GE 0106 Basic Engineering II

8/4/2019 GE 0106 Basic Engineering II

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BASIC ENGINEERING II

Part A ELECTRICAL ENGINEERING


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Electrical machines

Definition of mmf, flux and reluctance, leakage flux, fringing, magnetic materials

and B-H relationship. Problems involving simple magnetic circuits. Faradays

laws, induced emfs and inductances, brief idea on Hysteresis and eddy currents.

Working principle, construction and applications of DC machines and AC

machines (1-phase transformers, 3-phase induction motors, single phase

induction motors split phase, capacitor start and capacitor start & run motors).


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AC C i it

Ci it t , O l , Ki ff l . Average and RMS val es,

ncept f phasor representation. RLC serious circuits and series resonance.RLCparallel circuits (includessi pleproblems in C ACcircuits). Introduction

to threephasesystem typesofconnection, relationshipbet een lineandphase

values. (qualitative treatment only).


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Wiring & lighting

Types of wiring, wiring accessories, staircase & corridor wiring, Working and

characteristics of incandescent, fluorescent, SV & MV lamps. Basic principles of

earthing, simple layout of generation , transmission & distribution of power.

TEXT BOOKS

1. Muthusubramanian R, Salaivahanan S, Muraleedharan K A, Basic Electrical, Electronics

and Computer Engineering, Tata McGraw Hill, 1999

2. Mehta VK, Principles of Electronics, S Chand & Co, 1980


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MAGNETIC CIRCUITS

Introduction

A substance, which when suspended freely, points in the direction of north and

south is called a MAGNET. Magnet attracts iron fillings. It is also called a

permanent magnet. A current passing though a conductor also can produce

magnetic effect and it is called as Electromagnet.

A permanent magnet has one north pole and one south pole. The imaginary lines

which travel from north pole to south pole outside the magnet are called

magnetic lines of force. They are drawn by plotting successive directions pointed

out by a small compass needle in the magnetic field. Magnetic lines of forces are

shown in Fig. 1 and they pass through the magnet.

S N

Fig. 1 Magnetic lines of forces


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Fl x, i Force el ctance

e agnetic lines of force in the agnetic fiel is called Fl x. Its nit is

Weber(Wb). Wb = 08 agnetic lines. Fl x isdenotedby . agnetic fl x er

nit cross sectional area is called Fl x density and it is ex ressed in Weber /

etre2. Fl x density isdenotedbyB.

agneto oti e Force (mmf) is the source of roducing flux in the magnetic

circuit. Itcanbeex lainedthrough lectromagnet. WhenacurrentofIampere is

passedthroughacoil of turns, results inammfof I . his I ampereturns

iscalledthemmfand itsunit isampereturns(AT).


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Reluctance is the property of ma netic circuit that opposes the setting of flux.

Reluctance, S =fluxmmf

Its unitisampereturns / weber.

hefollowing tableshowsthesimilarities between ma neticandelectriccircuits.

Sl.

No.Ma neticcircuit Electriccircuit

1 Ma neticflux, webers Electriccurrent, I ampere

Ma neto motiveforce, AT EMF, E volts

3 Reluctance, S AT / Wb Resistance, R ohm

4 =reluctance

mmf Current =resistance

emf


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Leakage flux and Fringing effect

The flux which do not follow the desired path in a magnetic circuit is known as

leakage flux.

Usually we assume that all the flux lines take path of the magnetic medium. But,

practically, some flux lines do not confine to the specified medium. It is because,

to prevent the leakage flux, there is no perfect magnetic insulator. Some flux lines

can pass through air also. All the magnetic flux which complete the desired

magnetic circuit are the useful flux.

To account for the leakage flux, leakage coefficient is introduced. Leakage

coefficient, denoted by is defined as follows.

Leakage coefficient, =

fluxuseful

fluxtotal '!

Usually, leakage factoris greaterthan unity.


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An air gap is often introduced in the magnetic circuit out of necessit . hen

crossing an air gap, the magnetic linesof force have a tendenc to bulge out.

his is because the magnetic lines of force repel each other hen the are

passing throughnon magneticmaterial. hisphenomenon iskno nas fringing.

It issho ninFig.

Fringingeffect increases theeffectiveareaofcrosssectionof theair gapandas

aresult the fluxdensit in theair gapisreduced.

S

Fig. Fringingeffect

Areaat

air gappath

Areaat

ironpath

N


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Probl m involving impl m gn ti ircuit

B fore doing problem involving m gneticcircuit it i necessary to no some

more termsassociated ith the magneticcircuit.

Magnetic field intensity, (also called as Magnetizing force) denoted as H, is the

mmf perunit length of magnetic flu path. hus,

HN

IN

Flu densityis proportional to magnetic field intensity. hus g . heconstant

of proportionalityiscalledpermeability, . hus B H or

B / H

Permeability of vacuum orfreespaceis denoted as 0. Its valueis

4 10-7. Permeability ofany othermedium is given by

0 r

here riscalled therelative permeability of the medium.


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An expression for Reluctance, S can be obtainedasfollo s.

rr aete

Nince;

aN

aN

N N

N!!!!!

Permeance, Pisthe reciprocal ofReluctance.

An iron corecoil ith asmall air gap issho n in Fig.3.

Coil has n turns.

