ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune ......are only occasionally used in instruments...

: 2 : ESE-offline Test Series - 2017


01. (a)Sol: Given data:

Rm = 5 , RSW = 4 Im = 5 mA I = 1 A

So, voltage across shunt coil= ImRm + ImRsh

= (0.005) [4+5]= 45103V

The shunt should carry Ish

= I Im

= 1 0.005= 0.995A

The shunt resistance995.0

1045 3

= 0.045

Compensation of Temperature Error:The temperature error can be eliminated when the shunt and the moving-coil are made of the samematerial and kept at the same temperature. This method, however, is not satisfactory in practice asthe temperature of the two parts are not likely to change at the same rate. Disadvantage of usingcopper shunt is that they are likely to be bulky as the resistivity of copper is small. Copper shuntsare only occasionally used in instruments with built – in shunts.

In this case, a “swamping resistance” of manganin (which has a negligible temperature co-efficient) having a resistance 20 to 30 times the coil resistance is connected in series with the coiland a shunt of manganin is connected across this combination.

01. (b) (i)Sol: The atomic number of silicon is 14. Its electronic configuration is 1s2 2s2 2p6 3s2 3p2.

It undergoes hybridization and the valance electronic configuration becomes 3s1 3p3. Thus itexhibits a valance of 4. Hence it forms covalent bonding with four silicon atoms. It has thediamond cubic structure. The reference atom occupies the centroid of a regular tetrahedron, withfour other silicon atoms located at the corners of the tetrahedron.

It is represented in a planner configuration as below:

I

Im

Meter (moving)

RSW Swampingresistance

Shuntresistance

Rsh

IIm= Ish

Si Si Si

Si Si Si

Si Si Si

: 3 : E&TE-Conventional-5 (Solutions)


Thus each silicon atom is bonded covalently to four other atoms. Similarly, the atomic number ofGermanium is 32. It has valence electron configuration as 4s1 4p3. Germanium also has the samediamond cubic structure.

01. (b) (ii)

Sol: The energy of a diatomic molecule is given as6

)(r

b

r

arU

To find the expression for the equilibrium separation of the atoms of the molecule.

The force of interaction is given by

dr

dUrF )(

72

6

r

b

r

a

At equilibrium separation ro , the resultant force is zero. i.e., F(ro) = 0

06

70

20

r

b

r

a

ro5 = (6b/a) or

51

0

6

a

br



01. (c)

Sol: Percentage regulation of transformer

= (%R) cos2 (%X) sin2

Max. regulation occurs at only lag pf load

Hence,

% regulation = (%R) cos2 + (%X) sin2

Condition for Max. Regulation

0

d

gRe%d

2

12

%Xtan lag

%R

= 1 02

02

Xtan lag

R

Impedance triangle at Max. Regulation is

from the above pf of load corresponding to Max.

cos2 =Z%

X%sin;

Z%

R%2

Max. regulation =

Z%

X%X%

Z%

R%R%

=

Z%

X%R% 22

Z%

2

X02

R02



02. (a) (i)Sol: (i) D’Arsonval type Galvanometers, which are dc galvanometers, are made in three general types:

(1) Portable point type(2) Laboratory rectifying type(3) Box type

02. (a) (ii)Sol: Given,

Height, l = 25 mm = 0.025 mWidth, b = 20 mm = 0.02 mB = 0.11 Wb/m2

Moment of inertia, J = 0.2810–6 kg-m2

Control spring constant, C=3110–6 N-m/rad

deflection, = 135

= 135 180

= 2.356 rad

current, I = 0.11 mA= 0.011 A

(P) Number of turns (N):For steady deflection, GI = C

Displacement constant, G =I

C

=011.0

356.21031 6

= 6.64 10–3 N-m/A

G = NB b

N =02.0025.011.0

1064.6

bB

G 3

= 120.7 say 121N = 121

(Q) Resistance of coil for critical damping, Rc:

CJ2

GR

2

c

66

3

1028.010312

1064.6

= 7.48



02. (b) (i)Sol: Conductivity: In a conductor, the electrons are moving about with random velocity which depends

upon the temperature. When an electric field is applied, the electrons acquire a systematic velocity,which results in a net flow of charge, called electric current. The current density J is directlyproportional to the applied electric field strength E.

