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.INDUCTORS, CAPACITORS AND AC
CIRCUITS
Jabatan Kejuruteraan Mekanikal
Politeknik Ibrahim Sultan
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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
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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSInductor
Inductor S"#'o%(
Variable core Iron coreFerrite core Air core
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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
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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
=
=
===
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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
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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 $etween the ends of the conductor
"the simple concept of (' generator#
Conductor cuts ux
Magnetic field
conductor
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EXPLAIN THE BASIC PRINCIPLESOF INDUCTORST) Dirction o* E+F Producd
2-F%u Cut( ConductorThe direction of an induced e.m.f. is always such thatit tends to set up a CURRENT OPPOSIN5 the+OTION OR THE CHAN5E OF FLUX responsi$le forinducing that e.m.f. "Ln6( La1#
Bar magnet move in and move out from a solenoid
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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/.
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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
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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
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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSS%*:Inducanc 7 +utua%:Inductanc
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 $e found with this formula
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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSS%*:Inducanc 7 +utua%:Inductanc
ii. +utua%:Inductanc +- The emf is induced into an ad=acent coil situated
within the same magnetic eld, the emf is said to$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.
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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSS%*:Inducanc 7 +utua%:Inductanc
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
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EXPLAIN THE BASIC PRINCIPLES OFINDUCTORSS%*:Inducanc 7 +utua%:Inductanc
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
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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 $ut pass through the second
inductor and the third and so on. Then, inductors inseries have a Co##on Currnt )owing throughthem, for e%ample
I089 I0C9 I0:9 I(D...etc.
Ltotal= L1+ L2+ L3+ ..... + Lnetc.
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INDUCTORS
Connctin! Inductor in Sri( 7 Para%%%
i. Inductor in Sri( CircuitEa#% 4Three inductors of 85m7, @5m7 and E5m7 areconnected together in a series com$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
Connctin! Inductor in Sri( 7 Para%%%
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
Connctin! Inductor in Sri( 7 Para%%%
ii. Inductor in Para%%% CircuitEa#% 43Three inductors of
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INDUCTORS
Connctin! Inductor in Sri( 7 Para%%%
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
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INDUCTORS
Connctin! Inductor in Sri( 7 Para%%%
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
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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.
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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#.
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
Circuit S"#'o%( *or a Caacitor
( $asic %ed value type of capacitor consists oftwo plates made from metallic foil, separated $y aninsulator. This may $e made from a choice ofdiNerent insulating materials, having good
I!0!'TJI' properties.
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
Caacitor( Ha& +an" U((
= Hi!) 8o%ta! !lectrolyticused in power supplies.= Aia% E%ctro%"tic lowervoltage smaller si3e for generalpurpose where largecapacitance values are needed.
= Hi!) 8o%ta! di(> cra#icsmall si3e and capacitancevalue, e%cellent tolerancecharacteristics.= +ta%i(d Po%"ro"%n
small si3e for values up toaround C?F good relia$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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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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
http://www.kpsec.freeuk.com/components/capac.htmhttp://www.kpsec.freeuk.com/components/capac.htmhttp://www.kpsec.freeuk.com/components/capac.htmhttp://www.kpsec.freeuk.com/components/capac.htm7/25/2019 Chapter 2 - Inductors, Capacitors and Alternating Current Circuits
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
T"( o* Caacitor
S C C S O C C O S
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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.
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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.
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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.
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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 $edepleted as the current )ows. The rate atwhich the capacitor voltage reduces
towards 3ero will depend on the amountof current )owing, and thus on the value
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
Caacitanc
a#The amount of energy a capacitor can store depends on thevalue or CAPACITANCEof the capacitor.$#Lym$ol R Cc#Unit R *arad("sym$ol F#d#8 Farad is the amount of capacitance that can store 8 'oulom$
"
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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 $eing
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
Factor( A@ctin! Caacitanc4- PLATE AREA
2- PLATE SPACIN5
;- PER+ITTI8IT DIELECTRIC +ATERIAL-
/Jelative/ permittivity "etelapan# means the permittivity of a
material, relative to that of a pure vacuum.An aroi#ation o* caacitanc *or an" air o*
(aratd conductor( can ' *ound 1it) t)i(*or#u%a
)*
d
*1
*
r
relativetypermittivir
mF
spacefreeoftypermittivi
=
=
=
=
/121085.8
0
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
C)ar! on a Caacitor Caacitanc-
a#The charge "S# on a capacitor depends on a com$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
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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(
===
===
++
=
++++=
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
Caacitor( in Circuit(4.Caacitor( in Sri(
If TK capacitors
V**
*VV
**
*V
+++
-.
**
***
**
(
(
+=
+=
==
+
=
21
12
21
21
21
21
21
;
capacit!;an"in#$ta%e&e
capacit!;e#e!"ca!%e&e
ecapacitanc&ta$
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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
capacit!;an"in#$ta%e&e
;;
capacit!;e#e!"ca!%e&e
ecapacitanc&ta$
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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.