Currentthrough coil I

Mean radius ofmagnetic path Rm

Crosssection ofcoreiscircular ith

diameterd

Length ofairgap g

II

RmRm

Fig.3 Iron corecoil


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Example 5

n iron rod of1cm radius is bent to a ring ofmean diameter3 cmand woun

with 5 turns of wire. ssume the relative permeabilit ofiron as . n air ga

of .1cmiscut across the bent ring. Calculate thecurrent required to produceauseful flux of linesif (a leakageis neglectedand (b) leakage factoris1.1.

Solution

Given r = 1cm; Dm= .3m; N = 5 ; r= ; g = .001m

Flux = 0000 / (108) = 0.2mWb

Leakage i neglected

rea ofcrosssection, a = x10-4 = 0.0003142m2

Reluctance ofair gap = Wb/10x2.532710x4x0.0003142

0.001a

70

!!

gN

Requiredair gap mmf = 0.0002x2.532 x106 = 506.54 T

Length ofiron path = ( x0.3)0.001 = 0.9415 m

Reluctance ofiron path =

T10x2.9 000x10x4x0.0003142

0.9415a

6

r0

!!

N


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Requiredironpat f=0.0002x2. 807x 06=596. AT

Total frequired=506.5 + 596. = 02.68AT

Currentrequired= 02.68 / 250= . 07A

L g ctori 1.1

Asinpreviouscase, requiredairgap f =0.0002x2.5327x 06=506.5 AT

To aintainuseful fluxof0.2 Wbint eairgap,

fluxrequiredint eironpat = . x0.2=0.22 Wb

Requiredironpat f=0.00022x2.9807x 06=655.75 AT

Total frequired=506.5 + 655.75 = 62.29 AT

Currentrequired= 62.29 / 250= .6 92A


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Solution

iven 1 10cm; 2 3 18cm;a1 6.25x10=4

m2

; a2=a3=3x10=4

m2

g=1mm; r=800; =600; 1 =100x10-6 b

Reluctance ofpath 1

Reluctance ofairgap = b10x1.2 3210x4x10x6.2510x1

a6

4

-3

0

!!

N

Length ofiron path =100.1=9.9cm


b/AT10x0.1

00x10xx10x.0.099

a r0

!

!

N

Thus R1=(1.2 32 + 0.15 6)x106=1.4308x106AT / b

MMF1=100x10-6x1.4308x106=143.08AT


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100 x 10-6 Wb 1.2732 x 10 AT/Wb

0. 968x 10 AT/Wb 172.92 AT

0.1 76 x 10 AT/Wb

0. 968x 10 AT/Wb

0 x 10- Wb0 x 10- Wb

Reluctance of path 2


Wb/AT10x0. 968

800x10x4x10x30.18

a6

74r0

!

!

N

Fluxin path 1 will di ide equally; Thus flux= 0 x 10-6 Wb

F2= 0 x 10-6x 0. 968 x 106= 29.8 AT

Since path 2 and path 3 are in parallel, it is required to consider f for only one

of the .

Total F = 1 3.08 + 29.8 = 172.92 ATExciting current required= 172.92 / 600 = 0.2882 A

Electrical equi alent of the agnetic circuit consideredis shown in Fig. .

Fig. Electrical equi alent


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H ax, B ax

-H ax, -B ax

H

B

O

R

PS

M

Q

Hysteresis and eddy current losses

Consider an iron bar hich can be agnetized as sho n in Fig. 6. Magnetizing

force, H can be varied by controlling the current through the coil.Corresponding

values of fluxdensity B can be noted. First the B-H curve ill follo OM sho n if

Fig. 7. o ifHis decreased gradually, B ill not decrease along MO. Insteadit

ill decrease along M . Even hen H is zero, B has a definite value O . This

implies that even on removing the magnetizing force, H, the iron baris not getting

demagnetized completely. The value of O measures the retentivity of the

material.

Fig. 6 Circuit forB-H curve Fig. 7 Hysteresis loop


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To demagnetize the iron bar, the magnetizing force has to be applied in the

reversedirection.Fluxdensity, B becomeszeroat P. Thevalueof H as measured

by OP isknownascoercive force. If H is further increased, thecurve will follow

thepath PQ. By taking H back from Hmax, asimilarcurve QRSM isobtained. It is

seen that B always lagsbehind H. This laggingcharacterof B withrespect to H iscalled hysteresis and the complete loop is called hysteresis loop. Different

magnetic material will havedifferent hysteresisFig.8shows thehysteresis loop

ofcast steel andalloyedsteel.

Fig.8 Hysteresis loopofcast steel andalloyedsteel


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Areaof ysteresis loopgives t e ysteresis lossperunit volumeof t ematerial.About 4% additionofSlican tosteel giverise toreductionin ysteresis looparea

and ence ysteresis loss.

eneveraconductingmaterial cuts t emagnetic flux (armaturecorein t ecas

of rotating machines) an emf is induced in t e core. This emf sets up largecurrent t rough t esolidmass.Suchcurrent isknownaseddycurrent.Flow o

eddycurrent resultsineddycurrent loss.

Theeddycurrent loss isproportional tos uareof t e t icknessof t ematerial.

This loss can be minimi ed by using a laminated core, w ich offers ighresistance fort e flow ofeddycurrent.


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A B

K

Faraday s Laws ofElectroma netic Induction

When a current flows in a conductor, ma netic field is produced. he re erse

phenomenon whereby an Electro Moti e Force (EMF) and hence current isproduced in an electric circuit by some action of ma netic field, is called

electroma netic induction.

Considerthe setup showninFi . 9.

