5 EM Induction

8/2/2019 5 EM Induction

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SF027 1

UNIT 7: ELECTROMAGNETICUNIT 7: ELECTROMAGNETIC

INDUCTIONINDUCTION

DefinitionDefinition is defined as the productionis defined as the production

of an inducedof an induced e.m.fe.m.f. in a conductor/coil. in a conductor/coil

whenever the magnetic flux through thewhenever the magnetic flux through the

conductor/coil changes.conductor/coil changes.

SF027 2Fig. 7.1bFig. 7.1b

Fig. 7.1aFig. 7.1a

Consider some experiments were conducted by Michael Faraday thatled to the discovery of the Faradays law of induction as shown infigures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e.

7.1 The Phenomenon of Electromagnetic

Induction

NNSS

No movementNo movement

0v =

Move towards the coilMove towards the coil

v

I


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SF027 3

NN

No movementNo movement

0v =

SS

Fig. 7.1dFig. 7.1d

Fig. 7.1cFig. 7.1c

Move away from the coilMove away from the coil

v

I

SF027 4

From the experiments :

When the bar magnet is stationary, the galvanometer not show any

deflection (no current flows in the coil). When the bar magnet is moved relatively towards the coil, the

galvanometer shows a momentary deflection to the right. When thebar magnet is moved relatively away from the coil, thegalvanometer is seen to deflect in the opposite direction (Fig.7.1d).Therefore when there is any relative motion between the coil andthe bar magnet , the current known as induced current will flowmomentarily through the galvanometer. This current due to aninduced e.m.f across the coil.

Fig. 7.1eFig. 7.1e

NN SSMove towards the coilMove towards the coil

v

I


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SF027 5

Conclusion :

When the magnetic flux through a coil changes (magnetic field linesbeen cut) thus the induced e.m.f. will exist across the coil.

The magnitude of the induced e.m.f. depends on the speed of therelative motion where when

7.2.1 Faradays law of induction

States the magnitude of the inducedthe magnitude of the induced e.m.fe.m.f. is proportional to the. is proportional to therate of change of the magnetic flux.rate of change of the magnetic flux.

Mathematically,

The negative sign indicates that the direction of induced e.m.f. alwaysoppose the change of magnetic flux producing it (Lenzs law).

vv increaseincrease inducedinduced e.m.fe.m.f. also increase. also increase

vvdecreasedecrease inducedinduced e.m.fe.m.f. also decrease. also decrease

vv is proportional to the inducedis proportional to the induced e.m.fe.m.f..

7.2 Faradays law and Lenzs law

dt

dB

where

fluxmagneticofchange:Bd

timeofchange:dte.m.f.induced:

ordt

dB= (7.2a)(7.2a)

SF027 6

For a coil of N turns, eq. (7.2a) can be written as

Since

From the definition of magnetic flux,

Note :

if the coil is connected in series to a resistor of resistance R and theinduced e.m.fexist in the coil as shown in figure 7.2a.

dt

N

if )( =

, then eq. (7.2b) can be written asifB d =

fluxmagneticfinal:ffluxmagneticinitial:i

where

dt

dN B= (7.2b)(7.2b)

cosBAB =

dtBAd )cos( =

then eq. (7.2a) also can be written as

Therefore the induced currentIisgiven by

Fig. 7.2aFig. 7.2a RII

IR=dt

dB= and

dt

dIR B=


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SF027 7

To calculate the magnitude of induced e.m.f., the negative sign canbe ignored.

If the coil has N turns, then each of turns will have a magnetic flux,

B ofBAcos through it, therefore the magnetic flux linkagemagnetic flux linkage(refer to the combined amount of flux through all the turns) is givenby

Example 1 :

A rectangular coil of sides 10 cm x 5.0 cm is placed between N and S

poles with the plane of the coil parallel to the magnetic field as shown

in figure below.

If the coil is turned by 90 about its rotation axis and the magnitude ofmagnetic flux density is 1.0 T, find the change in the magnetic flux

through the coil.