J Eor J = E

Where is a constant for a given material of the conductor, called the conductivity. It is related tothe microscopic properties of the material as

=m

ne2

Wheren = number of electrons per unit volumee = electron charge = relaxation timem = mass of electron

Mobility (): When a conductor is subjected to an electric field, the electrons experience a force of

eE. As a result they acquire a drift velocity d = m

eE .

Then the velocity acquired by the electron is directly proportional to the electric field strength.The drift velocity acquired per unit electric field strength is called the mobility of the chargecarrier.

Ed

Thus the conductivity is related to mobility as = ne

02. (b) (ii)Sol: Conductivity nie(p+e)

ResistanceAenA

Repi)(

Givenl = 1 cm = 110-2mA = 10-310-3= 10-6m2

ni = 2.51019/m3

e= 0.39 m2V-1s-1

p= 0.19 m2V-1s-1

Hence

19619

2

1061101903901052

10R

.)..(. =

6.158.05.2

1010 62

= 4.31 103



03. (a) (i)Sol: Factors affecting the resistivity of electrical materials are listed below -

1. Temperature.2. Alloying.3. Mechanical stressing.4. Age Hardening.5. Cold Working.

1. Temperature The resistivity of materials changes with temperature. Resistivity of most of themetals increase with temperature. The change in the resistivity of material with change intemperature is given by formula given below-

121tt tt112

Where,

1t is the resistivity of material at temperature of t1

o C

2t is the resistivity of material at temperature of t2oC

α1 is temperature coefficient of resistance of material at temperature of t1o C. If the value of α1 is

positive, the resistivity of material is increase. The resistivity of metals increase with increase oftemperature, means the metals are having positive temperature coefficient of resistance. Severalmetals exhibit the zero resistivity at temperature near to absolute zero. This phenomenon is “calledthe superconductivity”. The resistivity of semiconductors and insulators decrease with increase intemperature, means the semiconductors and insulators are having negative temperature coefficientof resistance.

2. Alloying: Alloying is a solid solution of two or more metals. Alloying of metals is used to achievesome mechanical and electrical properties. The atomic structure of a solid solution is irregular ascompared to pure metals. Due to which the electrical resistivity of the solid solution increases morerapidly with increase of alloy content. A small content of impurity may increase the resistivitymetal considerably. Even the impurity of low resistivity increases the resistivity of base metalconsiderably. For example the impurity of silver (having lowest resistivity among all metals) incopper increase the resistivity of copper.

3. Mechanical stressing: Mechanical stressing of the crystal structure of material develops thelocalized strains in the material crystal structure. These localized stains disturb the movement offree electrons through the material. This results in an increase in resistivity of the material.Subsequently, annealing, of metal reduces the resistivity of metal. Annealing of metal, relieve themechanical stressing of material due to which the localized stains got removed from the crystalstructure of the metal. Due to which the resistivity of metal decrease. For example, the resistivity ofhard drawn copper is more as compared to annealed copper.

4. Age Hardening: Age hardening is a heat treatment process used to increase the yield strength andto develop the ability in alloys to resist the permanent deformation by external forces. Agehardening is also called “Precipitation Hardening”. This process increases the strength of alloys bycreating solid impurities or precipitate. These created solid impurities or precipitate, disturb the



crystal structure of metal which interrupts the flow of free electrons through metal/Due to whichthe resistivity of metal increases.

5. Cold Working: Cold working is a manufacturing process used to increase the strength of metals.Cold working is also known as “Work hardening” or “Strain hardening”. Cold working is used toincrease the mechanical strength of the metal. Cold working disturbs the crystal structure of metalswhich interfere with the movement of electrons in metal, due to which the resistivity of metalincreases

03. (a) (ii)Sol: Given the resistivity = 1.54 10–8

There are n = 5.8 1028 electron per m3. To Calculate the mobility of electron, we use theformula,

