Chapter 2 - Inductors, Capacitors and Alternating Current Circuits

7/25/2019 Chapter 2 - Inductors, Capacitors and Alternating Current Circuits

1/101

.INDUCTORS, CAPACITORS AND AC

CIRCUITS

Jabatan Kejuruteraan Mekanikal

Politeknik Ibrahim Sultan


2/101

EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSIntroduction

i. Inductors is a passive element.ii. The function - to store electrical charge in magnetic

elds.

iii. Used in power supplies, transformers radios, TVs,radars & electric motors.

iv. Inductors are usually made with coils of wiresometime called a coil or choe.

v. The wire coils are wound around iron cores, ferritecores, or other materials.

vi. Increase the current in the conductor will create achanging magnetic eld to generate a voltage in theconductor.

vii. !lectrical circuits that have properties againstchanges in current is called inductanc"earuhan#.

viii. Inductance is the a$ility of an inductorto store
http://en.wikipedia.org/wiki/Inductorhttp://en.wikipedia.org/wiki/Magnetic_fieldhttp://en.wikipedia.org/wiki/Energyhttp://en.wikipedia.org/wiki/Inductor


3/101

EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSInductor

Inductor S"#'o%(

Variable core Iron coreFerrite core Air core


4/101

EXPLAIN THE BASIC PRINCIPLES OFINDUCTORST) Function o* Inductor( in E%ctronic

Circuit( To smooth out the ripples in a wave of ' circuits

To improve the characteristics of waves in thetelephone line

To store the electrical energy

To improve the current in circuit


5/101

EXPLAIN THE BASIC PRINCIPLES OFINDUCTORST) E+F %ctro#oti& *orc- Inducd

*ro# F%u Cuttin!a# (n inductoris simply a coil of wire. t) currnt, " i #)owing through the coil produces a #a!ntic /u," *+# that is proportional to this )ow of electricalcurrent.

$# hen a conductor is moved across a magnetic eld soas to cut through the )u%, an electromagnetic force"e.m.f.# is produced in the conductor.

c- Farada"( La1 a &o%ta! i( inducd in aconductor 1)n t)at conductor i( #o&d

t)rou!) a #a!ntic $%d, or 1)n t) #a!ntic$%d #o&( a(t t) conductor3

sec/

;

)(

ampsinchangeofratecurrentsthedt

di

Webersinfluxofamount

wheredt

di

Ldt

dLi

dt

d

V tL

=

=

===


6/101

EXPLAIN THE BASIC PRINCIPLES OFINDUCTORST) E+F %ctro#oti& *orc- Inducd

*ro# F%u Cuttin!e# Induced e.m.f. on the conductor could $e produced$y two methods i. )u% cuts conductorii. conductor cuts )u%

4-F%u Cut( Conductorhen the magnet is moved towards the coil, a

de)ection is noted on the galvanometer showingthat a current has $een produced in the coil.

Coil (conductor)

Magnetic field

Flux cuts conductor


7/101

EXPLAIN THE BASIC PRINCIPLESOF INDUCTORST) E+F %ctro#oti& *orc- Inducd

*ro# F%u Cuttin!2-Conductor Cut( F%u

hen the conductor is moved through a magnetic ane.m.f. is induced in the conductor and thus a source ofe.m.f. is created $etween the ends of the conductor

"the simple concept of (' generator#

Conductor cuts ux

Magnetic field

conductor


8/101

EXPLAIN THE BASIC PRINCIPLESOF INDUCTORST) Dirction o* E+F Producd

2-F%u Cut( ConductorThe direction of an induced e.m.f. is always such thatit tends to set up a CURRENT OPPOSIN5 the+OTION OR THE CHAN5E OF FLUX responsi$le forinducing that e.m.f. "Ln6( La1#

Bar magnet move in and move out from a solenoid


9/101

INDUCTORS

Currnt 7 8o%ta! in

an Inductor

The voltage induced in theinductor depends therate of current change.Ln69( La1 stated that/the direction of aninduced emf is such thatit will always opposesthe change that is

causing it/.


10/101

INDUCTORS

Po1r in an Inductor

The instantaneous power used in forcing the current, " i #against this self-induced emf, " V0# is given as

1ower in a circuit is given as

dt

diL

dt

dLi

dt

dV tL ===

)(

==

== 2

2

2

1

2

1Li

dt

d

dt

diLi

dt

diLivP


11/101

INDUCTORS

Enr!" in an Inductor

hen power )ows into an inductor, energy is stored in itsmagnetic eld. hen the current )owing through theinductor is increasing and di2dt $ecomes greater than3ero, the instantaneous power in the circuit must also $e

greater than 3ero, " 1 4 5 # ie, positive which means thatenergy is $eing stored in the inductor.