BN=linkagefluxmagnetic

SSNNPP

QQRR

SS

I I

SF027 8

Solution:A=(10x10-2)(5.0x10-2)=50x10-4 m2,B=1.0 T

Initially,

Finally,

Therefore the change in magnetic flux through the coil is

From the figure, =90 thus theinitial magnetic flux through the coil is

BAi cos=0i =

Ar

Br

Br

Ar

ifB =Wb10x50 4B

=

From the figure, =0 thus the finalmagnetic flux through the coil is

BAf cos=

Wb10x50 4f =


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SF027 9

Example 2 :

The magnetic flux passing through a coil of 1000 turns is increased

quickly but steadily at rate of 2.0 x 10-2 Wb s-1. Calculate the induced

e.m.f. in the coil.

Solution:N=1000 turns,

By applying the Faradays law equation for a coil of N turns , thus the

induced e.m.f. is

Example 3 :

A circular shaped coil 3.0 cm in radius, containing 20 turns and have a

resistance of 5.0 is placed perpendicular to a magnetic field of fluxdensity of 5.0 x 10-3 T. If the magnetic flux density is reduced steadily to

zero in time of 2.0 ms, calculate the induced current flows in the coil.

Solution:N=20 turns, r=3.0x10-2 m, R=5.0 , Bi=5.0x10-3 T

, Bf=0, dt=2.0x10

-3

sThe area of the circular shaped coil is

1-2B Wb s10x02dt

d. =

dt

dN B=

V20=

2rA =23 m10x82A = .

SF027 10

initially,

By applying the Faradays law equation for a coil of N turns , thus

Example 4 : (exercise)

A flat coil having an area of 8.0 cm2 and 50 turns lies perpendicular to a

magnetic field of 0.20 T. If the flux density is steadily reduced to zero,

taking 0.50 s, find

a. the initial flux through the coil.

b. the initial flux linkage.

c. the induced e.m.f. (Lowe&Rounce,pg.206,no.1)

Ans. : 1.6 x 10-4 Wb, 80 x 10-4 Wb, 16 mV

A10x03I 2. =

ifB d =

Br

Ar

From the figure, =0 thus thechange in magnetic flux through the

coil is

ABABd ifB coscos =ABd iB =

dt

dN B= and IR =

( )

dt

ABNIR i

=


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SF027 11

In figure 7.2b the magnitude of the

magnetic field at the solenoid increases as

the bar magnet is moved towards it.

An e.m.f is induced in the solenoid and

galvanometer indicates that a current is

flowing.

To determine the direction of the current

through the galvanometer which

corresponds to a deflection in a particular

sense, then the current through the

solenoid seen is in the direction that makethe solenoid upper end becomes a north

pole. This opposes the motion of the bar

magnet and obey the lenzs law.

7.2.2 Lenzs law

States an induced electric current always flows in such aan induced electric current always flows in such a

direction that it opposes the change producing it.direction that it opposes the change producing it.

This law is essentially a form of the law ofconservation of energyconservation of energy.

An illustration of lenzs law can be shown by using the experimentsbelow.

First experiment : (figure 7.2b)

Fig. 7.2bFig. 7.2b

I

INN

Direction ofDirection of

induced currentinduced current

Right hand gripRight hand grip

rule.rule.

North poleNorth pole

SF027 12

Fig. 7.2dFig. 7.2d

When the conductor move to the left thus theinduced current needs to flow in such a wayto oppose the change which has induced itbased on lenzs law. Hence galvanometershows a deflection.

To determine the direction of the inducedcurrent (e.m.f.) flows in the conductor PQ,the Flemings right hand (Dynamo) rule isused as shown in figure 7.2d.

Second experiment : Consider a straight conductor PQ is placedperpendicular to the magnetic field and move

the conductor to the left with constant velocity vas shown in figure 7.2c.

vr

Fig. 7.2cFig. 7.2c

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XXPP

QQ

Br

e.m.f.induced orI

)(motionr

ThumbThumb direction ofdirection ofMotionMotion

First fingerFirst finger direction ofdirection ofFieldField

Second fingerSecond finger direction ofdirection ofInduced current or InducedInduced current or Induced e.m.fe.m.f..

Important

Therefore the induced current flows from Q to Pas shown in fig. 7.2c.

Since the current flows in the conductor PQ andis placed in the magnetic field then this conductorwill experience magnetic force.