A

G

Fi . 9 Static induced emf


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When the switch, K is closed from the open position, there will be induced

voltageand hencecurrent in coil asindicated by thegalvanometer .When the

key is opened from theclosed position, thecurrent flowwill be in the opposite

direction. hisillustrates the production ofstaticinducedemf.

onsider thesetup shown in Fig.10.

SA

Fig.10 ynamicinducedemf

G


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Induced emf

An emf is induced in a coil or conductor whenever there is a change influxlinkages. he change in flux linkages canoccurin two ways.

(i) he coil is stationary and the magnetic field is changing. Resulting

induced emf is known as static induced emf. ransformerworks on this

principle.(ii) he conductor is moved in a stationary magnetic field in such a way

that there is change in flux linkage. Resulting induced emf is known as

dynamic induced emf. eneratorworks on this principle.

Static induced emf

In this case, the coil is held stationary and the magnetic field is varied. he self

induced emfmaybe self inducedormutually induced.


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A B

Two coils are wound over a magnetic specimen. Coil A is energi ed using a

battery. Ifswitch is initiallyclosed, thenasteadycurrent of Iamperewill flow

through thecoil A. It producesa fluxof Wb. Let usassume that theentire flux

linkscoils A and B. When theswitch issuddenlyopened, thecurrent reduces toeroand the flux linkingboth thecoilsbecomes ero. AsperFaradays law, emf

is induced in both the coils A and B. Such emfs are known as static induced

emfs. Static induced emf can be classified into two categories, namely self

inducedemfand mutually inducedemf.

Fig. 9Static inducedemf

A

G


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Se ce em

Ifasinglecoil carriesacurrent, flux will beset up in it. If thecurrent changes,

the flux will change. hischange in flux will inducean emf in thecoil. hiskind oemf isknown asself induced emf. In other words, self induced emf is theemf

induced in a circuit when the ma netic flux linking it changes because of the

current changes in thesamecircuit.

he ma nitude of thisself induced emfedtd

M t a y ce em

Mutually induced emf is the emf induced in one circuit due to change of flu

linking it, the flux being produced by thecurrent in anothercircuit.

eferring to Fig. 9, when achange in current though coil A occurs, we find the

flux linking coil changes. ence, an emf is induced in coil and it iscalled as

mutually induced emf.


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SA

B

D namicinducedemf

Consider the experiment set shown in Fi . 10.The magnetic poles, produce a

stationar fluxdensit ofB Wb. / m2. Let theconductor lengthbe meters.The

conductoris movedat ri ht angle tohe field. Let thedistance movedin dtsecondbedxmeters.

Areaswept b theconductorindtsec. = dx m2

Magnetic fluxcut b theconductor= B dx Wb.

G

Fi .10D namicinducedemf


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Takingtheconductorhasoneturn, corresponding

flux linkage, B dx WbTurn

Rateofchangeofflux linkage B dtdx

AccordingtoFarada s Law, this isthe inducedemf, e intheconductor.

Thus inducedemf, e B olts

where linear elocit dtdx

Lettheconductorbe mo edwith elocit m / sec. inan inclineddirection,

makinganangle tothedirectionoffield. Then

Inducedemf, e B sin olts

This isthebasicprincipleofworkingofagenerator.


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S N

A

B

FI

Forceoncurrentcarry ngconductor

Consi er the setup shown in Fi . 11. When a current ofI ampere fows in the

conductorfromAtoB itwi experienceaforce F gi enby

F= B I Newton

Thisre ationistrueiftheconductorisatrightangletothemagneticfield. Incase

if the conductor is an inclined direction making an angle tothe direction of

field, then

F= B I sin Newton

Thisisthebasicprincipleofworking ofamotor.

Fig.11Forceoncurrentcarrying conductor


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I

Self inductance,

Self induct nce of a coil, L is the rate of change of flux linkages with respectto

the currentin itIts

un

it isen

r .Thus

=Id

d =Id

d enr

Equation forself inductance

onsideramagneticcircuit shown in ig. 2.

ig. 2 Self inductance


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Wit al tation

a neti in force, H =NIN AT / m

l x en it , B= rH = r(N

IN)W . / m

a neticfl x, = r( NIN

)a W .

l x linka e= N = r(N

I2

)a W . T rn

Selfin uctance, = Idd

=N I

= N

a 2r0

= a/( r0

2

N

=el ctance

N2


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Expressionforselfinducede finter sofselfinductance

The agnitudeofselfinducede f, e=Ndtd

Thusselfinducede f, e=Ndtdx

d

d I

I

=LI

Mutual inductance

Whent erearetwoare orecoils, t erewill be utual inductancebetweenany

twocoils. Ift etwocoilsarefarapart, t ent et erewill notbeanyco onflux

lin ingbot t ecoilsand ence utual inductancewill be zero.


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21

21NN

I1I2

Consider two aircorecoils havingself inductancesL1and L2 that arecloser to

each otherasshown in Fig. 12. When current passes through coil 1, flux11

is

produced in coil 1. Onl a part of this flux linkswith coil 1and theremaining flux

links both thecoils1and 2. Generall , the flux linking both thecoils is useful and

it iscalled mutual fluxand represented b21

. The otherpart of the flux iscalled

leakage fluxrepresented b1. When thecoil 2carriescurrent, flux produced in

it is22

and leakage flux is2and themutual flux is

12. Fluxes

1,

21,

2

and12

areshown in Fig. 13.

12

Fig. 13Two coils in proximit


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Example

coil ofresistance150 isplacedinamagnetic fluxof0.1mWb.Thehas 500

turnsandagalvanometerof 450 resistanceisconnectedinserieswithit.The

coil ismoved from thegiven field toanotherfieldof0.3mWb. In0.1sec.Find the

averageinducedemfand theaveragecurrent through thecoil.

