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DiplomSubject Code :17404 (EEN)

Important Instructions to examine

1) The answers should be examine

answer scheme.

2) The model answer and the answ

the understanding level of the can

3) The language errors such as gra

applicable for subject English and

4) While assessing figures, exami

The figures drawn by candidate an

equivalent figure drawn.

5) Credits may be given step wise

may vary and there may be some d

6) In case of some questions credit

based on candidate’s understandin

7) For programming language pap

concept.

RA STATE BOARD OF TECHNICAL EDU(Autonomous)

(ISO/IEC-27001-2005 Certified)

a in Engineering Summer – 2015 Examinatio Model Answers

s:

d by key words and not as word-to-word as

er written by candidate may vary but the exa

idate.

mmatical, spelling errors should not be given

Communication Skills).

er may give credit for principal components

d model answer may vary. The examiner ma

for numerical problems. In some cases, the a

ifference in the candidate’s answers and mo

may be given by judgement on part of exam

g.

rs, credit may be given to any other progra

ATION

sPage No :

iven in the model

miner may try to assess

more importance (Not

indicated in the figure.

y give credit for any

sumed constant values

el answer.

iner of relevant answer

based on equivalent

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1 Attempt any TEN of the follo

1 a) Define RMS value of an alterAns- RMS Value: The RMSin terms of the direct currentwhen same time is provided.

The RMS value or effedirect current which will doesthe same real effect as when t

1 b) Draw the neat labeled diagra

Or equivalent diagram.1 c) State any two applications of

Ans

1) To measure direct & altern2) To measure direct & altern3) To measure unknown resist4) To measure capacitances

1 d) State the function of commutaAns-

The ends of the armature coibrushes supplies the DC poreverses the direction of currtravel alternate. This results in

1 e) Define transformer.Ans-Transformer is a static (or statcircuit is transferred to anothevoltage in a circuit with corres

1 f) What is transformation ratio oAns: Transformation ratio oinduced emfs (voltage transfoAt substantial values of load c




ing.

ating quantity.alue or effective value of an alternating cuthat produces the same heating effect in the

Ortive value of an alternating current is definethe same amount of work in the same timee alternating quantity is applied.

of PMMC instrument.

igital multimeter.

ting voltages.ting currents.ance.

5) To check continuity of elelements as resistances, c

6) To test Diodes.7) To test the transistor

tor in D.C. motor.

ls are connected to the commutator, whicher to the armature windings of motor fornt in the armature conductors as the polesunidirectional torque production in the mot

ionary) device by means of which electricalr circuit with frequency unchanged. It can raiponding increase or decrease in current.

f transformer?turns ratio is equal to the ratio of prim

mation).urrents near full load values the ratio of curr

ATION

sPage No :

2 x 10 = 20

rent is measuredsame resistance

as that value ofr would produce

02 mark

Neat

diagram 01mark

Correctlabeling

carries 01mark

01 markeach point

any 02points = 2

marks

ctrical circuits /oils etc.

with the help ofits operation. Itnder which theyr armature.

1 mark

1 mark

ower in onese or lower the

02 marks

ry to secondary

ents is inverse of 1 mark

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the turns ratio.

Mathematically,

1 g) What is slip of an induction mAns-

The difference between synexpressed as fraction or percei.e. slip, s = [(Ns- N)/N S] p.u

OR percentage slip =

At standstill (N = 0) hence slii.e. s=1 or s= 100 %

1 h) Name any four safety tools usAns:1) Hand gloves2) Goggles3) Rubber mats4) Fire extinguishers

5)

Danger plates (Danger n6) Search lights7) Safety shoes or Gumboo8) Ear plugs

1 i) What is energy audit?Ans:Energy audit is inspection, suoccupancy, process or systemthe energy efficiency without

1 j) Draw the speed torque characAns:NS, Nr & S are synchronous sTm & TSt are maximum & star




otor? What is its value at standstill condition

hronous speed (Ns) and actual speed (N) oftage of the synchronous speed is known as s

[(Ns- N)/N S] x 100.

is equal to one p.u. or 100 per cent.

d in industry.

otice plates)ts

9) Fall arresters10) Life line rope11) Safety helmets12) Safety belts

13)

Safety mask14) Fire buckets15) First aid box16) Insulating stick o

rvey and analysis of energy flows in an eneto identify, suggest means to conserve ene

negatively impacting the output(s).

eristics of 3-phase induction motor.

eed, rotor speed & slip respectively;ting torques respectively.