Its direction is in opposite direction of the motion.

I

BFr

Only for theOnly for the

straightstraight

conductor.conductor.


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SF027 13

Third experiment : Consider two solenoids P and Q arranged coaxiallyclosed to each other as shown in figure 7.2e.

At the moment when the switch S is closedswitch S is closed, currentIbegins toflow in the solenoid P and producing a magnetic field inside thesolenoid P. Suppose that the field points towards the solenoid Q.

The magnetic flux through the solenoid Q increases with timeincreases with time.According to Faradays law ,an induced current due to inducede.m.f. will exist in solenoid Q.

The induced current flows in solenoid Q must produce a magneticfield that oppose the change producing it (increase in flux). Hencebased on Lenzs law, the induced current flows in circuit consists ofsolenoid Q is anticlockwiseanticlockwise (fig. 7.2e) and galvanometer shows adeflection.

SSwitch,PP QQ

Fig. 7.2eFig. 7.2e

NNSS SSNN

indI indI--++

ind

SF027 14

Fig. 7.2fFig. 7.2f

SSwitch,PP QQ

NNSS

II

At the moment when the switch S is openedswitch S is opened, the currentIstartsto decrease in the solenoid P and magnetic flux through thesolenoid Q decreases with timedecreases with time. According to Faradays law ,aninduced current due to induced e.m.f. will exist in solenoid Q.

The induced current flows in solenoid Q must produce a magneticfield that oppose the change producing it (decrease in flux). Hencebased on Lenzs law, the induced current flows in circuit consists ofsolenoid Q is clockwiseclockwise (fig. 7.2f) and galvanometer seen to

deflect in the opposite direction of fig.7.2e.

SS NN

indI indI

-- ++

ind


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SF027 15

Example 5 :

A single turn circular shaped coil has resistance of 10 ohm and area of

its plane is 5.0 cm2. It moves towards the north pole of a bar magnet as

shown in figure below.

If the average rate of change of magnetic flux density through the planeof the coil is 0.50 T s-1, determine the induced current in the coil andstate the direction of the induced current observed by the observershown in figure above.

Solution:N=1 turn, R=10 , A=5.0x10-4 m2,

By applying the Faradays law equation for a coil of N turns , thus

1-T s500dt

dB.=

dtdN B=

( )dt

BAdNIR

=

whereo180BAB cos= and IR =

A10x52I 5 . =

=dt

dB

R

NAI

SF027 16

Based on the lenzs law, hence the direction of induced current is

clockwiseclockwise as shown in figure below.


A bar magnet is held above a loop of wire in a horizontal plane, asshown in figure below.

NNSSindI

The south end of the magnet is toward the loop of thewire. The magnet is dropped toward the loop. Find thedirection of the current through the resistor

a. while the magnet falling toward the loop and

b. after the magnet has passed through the loop and

moves away from it.

(Serway&Jewett, pg.991, no.15)


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SF027 17

Consider a linear (straight) conductor PQ of lengthL is movedperpendicular with velocity v across a uniform magnetic fieldB asshown in figure 7.3a.

When the conductor moved through a distance x in time t, the areaswept out by the conductor is given by

Since the motion of the conductor is perpendicular to the magnetic field

B hence the magnetic flux cut by the conductor is given by

7.3 Induced E.m.f. in a linear conductor.

Fig. 7.3aFig. 7.3a

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

XX XX XX XX

PP

QQ

L

x

Br

vr

o0BAB cos=

LxA =

BLxB =

SF027 18

According to Faradays law, the e.m.f. is induced in the conductor andits magnitude is given by

In general, the magnitudemagnitude of the induced e.m.f. in a linear conductor isgiven by

In vector formvector form,

The induced e.m.f. exist in the linear conductor when cutting themagnetic flux is also known as motional inducedmotional induced e.m.fe.m.f..

The direction of the induced currentdirection of the induced current due to induced e.m.f. flows inthe linear conductor can be determine by using FlemingFlemings right hands right handrule (based on lenzs law).In case of figure 7.3a, the induced currentflows from P to Q.