Solution

iven Rc=150 ; 1 =0.1x10-3Wb.; N = 500 turns; Rg= 450 ; =0.3x10-3Wb.;

t =0.1sec.

Inducedemf, e= Ndtd = 500x

0.110x0.110x(0.3 33

= 500x x10-3=1.0Volt

Current, I=inducedemf / total resistance=1.0 / (150 + 450) =0.00166


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Example8conductor oflength 100cmmovesat rightangleto a uniformmagneticfield of

fluxdensity1.5 b. / m2with avelocity of30m / sec. Calculatetheemfinducedin

it. Findalso thevalue of inducedemfwhen theconductormovesatan angle of

600to thedirection ofthemagneticfield.

Solution

Given = 1.0m; B = 1.5 b. / m2; v = 30m / sec.; = 600

Inducedemf, e = B v = 1.5x1.0x30 = 45 V

ith = 600. Inducedemf, e = B vsin = 45xsin 600 = 38.9 11 V


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Example9

Aconductorof10cm long lies perpendicular to amagnetic field ofstrength 1000

AT / m., Find the forceacting on it when it carriesacurrent of60A.

Solution

Given =0.1m; H=1000AT / m; I=60A

Flux density, B= 0H= x10-7x1000=0.0012 7Wb. / m2

Force, F =B I=0.0012 7x0.1x60=0.007 Newton


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Example 10

n air cored toroidal coil has 80 turns, a mean length of 30 cm and a cross-

sectional area of cm2. Calculate (a) the inductance of the coil and (b) the

average induced emf, if a current of is reversed in 60 msec.

Solution

Given N = 80 turns; = 0.3 m; a = x 10- m2; dI= 8 ; dt = 60 x 10-3sec.

Inductance, L= N2 / Reluctance

Reluctance, S= / ( a 0) = 97 10x0.47710x4x10x50.3

!

Inductance, L= 0. 8250x0. 8250x0. 775

80 392

!!

Induced e f, e =L 0.0 33x 008

x0x0. 825dtd

3

3!!

I


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Example11

current of5 when flowing through acoil of1000 turnsestablishesa flux of

0.3m Wb. Determine theself inductance if thecoil.

Solution

Gi en I=5 ; N=1000 turns; =0.3x10-3 Wb.;

Self inductance, L= 0.05

10x0.3x1000

dd 3

!!

I


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Example12

A coil has a self inductance of 30 mH. Calculate the emf in the coil w en the

current in thecoil (a) increasesat the rate of300 A / sec. (b) raises from 0 to 10 A

in 0.06sec.

Solution

Given L=30x10-3 H

(a) Induced emf, e=L 300x10x30dtd 3

!!

I

(b) Induced emf, e=Ldd

!!

I

Example 13


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Example13

Thenumberof turns inacoil is250. Whenacurrent of2 flows in thiscoil, theflux in thecoil is0.3mWb. When thiscurrent is reduced tozero in2msec., thevoltage induced inanothercoil is63.75 V. If thecoefficient ofcouplingbetweenthe twocoils is0.75, find theself inductancesof the twocoils, mutual inductanceand thenumberof turns in thesecondcoil.

Solution

ivenN1=250;I1=2 ; 1 =0.3x10-3Wb.;dI1=2 ; dt1=2msec;e2=63.75 V;

k=0.75

Self inductance, L1= H0.0375

2

10x0.3x250

d

d 3!!

I

Inducedemf incoil 2, e2=M2

2M

d

d

1

!!1

Thusmutual inductance, M=63.75mH

SinceM=k

0.063752

=0.752

x0.0375x 2Thusself inductanceofcoil 2, 2=0. 927H

Flux

k!2

= 0.75x0.3x10-3Wb=0.225x10-3Wb

lso, e2=N2x 63.7510x2

10x0.225xN

dt

d3

3

22

!!

ThusN2=567


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Working principle, construction andapplications of enerator

he dc generator is rotating electrical achine which converts echanical

energy into electrical energy. hegenerator is usually driven by astea turbine

orwaterturbinewhich iscalledas pri e over.

he dc generator operates on the principle based on the Faradays Law of

electro agnetic induction. he generator should have (i) agnetic field (ii)

conductors capable of carrying current (iii) ove ent of conductors in the

agnetic field. ecessary agnetic field is produced by field coil. he set of

conductorsiscalledthear ature.

Fig.14 Principle ofoperation of enerator


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The oltage inducedin the coil will be as shown in Fig. 1 .

Fig. 1 EMF inducedin an armature coil


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YYY

YYY

Z

ZZ

Z

ZZ

Y

YY

epending on ho the rmature and Field inding are connected, e ha e

different type ofdcgenerator .Theyare ho ninFig.18.

Fig.18 (b) Shunt generatorFig.18 (a) Serie generator

Fig.18 (c) Short hunt compoundedgenerator

Fig.18 (d) Long hunt compoundedgenerator

L

O

A

D

G

L

O

A

D

G

G

L

O

A

D

L

O

A

D

AA

G

A


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Application o dcgenerators

Shunt eneratorsareuse insuppl ingnearl constant loa s. The areuse fo

char ing atteriesandsuppl ingthefiel sofs nchronousmachines.

Se

ries

ene

rato

rs

a

re

use

to

ooste

rs

fo

ra

ing

vo

lta e

to

transm

iss

ion

lines

tocompensateforthe line rop.

umulativecompound eneratorsareuse for rives hich requireconstant c

volta esuppl .

ifferential compound eneratorsareuse inarc el ing.