Where;W- Energy (joules, J)

L- Inductance (Henry, H)i- Current (Ameres, A)

2)()(

2

1tt LiW =


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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSInductanc

This a$ility of aninductor to resistchanges in current andwhich also relatescurrent, i with its

magnetic )u% linage,*+ as a constant ofproportionality is calledInductancwhich isgiven the sym$ol L

with units of Hnr""H#after 6oseph 7enry.


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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSUnit o* Inductanc

The unit of inductance is Hnr" H-The sym$ol L

8 7enry is the amount $y the winding inductancewhen it( currnt c)an!( at a rat o* 4 a#rcurrnt and inducd &o%ta! roducd '" 4

&o%t.

8 mili7enry "m7# 9 828555 atau 85-:7enry "7#

8 miro7enry ";7# 9 82 8,555,555 or 85-


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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSS%*:Inducanc 7 +utua%:Inductanc

Two type of Inductance i. S%*:Inductanc L-

the induction of a voltage in a current-carryingwire when the current in the wire itself ischanging.

In the case of self-inductance, the magnetic eldcreated $y a changing current "ac# in the circuititself induces a voltage in the same circuit.

Therefore, the voltage is self-induced


15/101


i. S%*:Inductanc L-:For#u%a

SELF:INDUCTANCE

SELF:INDUCTANCE

di

d

NL

=

sec)/(

)(

)@(tan;

;')

;)

;)

21

2

1

ampcurrentofchangeofratetime

dt

di

fluxsinchangesdt

d

voltsvoltageinducede

circuitinturnofnumberN HenryHceinducselfLwhere

di

dt

dt

dNL

dt

diL

dt

dN

ee

LawsFaradaycdtdiLe

currentofchangetocausedgeneratedEMFbdt

dNe

fluxofchangetocausedgeneratedEMFa

=

=

== =

==

=

=

=


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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSS%*:Inducanc 7 +utua%:Inductanci. S%*:Inductanc L-

(n appro%imation of inductance for any coil of wirecan $e found with this formula


17/101


ii. +utua%:Inductanc +- The emf is induced into an ad=acent coil situated

within the same magnetic eld, the emf is said to$e induced magnetically, inductively or $y +utua%induction, sym$ol " > #.

Then when two or more coils are magneticallylined together $y a common magnetic )u% theyare said to have the property of +utua%

Inductanc. >utual induction as the current )owing in one coil

induces an emf in an ad=acent coil.


18/101


ii. +utua%:Inductanc +-Mutual Inductance between Coils

(ssuming a perfect )u%linage $etween the two

coils the mutualinductance that e%ists$etween them can $egiven as.

here?ois the permea$ility of free

space "@.A.85-B#?ris the relative permea$ility

of the soft iron core* is in the num$er of coil


19/101


ii. +utua%:Inductanc +-Mutual Induction

7ere the current )owing in coil one, 08 sets up a

magnetic eld around itself with some of thesemagnetic eld lines passing through coil two, 0

C

giving us mutual inductance. 'oil one has a current

of I8 and *8 turns while, coil two has *C turns.

Therefore, the mutual inductance, >8C

of coil two

that e%ists with respect to coil one depends on theirposition with respect to each other and is given as


20/101

INDUCTORS

Connctin! Inductor in Sri( 7 Para%%%

i. Inductor in Sri( Circuit

The current, " I # that )ows through the rst inductor,08has no other way to go $ut pass through the second

inductor and the third and so on. Then, inductors inseries have a Co##on Currnt )owing throughthem, for e%ample

I089 I0C9 I0:9 I(D...etc.

Ltotal= L1+ L2+ L3+ ..... + Lnetc.


21/101

INDUCTORS


i. Inductor in Sri( CircuitEa#% 4Three inductors of 85m7, @5m7 and E5m7 areconnected together in a series com$ination with nomutual inductance $etween them. 'alculate the total

inductance of the series com$ination.

Ea#% 23

'alculate the total inductance.


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INDUCTORS


ii. Inductor in Para%%% Circuit

Inductor(are said to $e connected together in

/Para%%%/ when $oth of their terminals arerespectively connected to each terminal of theother inductor or inductors. The voltage dropacross all of the inductors in parallel will $e thesame. Then,

Inductor( in Para%%% have a

Co##on 8o%ta! across them and in oure%ample $elow the voltage across the inductorsis given as VL1= VL2= VL3= VAB...etc

Tota%Inductanc3


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INDUCTORS


ii. Inductor in Para%%% CircuitEa#% 43Three inductors of


24/101

INDUCTORS


ii. Inductor in Para%%% CircuitEa#% ;3Find the total inductance $etween points ( andD.