ATION

sPage No :

01 mark

?

he rotorlip (s). 01 marks

01 marks

½ markeach point

student canwrite any 04

points = 02marks

discharge rod

rgy consumptionrgy and increase

02 mark

02 mark

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1 k) State the method to reverse th

Ans- Direction of rotation ofphase sequence of the three pi.e. R-Y-B ---- clockwise dire R-B-Y or Y-R-B --- antic

1 l) State any two applications ofAns:

1) Submersible Water pu2) Water pumps on the bo3) Crop Cutting machines4) Diesel generator5) Electrical oven and spa

2 Attempt any FOUR of the foll2 a) Draw the single line diagram

Ans:

Or equivalent diagram




direction of rotation of 3-phase induction m

three phase induction motor can be reversedase supply.ctionlockwise direction

lectrical machines in electro agro system.

pewells.

e heaters.

owing.f electrical power system and mark the volt

ATION

sPage No :

otor.

by changing the 1 mark

1 mark

01 mark each,

any 02points = 2

marks

4 X 4 = 16 ges.

Correctdiagram 02

marks

and

CorrectVoltage

levels 02marks

Total = 4marks

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2 b) Define the following terms as

(i) Cycle(ii) Frequency(iii) Maximum value(iv) Average value

Ans-(i) Cycle: A complete

(ii) Frequency: Numbefrequency. Unit He

(iii) Maximum value:its maximum value

(iv) Average value: Avall instantaneous v

2 c) State any four advantages of tAns:

(i) In a three phase systemstable balanced conditiomaintenance problems

(ii) Three phase transformer(iii) Three phase induction

motors are not self-starti(iv) Three phase motors hav(v) Three phase motors hav(vi) Single phase motors co

motors are useless for inhigh.

(vii) The size of three phaserating.

(viii) Three phase system istransmission and distrib

2 d) What is line voltage and phasmark line voltage and phase vAns-Line voltage- The voltage avavoltage (V L).Phase voltage- the voltage ind

Lin




referred to an alternating quantity:

set of positive and negative values of an alte

r of repetitions of one cycle in one second isrtz or cycles/second.

he maximum instantaneous value of alternat

or its peak value.

erage value of an alternating quantity is defilues over a period of half cycle.

ree phase system over single phase system.

ith balanced loads the supply systems can hns which increases the life of the system and

unwarranted conditions.s are more economical (power capacity/weigotors are self-starting, whereas single phaseng machines unless provided with an extra shigher efficiency as compared to single phbetter power factor.monly designed for fractional horse power r

dustrial applications, where the power requir

otors is small as compared to single phase

much cheaper with regards to generation oftion compared to a single phase system.

voltage? Draw star connected 3- phase supltage.

ilable between any pair of terminals (or oute

uced in each winding is called Phase voltage

e Voltages = V RY , VYB, VBR, Phase Voltage

ATION

sPage No :

rnating quantity.

called as

ing quantity is

ed as average of

01 mark

01 mark

01 mark

01 mark

old on to theirleads to less

ht ratio is more)inductionarting winding.se motors.

ating so theseement is quite

otors of same

ower and its

01 markeach point

any 04points = 04

marks

ly system and

s) is called line

(Vph).