The resistivity =ne

1

Where = mobility

=ne

1

=81928 1054.1106.1108.5

1

= 0.007 m2/v-s



The relaxation time is related to resistivity as

=2ne

m

=2ne

m

m = mass of electron = 9.1 10–31 kg

= resistivity = 1.54 10–8

n = 5.8 1028/m3

=83828

31

1054.11056.2108.5

101.9

= 3.98 10–14 sec

03. (b)

Sol: Given data, Psc = PI + Pm, rc sc

1P P

3 ,

= 79% and PI = Pm

output

output losses

Output = 0.79(output + losses)

Output = 3.76 losses

PGross – Pm = 3.76 [losses]

rc I sc I rc m

(1 s)P P 3.76(P P P P )

s

rc I I I I I

1 s 2P P 3.76 2P P P P

s 3

I I

(1 s) 2 141 P 3.76 P

s 3 3

= 1 s

27.82s



128.82

s

s = 0.03469

04. (a)Sol: Power = 4 kVA

W0 = V1I0cos0

cos0 =7.0200

60

= 0.428

(i) Iw = I0cos0 = 0.3 A

I = I0sin0 = 0.632 A(ii) Efficiency = ?

% =cu

2i WxWcoskVAx

coskVAx

Wi = 60 W

Wcu I2

Ifull-load current =400

4000 = 10 A

2

2cu

6

10

6.21

W

Wcu2 = 60 W

% = 1006060104

11043

3

= 97.08%

(iii) cos = 1From short circuit data of high voltage side

ZSc =6

9

I

V

SC

SC = 1.5

RSC =2SC

SC

I

W = 0.6

XSC = 1.374 Voltage Drop = I2R02cos I2X02sin2

I2 =400

4000

V

P = 10A

Voltage drop for UPF

= 10 0.6 1 + 10 1.374 0= 6 V



Secondary terminal voltage = E2 Drop

= 400 6 = 394 V

Voltage drop for 0.8 lag

= 10 0.8 0.6 + 10 1.374 0.6

= 13.04 V

Secondary terminal voltage (V2)

= E2 Drop

= 386.96 V

Voltage drop for 0.8 lead

= 100.80.6 01.3740.6

= –3.44V

Secondary terminal voltage (V2)

= E2 – Drop

= 403.44V

04. (b) (i)Sol: Dynamic error: It is the difference between the true value of the quantity (under measurement)

changing with time and the value indicated by the measurement system if no static error isassumed. It is also called Measurement Error.

Resolution: The smallest voltage that can be measured in lowest voltage range is known asresolution of an instrument.

Scale resolutionN10

1

Where N = no. of full digits

Resolution in a selected voltage range is given as =countsTotal

rangevoltageSelected

(or)

countMaximum

rangethatreadinginscaleFull



04. (b) (ii)Sol:

W = 400 watt.

W = Vph Iph cos

Vph Iph = 400/0.8

This type of connection gives reactive power

W = sinIV3 pp

= 6.08.04003

= 519.6 VAR

05. (a)

Sol: Error: The amount of deviation of measured value from true value is called error.

Error = measured value – true value.

= AM At

Types of Errors:

In any measuring instrument, there are three types of errors.

R

Y

B

W

R

Y

B

W



1. Gross error

All human negligence errors, while using the instrument and operating the calculators, are

called gross errors.

2. Systematic error: Systematic errors are of 3 types.

i) Instrumental error:- These errors are due to ageing, manufacturing defects.

ii) Environmental errors: These errors are due to the changes in environment

Ex: change in temperature.

iii) Observational errors: - These errors are due to parallax error i.e. observing the reading from

sides.

To avoid parallax errors, readings must be observed exactly at 900 to the display.

3. Random errors:

There is no particular reason for occurrence of these errors.

All instruments will have random errors. Random errors may be +Ve (or) Ve.

Random errors can be solved by using mathematical tool statistics like arithmetic mean and

arithmetic mode and standard deviation.

Sources of errors:

i) Gross error: These type of errors mainly cover human mistakes in reading instruments and

recording and calculating measurement results. The responsibility of the mistake normally lies

with the experimenter.

ii) Systematic errors: These type of errors are of three types.

a) Instrumental errors: These errors are due to three main reasons

i) Due to inherent shortcomings in the instrument

ii) Due to misuse of the instruments and

iii) Due to loading effects of instruments

b) Environmental errors: These errors are due to conditions external to the measuring device,

including conditions in the area surrounding the instrument. These may be effects of

temperature, pressure, humidity, dust, vibrations or of external magnetic or electrostatic fields.