( )dt

BLxd=

vdt

dx=and

dt

dxBL=

dt

dB=

BLv=

BLv sin=Bvrr

andbetweenangle:where

( )BvLrr

=

(7.3a)(7.3a)

(7.3b)(7.3b)


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SF027 19

Note that the eq. (7.3a) also can be used for the rectangular coil of oneturn moved across the uniform magnetic field.

For a rectangular coil of N turns,

Example 7 :

A 20.0 cm long metal rod PQ is moved at speed of 100 m s-1 across a

uniform magnetic field of flux density 100 mT. The motion of the rod is

perpendicular to the magnetic field as shown in figure below.

iii. the electrical energy dissipated through the resistor in one minute.

Solution:L=20.0x10-2

m, v=100 m s-1

,B=100x10-3

T, =90a. By applying the equation of motional induced e.m.f in the linear

conductor, thus the induced e.m.f. is

PP

QQ

Br

1sm100

a. Calculate the motional induced e.m.f

in the rod.

b. If the rod is connected in series to the

resistor of resistance 10.0 , determine

i. the induced current and its direction.

ii. the total charge passing through the

resistor in one minute.

V00.2 =BLv sin=

NBLv sin= (7.3c)(7.3c)

SF027 20

b. GivenR=10.0 i. From the Ohms law , thus

Direction : using FlemingFlemings right hands right hand rule

ii. Given t=60.0 s

The total charge flows through the resistor is

iii. Given t=60.0 sBy using the equation of electrical energy, thus


A linear conductor of length 20 cm moves in a uniform magnetic field of

flux density 20 mT at a constant speed of 10 m s-1. The velocity makes

an angle 30 to the field but the conductor is perpendicular to the field.Determine the induced e.m.f. across the two ends of the conductor.

Ans. : 2.0 x 10-2 V

A2000I .=IR =

From P to QFrom P to Q

C012Q .=ItQ =

J024E .=ItE= RtIE 2=or


11/26


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SF027 23

Note :

This phenomenon was the important part in the development ofthe electric generator or dynamo.

Eq. (7.4a) also can be written as

Example 9 :

A rectangular coil of 200 turns has size 10 cm x 15 cm. It rotates at a

constant angular velocity of 600 r.p.m. in a uniform magnetic field of

flux density 20 mT. Calculate

a. the maximum e.m.f. produced by the coil.

b. the induced e.m.f. at the instant when the plane of the coil makes an

angle of 60 with the magnetic field.

Solution:N=200 turns, A=(10x10-2)(15x10-2)=150 x 10-4 m2

, B=20x10-3 T ,

a. By applying the equation of maximum induced e.m.f. for rotating coil,

thus

V77.3 =max

1-rad s20

60

2x600 ==

NBA =max

NBA sin=

BA

rr

andbetweenangle:where

(7.4b)(7.4b)

SF027 24

b.


A coil of area 0.100 m2 is rotating at 60.0 rev s-1 with the axis of rotation

perpendicular to a 0.200 T magnetic field.

a. If the coil has 1000 turns, find the maximum e.m.f. generated in it.

b. What is the orientation of the coil with respect to the magnetic fieldwhen the maximum induced e.m.f. occurs?

(Serway&Jewett, pg.991, no.15)

Ans. : 7.54 kV


A circular coil has 50 turns and diameter 1.0 cm. It rotates at a constant

angular velocity of 25 rev s-1 in a uniform magnetic field of flux density

50 T. Determine the induced e.m.f. when the plane of the coil makesan angle 55 to the magnetic field.Ans. : 1.77 x 10-5 V

Ar

Br

o60

From the figure, =90-60=30Hence the induced e.m.f. is

V88.1 =

NBA sin= and maxNBA = sinmax=


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SF027 25

7.5.1 Self-induction

Consider a solenoid which is connected to a battery , a switch S andvariable resistor R, forming an open circuit as shown in figure 7.5a.

According to the Faradays law, an e.m.f. has to be induced in thesolenoid itself since the flux linkage changes.

In accordance with Lenzs law, the induced e.m.f. opposes the changethat has induced it and it is therefore known as a backback e.m.fe.m.f.

For the currentIincreases :

7.5 Self-induction and Self-inductance

NNSS

Fig. 7.5aFig. 7.5a

SS RR

When the switch S is closed, a current

Ibegins to flow in the solenoid.

