Introduction to Capacitors, Inductors & Magnetic Circuits

Introduction to Capacitors, Inductors& Magnetic Circuits

Related BookBasic Circuit Analysis

6thEdby David Erwin

1

Chap. 6: Capacitors and Inductors

IntroductionCapacitorsSeries and Parallel CapacitorsInductorsSeries and Parallel Inductors

2

6.1 Introduction

Resistor: a passive element which dissipates energy only

Two important passive linear circuit elements:1) Capacitor2) Inductor

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

3

Michael Faraday (1971-1867)

4

6.2 Capacitors

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

(F/m)10854.8

ε

120

0

r

dAC

5Engineering

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.

6

dAC ε

Fig 6.4

7

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

Fig 6.5

8

Variable capacitors

Fig 6.3

9

Fig 6.2

10

Charge in Capacitors

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

11

CvqC/V1F1

v

q LinearNonlinear

Voltage 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.

12

I-V Relation of Capacitor

dtdvC

dtdqiCvq ,

13

+

-

v

i

C

Physical Meaning

dtdvCi

14

• 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 6.7

A capacitor is an open circuit to dc.

The voltage on a capacitor cannot change abruptly.

15

Abrupt change

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

16

dtdvCi

tidt

Ctv 1)(

t

to

otvidt

Ctv )(1)(

0)( v

Ctqtv oo /)()(

+

-

v

i

C

Energy Storing in Capacitor

17

dtdvCvvip

t tvv

tv

v

t CvvdvCdtdtdvvCpdtw )(

)(2)(

)( 21

)(21)( 2 tCvtw

Ctqtw 2)()(

2

)0)(( v +

-

v

i

C

Model of Practical Capacitor

18

Example 6.1

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

(b)Find the energy stored in the capacitor.

19

Example 6.1

Solution: (a) Since

(b) The energy stored is

20

pC6020103 12 q

pJ60040010321

21 122 Cvw

,Cvq

Example 6.2

The voltage across a 5- F capacitor is

Calculate the current through it. Solution: By definition, the current is

21

V 6000cos10)( ttv

6105 dtdvCi

6000105 6

)6000cos10( tdtd

A6000sin3.06000sin10 tt

Example 6.3

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

Assume that the initial capacitor voltage is zero.

Solution: Since

22

mA6)( 3000teti

t tev 0

30006 6102

10

30003

3000103 tte

t vidt

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

310dt

V)1( 3000te

Example 6.4

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

23

Example 6.4

Solution:The voltage waveform can be described mathematically as

24

otherwise043V 5020031V 5010010V 50

)( tttttt

tv

Example 6.4

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.

25

otherwise043mA1031mA1010mA10

otherwise0435031501050

10200)( 6ttt

ttt

ti

Example 6.4

26

Example 6.5

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

27

Example 6.5

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

28

mA2)mA6(4233

i

,V420001 ivmJ16)4)(102(2

121 232

111 vCw

mJ128)8)(104(21

21 232

222 vCw

V840002 iv

Fig 6.14

Neq CCCCC ....321

29

6.3 Series and Parallel Capacitors

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

30

Niiiii ...321

dtdvC

dtdvC

dtdvC

dtdvCi N ...321

dtdvC

dtdvC eq

N

kK

1

Neq CCCCC ....321

Fig 6.15

Neq CCCCC1...1111

321

31

Series Capacitors

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

32

Neq

t

N

t

eq

Ctq

Ctq

Ctq

Ctq

idCCCC

idC

)()()()(

)1...111(1

21

321

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

21

111CCCeq

21

21CCCCC eq

Summary

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

33

Example 6.6

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

34

Example 6.6

Solution:

35

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

Example 6.7

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

36

Example 6.7

37

Example 6.7

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) 38

mF10mF1201

301

601

eqC

C3.0301010 3 vCq eq

Example 6.7Therefore,

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. 39

,V1510203.0

31

1

Cqv

V1010303.0

32

2

Cqv

V530 213 vvv

V510603.0

mF60 33

qv

Joseph Henry (1979-1878)

40

6.4 Inductors

An inductor is made of a coil of conducting wire

lANL 2

41

Fig 6.22

42

(H/m)104 70

0

2

r

lANL

turns.ofnumber:Nlength.:l

area. sectionalcross: Acoretheoftypermeabili:

Fig 6.23

43

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

Flux in Inductors

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

44

LiWeber/A1H1

i

ψ LinearNonlinear

Energy 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.

45

I-V Relation of Inductors

An inductor consists of a coil of conducting wire.

46

dtdiL

dtdv

+

-

v

i

L

Physical 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

47

Fig 6.25

An inductor are like a short circuit to dc.

The current through an inductor cannot change instantaneously.

48

49

t

to

otidttv

Li )()(1

tdttv

Li )(1

memory. hasinductor The

vdtL

di 1

+

-

v L

Energy Stored in an Inductor

The energy stored in an inductor

idtdiLviP

50

t t idtdtdiLpdtw

)()(

22 )(21)(2

1ti

iLitLidiiL ,0)( i

)(21)( 2 tLitw

+

-

v L

Model of a Practical Inductor

51

Example 6.8

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:

52

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

J5100)1.0(21

21 1021022 tt etetLiw

,H1.0andSince LdtdiLv

isstoredenergyThe

Example 6.9

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:

53

0,00,30)(

2

ttttv

.H5and L)()(1 Since0

0 t

ttidttv

Li

A236 33

tt t dtti 0

2 03051

Example 6.9

54

50

65 kJ25.1560

566060 tdttpdtw

thenisstoredenergytheand,60powerThe 5tvip

before.obtainedas

usingstoredenergytheobtaincanweely,AlternativwritingbyEq.(6.13),

)0(21)5(2

1)0()5( 2 LiLiww

kJ25.1560)52)(5(21 23

Example 6.10

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.

55

Example 6.10

Solution:

56

,25112 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 Series

Neq LLLLL ...32157

Inductors in Parallel

Neq LLLL1111

21

58

6.5 Series and Parallel Inductors

Applying KVL to the loop,

Substituting vk = Lk di/dt results in

59

Nvvvvv ...321

dtdiL

dtdiL

dtdiL

dtdiLv N ...321

dtdiLLLL N)...( 321

dtdiL

dtdiL eq

N

KK

1

Neq LLLLL ...321

Parallel Inductors

Using KCL,But

60

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

21 0tititivdt

LLL Nt

tN

t

teq

N

kk

t

t

N

k ktivdt

Ltivdt

L 00)(1)(1

01

01

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

61

Table 6.1

62

Example 6.11

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

63

Example 6.11

Solution:

64

10H12H,,H20:Series

H6427427

: Parallel

H18864 eqL

H42

Practice Problem 6.11

65

Example 6.12

Find the circuit in Fig. 6.33,

If find :

66

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

1i

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

Example 6.12

Solution:

67

.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)()()( 1012tetvtvtv

mV80mV)10)(4(22)( 10101

tt eedtditv

isinductanceequivalentThe)(b

Example 6.12

68

t t t dteidtvti 0 010

121 mA54120)0(4

1)(

mA38533mA503 101010 ttt eete

t tt dteidtvti 010

20 22 mA112120)0(12

1)(

mA11mA10101010 ttt eete

)()()(thatNote 21 tititi

t idttv

Li 0 )0()(1)(c