(c) Observational errors: There are many sources of observational errors. As an example, thepointer of a voltmeter rests slightly above the surface of the scale. Thus, an error on account ofparallax will be incurred unless the line of vision of the observer is exactly above the pointer.

iii) Sources of Random error: These errors are due to a magnitude of small factors, which changeor fluctuate from one measurement to another and are due surely to chance.Given data

R1 = 100 2%

R2 = 200 2.5%

R1 = 100 100

2 100 = 100 2

R2 = 200 100

5.2 200 = 200 5

The limiting value of resistance

R = (100 + 200) (2 + 5)

= (300 7).

Magnitude of resistance = 300

Error in ohm = 7.

Percent limiting error of series combination of resistance =300

7 100 = 2.333%

Standard deviation:

R1 = 100 2%

R2 = 200 2.5%

R3 = R1 + R2

R3 = 100 + 50 = 150

R1 = 100100

2 = 2

R2 = 5.150100

3

R3 = 2R

2

2

32R

2

1

321 R

R

R

R



1R

R

1

3

, 1R

R

2

3

2

2223 2

312.1R

= 5.24

25

%667.1100150

5.2R 3

R3 = 150 1.667%

05. (b) (i)

Sol: Ceramic materials:

Ceramic materials are inorganic, non-metallic materials made from compounds of metal andnon-metals.

They may be crystalline or non-crystalline and are formed by action of heat and subsequent heattreatment.

They do not have free electrons.

They possess both covalent and ionic bonds in different proportions.

Due to presence of ionic & covalent bond, ceramic materials are solid, inert hard, brittle, strongin compression, weak in shearing and tension.

Other characteristics are low thermal expansion, low water absorption and good electricalproperties.

They are classified in two groups based on permittivity.

Permittivity less than 12 (r < 12)

Ex. Porcelain, Alumina, fused silica etc

Porcelain informally referred as ‘china clay’ is used as an insulator in transmission anddistribution, also used in fuses, plugs and sockets.

Most common use of porcelain is for utilitarian wares and artistic objects.

Alumina is used in some sodium vapour lamps and also in preparation of coating suspensions inCFLs.

Permittivity greater than 12 (r>12)

Ex: Barium titanate (BaTiO3) calcium copper titanate



Barium titanate is used for capacitors, microphones and other transducers.

CCTO has permittivity of approximately 12000 at room temperature and is used for reducingsize of capacitors.

High permittivity materials mainly used in semi conductor manufacturing process to replacesilicon dioxide.

05. (b) (ii)

Sol: It is to increase the tensile strength and tensile modulus of the composite.

i) In case of some insulators they can provide conducting path.

ii) It reduces the overall weight of the composite material and thereby reducing the cost of thecomposite.

iii) Fibre reinforced plastics have very good shear strength and can replace heavier metallicparts.

iv) Wires are used as a radial steel reinforcement in automobile tyres and also in high pressurehose pipes.

06. (a) (i)

Sol: Superconductivity:

The resistivity of some materials abruptly becomes zero below a specific critical temperature. It iscalled Superconductivity. It was first observed in pure mercury at the critical temperature of4.2 K. Different materials have different critical temperatures.

B = H

= orH

i. e. B = orH + oH oH

= oH + oH (r 1)

= oH + oM

Where the magnetisation M is equal to H

(r 1)

i.e. B = o(H + M) ------(1)

The first term on the right side of Eq. (1) is due to external field. The second term is due to themagnetisation.