L= 200 mH L!= 1 H

L"= 1 HL#= 2 HL$= 3 H

L%= 800 mH L&= 0.8 H L'= 2 H

A

B


25/101

INDUCTORS


ii. Inductor in Para%%% Circuit

HLLL 3218778 =+=+=

HLL

LLL 2.1

32

)3)(2())((

876

876

678 =

+

=

+

=

HLLLL 32.18.016785445678 =++=++=

HLL

LLL 5.1

33

)3)(3())((

845673

845673345678 =+=+=

HLLLL( 5.25.11080010200 33

34567821 =++=++=

Solution 3;

Total inductance between points A and B, LT


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INDUCTORS


27/101

INDUCTORS

An(1rCalculate the first inductor branch, L

A

Calculate the second inductor branch, LB

Calculate the equivalent circuit inductance, LEQ

Then the equivalent inductance is 15 mH.


28/101

EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS

Introduction

a#( capacitors are one of the most useful components inelectronics, and after resistors are the most numerouscomponents in circuits.$#'apacitors a'i%it" to (tor %ctric c)ar!, as an!0!'TJKLT(TI' FI!0 created $etween two metal /plates/.

c#Limple capacitors consist of two plates made of anelectrically conducting material "e.g., a metal# andseparated '" a nonconductin! #atria% or di%ctric

"e.g., glass, paraMn, mica, oil, paper, tantalum, or

air#.


29/101


Circuit S"#'o%( *or a Caacitor

( $asic %ed value type of capacitor consists oftwo plates made from metallic foil, separated $y aninsulator. This may $e made from a choice ofdiNerent insulating materials, having good

I!0!'TJI' properties.


30/101


Caacitor( Ha& +an" U((

= Hi!) 8o%ta! !lectrolyticused in power supplies.= Aia% E%ctro%"tic lowervoltage smaller si3e for generalpurpose where largecapacitance values are needed.

= Hi!) 8o%ta! di(> cra#icsmall si3e and capacitancevalue, e%cellent tolerancecharacteristics.= +ta%i(d Po%"ro"%n

small si3e for values up toaround C?F good relia$ility.= Su' #iniatur +u%ti %a"rcra#ic c)i "surface mount#capacitor. Jelatively highcapacitance for si3e achieved

$y multiple layers, eNectively


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T"( o* Caacitor

There are many types of capacitor $ut they can $e split into twogroups, polarisedand unpolarised.

8.1olarised capacitors "largevalues, 8?F O#

!lectrolytic 'apacitors

Tantalum Dead 'apacitors

C.Unpolarised capacitors "smallvalues, up to 8?F#

1olystyrene 'apacitors"color code P no. code#

Varia$le capacitors

Trimmer capacitors
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T"( o* Caacitor

S C C S O C C O S


33/101


T) Function a Caacitor in E%ctronic Circuit

a# 'apacitors store electric charge.$# 'apacitors improve electric current.

c# They are used with resistors in timing circuits$ecause it

taes time for a capacitor to ll with charge.

d# They are used to smoothvarying ' supplies $y acting

as a reservoir of charge.

e# They are also used in lter circuits $ecause capacitors

easily pass (' "changing# signals $ut they $loc '

"constant# signals.

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C)ar! 7 Di(c)ar!Ho1 a caacitor !t( it( c)ar!

a#( capacitor is connected in a ' circuit,current )ow, $ut only for a short time.$#hen the switch is closed to contact ( andelectrons $egin to )ow from the negative

$attery terminal, and appear to $e )owingaround the circuit. Kf course they canQt$ecause the capacitor has a layer ofinsulation $etween its plates, so electronsfrom the negative $attery terminal crowdonto the right hand plate of the capacitor

creating an increasingly strong negativecharge.c#The very thin insulating "dielectric# layer$etween the plates is a$le to eMcientlytransfer this negative charge from the

electrons, and this charge repels the samenum$er of electrons from the left hand plate



35/101


C)ar! 7 Di(c)ar!?)" t) currnt *a%%(

a#(fter a short time, a largenum$er of electrons havegathered on the right hand plateof the capacitor, creating a

growing negative charge, maingit increasingly diMcult forelectrons )owing from thenegative $attery terminal to reachthe capacitor plate $ecause of therepulsion from the growing

num$er of negative electronsgathered there.