= ER, EY, EB

01 mark

01 mark

Neat

labellleddiagram 02mark

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2 e) The current flowing through a

(i) Frequency(ii) Phase difference(iii) RMS value of curr(iv) Amplitude

Ans:(i) Frequency –

ῳt = 314tῳ = 2πf = 314

∴ 3142 49.9

(ii) Phases differenceSince,∅ ∴ ∅ 30° or ( π /6

(iii) RMS value of currMaximum value =

√ 2 14.√

(iv) Amplitude 14.14

2 f) Explain the principle of workiAns-

Working of clip on ammeter-

The principle of opclamp ‐on jaws of the instrconductor acts as a primarythe instrument (taking the t




circuit is i = 14.14 sin ( 314t – π /6) . find :

nt

7 50

) radians

nt 14.14

4 9.99 10

ng of clip-on ammeter. Where it is used?

ration is shown in above figure, where it cament act as a transformer core and theinding. The resulting secondary current is trns-ratio of the current transformer into

ATION

sPage No :

01 mark

01 mark

01 mark

01 mark

be seen that thecurrent ‐carryingen measured by

account).Current

Diagram 01mark

Workingprinciple 02

mark

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induced in the secondary win

Use:Clip-on ammeter is used

without opening the circuit.

3 Attempt any FOUR of the foll

3 a) Explain the working principleAns:

Working Principle of D.C. MA machine that con

motor. Its operation is basedplaced in a magnetic field, ththis force is given by Fleming

F = BIL newtonsWhere,B= Flux densityI = Current through the conduL = length of the conductor

Significance of back EMF -The presence of back e.

makes the motor to draw as mtorque required by the load.Armature current (i) When the motor is runningand windage losses. Thereforeequal to the applied voltage.

(ii) If the motor is suddenly loslow down. Therefore, the sp

is reduced and hence the baccurrent to flow through theThus, the driving torque incrdown when the armature currby the load.

(iii) If the load on the motor isrequirement so that armaturee.m.f. Eb also increases and caccelerating when the armatrequired by the load.

It follows, therefore, tcurrent i.e., it automatically c




ing is rectified and applied to a moving ‐coil

to measure the AC current without breaki

owing.

of DC motor and explain the significance of

torverts d.c. power into mechanical power ison the principle that when a current carryconductor experiences a mechanical force.

’s left hand rule and magnitude is given by;

tor

.f. makes the d.c. motor a self-regulating much armature current as is just sufficient to d

on no load, small torque is required to over, the armature current Ia is small and the bac

aded, the first effect is to cause the armatureed at which the armature conductors move

k e.m.f. Eb falls. The decreased back e.m.frmature and larger current means increaseases as the motor slows down. The motornt is just sufficient to produce the increase

decreased, the driving torque is momentarilis accelerated. As the armature speed incuses the armature current Ia to decrease. Thre current is just sufficient to produce th

at back e.m.f. in a d.c. motor regulates theanges the armature current to meet the Load

ATION

sPage No :

meter.

g the circuit or01 mark

4 X 4 = 16

back e.m.f.

known as a d.c.ing conductor isThe direction of

chine i.e., itvelop the

ome the frictionk e.m.f. is nearly

tothrough the field

. allows a largerdriving torque.

ill stop slowingtorque required

in excess of thereases, the back

motor will stopreduced torque

flow of armaturerequirement.

02 mark

02 mark

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Impedance-

2

(i) Current:

100340.17 (ii) Impedance: 340.17

(iii) Phase difference between

Since, given circuit is seri

∅ a (iv) Power – c

3 e) A balanced three-phase delta

connected to a 400V, 3-phase(i) Phase voltage(ii) Phase current(iii) Line current(iv) Power Consumed

Ans:Given data- R= 4 /phase, V L

(i) Phase voltage - :

For delta connecti∴ 400

(ii) Phase current-

∴

(iii) Line current -

√ 3∗

(iv) Power consumed:

√ 3 c




1 12 ∗50 ∗ 10 ∗10 318.31Ω

120 318.30 340.17Ω

0.2939

current and supply voltage-

es RC circuit,

a318.30120 69.34°

∅ 100 ∗ 0.2939 ∗ 69.34° 10.36onnected load consists of three resistances e

50 Hz supply. Find:-

= 400 Volts, delta connected load

n;