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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
EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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 $efore and after the capacitors
are connected in parallel( )( ) **V+a 01.02001050) 6 ===
hen the capacitors are connected in parallel, totalcapacitors is G5 ?F
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
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EXPLAIN THE BASIC PRINCIPLES OF CAPACITORS
Enr!" Stord in a CaacitorErci(3
Three capacitors of C, : and < ?F respectively areconnected in series across a E55V d.c supply. 'alculatea#The charge on each capacitor$#The p.d. across each capacitorc#The energy stored in the < ?F capacitor
An(1ra# E55 ?'$# CE5V 8
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ALTERNATIN5 CURRENT
Introduction
a) &irect current "'# is electricity )owing in aconstant direction, and2or possessing a voltagewith constant polarity "$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
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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
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ALTERNATIN5 CURRENT
a# The sine waveoutput is theresult of oneside of thegenerator loopcutting lines offorce.
$# In the rst halfturn of rotationthis produces apositive currentand in thesecond half ofrotationproduces anegativecurrent. This
D&%o#nt o* aSin:?a& Outut
ALTERNATIN5 CURRENT
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ALTERNATIN5 CURRENT
D&%o#nt o* a Sin:?a& Outut
'oil (ngle
" W #5 @E X5 8:E 8G5 CCE CB5 :8E :
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ALTERNATIN5 CURRENT
Radian(
Radian, "rad# is a Huadrant ofa circle where the distancesu$tended on thecircumference eHuals theradius "r# of the circle.
Lince the circumference ofa circle is eHual toCA % radius, there must $eCA radians around a :
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ALTERNATIN5 CURRENT
Radian( : Ea#%
The conversion $etween degrees and radians
ALTERNATIN5 CURRENT
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ALTERNATIN5 CURRENT
Sinu(oida% ?a&*or#
ALTERNATIN5 CURRENT
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ALTERNATIN5 CURRENT
tVtv m sin)( =
t1ti m sin)( =
ALTERNATIN5 CURRENT
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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 $ottom of the wave.
e# (verage Value "Va 9 average value95.
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ALTERNATIN5 CURRENT
T"( o* Priodic ?a&*or#
R%ation()i Bt1n FrGunc" andPriodic Ti#
!ei* +einitin ,!itten as e!i-ic &ime
i$ &'/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
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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
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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
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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
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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.html7/25/2019 Chapter 2 - Inductors, Capacitors and Alternating Current Circuits
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Ea#% 4'alculate the current and power consumed in a single
phase C@5V (' circuit $y a heating element whichhas an impedance of
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Ea#% 2( sinusoidal voltage supply dened as
V"t# 9 855 % cos"\t O :5o
# is connected to a pureresistance of E5 Khms. etermine its impedance andthe value of the current )owing through the circuit.raw the corresponding phasor diagram.So%ution 2
INDUCTI8E REACTANCE
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Introduction
a# Inductive Jeactance is an inductors resistance in an(' circuit.$# 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.
$# The ac voltage, V is
increasing in a positivedirection at its ma%imumrate as it crosses the 3eroreference a%is at aninstant in time given as
5o
.
CAPACITI8E REACTANCE
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Sinusoidal Waveforms for AC Inductance
CAPACITI8E REACTANCE
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Purely Capacitive
V#$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.
CAPACITI8E REACTANCE
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Capacitive eactance a!ainst Fre"uency
RL SERIES AC CIRCUIT
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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
RL SERIES AC CIRCUIT
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The voltage triangle is derived from 1ythagorasQstheorem and is given as
RL SERIES AC CIRCUIT
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#$e Impedance #rian!le
RL SERIES AC CIRCUIT
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%&le 'o( );
A solenoid coil )as a resistance of 3( *)s and an inductance of (.-. %f t)e current
flowing t)roug) t)e coil is aps. #alculate,
a" T)e /oltage of t)e suppl& if t)e fre0uenc& is (-.
b" T)e p)ase angle between t)e /oltage and t)e current.
RL SERIES AC CIRCUIT
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Power #rian!le
RC SERIES AC CIRCUIT
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1" A resistance, and a capacitance,
# in series.
2" T)e current flowing t)roug) t)e
resistance and capacitance, w)ilet)e /oltage is ade up of t)e two
coponent /oltages, Vand V#.
3" T)e resulting /oltage of t)ese two
coponents can be found
at)eaticall& but since /ectors
V and V# are '(o outofp)ase,
t)e& can be added /ectoriall& b&
constructing a /ector diagra.
" T)e indi/idual /ector diagras for
a pure resistance and a pure
capacitance are gi/en as
RC SERIES AC CIRCUIT8ctor Dia!ra# o* t) R(u%tant
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8ctor Dia!ra# o* t) R(u%tant8o%ta!
RC SERIES AC CIRCUIT
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#$e C Impedance #rian!le
RC SERIES AC CIRCUITE %
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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
RC SERIES AC CIRCUIT
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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 ".
RCL SERIES AC CIRCUIT
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Individual Volta!e Vectors
P$asor +ia!ram for a Series *C Circuit
RCL SERIES AC CIRCUIT
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Volta!e #rian!le for a Series *C Circuit
B& substituting t)ese /alues into 5&t)agoras6s e0uation abo/e for t)e
/oltage triangle will gi/e us
RCL SERIES AC CIRCUIT
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#$e Impedance #rian!le for a Series *C Circuit
RCL SERIES AC CIRCUIT
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%&le ;
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.
RCL SERIES AC CIRCUIT
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Solution ;
%nducti/e eactance, 9L.
#apaciti/e eactance, 9#.
#ircuit %pedance, :.
#ircuits #urrent, %.
RCL SERIES AC CIRCUIT
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Solution ;
Voltages across t)e ;eries L# #ircuit, V, VL, V#.
#ircuits 5ower factor and 5)ase Angle,
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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-
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