36/101


C)ar! 7 Di(c)ar!Fu%% C)ar!

a#The repulsion from the electronson the capacitorQs right hand plateis appro%imately eHual to the forcefrom the negative $attery terminal

and current ceases.$#Knce the $attery and capacitorvoltages are eHual we can say thatthe capacitor has reached itsma%imum charge.c#If the $attery is now

disconnected $y opening theswitch, the capacitor will remain ina charged state, with a voltageeHual to the $attery voltage, andprovided that no current )ows, it

should remain charged indenitely.



37/101


C)ar! 7 Di(c)ar!Di(c)ar!in! t) caacitor

a#Luppose that with the capacitor fullycharged, the switch is now closed inposition D the circuit is complete oncemore, $ut this time consisting of a resistor

and capacitor.

$#!lectrons will now )ow around the

circuit via the resistor as the charge oncapacitor acts as the source of currentThe charge on the capacitor will $edepleted as the current )ows. The rate atwhich the capacitor voltage reduces

towards 3ero will depend on the amountof current )owing, and thus on the value



38/101


Caacitanc

a#The amount of energy a capacitor can store depends on thevalue or CAPACITANCEof the capacitor.$#Lym$ol R Cc#Unit R *arad("sym$ol F#d#8 Farad is the amount of capacitance that can store 8 'oulom$

"


39/101


Factor( A@ctin! Caacitanc

Three $asic factors of capacitor construction determining theamount of capacitance created. These factors all dictatecapacitance $y aNecting how much electric eld )u% "relativediNerence of electrons $etween plates# will develop for a givenamount of electric eld force "voltage $etween the two plates# 4- PLATE AREA (ll other

factors $eing eHual, greater

plate area gives greatercapacitance less plate areagives less capacitance.

2- PLATE SPACIN5(ll otherfactors $eing eHual, further

plate spacing gives lesscapacitance closer platespacing gives greatercapacitance.

;- PER+ITTI8ITDIELECTRIC +ATERIAL-

(ll other factors $eing



40/101


Factor( A@ctin! Caacitanc4- PLATE AREA

2- PLATE SPACIN5

;- PER+ITTI8IT DIELECTRIC +ATERIAL-

/Jelative/ permittivity "etelapan# means the permittivity of a

material, relative to that of a pure vacuum.An aroi#ation o* caacitanc *or an" air o*

(aratd conductor( can ' *ound 1it) t)i(*or#u%a

)*

d

*1

*

r

relativetypermittivir

mF

spacefreeoftypermittivi

=

=

=

=

/121085.8

0



41/101


C)ar! on a Caacitor Caacitanc-

a#The charge "S# on a capacitor depends on a com$inationof the a$ove factors, which can $e given together as the'apacitance "'# and the voltage applied "V#.$#For a component of a given capacitance, the relationship$etween voltage and charge is constant.c#Increasing the applied voltage results in a proportionallyincreased charge. This relationship can $e e%pressed in theformula

< C8 or C


42/101


Ea#% 3

8#( capacitor having a capacitance of G5 ?F is connectedacross a E55V d.c. supply. 'alculate the charge.

C#( capacitor is made with metal plates and separated $ysheets of mica having a thicness of 5.: mm and relativepermittivity of


43/101


Ea#% 3

8#( capacitor having a capacitance of G5 ?F is connectedacross a E55V d.c. supply. 'alculate the charge.

C#( capacitor is made with metal plates and separated $ysheets of mica having a thicness of 5.: mm and relativepermittivity of


44/101


Caacitor( in Circuit(4.Caacitor( in Sri(

'onnected in series. Increases the thicness of the dielectric, and soreduces the total capacitance. The total capacitance is inversely proportional to the

distance $etween the plates. The formula we use for capacitors in L!JI!L "morethan C capacitors #

3

3

2

2

1

1

321

323121

321

321

;;

capacit!;an"in#$ta%e&e

capacit!;e#e!"ca!%e&e

11111

ecapacitanc&ta$

*

+V

*

+V

*

+V

++++

******

****

-.*****

(*

(*

(*

(

(

n(

===

===

++

=

++++=



45/101


Caacitor( in Circuit(4.Caacitor( in Sri(

If TK capacitors

V**

*VV

**

*V

+++

-.

**

***

**

(

(

+=

+=

==

+

=

21

12

21

21

21

21

21

;



ecapacitanc&ta$



46/101


Caacitor( in Circuit(2.Caacitor( in Para%%%

'onnecting capacitors in parallel eNectivelyincreases the area of the plates.

(***

(*(*(*

(

VVVV

*V+

*V+

*V+

****

===

===

++=

321

3

3

2

2

1

1

321


;;


ecapacitanc&ta$



47/101


Caacitor( in Circuit(:Ea#% 3

8# hat is the appro%imate total capacitance of thisparallel circuit

a# :C5pF $# 8@BpF c# :.CnF d# 8.@BnF

C# hat is the appro%imate total capacitance of thisseries circuit

a# 8.