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Working principle, construction andapplications of enerator

Wheneveracurrent carryingconductoriskept in astationary magnetic field, an

electromotive force is produced. This force is exerted on the conductor and

henceis movedaway from the field.Thisis the principle usedin dc motors.

onstruction ofdc motorisexactlysimilar to dcgenerator.

In a dc motor, both the armature and the field windings are connected to a dc

supply. Thus, we have current carrying armature conductors placed in a

stationary magnetic field. ue to electromagnetic torqueexerted on thearmature

conductors, the armature starts revolving. Thus, electrical energy is convertedinto mechanical energyin thearmature.


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When the armature is in motion, we have revolving conductors in a stationarymagnetic field. As per Faradays Law of electromagnetic induction, an emf is

induced in thearmatureconductors. As perLenzs law, this induced emf opposes

the voltage applied to the armature. Hence it is called back emf. There will be

small voltage drop due to armature resistance. Thus, theapplied voltagehas to

overcome the backemf in addition to supplying thearmature voltage drop. The

input power is used to produce necessary tor ue for thecontinuous rotation of

thearmature.

Depending on how the Armature and Field windings are connected, we have

different types of dcmotors. They areshown in Fig. 19.


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YY

upply

olta e AA

A

Y

-

+

A

AA

+

+

upplyolta e

YYY

AA

A

-

upplyolta e

Y

YY

+

-

upplyolta e

MM

Fi .19 (a) erie motor Fi .19 (a) unt motor

G

Fi .19 (c) ort unt compoun e motor

AA

G

A

Fi .19 ( ) on unt compoun e motor


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Appli ationofdcmotors

DC series motorsareused inelectric trains, cranes, hoists, conveyorsetc. where

highstarting torque isrequired.

Shunt motors are used where the speed has to remain constant under loaded

condition.

Compound motorsareused fordrivingheavy tools for intermittent heavy loads

suchasrolling mills, printing machinesetc.


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EMF induced in primary side E1 = N1dd

Since same flux is linking bo h he primary and secondary coils

EMF induced in primary side E2 = N2dd

Vol age ra io2

1

2

1

N

N

EE

!

Since losses in he ransformer are very less

E1I1 = E2I2

hen the current ratio1

2

1

2

2

1

EE

!!

I

I


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The transformer mainly consists of a good magnetic core and primary and

secondary windings.

The transformercore isgenerally laminatedand is madeout ofagood magnetic

material such as transformer steel or silicon steel. Such a material has high

relati epermeabilityand low hysteresis loss. Thereare two typesof transformer

cores. They are known as Core Type and Shell type. In core type, L shaped

stampingsasshown inFig. 21areused. Onecore type transformer isshown in

Fig. 22.

Fig. 21 L typestampings


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Laminate core of a s ell type transformer is s own inFi . 23. In t is E type

an I type laminations are use .Fi . 24 s ows a s ell type transformer.

Fi . 23 Laminate core of s ell type transformer


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Workingprinciple, construction and applications of 3- phase inductionmotor

When a three phase alanced oltage is applied to a three phase alanced

winding, a rotating magnetic field is produced. his field has a constant

magnitude androtates in space with a constant speed. If a stationar conductor

is placed in this field, an emfwill e induced in it. B creating a closedpath for

the current to flow, an electromagnetic torque can e exertedon the conductor.

hus the conductoris put inrotation.

he important parts of a three phase induction motor are schematicall

represented in Fig. 25. Broadl classified, the are stator and rotor which are

descri ed elow.


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Stator is t e stationary part oft e motor. The statorcore consist ofhigh grade,low loss electrical sheet-steel stampings assembled in t e frame. Slots are

provided on t e inner periphery of t e stator to accommodate t e stator

conductors. equired numbers of stator conductors are housed in t e slots.

These conductors are arranged to form a balanced t ree phase winding. The

statorwinding may be connectedin starordelta.

otor is t e rotating part oft e induction motor. The airgap between t e stator

and rotor is as minimum as possible. The rotor is also in t e form of slotted

cylindrical structure. There are to types ofrotors, namely Squirrel age rotorand

Slip-ring or oundrotor.

f


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Fig.26 ho thecon tructionofa uirrel cage rotor.

In thi type, each rotor lot accommodate a rodorbarmadeofgoodconductingmaterial.The e rotorbar are hort circuitedat bothend bymean ofend ring

madeof the amemetal a that of rotorconductor .Thu the rotorcircuit form a

clo edpath foranycurrent to flo through.

Fig.26Squirrel cage rotorof threepha einduction


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Fig.2 s ows the rotor ofslip-ring induction motor. In thiscaseconductorsar

housed in rotorslots. heseconductorsareconnected to form astarconnected

balanced three phasewinding. he rotoriswound to givesame number of poles

as thestator. he threeends of the rotorwindingareconnected to the brus e riding over the slip-rings. Slip-rings are s ort circuited at the time of starting.

External resistances can be connected to control the speed of the motor.

Althoug thewound rotor motorcosts more than asquirrel cage motor, it has the

features ofcontrolling the torqueand thespeed.

Fig.2 Rotor ofslip-ringinduction motor

Startingresistanceandspeedcontroller


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A three phase balanced voltage is applied across the three phase balanced stator

winding. A rotatingmagnetic field is produced. This magnetic field completes itspath through the stator, the airgap and the rotor. The rotorconductors, which are

stationar at the time of starting, are linkedb time var ing start magnetic field.