The current produces a magnetic fieldwhose field lines through the solenoidand generate the magnetic flux linkage.

If the resistance of the variable resistorchanges, thus the current flows in thesolenoid also changed, then so toodoes magnetic flux linkage .

Fig. 7.5b : initialFig. 7.5b : initial

NNSS

I

ind

SSNN

-- ++

NNSS

Fig. 7.5c :Fig. 7.5c :IIincreasesincreases

IindI

indI

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For the currentIdecreases :

This process is known as self-induction.

SelfSelf--inductioninduction is defined as the process of producing an inducedthe process of producing an inducede.m.fe.m.f. in the coil due to a change of current flowing through the. in the coil due to a change of current flowing through the

same coil.same coil.

This effect can be shown by the currentIagainst time tgraph forresistor and solenoid in figure 7.5d.

Direction of the inducedDirection of the induced e.m.fe.m.f. is in. is in

the opposite direction of the currentthe opposite direction of the currentII..

Fig. 7.5d : initialFig. 7.5d : initial

NNSS

I

NNSS

++ --ind

NNSS

IFig. 7.5e :Fig. 7.5e :IIdecreasesdecreases

I

Direction of the inducedDirection of the induced e.m.fe.m.f. is in. is in

the same direction of the currentthe same direction of the currentII..

indI

indI


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SF027 27

7.5.2 Self-inductance,L

From the process of the self-induction, we get

From the Faradays law,

SolenoidSolenoid

ResistorResistor

SelfSelf--inductioninduction

effecteffect

t0 Fig. 7.5dFig. 7.5d

LIB =IB

coiltheofinductance-self:Lwhere

dt

dIL=

dt

d

B=

(7.5a)(7.5a)

( )dt

LId

=

(7.5b)(7.5b)

current:I

SF027 28

From the eq. 7.5b, SelfSelf--inductanceinductance is defined as the ratio of the self inducedthe ratio of the self induced e.m.fe.m.f..

to the rate of change of current in the coil.to the rate of change of current in the coil.

If the coil has N turns, hence

Self-inductance is a scalar quantityscalar quantity and its unit is henryhenry (H)(H).

Unit conversion :

The value of the self-inductance depends on the size and shape of the coilthe size and shape of the coil

the number of turn (the number of turn (NN))

the permeability of the medium in the coil (the permeability of the medium in the coil ())..

I

NL B=

anddt

dIL=

dt

dN B=

dt

dN

dt

dIL B=

(7.5c)(7.5c)

= BdNdILBNLI=

1-21- AT m1Wb A1H1 ==


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SF027 29

A circuit element which possesses mainly self-inductance is known asan inductorinductor. It is used to store energy in form of magnetic fieldstore energy in form of magnetic field.

The symbol of inductor in the electrical circuit is shown in figure 7.5e.

7.5.3 Self-inductance of a Solenoid The magnetic flux density at the centre of the air-core solenoid is given

by

The magnetic flux passing through the solenoid is given by

Therefore the self-inductance of the solenoid is given by

Fig. 7.5eFig. 7.5e

andl

Nn =nIB 0=

l

NIB 0=

o0BAB cos=

Al

NI

0B

=l

NIA

0B =

I

NL B=

=l

NIA

I

NL 0

l

ANL

2

0= or AlnL 20=

turnsofnumber:Nwheresolenoidtheoflength:l

SF027 30

For the medium-core solenoid :

Example 12 :

At an instant, the current in an inductor increases at the rate of

0.06 A s-1 and back e.m.f. of 0.018 V was produced in the inductor.

a. Calculate the self-inductance of the inductor.

b. If the inductor is a solenoid with 300 turns, find the magnetic flux

through each turn when the current of 0.80 A flows in it.