48/101


Enr!" Stord in a Caacitor

Luppose the p.d. across a capacitor of capacitance, 'farads to $e increased from !to (!%d!) volts in dtseconds.The charging current, iamperes, given $y dt

dv*i=

Instantaneous value of power to capacitor is watts

dt

dvv*iv=

(nd energy supplied to capacitor during interval dt is dv*vdtdt

dv

*vdtdt

dv

v* =

=

7ence total energy supplied to capacitor when p.d. isincreased from 5 R V volts is [ ]

V

+*where+VWalso

*

+

*

+*Walso

*VW

/oules*Vv*dv*vEnergy VV

==

=

=

=

===

;

2

1

2

1

2

1

21

;2

1

2

1

22

2

2

0

2

0



49/101


Enr!" Stord in a CaacitorEa#%( E5 ?F capacitor is charged from C55V supply. (fter $eingdisconnected it is immediately connected in parallel with a:5 ?F capacitor which is initially uncharged. Finda#The p.d. across the com$ination$#The electrostatic energies $efore and after the capacitors

are connected in parallel( )( ) **V+a 01.02001050) 6 ===

hen the capacitors are connected in parallel, totalcapacitors is G5 ?F

( )( )

( )( ) 0ouleWparallel1n

0oule*VWb

625.012510802

1;

120010502

1

2

1)

26

262

==

===

An(1r

( )

Vcapacitorsacrossdp

dp

*V+

125..

..108001.0 6

=

==



50/101


Enr!" Stord in a CaacitorErci(3

Three capacitors of C, : and < ?F respectively areconnected in series across a E55V d.c supply. 'alculatea#The charge on each capacitor$#The p.d. across each capacitorc#The energy stored in the < ?F capacitor

An(1ra# E55 ?'$# CE5V 8


51/101

ALTERNATIN5 CURRENT

Introduction

a) &irect current "'# is electricity )owing in aconstant direction, and2or possessing a voltagewith constant polarity "$attery-positive andnegative terminals.

$# 'ertain sources of electricity "most nota$ly,rotary electro-mechanical generators# naturallyproduce voltages alternating in polarity,reversing positive and negative over time.

c# !ither as a voltage switching polarity or as a

current switching direction $ac and forth, thisind of electricity is nown as (lternating'urrent "('#.

ALTERNATIN5 CURRENT


52/101

ALTERNATIN5 CURRENT

Ba(ic AC5nrator

a# (' generatorconsists of aconductor, or loop ofwire in a magnetic

eld that isproduced $y anelectromagnet.

$# The two ends of theloop are connected

to slip rings, andthey are in contactwith two $rushes.

c# hen the looprotates it cuts

magnetic lines of

ALTERNATIN5 CURRENT


53/101

ALTERNATIN5 CURRENT

a# The sine waveoutput is theresult of oneside of thegenerator loopcutting lines offorce.

$# In the rst halfturn of rotationthis produces apositive currentand in thesecond half ofrotationproduces anegativecurrent. This

D&%o#nt o* aSin:?a& Outut

ALTERNATIN5 CURRENT


54/101

ALTERNATIN5 CURRENT

D&%o#nt o* a Sin:?a& Outut

'oil (ngle

" W #5 @E X5 8:E 8G5 CCE CB5 :8E :


55/101

ALTERNATIN5 CURRENT

Radian(

Radian, "rad# is a Huadrant ofa circle where the distancesu$tended on thecircumference eHuals theradius "r# of the circle.

Lince the circumference ofa circle is eHual toCA % radius, there must $eCA radians around a :


56/101

ALTERNATIN5 CURRENT

Radian( : Ea#%

The conversion $etween degrees and radians

ALTERNATIN5 CURRENT


57/101

ALTERNATIN5 CURRENT

Sinu(oida% ?a&*or#

ALTERNATIN5 CURRENT


58/101

ALTERNATIN5 CURRENT

tVtv m sin)( =

t1ti m sin)( =

ALTERNATIN5 CURRENT


59/101

ALTERNATIN5 CURRENT

C)aractri(tic( o* a Sin ?a&

The main characteristics of an (' waveform are denedas

a# The Priod, T# is the length of time in seconds thatthe waveform taes to repeat itself from start tonish.

$# The FrGunc", - is the num$er of times thewaveform repeats itself within a one second timeperiod. FreHuency is the reciprocal of the time period," Y 9 82T # with the unit of freHuency $eing the Hert',"73#.

c# The A#%itud A- is the magnitude or intensity ofthe signal waveform measured in volts or amps.

d# The 1!(Z-TK-1!(Z "Vp-p 9C Vp-p # value is the vertical

distance $etween the top and $ottom of the wave.

e# (verage Value "Va 9 average value95.