100 …………………… ,

√ 3∗ 100 173.20 if V & I are line values)

∅ √ 3 ∗400 ∗ 173.20 ∗1 120000

ATION

sPage No :

½ mark

½ mark

1 mark

01 mark

01 mark

ach of four ohms

120

01 mark

01 mark

01 mark

01 mark

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3 f) Derive e.m.f. equation of tran

Ans: EMF equation of transfo

Fig.(i)Let,N1 = No. of primary turns, N 2 Øm = Max. flux in core = B m

As shown in above fig (ii), fluone quarter of cycle i.e in (¼ ) Therefore Average rate o

Now, rate of change of flux pTherefore Average e.m.f/turnIf flux Ø varies sinusoidally, tthe average value with the for

F fac . . Therefore value of e.m.f/turnNow, r.m.s. value of the induc

ESimilarly r.m.s. value of the i

EEquation (i) & (ii) are e.m.f. e

4 Attempt any FOUR of the foll4 a) Classify the transformers acco

No. of phase

Construction Voltage Application

Ans –(i) By number of phases: a) Si




former.

rmer-

Fig. (ii)

= No. of secondary turns, f = Frequency ofA (webers).

x increases from its zero value to maximumf second.change of flux

Ø

/

4fØ Wb/ r turn means induced emf in volts. 4fØ

hen r.rm.s. value of induced e.m.f is obtainedfactor.

1.111.11 X

4fØ

4.44fØ

ed e.m.f. in the whole of primary winding= (induced e.m.f./turn) X (No. of primary tu1 4.44 f ………………(i)duced e.m.f. in the whole of secondary wind

4.44 f ………………(ii)uation of transformer.

owing.rding to

ngle phase, and b) Three phase

ATION

sPage No :

.C. supply in Hz

value of Ø m in

by multiplying

s)

ing

1 mark

1 mark

1 mark

1 mark

4 x 4 = 16

01 mark

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(ii) By Construction: a) Core t(iii) By Voltage: a) Step up, b(iv) By Application: a) Instru

for rele)Aut

4 b) Draw a neat labeled diagram

Ans-

Advantage –(i) This method is che(ii) Power loss in the s

current as per relat(iii) No external resista

4 c) How speed control of inducti

Ans-

Or




ype, b) Shell type, c) Berry type.Step down.ent transformer for measurement, b) Instruying, c) Power transformer, d) Distribution-transformers.

f star-delta starter and write any one advanta

ap and effective.tarter can be reduced considerably by limitinon IL(star) = 1/3 I L (delta).

nces required to achieve the reduced voltage

n motor is done by VFD.

ATION

sPage No :

ent transformerransformer

01 mark 01 mark Any two =

01 mark

ge of this starter.

the starting

starting.

(schematicconnection

changingarrangementfrom star to

deltaexpected)

Neat labeleddiagram 03

mark,unlabeled or

partiallylabeled 1 to

2 marks.

Any oneAdvantage

01 mark

Diagram 02mark

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Explanation-The synchronous sp

synchronous speed of an ifrequency (f). Variable frequdrives).

A basic block

frequency source is shown incontrolled DC. This DC voltby pulses from voltage to frethe circuit to filter the control

4 d) State any four advantages of eAns-Advantages of electric drive o

1. These drives are available2. The control characteristic

these can be shaped to stecontrol, electric braking,

3. They are adaptable to anyor rough it is.

4. They can operate in all thapplicable for other prime

5. They do not pollute the e6. They do not need refuelin

loaded immediately.7. They are powered by elec

of power

4 e) What is universal motor? WhAns:




ed of the induction motor is given by,nduction motor can be changed by channcy can be obtained from solid state equip

iagram of speed control of induction mot

above fig. Three phase supply at input is firge is applied to inverter circuit whose frequeuency controller unit. A smoothing reactor,led DC.lectric drive over any other type of drives.

ver any other type of drivesin wide range torque, speed and power.

s of these drives are flexible. According to loady state and dynamic characteristics. As weearing, starting many things can be accompl

type of operating conditions, no matter how

four quadrants of speed torque plane, whicmovers.vironment.g or preheating, they can be started instantly

trical energy which is atmosphere friendly a

re is it used? State its types.


