Capacitors and Inductors 1 Eastern Mediterranean University.

Capacitors and Capacitors and InductorsInductors

11Eastern Mediterranean UniversityEastern Mediterranean University

Chap. 6, Capacitors and InductorsChap. 6, Capacitors and Inductors

Introduction

Capacitors

Series and Parallel Capacitors

Inductors

Series and Parallel Inductors

22EEastern astern MMediterranean editerranean UUniversityniversity

6.1 Introduction6.1 Introduction

Resistor: a passive element which dissipates energy only

Two important passive linear circuit elements:1) Capacitor

2) Inductor

Capacitor and inductor can store energy only and they can neither generate nor dissipate energy.


Michael Faraday (1971-1867)Michael Faraday (1971-1867)


6.2 Capacitors6.2 Capacitors

A capacitor consists of two conducting plates separated by an insulator (or dielectric).

(F/m)10854.8

ε

120

0

r

dAC


Three factors affecting the value of capacitance:1. Area: the larger the area, the greater the capacitance.

2. Spacing between the plates: the smaller the spacing, the greater the capacitance.

3. Material permittivity: the higher the permittivity, the greater the capacitance.

66

dAC ε

EEastern astern MMediterranean editerranean UUniversityniversity

Fig 6.4Fig 6.4

77

(a) Polyester capacitor, (b) Ceramic capacitor, (c) Electrolytic capacitor


Fig 6.5Fig 6.5

88

Variable capacitors


Fig 6.3Fig 6.3


Fig 6.2Fig 6.2


Charge in CapacitorsCharge in Capacitors

The relation between the charge in plates and the voltage across a capacitor is given below.

1111

Cvq

C/V1F1

v

q LinearNonlinear


Voltage Limit on a CapacitorVoltage Limit on a Capacitor

Since q=Cv, the plate charge increases as the voltage increases. The electric field intensity between two plates increases. If the voltage across the capacitor is so large that the field intensity is large enough to break down the insulation of the dielectric, the capacitor is out of work. Hence, every practical capacitor has a maximum limit on its operating voltage.


I-V Relation of CapacitorI-V Relation of Capacitor

dtdvC

dtdqiCvq ,

1313

+

-

v

i

C


Physical MeaningPhysical Meaning

dtdvCi

1414

• when v is a constant voltage, then i=0; a constant voltage across a capacitor creates no current through the capacitor, the capacitor in this case is the same as an open circuit.

• If v is abruptly changed, then the current will have an infinite value that is practically impossible. Hence, a capacitor is impossible to have an abrupt change in its voltage except an infinite current is applied.

+

-

v

i

C


Fig 6.7Fig 6.7

A capacitor is an open circuit to dc.

The voltage on a capacitor cannot change abruptly.

1515

Abrupt change


The charge on a capacitor is an integration of current through the capacitor. Hence, the memory effect counts.

1616

dtdvCi

tidt

Ctv 1)(

t

to

otvidt

Ctv )(1)(

0)( v

Ctqtv oo /)()(

+

-

v

i

C


Energy Storing in CapacitorEnergy Storing in Capacitor

1717

dtdvCvvip

t tvv

tv

v

t CvvdvCdtdtdvvCpdtw )(

)(2)(

)( 21

)(21)( 2 tCvtw

Ctqtw

2)()(

2

)0)(( v +

-

v

i

C


Model of Practical CapacitorModel of Practical Capacitor


Example 6.1Example 6.1

(a) Calculate the charge stored on a 3-pF capacitor with 20V across it.

(b) Find the energy stored in the capacitor.



Solution:

(a) Since

(b) The energy stored is

2020

pC6020103 12 q

pJ60040010321

21 122 Cvw

,Cvq



The voltage across a 5- F capacitor is

Calculate the current through it.

Solution:

By definition, the current is

2121

V 6000cos10)( ttv

6105 dtdvCi

6000105 6

)6000cos10( tdtd

A6000sin3.06000sin10 tt



Determine the voltage across a 2-F capacitor if the current through it is

Assume that the initial capacitor voltage is zero.

Solution:

Since

2222

mA6)( 3000teti

t tev0

30006 6

1021

03000

3

3000103 tte

t vidt

Cv

0)0(1 ,0)0(and v

310dt

V)1( 3000teEEastern astern MMediterranean editerranean UUniversityniversity


Determine the current through a 200- F capacitor whose voltage is shown in Fig 6.9.