Solution: =0.018 V,

a. By applying the equation below, thus

and 0r =l

ANL

2

=

or

l

ANL

2

0r= typermeabilirelative:rwhere

spacefreeoftypermeabili:0mediumoftypermeabili:

solenoidtheofarea:A

1-A s060dt

dI.=

dt

dIL= H300L .=


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SF027 31

b. GivenN=300 turns, I=0.80 A

By using the equation below, thus


An e.m.f. of 24.0 mV is induced in a 500 turns coil at an instant whenthe current is 4.00 A and is changing at the rate of 10.0 A s-1. Find the

magnetic flux through each turn of the coil. (Serway&Jewett, pg.1025, no.6)

Ans. : 19.2 Wb


A 40.0 mA current is carried by a uniformly wound air-core solenoid

with 450 turns, a 15.0 mm diameter and 12.0 cm length. Calculate

a. the magnetic field inside the solenoid.

b. the magnetic flux through each turn.

c. the inductance of the solenoid.

(Given 0

= 4 x 10-7 H m-1)

Ans. :188 T, 33.3 nWb, 0.375 mH

I

NL B= Wb10x0.8 4B

=

SF027 32

Consider a coil of self-inductanceL. Suppose that at time tthe currentin the coil is in the process of building up to its stable valueIat a rate

dI/dt. The magnitude of the back e.m.f. is given by

The powerPin overcoming this back e.m.f. is given by

The total energy stored in the inductor, U, as the current increasesfrom 0 toIcan be found by integrating the eq. (7.6a). Thus

7.6 Energy Stored in an Inductor

dt

dIL=

dt

dILIP=IP=

LIdIPdt= dUPdt=andLIdIdU= (7.6a)(7.6a)

2LI2

1U=

=I

0IdILdU

(7.6b)(7.6b) analogous to2

CV2

1U=


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SF027 33

For a long air-core solenoid, the self-inductance is

Therefore the energy stored in the solenoidenergy stored in the solenoid is given by

Note :

a.

b.

c.

l

ANL

2

0=

=

l

AIN

2

1U

22

02

LI2

1

U=(7.6c)(7.6c)

BconstantI

0 =0VAB =

BincreasesI

0VAB >

BdecreasesI

0VAB NSthe transformer isa stepstep--down transformerdown transformer.

IfNP< NSthe transformer isa stepstep--up transformerup transformer.

The symbol of transformer in

circuit is shown in figure 7.8b.

laminated

iron core

primary coilsecondary coil

NPturns

NSturns

Fig. 7.8aFig. 7.8a

Fig. 7.8bFig. 7.8b

SF027 44

By referring to mutual inductance, the induced e.m.f. in the primary andsecondary coil is given by

For an ideal transformer, there is no flux leakage so that

By dividing eq. (7.8a) with (7.8b),

For an ideal transformer, the electrical power is given by

dt

dN SSS =

dt

dN PPP =

and

(7.8a)(7.8a)

(7.8b)(7.8b)

dt

d

dt

d SP = same for both primary andsame for both primary andsecondary coils.secondary coils.

S

P

S

P

N

N

=

SP PP =

II SSPP =

P

S

S

P

I

I

=

primaryofpower:PPsecondaryofpower:PS

where


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SF027 45

Fig. 7.8dFig. 7.8dFig. 7.8cFig. 7.8c

In general:

7.8.1 Energy losses in transformers

Although transformers are very efficient devices, small energy losses do

occur in them owing to four main causes: Resistance of coils

The wire used for the primary and secondary coils has resistance and

so ordinary (I2R) heat losses occur.Overcome :Overcome : The transformer coils are made ofthick copper wirethick copper wire.

Eddy current

The alternating magnetic flux induces eddy currents in the iron core.This current causes heating and dissipation of power in the core.

Overcome :Overcome : The effect is reduced by using laminated corelaminated core as shownin figure 7.8c and 7.8d.

P

S

S

P

S

P

S

P

I

I

N

N

V

V

=== (7.8c)(7.8c)

SF027 46

Hysteresis

The magnetization of the core is repeatedly reversed by thealternating magnetic field. The resulting expenditure of energy inthe core appears as heat.

Overcome :Overcome : By using a magnetic materialmagnetic material (such as Mumetal) whichhas lowlow hysteresishysteresis lossloss.

Flux leakage

The flux due to the primary may not all link the secondary. Someof the flux loss in the air.

Overcome :Overcome : By designing the iron core suitably.