ATION

sPage No :

Ns=120f/P. Theging the supplyent or (i.e. VFD

r using variable

st converted intoncy is controlledL is connected in

Explanation

02 mark

ad requirementsll as speedished.much vigorous

is not

and can be

d cheap source

01 mark toeach point04 correct

points = 04marks

Diagram 01mark

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Universal motor is one

The stator and rotor windincommutator. The universalmotor.

Universal motor applications(i) Vacuum cleaners, (ii) DrinPortable drills, (v) Blenders et Types of universal motor-

1. Non-compensated typ2. Compensated type uni

4 f) Explain the working principle

Working Principle:

Fig 1OR

Fig3


Figures show thealternator. Operates on the pwinding and a magnetic fistationary) are cut by magnBecause the magnetic poles ain armature conductors, whicalternating emf is producednumber of N and S poles and

A schematic dia




hat can be used with a single phase AC sour

s of the motor are connected in series totor is also known as AC series motor or

and food mixers, (iii) Domestic sewing mac.

universal motor.ersal motor.

and operation of self-excited alternator.

Fig 2

stationary armature, two pole rotating fieldrinciple of electromagnetic induction. Con

ld. When the rotor rotates, the stator ctic flux, hence they have induced emf pr

re alternately N and S, they induce an emf ah first flows in one direction and then inn the stator conductors (i) whose frequenii) whose direction is given by Flemings Rigram of one type of self excited alternator

ATION

sPage No :

e as well as DC.

rough the rotorAC commutator

hine etc. (iv)

1 mark

explanation

Any two ½mark each =

1 mark

01 mark

type three phaseists of armaturenductors (being

oduced in them.nd hence currentther. Hence, an

y depends uponht Hand rule.orking is shown

Any simplediagram1 mark

Explanation03 marks

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in Fig 2. The three phase gene

to the field winding. This DCis maintained at the required vit is called self excited alterna

Students may drmust be assessed.

5 Attempt any four of the follo5 a) Explain the construction of th

Ans:

OR Fig BMain parts:Stator: laminated electromagnwindings spaced 120 o apart inframe has terminal box for the

Rotor: laminated constructionwindings in slip ring rotors.end covers. Slip rings providtype). Fan for cooling mounte




rator output is fed to the rectifier whose out

output is regulated such that the induced emalue. As the field is fed from alternator’s oor.aw a simple diagram as below or its equiv

ing.ee phase induction motor with diagram.

etic core with slots to accommodate 3 phasespace. Stator core housed in standard sized fwindings. Windings may be star or delta co

with slots to house conductor bars in squirreounted on steel shaft supported by bearin

d to connect to the rotor windings in wounon shaft.

ATION

sPage No :

ut (DC) is given

in the alternatorn generated emf

alent which also

04 x 4= 16

insulatedrames. Thenected.

l cage rotors andgs housed in the

rotor (slip ring

Diagram 02marks

1 mark

1 mark

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5 b) Explain the working of single

Ans:

Fig(i)

The capacitor-start motowinding has as many turns asseries with the starting windithat Is leads Im by about 80°split-phase motor. Consequenof a split-phase motor Again,the motor attains about 75%phase induction motor and co

5 c) What is (i) Resistance heatingAns:

(i) Resistance heating :

It is based on theelement I 2 R loss takes place

In this method the materithrough it. The charge may belectrodes are inserted in thecharge in the form of small pthe surface of the charge to av

through it. This method of hecharge itself.

(ii) Induction heating :Induction heating: high freq




phase capacitor start induction motor.