Therefore emf is induced in the rotorconductors. Since the rotorcircuit forms a

closedpath, rotorcurrent is circulated. Thus the current carr ing conductors are

placed in a rotatingmagnetic field. Hence an electromotive force is exertedon the

rotorconductors and the rotorstarts rotating.

According to Lenzs law, the nature of the induced current is tooppose the cause

producing it. Here the cause is the relative motionbetween the rotorconductors

and the rotatingmagnetic field. Hence the rotorrotates in the same direction as

hat of the rotatingmagnetic field.


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In practice, the rotor speed never equals to the speed of the rotating magnetic

field. The difference in the two speeds is called the slip. The current drawn by the

stator gets adjusted according to the load on the motor.

Three phase induction motors are used in industry forvery many purposes. Theyare used in lathes, drilling machines, agricultural and industrial pumps,

compressors and industrial drives. They are also used in lifts, crane and

conveyors.


72/106

Workingprinciple, construction and applications of single phase induction otor

Single phase induction otors are used in variety of applications at ho e,factory, office andbusiness establish ents. Single phase induction otor is not

self starting. dditional arrange ent has tobe ade to ake it self-starting. his

could be achieved by using t o indings, ain inding and starting inding,

ith large phased

ifference bet een the currents carried

by the . his kind

ofsplit-phase otor produces a revolving flux and hence akes the otor self

starting. ependingon the circuit ele ent connected in series ith the starting

inding, the split-phase otors are classified into

(i) esistance-start induction otor(ii) apacitance-start induction otor

(iii) apacitance-start-and-run otor


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Resi tance-startinduction otor

Mainwinding

otor

Startingwinding

Singlephasea.c.supply

Is

S

I

i .28 esistancestartinduction otor


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Resistancestart inductionmotor is shown in ig. 28. he startingwindinghas a

high resistance connecte in series with it. he current flowing through it is given

Is. he centrifugal switchS isconnects the startingwindingwhen the motorspee reaches 80%of full loa spee . he mainwindinghas low resistance and

high reactance and it carries current Im. urrent in starting winding is Is. he

torque evelope the motor is proportional to sinwhere is the angle

etween Im and Is as shown in ig. 29. orobtaininghigh torque, shoul be as

high as possible. Here is the power factor angle.

V

Im

Is

Iig. 29Phasor iagramof Resistance start inductionmotor

Ca acit r-starti ducti t r


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In thecapacitor-start induction motor, acapacitorisconnected in serieswith the

starting winding asshown in Fig. 30.Im Main winding

Starting winding

C

Rotor

Single phasea.c. suppl

Is

S

Fig. 30 Capacitorstart-induction motor

Is

I

Fig. 31Phasordiagram ofcapacitor-start induction motor

V

Im


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The phasor diagram of capacitor-start induction motor is shown in Fig. 31.

The following are the advantages ofcapacitor-start induction motor(i) Increase in starting torque(ii) etter starting power factor

Ca acitor-start-a d- runmotor

Capacitor-start-and-run motor is similar to that of the capacitor-start motorexcept that the capacitor in the starting winding circuit remains there through out

the operation of the motor. The advantages of this t pe of motor are

(i) Low noise in the motorwhile running

(ii) igher power factor(iii) igher efficienc

(iv) Improved over-load capacit


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ELECTRIC CIRCUITS

Electric circuits are broadl classified as Direct Current (D C ) circuits and


78/106

Electric circuits are broadl classified as Direct Current (D.C.) circuits and

Alternating Current (A.C.) circuits. he following are the various ele ents that

for electriccircuits.

D.C. Circuits A.C. Circuits

Ele ents Representation Ele ents Representation

oltagesource oltagesource

Current source Current source

Resistor Resistor

Inductor

Capacitor

+- +-~


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I

V

First we shall discuss about the anal sis of DC circuit. The voltage across an

element is denoted as Eor V. The current through the element is I.

Conductoris used to carr current. When a voltage is applied across a conductor,current flows through the conductor. If the applied voltage is increased, the

current also increases. The voltage current relationship is shown inFig. 1.

It is seen that Ig V. Thus we can write

I = G V (1)

where G is called the conductance of the conductor.

Fig. 1 Voltage current relationship

Very often we are more interested on RESISTANCE R of the conductor than the


80/106

Very often wearemoreinterested on RESISTANCE, R oftheconductor, than the

conductance of the conductor. Resistance is the opposing property of the

conductoranditisthe reciprocal oftheconductance, Thus

R = or =R

(2)

Therefore

I = RV (3)

Theabove relationship isknown as OHMs law.Thus Ohm lawcan bestatedas

the current flows through a conductor is the ratio of the voltage acro ss the

conductorandits resistance. Ohms lawcan also bewritten as

V = RI (4)

R =I

V (5)

The resistance of a conductor is directl proportional to its length, inversel


81/106

proportional to its area of cross section. It also depends on the material of the

conductor. Thus

R = N (6)

where is called the specific resistance of the material b which the conductor is

made of. The unit of the resistance is Ohm and is represented as . Resistance of

a conductor depends on the temperature also. The power consumed b the

resistor is givenb

P =VI ( )

hen the voltage is in volt and the current is in ampere, power will be in watt.

lternate expression forpower consumedb the resistors are givenbelow.