60/101

ALTERNATIN5 CURRENT

T"( o* Priodic ?a&*or#

R%ation()i Bt1n FrGunc" andPriodic Ti#

!ei* +einitin ,!itten as e!i-ic &ime

i$ &'/san- H 1m

e%a i$$in H 1/

i%a Bi$$in H 1n

&e!!a &!i$$in &H 1p

sec11

)(

11)(

fFre2uency((imePeriodicor

Hert3((imePeriodic

fFre2uency

==

==

ALTERNATIN5 CURRENT


61/101

ALTERNATIN5 CURRENT

Ea#% 4The freHuency of a 8C5 V (' circuit is F at that pointSo%ution 4

( ) ( )sec

8.3766014.322) radianH3fa ===

( )( ) radiansradtb 3768.0sec001.0sec

8.376) ===

( ) ( ) VVtVvc m 15.443768.0sin120sin) ===

Ea#% 2

If a sine wave has a J>L voltage of 8Cvolts, what will $eits 1ea-to-1ea voltage

a# ::.XV$# G.@G@V

c# 8


62/101

ALTERNATIN5 CURRENT

Ea#% ;hat is the pea value of a sine wave whose V(V value is

8EV

a# 8XV$# X.EVc# C8.CV

d# C:.EVEa#% If an (' waveform has a periodic time of Cms, what will

$e its freHuency

a# C73$# E5573c# C>73d# E573

ALTERNATIN5 CURRENT


63/101

ALTERNATIN5 CURRENT

Ea#% ith reference to Fig 8.:.8, what is the value la$elled (

a# 1eriodic time$# (mplitudec# FreHuencyd# J>L value

Ea#% Jith reference to Fig 8.:.8, if the level la$elled [ has a

value of CV what is the value la$elled D

a# The Joot >ean LHuared value.$# The (mplitude.c# The (verage value.d# The 1ea value.

ALTERNATIN5 CURRENT


64/101

ALTERNATIN5 CURRENT

Ea#% In Fig 8.:.C, how many complete cycles are shown

a# C $# : c# @ d# B

Ea#% Khat value is given $y the formula V1Z% 5.L $# V>([ c# The Form Factord# V(V

Ea#% 'omplete the sentence /( sine wave...

a# ...has many harmonics$# ...is a comple% wavec# ...consists of a fundamental onlyd# ...always has the same freHuency/

ALTERNATIN5 CURRENT


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ALTERNATIN5 CURRENT

Ea#% 40hat is the direction of an (' current

a# It is %ed $# It eepschanging

c# It eeps reversing d# It cannot $efound

Ea#% 44(ll the following statements on alternating current is

true EXCEPT

i. It carries electric charges in the form of sinewave.

ii. The movement of electric charge reverses

direction repeatedly.iii. !lectric charge )ows in two directions.iv. 'urrent always )ows in the same direction

$etween positive and negative terminal.

a# iv only

ALTERNATIN5 CURRENT


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ALTERNATIN5 CURRENT

T1o Sinu(oida% ?a&*or# M in:)a(

a# Two alternating Huantities such as a voltage, v and acurrent, i have the same freHuency Y in 7ert3.

$# The freHuency of the two Huantities is the same theangular velocity, \ must also $e the same.

c# The angle of rotation within a particular time period

will always $e the same and the phase diNerence$etween the two Huantities of v and i will therefore $e3ero and + 9 5.

ALTERNATIN5 CURRENT


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ALTERNATIN5 CURRENT

P)a( Di@rnc o* a Sinu(oida% ?a&*or#

a# The voltage, v and the current, i have a phasediNerence $etween themselves of :5o, so "+ 9 :5oor A2< radians#.

$# (s $oth alternating Huantities rotate at the samespeed, i.e. they have the same freHuency, this phase

diNerence will remain constant for all instants in time,then the phase diNerence of :5o $etween the twoHuantities is represented $y phi, + as shown $elow.

ALTERNATIN5 CURRENT


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ALTERNATIN5 CURRENT

a# The current, i is lagging the

voltage, v $y angle + ":5o#.

$# The current phasor lags thevoltage phasor $y the angle,+, as the two phasors rotate

in an anticlocisedirection.

P)a(or Dia!ra# o* a Sinu(oida%?a&*or#

P)a(or Addition o* T1o P)a(or(

'onsider two (' voltages,V8having a pea voltage of

C5 volts, and VC having a

pea voltage of :5 voltswhere V8 leads VC $y


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a# (ny ideal resistor can $e

descri$ed in terms of itsvoltage and current, thevoltage across a pureohmic resistor is linearlyproportional to the

current )owing through itas dened $y *hm+s La.