Fig(ii)

r is identical to a split-phase motor exceptthe main winding. Moreover, a capacitorg as shown in Fig. (i). The value of capac

(i.e., α ~80°) which is considerably greater tly, starting torque (Ts = k Im Is sin α) is muthe starting winding is opened by the centrifof synchronous speed. The motor then opetinues to accelerate till it reaches the normal

(ii) Induction heating (iii) Dielectric heati

I2 R effect. When current is passed throhich produce heat.

l to heated is treated as a resistance andin the form of powder, small solid pieces o

charge and connected to either AC or DC sieces, a powder of high resistivity materialoid direct short circuit. Heat is produced wh

ating has high efficiency because the heat is

ency supply passing through a coil (wor

ATION

sPage No :

that the startingis connected in

itor is so chosenhan 25° found inh more than thatgal switch whenates as a single-

speed.

Diagram 02mark

Working 02marks

g ?

gh the resistive

urrent is passedr liquid. The twoupply. When theis sprinkled overn current passes

produced in the

coil) generates

Diagram orequivalent ½

mark

½ mark

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strong magnetic field that ind

heated. As the path of induceleading to large amount of hecurrent paths leading to larmelting metal charge etc.

High freqSupply

(iii) Dielectric heating :Dielectric heating: heating ofconductivity) due to the leakahigh freq. ac supply. The freq20 kV. Material to be heatedapplied.Applications: 1) drying of indetc. 2) rubber vulcanizing, 3)

5 d) A six pole 50 Hz induction mAns:Given f =50 Hz, s=2.5 % = 0.

120 120 ∗ 506 10

; 1 1000 975

10

5 e) Classify electric drive. ExplaiAns:Classification of electric drive

(i) Group drive(ii) Individual Drive

(iii)

Multi-motor Driv (i) Group drive

• If several group of meor actuated by one mo




uces emf and hence current in the metallic

currents (eddy currents) is short circuit tht production. Also skin effect causes high rer amount of heat produced. Used for su

Enclosure contains charge to be heate

Work coil

nsulators (or non-metallic material with extre current (and hence heat losses) in them w

is normally between 12 kHz to 30 kHz and vs placed between two metal electrodes and t

strial products as textiles, agricultural produuring of rubber and plastics, 4) sterilization.

tor has a slip of 2.5%. Find its actual speed

025

00

0.025 0 975 25 each type in brief.

s-

hanisms or machines are organized on one sor, the system is called a group drive or shaf

ATION

sPage No :

job/charge to be

se are very highsistance in eddyrface hardening,

d

mely pooren subjected tooltage is aboute supply is

ce as tobacco

1 mark


mark

1 mark


mark

nd slip speed.

01 mark

01 mark

01 mark

01 mark

haft and drivendrive.

List 1 mark

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• Single motor drives a

Advantage : This is most econDisadvantage:1. Any Fault that occurs in the2. Efficiency low because of l(Power loss).3. Not safe to operate.4. Noise level at the working s5. Low Flexibility.

(ii) Individual Drive• A single motor drives• Each machine load op

Advantage: specialized for giDisadvantage: Costly, invest

(iii) Multi Motor Drive:• Each operation of the• The System contains s

operate its own mech• Separate motors are p

Advantage :1. Each Machine is driven by2. Machines not required can3. There is flexibility in the in4. In the case of motor fault, ocontinue working undisturbed5. Absence of belts and line spersonnel.

Disadvantage:

Initial high cost.

5 f) Explain the concept and princ

Electroplating is the applicatiby an electrochemical process




o of machines through belt from common s

omical type of drive.

driving motor renders all the driving equipsses occurring in the energy transmitting m

pot is high.

or actuates a given mechanism.erating system is driven by its own separate

en application hence increases productivityent wasted if drive remains idle for longer ti

mechanism is taken care of by a separate mo everal individual drives (motors) each of wh

nism.ovided for actuating different parts of the dri

separated motor it can be run and stopped ae shut down and replaced with a minimumtallation of different machines.

nly its connected machine will stop where as.afts greatly reduces the risk of accidents to t

ple of electroplating with neat sketch.

n of a metal coating to a metallic or other c.