P = R IxI=I2 R (8)

P =Vx = (9)


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KI FFs WS

here are two Kirchhoffs laws. he first one is calle Kirchhoffs current law,

K andthe secondone is Kirchhoffs voltage law, K . Kirchhoffs current law

eals withthe element currents meeting at a junction, which is a meetingpointof

two are more elements. Kirchhoffs voltage law eals with element voltages in a

close loop also calle as close circuit.

irchhoff current law


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I4

I

I2

I1

P

irchhoff current law tate that thealgebraic umofelement current meeting

at a junctioni zero.

Con idera junctionP wherein fourelement , carryingcurrent I1, I2, I andI4, aremeetinga howninFig.2.

Note that current I1andI4are flowingout from the junction while thecurrent I2

andI are flowinginto the junction. According to CL,

I1 I2 - I + I4= (10)

Theabo eequationcanbe rearrangeda

I1 + I4 =I2 + I (11)

From equation (11), CL can al o tated a at a junction, the um of element

current that flow out i equal to the umofelement current that flow in.

Fig.2Current meetingat a junction

irchhoffs voltage law


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V3

V1+ -

V2

V4

+

+

+

-

-

-

irchhoffs voltage law states that the algebraic sum of element voltages around

a closed loop is zero.

Consider a closed loop in a circuit wherein four elements with voltages V1, V2

, V3and V4, are present as shown in Fig. 3.

Assigning positive sign for voltage drop and negative sign for voltage rise, when

the loop is traced in clockwise direction, according to VL

V1 - V2 - V3+ V4=0 (12)

The above equation can be rearranged as

V1 + V4 = V2+ V3 (13)

From equation (13), VL can also stated as, in a closed loop, the sum of voltage

drops is equal to the sum of voltage rises in that loop.

Fig. 3 Voltages in a closed loop

Resistorsconnecte inseries


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R3R2R1

+

+++

-

--- V3V2V1

I

E

Tworesistorsaresai tobeconnecte inserieswhen there isonlyonecommo

point between them andnoother element is connecte in that commonpoint.

Resistorsconnecte inseriescarrysamecurrent. Consi erthreeresisters R1, R2and R3 connecte in series as shown in i . 4. With the supply volta e of E,

volta esacross the threeresistorsare V1, V2and V3.

AsperOhms law

V1= R1I

V2= R2I (14)

V3= R3I

i . 4 Resistorsconnecte inseries


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Voltagedivision rule


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++ V1 - -V2

R2R1 I

+ -E

Consider two resistorsconnectedin series. hen

V1 = R1I

V2 = R2I

E = (R1+R2) Iand hence I= E / (R1+R2)

otal voltage of E isdroppedin two resistors.Voltageacross the resistorsare given by

V1 =21

1

RRR

E and (20)

V2 =2

2

RR

R

E (2 )

Resistors connectedin parallel

T i t id t b t d i ll l h b th t d


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(22)

I2

R1

R2

+ -E

I1

I

A

Two resistors are said to be connected in parallel when both are connected

across same pair of nodes. Voltages across resistors connectedin parallel will be

equal.

Consider two resistors R1 andR2 connectedin parallel as shown in Fig.5.

As per Ohms law,

I1=1

I2=2

Fig.5 esistors connectedin parallel

I1 1


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E

I1

s per h s law

I1 =1

E

I2 =2

E

I2

1

2

+ -

I

Appl ing K Lat nodeA

I = I1+I2 = E )11

21

=

eR(23)

Thus forthecircuit shown in Fig. 5

I =

eR(24)

where is the circuit voltage, I is the circuit current and Re is thee uivalent

resistance. ere

2

qe RRR

(25)


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21qe R

1

R

1

R

1!

(25)

From the above21

21

qe RR

RR

R

1 !

hus21

21

RR

RRR

!qe (26)

When n numbers of resistors are connected in parallel, generalizing e . (25), Re

can be obtained from

n2e R................

RRR (2 )

Currentdivisionrule


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(29)

(30)

I2

R1

R2

+ -E

I1

I

A

Referring to Fig. 5, it is noticed the total current gets divided as I1 and I2. The

branchcurrentsareobtainedasfollows.

Fromeq.(23)

E = I21

21

RRRR

Substitutingtheaboveineq.(22)

I1 = I

1

I2 = I2

Fig. 5 Resistorsconnectedinparallel

Exa ple


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30200

+

+++

-

--- V3V2V

I

00V

Exa ple

Threeresistors 0, 20and30areconnected inseriesacross 00Vsupply.

Findthevoltageacrosseachresistor.

Solution

Current I= 00 / ( 0 + 20 + 30) = .6667A

Voltageacross 0= 0x .6667 = 6.67V

Voltageacross20=20x .6667 = 33.33V

Voltageacross30=30x .6667 = 50V

Example 2


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I2

4

6

I1

30A

Two resistors of4 and6 are connected inparallel. If the suppl current is 30A,

find the current in each resistor.

Solution

Using the current division rule

urrent through4 = A1830x64

6!

urrent through6 = A1230x64

4!

Example3


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Fourresistors of2 ohms, 3 ohms, 4 ohmsand 5 ohmsrespectivelyareconnected

in parallel. What voltagemust beapplied to the group in orderthatthetotal powe

of100W isabsorbed?

Solution

Let Tbethetotal equivalentresistor. Then

12015

4120 24304060514131211T!

!!

esistance T= 0. 2154120

!

Let E bethesupply voltage. Then total currenttaken = E / 0. 2 A

Thus 1000. 2x)0. 2

E( 2 ! and hence E2=100x0. 2= . 2

equired voltage= . 2 2. 2 !