Pur%" R(i(ti& Circuit

Pur%" R(i(ti& M Sinu(oida%?a&*or#(

Pur%" R(i(ti& MP)a(or Dia!ra#

Pur%" R(i(ti& MI#danc

RESISTI8E REACTANCE
http://www.electronics-tutorials.ws/dccircuits/dcp_2.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_2.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_2.html


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Ea#% 4'alculate the current and power consumed in a single

phase C@5V (' circuit $y a heating element whichhas an impedance of


71/101

Ea#% 2( sinusoidal voltage supply dened as

V"t# 9 855 % cos"\t O :5o

# is connected to a pureresistance of E5 Khms. etermine its impedance andthe value of the current )owing through the circuit.raw the corresponding phasor diagram.So%ution 2

INDUCTI8E REACTANCE


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Introduction

a# Inductive Jeactance is an inductors resistance in an(' circuit.$# Unit KhmQs !"c# Lym$ol XL, is the property in an (' circuit which

opposes the change in the current.

where f9freHuency "73#

09Inductor2Inductance "7enry#L 9Inductive

reactance "#

9angular freHuency"radians#

L

1

V4

L

LL ==

fL4L 2=

INDUCTI8E REACTANCE


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Pur%" Inducti&

a# Inductor is connected directly across the (' supplyvoltage.

$# The supply voltage increases and decreases with thefreHuency, the self-induced $ac emf also increasesand decreases in the coil with respect to this change.

c# In self-induced emf is directly proportional to the rateof change of the current.

d# The voltage and current waveforms show that for apurely inductive circuit the current lags the voltage $yX5o or the voltage leads the current $y X5o.

e# e can dene the value of the current at any point intime as $eing

INDUCTI8E REACTANCE


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Inducti& Ractanc A!ain(t FrGunc"

The inductive reactance of an inductor increases as the

freHuency across it increases therefore inductivereactance is proportional to freHuency " [0] Y #

INDUCTI8E REACTANCE


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8# ( coil of inductance 8E5m7 and 3ero resistance isconnected across a 855V, E573 supply. 'alculate theinductive reactance of the coil and the current )owingthrough it.

C# 'alculate the supply voltage "VL# needed to cause a

current of 85m( to )ow through a 8Em7 inductor at a

supply freHuency of @5573.

Erci( 3

INDUCTI8E REACTANCE


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So%ution 3

mVV41V LL5 377377.0)7.37)(1010( 3

====

=== 7.37)1015)(400)(14.3(22 3fL4L

$# Voltage supply

a# Voltage supply

:# hich of the following graphs illustrates the formulaCAY 0

@# 'alculate the appro%imate supply voltage needed tocause a current of 85m( to )ow through a 8Em7

inductor at a supply freHuency of @73 .

a# :.GV$# 5.@Vc# :B.BVd# :B


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E# hat is the reactance of the inductorin Fig


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c# Lince the rate of change of the potential diNerenceacross the plates is now at its ma%imum value, the)ow of current through the capacitor will also $e at

its ma%imum rate as the ma%imum amount ofelectrons are moving from one plate to the other.d# (s the sinusoidal supply voltage reaches its X5opoint

on the waveform it $egins to slow down and for avery $rief instant in time the potential diNerence

across the plates is neither increasing nor decreasingtherefore the current decreases to 3ero as there is no

Introduction

a# hen the switch isclosed, a high current willstart to )ow into thecapacitor.

$# The ac voltage, V is

increasing in a positivedirection at its ma%imumrate as it crosses the 3eroreference a%is at aninstant in time given as

5o

.

CAPACITI8E REACTANCE


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Sinusoidal Waveforms for AC Inductance



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Purely Capacitive

V#$lags$ %#b& '(o, or we can sa& t)at %#$leads$ V#b& '(

o

CAPACITI8E REACTANCEi. Unit KhmQs "#


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ii. Lym$ol " ['#

iii. The voltage changes the less current they will pass.This means then that the reactance of a capacitor is

/inversely proportional/ to the freHuency of thesupply as shown.

here [' is the 'apacitive Jeactance in

Khms,Y is the freHuency in 7ert3' is the capacitance in Farads, sym$ol

F

iv. e can also dene capacitive reactance interms of radians, where Kmega, \ eHuals CAY.



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Capacitive eactance a!ainst Fre"uency

RL SERIES AC CIRCUIT


83/101

a# In an (' circuit, inductance,0 and resistance, J the

voltage, V will $e the phasorsum of the two componentvoltages, VJand V0.

