capacitors and inductors 1 svbitec.wordpress.com vishal jethva

Capacitors and Inductors

1svbitec.wordpress.com

Vishal Jethva

Chap. 6, Capacitors and Inductors

• Introduction• Capacitors• Series and Parallel Capacitors• Inductors• Series and Parallel Inductors


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.


Michael Faraday (1971-1867)


6.2 Capacitors

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

(F/m)10854.8

ε

120

0

r

d

AC


• 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

dA

Cε

svbitec.wordpress.com

Fig 6.4

7

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


Fig 6.5

8

Variable capacitors


Fig 6.3


Fig 6.2


Charge in Capacitors

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

11

Cvq

C/V1F1

v

q Linear

Nonlinear


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.


I-V Relation of Capacitor

dt

dvC

dt

dqiCvq ,

13

+

-

v

i

C


Physical Meaning

dtdv

Ci

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

dtdv

Ci tidt

Ctv

1)(

t

to

otvidt

Ctv )(

1)(

0)( v

Ctqtv oo /)()(

+

-

v

i

C


Energy Storing in Capacitor

17

dt

dvCvvip

t tvv

tv

v

tCvvdvCdt

dtdvvCpdtw )(

)(2)(

)( 21

)(2

1)( 2 tCvtw

C

tqtw

2

)()(

2

)0)(( v +

-

v

i

C


Model of Practical Capacitor


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.


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 dt

dvCi

6000105 6

)6000cos10( tdt

d

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 tev0

30006

61021

0

30003

3000103 tte

t

vidtC

v0

)0(1

,0)0(and v

310dt

V)1( 3000tesvbitec.wordpress.com

Example 6.4

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


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)( 6

ttt

ttt

ti


Example 6.4


Example 6.5

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


Example 6.5

Solution:• Under dc condition, we replace each capacitor

with an open circuit. By current division,

28

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 6.14

Neq CCCCC ....321


6.3 Series and Parallel Capacitors

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

30

Niiiii ...321

dtdv

Cdtdv

Cdtdv

Cdtdv

Ci N ...321

dtdv

Cdtdv

C eq

N

kK

1

Neq CCCCC ....321


Fig 6.15

Neq CCCCC

1...

1111

321


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

C

tq

C

tq

C

tq

C

tq

idCCCC

idC

)()()()(

)1

...111

(1

21

321

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

21

111CCCeq

21

21

CCCC

Ceq


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 △


Example 6.6

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


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.


Example 6.7


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

mF10mF1

20

1

30

1

60

1 eqC

C3.0301010 3 vCq eq


Example 6.7• 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.

39

,V1510203.0

31

1

Cq

v

V1010303.0

32

2

Cq

v

V530 213 vvv

V510603.0

mF60 33

qv


Joseph Henry (1979-1878)


6.4 Inductors

• An inductor is made of a coil of conducting wire

lAN

L2


Fig 6.22

42

(H/m)104 70

0

2

r

l

ANL

turns.ofnumber:N

length.:larea. sectionalcross: A

coretheoftypermeabili:


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 Linear

Nonlinear


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.


I-V Relation of Inductors

• An inductor consists of a coil of conducting wire.

46

dt

diL

dt

dv

+

-

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.

dt

diL

dt

dv


Fig 6.25

• An inductor are like a short circuit to dc.• The current through an inductor cannot

change instantaneously.


49

t

to

otidttv

Li )()(

1

t

dttvL

i )(1

memory. hasinductor The

vdtL

di1

+

-

vL


Energy Stored in an Inductor

• The energy stored in an inductor

idt

diLviP

50

t tidt

dtdiLpdtw

)(

)(

22 )(21

)(21ti

iLitLidiiL ,0)( i

)(2

1)( 2 tLitw

+

-

vL


Model of a Practical Inductor


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 teetetedtd

v 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

A23

6 33

tt

tdtti

0

2 03051


Example 6.9

54

5

0

65 kJ25.156

05

66060t

dttpdtw

thenisstoredenergytheand,60powerThe 5tvip

before.obtainedas

usingstoredenergytheobtaincanweely,AlternativwritingbyEq.(6.13),

)0(21

)5(21

)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.


Example 6.10

Solution:

56

,251

12Aii 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 ...321


Inductors in Parallel

Neq LLLL

1111

21


6.5 Series and Parallel Inductors

• Applying KVL to the loop, • Substituting vk = Lk di/dt results in

59

Nvvvvv ...321

dtdi

Ldtdi

Ldtdi

Ldtdi

Lv N ...321

dtdi

LLLL N )...( 321

dtdi

Ldtdi

L eq

N

KK

1

Neq LLLLL ...321


Parallel Inductors

• Using KCL,• But

60

Niiiii ...321

t

t kk

k otivdt

Li )(

10

t

t

t

t sk

tivdtL

tivdtL

i0 0

)(1

)(1

02

01 t

t NN

tivdtL 0

)(1

... 0

)(...)()(1

...11

0020121

0

tititivdtLLL N

t

tN

t

teq

N

kk

t

t

N

k k

tivdtL

tivdtL 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.


Table 6.1


Example 6.11

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


Example 6.11

• Solution:

64

10H12H,,H20:Series

H6427427 : Parallel

H18864 eqL

H42


Practice Problem 6.11


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 eedtdi

Ltv

mV120)()()( 1012

tetvtvtv

mV80mV)10)(4(22)( 10101

tt eedtdi

tv

isinductanceequivalentThe)(b


Example 6.12

68

t t t dteidtvti0 0

10121 mA5

4120

)0(41

)(

mA38533mA50

3 101010 ttt eet

e

t ttdteidtvti

0

1020 22 mA1

12120

)0(121

)(

mA11mA10

101010 ttt eet

e

)()()(thatNote 21 tititi

t

idttvL

i0

)0()(1

)(c


capacitors and inductors 1 svbitec.wordpress.com vishal jethva

Documents