95/106

R1

230V

12A

16A

R1

R2

230V

12A

4A

-+

-+

Example4

W ena resistor isplace acrossa230Vsupply, t ecurrent is12A. W at ist evalueoft e resistort at must eplace inparallel, to increaset e loa to16A

Solution

o maket e loa current16A, currentt rou t esecon resistor=1612=4A

Valueofsecon resistorR2=230/4= 7.

Example5


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50

100

E

7.2A

-

A50resistor is in parallel with a100resistor. Thecurrent in 50resistor is

7.2A. What is thevalue of third resistor to beadded in parallel to make the line

current as12.1A?Solution

Suppl voltage E = 50x7.2 = 360V

Current through 100 = 360/100 = 3.6AWhen the linecurrent is12.1A, current through third resistor= 12.1 (7.2 3.6)

= 1.3A

Value of third resistor= 360/1.3 = 276.9230

Example 6


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3.6 4.56

RT= 6

R1

p

A resistor of 3.6 ohms is connected in series with another of 4.56 ohms.Wha

resistance must be placed across 3.6 ohms, so that the total resistance of th

circuit shall be 6 ohms?

Solution

3.6 R1= 6 4.56 = 1.44

Thus 1.5R3.6

.5;.4

1R

R3.6Therefore1.44;

R3.6Rx3.6

11

1

1

1!!!!

Requiredresistance R1= 3.6/1.5= 2.4

Example


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8

+ -22 V

12 R

A resistance R is connected in series with a parallel circuit comprising two

resistors12 and8 respectively.Total powerdissipatedin thecircuitis 0 W

when theappliedvoltageis22 V. Calculatethevalue oftheresistorR.

Solution

Total currenttaken = 0 / 22=3.1818 A

Equivalent of12 8 =96/20= 4.8

Voltageacross parallel combination = 4.8x3.1818=15.2 26 V

VoltageacrossresistorR=2215.2 26=6. 2 4 V

Value ofresistorR=6. 2 4/3.1818=2.1143

Example8

The resistors 12 and 6 are connected in parallel and this combination is


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Voltageacross6 = 6 V

12 6.25

6

0.25 E

The resistors 12 and 6 are connected in parallel and this combination is

connectedinserieswitha6.25 resistanceandabatter whichhasaninternal

resistanceof 0.25 . Determine theemfof thebatter if thepotential differenc

across6 resistanceis6 V.

Solution

Currentin6 = 6/6 = 1 A

Currentin12 = 6/12 = 0.5 A

Thereforecurrentin25 = 1.0+0.5 = 1.5 AVoltagedropin6.25 and0.25 puttogether= 6.5 x1.5 = 9. 5 V

Total voltagedropinresistors = 9. 5 +6 = 15. 5 V

AsperKVL, inaclosed loop, voltagedropisequal tovoltagerise.

Thereforebatter emfE = 15. 5 V

Example9

A circuit consist of three resistors 3 4 and 6 in parallel and a fourth


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44

6

6 12V

3

A circuit consist of three resistors 3 , 4 and 6 in parallel and a fourth

resistorof 4 in series. A batter of12Vand an internal resistance of6 is

connected acrossthecircuit. Find thetotal current in thecircuitand theterminal

voltageacrossthe batter .

Solution

4 6 = 24/10=2.41.4 3 = 7.2/ .4=1.3333

Total circuitresistance=4+6+1.3333=11.3333

Circuitcurrent=12/11.3333=1.0 88A

Voltage drop in internal resistance=6x1.0 88=6.3 28V

Terminal voltageacrossthe batter =126.3 28= .6472V

Exampl 10

n l tri al n t or i arrang d assho n. Find i)the urrentin bran h F ii)


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F ED

CB

24

13

1

22

18

14

11

5

F ED

CB

24

13

1

22

18 14

11

14

F E

CB

24

13

1

22

18 7

11

the po erabsorbedin bran hBEand iii) potential differen eacrossthe branch

CD.

Solution

ariousstages ofreduction aresho n.

1

2


102/106

FE

24V

1

22

187

F E

CBA

24V

13

1

22

18

18

2

3


103/106


104/106

Currentin branch AF =24/12=2Afro F to A

sing currentdivision rulecurrentin 13in Fig. 4=1A

eferring Fig.3, currentin branch E =0.5A

Powerabsorbedin branch E =0.52x18= 4.5W

Voltageacross E =0.5x18=9VCurrentin 11resistorin Fig.1=9/18=0.5A

VoltageacrossCE in Fig.1=9 11x0.5)=3.5V

eferring Fig. given in the proble , using voltagedivision rule, voltageacrossin

branch CD= V1.253.5x145 !

Example 11


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6 3

25V 45V4

6 3

25V 45V4

I1

I2

I1-I2

A

B C D

Using Kirchhoffs laws, find the current in various resistors in the circuit shown.

Solution

Let the current suppliedb the batteries be I1 andI2

6 3B C D


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A

2 4 4

I1

I2

I1-I2

Using CL, currentin4 resistor=I1 I2

Consideringthe loopABCA, L yields

6 I1 + 4(I1 I2)= 2

Forthe loopCDAC, L yields

3 I2 + 4(I2 I1)=4

Thus 10 I1 -4I2 = 2

-4I1 + 7 I2=4

On solvingthe above I1= 6. 74A; I2 = 10.18 2 A

Currentin4 resistor=I1 I2 = 6. 74 10.18 2 =- 3.6112 A

GE 0106 Basic Engineering II

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Transcript of GE 0106 Basic Engineering II