$# The current )owing throughthe coil will still lag the

voltage, $ut $y an amountless than X5o dependingupon the values of VJand V0.

c# The new angle $etween thevoltage and the current is

nown as the phase angle ofthe circuit and is given the_ree sym$ol phi, +.

d# In a series connected J-0circuit the current is common

as the same current )ows



84/101

The voltage triangle is derived from 1ythagorasQstheorem and is given as



85/101

#$e Impedance #rian!le



86/101

%&ample 'o( );

A solenoid coil )as a resistance of 3( *)s and an inductance of (.-. %f t)e current

flowing t)roug) t)e coil is aps. #alculate,

a" T)e /oltage of t)e suppl& if t)e fre0uenc& is (-.

b" T)e p)ase angle between t)e /oltage and t)e current.



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Power #rian!le

RC SERIES AC CIRCUIT


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1" A resistance, and a capacitance,

# in series.

2" T)e current flowing t)roug) t)e

resistance and capacitance, w)ilet)e /oltage is ade up of t)e two

coponent /oltages, Vand V#.

3" T)e resulting /oltage of t)ese two

coponents can be found

at)eaticall& but since /ectors

V and V# are '(o outofp)ase,

t)e& can be added /ectoriall& b&

constructing a /ector diagra.

" T)e indi/idual /ector diagras for

a pure resistance and a pure

capacitance are gi/en as

RC SERIES AC CIRCUIT8ctor Dia!ra# o* t) R(u%tant


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8ctor Dia!ra# o* t) R(u%tant8o%ta!



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#$e C Impedance #rian!le

RC SERIES AC CIRCUITE %


91/101

Ea#%( capacitor which has an internal resistance of 85^Qs anda capacitance value of 855uF is connected to a supply

voltage given as V"t# 9 855 sin ":8@t#. 'alculate thecurrent )owing through the capacitor. (lso construct avoltage triangle showing the individual voltage drops.

T)e capaciti/e reactance and circuit ipedance is calculated as



92/101

T)en t)e current flowing t)roug) t)e capacitor is gi/en as

T)e p)ase angle between t)e current and /oltage is calculated fro t)e ipedance

triangle abo/e as

T)en t)e indi/idual /oltage drops around t)e circuit are calculated as

T)en t)e resultant /oltage triangle will be.

RCL SERIES AC CIRCUIT


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#$e Series *C Circuit

a"T)ree coponents, , L and # )a/e /er& different p)ase relations)ips to

eac) ot)er w)en connected to a sinusoidal A# suppl&.b"%n a pure o)ic resistor t)e /oltage is $inp)ase$ wit) t)e current, in a pure

inductance t)e /oltage $leads$ t)e current b& '(o, ! 4L% " and wit) a pure

capacitance t)e /oltage $lags$ t)e current b& '(o, ! %#4 ".



94/101

Individual Volta!e Vectors

P$asor +ia!ram for a Series *C Circuit



95/101

Volta!e #rian!le for a Series *C Circuit

B& substituting t)ese /alues into 5&t)agoras6s e0uation abo/e for t)e

/oltage triangle will gi/e us



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#$e Impedance #rian!le for a Series *C Circuit



97/101

%&ample ;

A series L# circuit containing a resistance of 127, an inductance of (.1- and

a capacitor of 1((u8 are connected across a 1((V, (- suppl&. #alculate t)etotal circuit ipedance, t)e circuits current, power factor and draw t)e /oltage

p)asor diagra.



98/101

Solution ;

%nducti/e eactance, 9L.

#apaciti/e eactance, 9#.

#ircuit %pedance, :.

#ircuits #urrent, %.



99/101

Solution ;

Voltages across t)e ;eries L# #ircuit, V, VL, V#.

#ircuits 5ower factor and 5)ase Angle,


100/101

1" ;tate t)e definition and define forulae of

a" %nducti/e reactance

b" #apaciti/e reactance

2" #alculate t)e following /alues>

1" T)e inducti/e reactance and capaciti/e reactance

2" T)e ipedance and current flow t)roug) t)e circuit

3" T)e power factor and p)ase angle

3" T)ree coils in a balanced t)ree p)ase s&ste are connected wit) delta

connection, w)ic) )as a series capacitor /alue '.? @8, inductor wit) (.?- /alue

and resistor wit) 2( *) per p)ase. %f t)is load is connected to a 2V, (-

suppl&, calculate p)ase current and reacti/e power for t)is connection.

L #

33 1((- 22(8

2( V, 1-


101/101

Chapter 2 - Inductors, Capacitors and Alternating Current Circuits

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Transcript of Chapter 2 - Inductors, Capacitors and Alternating Current Circuits