tothe circuit.

i "=−i *

v *=v "= "i "

∴ *di *

dt =− "i *

/i *

di *=− " *

dt

∫ I

i *t /

i *

di *=∫t

t

− "

*dt



∫ I

ic

t

/i *

di *=∫t

t

− " *

dt

lni *t

I =−

t −t

*/ "

i *t = I e−

t −t

* / " , t ≥

i *t = I e

−t −t

, t ≥

v *t =v

"t =−i

*t "=− I

( "e

−t −t

(

, t ≥(




Applications", #th $dition, %rentice &all, ', page 5'.

t =



Example:Example:

Initially, the inductor is fullydeenergi;ed and the switch ! isat position '. 4t time t s, theswitch is switched to position a,and at t /s, the switch isswitched back to position '.

8iven that E *6, "5Ω and *'&. Braw the current andvoltage curves for the followingcircuit from t s to t 's.

"

*

E

a

b

i

F

v *

3

F v " 3

!



Solutio:Solutio:



v i

(

(

E

v *

i *

t

"

E

3 E

/s

$nergi;ing Be3energi;ing

's /'s

.#4*6

3*6

Lecture DrHaidi CapacitorAndInductor

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