Solution:

The voltage waveform can be described mathematically as

2424

otherwise043V 5020031V 5010010V 50

)( tttttt

tv



Since i = C dv/dt and C = 200 F, we take the derivative of to obtain

Thus the current waveform is shown in Fig.6.10.

2525

otherwise043mA1031mA1010mA10

otherwise0435031501050

10200)( 6

ttt

ttt

ti



Obtain the energy stored in each capacitor in Fig. 6.12(a) under dc condition.



Solution:

Under dc condition, we replace each capacitor with an open circuit. By current division,

2828

mA2)mA6(423

3

i

,V420001 iv

mJ16)4)(102(21

21 232

111 vCw

mJ128)8)(104(21

21 232

222 vCw

V840002 iv


Fig 6.14Fig 6.14

Neq CCCCC ....321


6.3 Series and Parallel Capacitors6.3 Series and Parallel Capacitors

The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitance.

3030

Niiiii ...321

dtdvC

dtdvC

dtdvC

dtdvCi N ...321

dtdvC

dtdvC eq

N

kK

1

Neq CCCCC ....321


Fig 6.15Fig 6.15

Neq CCCCC1...1111

321


Series CapacitorsSeries Capacitors

The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.

3232

Neq

t

N

t

eq

Ctq

Ctq

Ctq

Ctq

idCCCC

idC

)()()()(

)1...111(1

21

321

)(...)()()( 21 tvtvtvtv N

21

111CCCeq

21

21

CCCCCeq


SummarySummary

These results enable us to look the capacitor in this way: 1/C has the equivalent effect as the resistance. The equivalent capacitor of capacitors connected in parallel or series can be obtained via this point of view, so is the Y- connection and its △transformation



Find the equivalent capacitance seen between terminals a and b of the circuit in Fig 6.16.



Solution:

3535

F4520520

F302064

F20F60306030 eqC

:seriesin are capacitors F5 and F20

F6 with the parallel in iscapacitor F4 :capacitors F20 and

withseriesin iscapacitor F30 capacitor. F60 the



For the circuit in Fig 6.18, find the voltage across each capacitor.



Solution: Two parallel capacitors:

Total charge

This is the charge on the 20-mF and 30-mF capacitors, because they are in series with the 30-v source. ( A crude way to see this is to imagine that charge acts like current, since i = dq/dt)

3838

mF10mF1

201

301

601

eqC

C3.0301010 3 vCq eq



Therefore,

Having determined v1 and v2, we now use KVL to determine v3 by

Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage v3 and their combined capacitance is 40+20=60mF.

3939

,V1510203.0

31

1

Cqv

V1010303.0

32

2

Cqv

V530 213 vvv

V510603.0

mF60 33

qvEEastern astern MMediterranean editerranean UUniversityniversity

Joseph Henry (1979-1878)Joseph Henry (1979-1878)


6.4 Ind6.4 Inducuctorstors

An inductor is made of a coil of conducting wire

lANL 2


Fig 6.22Fig 6.22

4242

(H/m)104 70

0

2

r

lANL

turns.ofnumber:Nlength.:l

area. sectionalcross: Acoretheoftypermeabili:


Fig 6.23Fig 6.23

4343

(a) air-core(b) iron-core(c) variable iron-core


Flux in InductorsFlux in Inductors

The relation between the flux in inductor and the current through the inductor is given below.

4444

LiWeber/A1H1

i

ψ LinearNonlinear


Energy Storage FormEnergy Storage Form

An inductor is a passive element designed to store energy in the magnetic field while a capacitor stores energy in the electric field.


I-V Relation of I-V Relation of Inductors Inductors

An inductor consists of a coil of conducting wire.

4646

dtdiL

dtdv

+

-

v

i

L


Physical MeaningPhysical Meaning

When the current through an inductor is a constant, then the voltage across the inductor is zero, same as a short circuit.

No abrupt change of the current through an inductor is possible except an infinite voltage across the inductor is applied.

The inductor can be used to generate a high voltage, for example, used as an igniting element.

dtdiL

dtdv


Fig 6.25Fig 6.25

An inductor are like a short circuit to dc.

The current through an inductor cannot change instantaneously.