Example 21:

The primary coil of a transformer has 1200 turns and the secondary coilhas 60 turns. The primary coil is connected to an a.c. supply of 240 V. Aresistor of resistance 3.0 is connected to the secondary coil. Assumethat there are no loss of power and magnetic flux, calculate the currentflows in the secondary circuit.

Solution:NP=1200 turns, NS=60 turns, P= 240 V, R = 3.0 For ideal transformer,

S

P

S

P

N

N

=

P

P

SS

N

N

= V12S =


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SF027 47

Thus the current flows in the secondary coil is

Example 22: (exercise)

A transformer, assumed to be 100% efficient, is used with a supplyvoltage of 120 V. The primary winding has 50 turns. The required outputvoltage is 3000 V. The output power is 200 W.

a. Name this type of transformer.

b. Calculate the number of turns in the secondary winding.

c. Calculate the current supplied to the primary winding

Ans. : 1250 turns, 1.67 A


A transformer with a 100 turns primary coil and a 500 turns secondarycoil is connected to a supply voltage of 2.0 V. Calculate the outputvoltage and the maximum current in secondary coil if the current inprimary coil is to be limited to 0.10 A.

Ans. : 10 V, 0.020 A

RISS =

A04IS .=

SF027 48

Fig 7.9a shows a simple d.c. motor.

As the motor speeds up, the back e.m.f.,B increases because it is

proportional to the frequency,f

7.9 Back e.m.f. in d.c. motor

( )f2NBANBAB ==Fig 7.9aFig 7.9a

FrF

r

When current,Iflows in the coilof the armature which is in the

magnetic field, magnetic force is

produced and will cause the coil

to rotate as shown in figure 7.9a.

As the coil rotates, its magnetic

flux changes and so an e.m.f. is

induced across the coil.

(Faradays law)

By Lenzs law this induced e.m.f.

opposes the current which is

making the coil turns. Therefore

it is called back e.m.f. (B

)

fB final

initial

final

initial

f

f

= (7.9a)(7.9a)


25/26

SF027 49

When the motor is first switched on, the back e.m.f. is zero: it rises as themotor speeds up.

When the motor is running freely, the back e.m.f is nearly equal to thesupply voltage and so there will not be much current drawn.

When a load is applied to the motor, the motor slows down, the backe.m.f. falls, and so the current in the coil increases.

Figure 7.9a also can be simplified into circuit shown in figure 7.9b.

IRV B =

RIIVI 2B =

Applying Kirchhoffs 2nd law:

where

VI: power supplied

RIIVI B2+=

Fig 7.9bFig 7.9b

I

B

V

R

MotorMotor

LL

Loop L: (7.9b)(7.9b)

Eq. (7.9b) xI :

(7.9c)(7.9c)

I2R : power lost as heat in coil

B

I: mechanical power

SF027 50

Example 24:

A motor rotates at a rate of 5000 rotations per minute. The supply voltageis 240 V and the resistance of the armature is 4.5 .a. Calculate the back e.m.f. if the current in the armature is 12 A.

A load is applied to the motor and the speed of the rotation is found todecrease to 4000 rotations per minute. Calculate

b. the back e.m.f. now.

c. the new current in the armature.

d. the mechanical power produced by the motor.

Solution: V =240 V, R =4.5 , initial= 5000 rpm

a. By using the equation below:

Given final= 4000 rpmb.

IRV B +=

final

initial

final

initial

=

IRVB =V186B =

V8148final .=


26/26

SF027 51

c. The new current in the armature is

d. The mechanical power is


The resistance of the armature of a d.c. motor is 0.75 . A supply of 240V is connected to this motor. When the motor rotates freely without load,the current in the armature is 4.0 A and the rate of rotation is 400 rpm.

Calculate

a. the back e.m.f. produced.

b. the mechanical power generated.

If a load is applied, the current increases to 60 A. Calculate

c. the back e.m.f. now.d. the mechanical power.

e. the rotation speed of the armature.

Ans. : 237 V, 948 W, 195 V, 11.7 kW, 329 rpm

IRV B +=

IpowerMechanical B=

I548148240 ).(. +=

A320I .=

).)(.( 3208148powerMechanical =W3020powerMechanical =

Date post:	06-Apr-2018
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