ATION

sPage No :

aft.

ent idle.chanisms

otor.

qualitye.

tor.ich is used to

ven mechanism.

s desired.f dislocation.

others will

e operating

01 mark

01 mark

01 mark

nducting surface

Diagram 01mark

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Working principle-The article to be plateimmersed into an electrolyticcoating is desired on the articl

The electrolyte is a solutiopassing electric current, the mequivalent amount of anode gSmooth and brighter deposits

At cathode Zn2 +(aq) + 2 e -

At anode Zn(s)

Faraday's laws of electrolysis

6 Attempt any FOUR of the foll6 a) Explain the principle of eddy

Ans –

By heating an article bycurrent-carrying coil. The altecurrents in the article which, ccoil or work coil & the materieddy current loss which is resalso contributes to some exten

The eddy current loss Hence, this loss can be controThis loss is greatest on surfac

6 b) Draw the wiring diagram for cAns-




(the work) is made as cathode (negativsolution. The anode is usually a pure plateis also suspended in electrolytic solution.

n of salt of metal with which article to be ele tal ions from electrolyte get deposited on th

t dissolved in a form of ions and passed intoare obtained. For zinc electroplating,

Zn (s)

Zn2+

(aq) + 2 e-

govern the amount of metal deposited.

owing.urrent heating in detail.

eddy currents, it is placed in-side a high freqrnating magnetic field produced by the coil sonsequently, gets heated up. Such a coil is kl to be heated is known as charge or load. Ponsible for the production of heat although

t in the case of non–magnetic material.

.lled by controlling flux density B and the suof the material but decreases as we go deep

ontrol of one lamp one fan by two different

ATION

sPage No :

electrode) andof metal whose

ctroplated. Onarticle. The

electrolyte.

01 mark

02 mark

4 x 4 = 16

ency a.c.ts up eddy-

own as heaterimarily. It is theysteresis loss

pply frequency f .inside.

Diagram 02mark

Explanation02 marks

witches.

Correctdiagram 04

marks

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MAHARASH


6 c) Write the function of (i) MCBAns-Function of MCB: Provide s

Function of ELCB: Providecurrents flowing to earth and

Function of Fuse: Provides son its size & related circuit rat

6 d) State the necessity of earthingAns-

Earthing is groundingdue to leakage current.equipment’s, it might

Earthing is necessary flightning strikes.

To provide safe path t To provide stable platf

Equipment earthing anSr.No. Equipment earthing

01Body earthing is paor leakage current tconductor connecte

02

Equipment earthingearthing of non-curmetallic parts thatduring earth fault.order to prevent anperson.




(ii) ELCB (iii) Fuse.

ort circuit protection & over load protection

protection by detecting the unsafe magnitudisconnecting the supply to the circuit.

ort circuit & occasionally over current prote ing.

. Compare equipment earthing and neutral ea

he body of the electrical equipment’s to avoiIf the leakage current keeps circulating in thesult in electrical shocksor better voltage regulation and protection fr

dissipate lightning and short circuit currents orm for operation of sensitive electronic equi

d neutral earthing.

Neutral earthing

th for fault currentrough solidto earth.

Neutral earthing is returcurrent by making earthfor star points of supply

is related toent carryingay get energizedo it is done inshock to any

Star point connected toknown as neutral earthiessential for preventingground effect duringfaulty conditions and mzero potential at neutral

ATION

sPage No :

to electric loads.

es leakage

ction depending

01 mark

02 mark

01 mark

rthing.

d the hazardse body of the

m surges and

.pment’s.

path foras conductorsystems.round isg. It isarcing

intaining

01 mark

01 mark toeach pointany three

pointsexpected=03 marks

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8/18/2019 17404S15 msbte

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17404S15 msbte