Thus the magnetic induction (B) in a solid is

B = o(H + M)



HenceMH

Bo

The relative permeability

H

M1

H

MH

MH/B

H/B

or

r = 1 +

06. (a) (ii)

Sol: Hc = H0 cc

2

c

rH2iandT

T1

Given T = 4.2 K

Tc = 7.18 K

Hc = 6.5 104

2

18.7

2.41

= 4.276 104 A/m

Substituting in the formula for the critical current

ic = 10–3 4.28 104

= 134.51 A

06. (b)Sol: Conditions to be satisfied for parallel operation of transformers:

Necessary conditions for possible parallel operation:1. Voltage ratings mentioned on the name plate of transformers to be connected in parallel must

be same.2. The transformers must be connected in parallel with correct polarity.

For this, dotted terminals should be connected to same bus bars and undotted terminals shouldbe connected to another bus bar.

3. The phase sequence of 3- transformers to be connected in parallel must be same.To get correct phase sequence, the terminals of both secondaries which attain their peak valuesimultaneously must be connected to same bus bar, so that potential difference between theterminals connected to same bus bars equal to zero and there will not be any circulatingcurrent.

4. A part from phase sequence matching the phase displacement between the secondaries of bothtransformers must be zero. This condition can be fulfil if the two transformers belongs to samePhasor group.



Desirable conditions for satisfactory parallel operation:1. The voltage ratios of transformers to be connected in parallel should be same to avoid

circulating currents.If voltage ratios are not same, there is a possibility of circulating current given by

BA

B2A2C ZZ

EEI

E2A=Secondary voltage of transformer AE2B =secondary voltage of transformer BZA and ZB are the equivalent leakage impedances in ohms referred to the secondary oftransformer A & B respectively.

2. The two transformers should share the common load proportional to their KVA ratings.More KVA then more sharing of load.Less KVA then less sharing of load.

3. TheRx

ratios of transformers to be connected in parallel should be equal to avoid operation

of transformers at different power factors.

B

B

A

A

R

X

R

X

A = tan-1

A

A

R

X, B = tan-1

B

B

R

X

A = B

cosA = cosB = cosL

07. (a)

Sol: Given data, 460 V, 100 Hp, 50 Hz, P = 4,s = 0.05

(1) rpm15004

50120

P

f120Ns

N = Ns(1 –s) = 1500 (1 – 0.05) = 1425 rpm

(2) Synchronous speed w.r.t stator = 1500 rpm(3) Rotor frequency is slip frequency

= sf= (0.05) 50= 2.5 Hz

(4) Slip rpm

=P

)sf(120= rpm75

50

5.2120



(5) (i) Speed of rotor field w.r.t rotor= 75 rpm

(ii) w.r.t stator = 1500 rpm(iii) w.r.t stator rotating field = ‘0’Since the relative velocity between stator field and rotor field is zero.

(6) Let us consider primary connection is ‘Y’ connection

1

5.0

ph/E

ph/E

ph/N

ph/NK

1

2

1

2

2

460 / 3E

2

At the operating speed, s = 0.05

2

230sE 0.05

3 = 6.63 V

07. (b)Sol: The relative permittivity is calculated from the measured value of capacitance and the dimensions

of the electrodes. For a parallel plate arrangement, relative permittivity

r =A

dC

0

S

Cs = measured value of capacitance with specimen as dielectric,d = spacing between electrodesA = effective area of electrodes0 = permittivity of free space

A method, which avoids the necessity for close contact between electrodes and specimen, uses apair of solid electrodes. The capacitance of the arrangement is measured. The specimen is thenremoved, and the spacing between the electrodes is adjusted, by means of micrometer adjustments,until the capacitance is the same as before.The relative permittivity of the specimen can becalculated from the thickness of the specimen and the alteration in electrode spacingC = capacitance with specimen between electrodesA = area of electrodes

electrodex

t

d

Specimen Electrode



d = thickness of specimen

t = gap between specimen and electrode

x = reduction in separation between the two measurements

CS = capacitance of specimen

C0 = capacitance due to space between specimen and electrode

r 0 0S 0

S 0 0 r 0

A / d A / tC CC

C C ( A / d) ( A / t)

r 0

r

A

t d

When the specimen is removed and the spacing readjusted to give the same capacitance, the

expression for capacitance is:

C =xdt

A0

Hencedt

A

xdt

A

r

0r0

r =xd

d

ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune ......are only occasionally used in instruments...

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Transcript of ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune ......are only occasionally used in instruments...