4949

t

to

otidttv

Li )()(1

t

dttvL

i )(1

memory. hasinductor The

vdtL

di 1

+

-

v L


Energy Stored in an InductorEnergy Stored in an Inductor

The energy stored in an inductor

idtdiLviP

5050

t t idtdtdiLpdtw

)(

)(22 )(

21)(

21ti

iLitLidiiL ,0)( i

)(21)( 2 tLitw

+

-

v L


Model of a Practical InductorModel of a Practical Inductor



The current through a 0.1-H inductor is i(t) = 10te-5t A. Find the voltage across the inductor and the energy stored in it.

Solution:

5252

V)51()5()10(1.0 5555 teetetedtdv tttt

J5100)1.0(21

21 1021022 tt etetLiw

,H1.0andSince LdtdiLv

isstoredenergyThe



Find the current through a 5-H inductor if the voltage across it is

Also find the energy stored within 0 < t < 5s. Assume i(0)=0.

Solution:

5353

0,00,30)(

2

ttttv

.H5and L)()(1 Since0

0 t

ttidttv

Li

A23

6 33

tt t dtti0

2 03051



5454

5

0

65 kJ25.1560

56

6060 tdttpdtw

thenisstoredenergytheand,60powerThe 5tvip

before.obtainedas

usingstoredenergytheobtaincanweely,AlternativwritingbyEq.(6.13),

)0(21)5(

21)0()5( 2 LiLiww

kJ25.1560)52)(5(21 23



Consider the circuit in Fig 6.27(a). Under dc conditions, find:

(a) i, vC, and iL.

(b) the energy stored in the capacitor and inductor.



Solution:

5656

,251

12 Aii L

,J50)10)(1(21

21 22 cc Cvw

J4)2)(2(21

21 22 iL Lw

)(a :conditiondcUnder capacitorinductor

circuitopencircuitshort

)(b

V105 ivc


Inductors in SeriesInductors in Series

Neq LLLLL ...321


Inductors in ParallelInductors in Parallel

Neq LLLL1111

21


6.5 Series and 6.5 Series and PParallel arallel IInductorsnductors

Applying KVL to the loop,

Substituting vk = Lk di/dt results in

5959

Nvvvvv ...321

dtdiL

dtdiL

dtdiL

dtdiLv N ...321

dtdiLLLL N )...( 321

dtdiL

dtdiL eq

N

KK

1

Neq LLLLL ...321


Parallel InductorsParallel Inductors

Using KCL,

But

6060

Niiiii ...321

t

t kk

k otivdt

Li )(1

0

t

t

t

t sk

tivdtL

tivdtL

i0 0

)(1)(10

201

t

t NN

tivdtL 0

)(1... 0

)(...)()(1...1100201

210

tititivdtLLL N

t

tN

t

teq

N

kk

t

t

N

k k

tivdtL

tivdtL 00

)(1)(10

10

1


The inductor in various connection has the same effect as the resistor. Hence, the Y-Δ transformation of inductors can be similarly derived.


Table 6.1Table 6.1



Find the equivalent inductance of the circuit shown in Fig. 6.31.



Solution:

6464

10H12H,,H20:Series

H6427427 : Parallel

H18864 eqL

H42


Practice Problem 6.11Practice Problem 6.11



Find the circuit in Fig. 6.33,

If find :

6666

.mA)2(4)( 10teti ,mA 1)0(2 i )0( (a)

1i

);(and),(),((b) 21 tvtvtv )(and)((c) 21 titi



Solution:

6767

.mA4)12(4)0(mA)2(4)()(a 10 ieti t

mA5)1(4)0()0()0( 21 iii

H53212||42 eqL

mV200mV)10)(1)(4(5)( 1010 tteq eedtdiLtv

mV120)()()( 1012

tetvtvtv

mV80mV)10)(4(22)( 10101

tt eedtditv

isinductanceequivalentThe)(b



6868

t t t dteidtvti0 0

10121 mA5

4120)0(

41)(

mA38533mA503 101010 ttt eete

t tt dteidtvti0

1020 22 mA1

12120)0(

121)(

mA11mA10101010 ttt eete

)()()(thatNote 21 tititi

t idttv

Li

0)0()(1)(c


Capacitors and Inductors 1 Eastern Mediterranean University.

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