1

Contents

Manual for K-Notes ................................................................................. 2

Error Analysis .......................................................................................... 3

Electro-Mechanical Instruments ............................................................. 6

Potentiometer / Null Detector .............................................................. 15

Instrument Transformer ....................................................................... 16

AC Bridges ............................................................................................. 18

Measurement of Resistance ................................................................. 21

Cathode Ray Oscilloscope (CRO) ........................................................... 25

Digital Meters ....................................................................................... 28

Q–meter / Voltage Magnifier ................................................................ 30

© 2014 Kreatryx. All Rights Reserved.

3

Error Analysis

Static characteristics of measuring system

1) Accuracy

Degree of closeness in which a measured value approaches a true value of a quantity under

measurement.

When accuracy is measured in terms of error :

Guaranteed accuracy error (GAE) is measured with respect to full scale deflation.

Limiting error (in terms of measured value)

GAE * Full scale deflectionLE

Measured value

2) Precision

Degree of closeness with which reading in produced again & again for same value of input

quantity.

3) Sensitivity

Change the output quantity per unit change in input quantity.

o

i

qS

q

4) Resolution

Smallest change in input which can be measured by an instrument

5) Threshold

Minimum input required to get measurable output by an instrument

6) Zero Drift

Entire calibration shifts gradually due to permanent set

4

7) Span Drift

If there is proportional change in indication all along upward scale is called span drift.

8) Dead zone & Dead time

The range of input for which there is no output this portion is called Dead zone.

To respond the pointer takes a minimum time is called dead time.

TYPES OF ERROR

a) Gross Error : Error due to human negligency, i.e. due to loose connection, reading the value

etc.

b) Systematic error : Errors are common for all observers like instrumental errors,

environmental errors and observational errors.

c) Random errors : Error due to unidentified causes & may be positive or negative.

Absolute Errors :

m rA A A

mA Measured value

rA True value

Relative Errors :

r =

A

T

AbsoluteErrors

Truevalue A

mm rT T

r

AA A A 1

1

5

Composite Error :

i) Sum of quantities

1 2

X X X

x x1 x2

ii) Difference of quantities

1 2

X X X

x x1 x2

So for sum & difference absolute errors are added.

iii) Multiplication of quantities

1 2 3

X X X X

31 2

1 2 3

XX XX

X X X X

iv) Division of quantities

1

2

XX

X

1 2

1 2

X XX

X X X

So, for multiplication & division, fractional or relative errors are added.

If m m

1 2

p

3

X XX

X

31 2

1 2 3

XX XXm n p

X X X X

Precision Index

Indicates the precision for a distribution

1h

2

6

Probable Error

r = 0.6745

0.4769r

h

Standard deviation of combination of quantities

2 2 2

1 2n1 2

n

2 2 2x x x x

X X X......

X X X

Probable Error

2 2 2

1 2 nn1 2

22 2x x x x

X X Xr r r ...... r

X X X

Electro-Mechanical Instruments 1) Permanent magnet moving Coil (PMMC)

Deflecting Torque

Td = nIAB

Where n = no. of turns

I = current flowing in coil

A = Area of coil

B = magnetic flux density

Deflection G

Ik

G = NBA & K = Spring constant

Eddy current damping & spring control torque in used.

For pure AC signal, the pointer vibrates around zero position.

It is used to measured DC or average quantity.

It can directly read only up to 50mV or 100mA.

7

Enhancement of PMMC

i) Ammeter

For using PMMC as an ammeter with wide range, we connect a small shunt resistance in

parallel to meter.

m

Im

I multiplication factor

Basically, ‘m’ is ratio of final range (as an ammeter) to initial range of instrument.

m

sh

RR

m 1

; mR = meter resistance

ii) Voltmeter

A series multiples resistance of high magnitude is connected in series with the meter.

M = multiplication factor

m

Vm

V

s mR R m 1

Sensitivity of voltmeter

s mv

fsd

R R1S / V

I V

Application of PMMC

1) Half wave rectifier meter

mavg

II I

8

RMS

s m f

2V

R R R

RMS

avg

s m f

0.45VI

R R R

; For Ac input

For DC input

DCavg

s m f

VI

R R R

avg AC DCavgI 0.45 I (Assuming

DC RMSV V )

AC DC(Sensitivity) 0.45(Sensitivity)

2) Full wave rectifier meter

RMS

ACf

avgs m

2 2VI

R R 2R

RMS

s m f

0.9V

R R 2R

DCavg DC

s m f

VI

R R 2R

avg DCACavgI 0.9 I (Assuring RMS DCV V )

AC DCSensitivity 0.9 Sensitivity

2) Moving iron meter

Deflecting torque, 2

d

1 dLT I

2 d

I = current flowing throw the meter

L = Inductance

= deflection

Under steady state

9

21 dLK I

2 d

2 I

MI meter measures both ac & dc quantities. In case of AC, It measures RMS value.

1T 2

2

RMS

0

1I i t dt

T

If 0 1 2i t I I sinwt I sin2wt .......

2 2 2

RMS 0 1 2

1I I I I .......

2

Air friction Damping is used

Condition for linearity

dLcons tant

d

MI meter cannot be used beyond 125Hz, as then eddy current error is constant.

3) Elector dynamometer

Deflecting Torque, d 1 2

dMT i i

d

For DC, 1 2i i I

2

d

dMT I

d

2 I

For AC, 1 m1i I sin t

2 m2i I sin t

1 2d avg

dMT I I cos

d

Where m11

II

2 &

2

I2I

2

10

Applications of dynamometer

1) Ammeter

Fixed coils are connected in series.

1 2I I I

0 (Angel between 1 2I & I )

2d

dMT I

d

At balance, c dT T

2 dMK I

d

2 I

It reads both AC & DC & for AC it reads RMS.

2) Voltmeter

sR Series multiplier resistance

2 1

s

VI I

R , 0

cos 1

2

d 2

s

V dMT

dR

At balance, cdT T

2

2s

V dMK

dR

2 V

It reads both AC & DC & for AC it reads RMS.

11

3) Wattmeter

Fixed coils carry same current as load & as called as current coils.

Moving coil is connected across voltage and thus current voltage, a high non-inductive

load is connected in series with MC to limit the current.

d 1 2

dMT I I cos

d

avg

s s

PV dM dMI cosR d R d

At balance, d

k T

avg P

Symbol :

Two wattmeter method

1 RY R RY RW V I cos V & I

L LV I cos 30

2 BY B BY BW V I cos V & I

L LV I cos 30

Where LV is line to line voltage

12

LI is line current

These expression remain same for -connected load.

1 23P W W

L L3V I cos

2 13Q 3 W W

L L3V I sin

2 1

1 2

3

3

Q 3 W Wtan

P W W

2 11

1 2

3 W Wtan

W W

for lag load

2 11

1 2

3 W Wtan

W W

for lead load

= Remember, In our case 1

W is wattmeter connected to R-phase and 2

W is wattmeter

connected to B-phase.

= If one of the wattmeter indicates negative sign, then pf < 0.5

Errors in wattmeter

a) Due to potential coil connection

2

T

cLr

I r% * 100

P

LI = load current

Cr = CC Resistance

TP = True Power

2

T

rs

V% *100

R P

V = voltage across PC

13

sR = Series multiplier resistance

TP = True Power

b) Due to self inductance of PC

If PC has finite inductance

p p sZ R R jwLp

p sR R p sZ R jwLp

r% tan tan *100

= load pf angle

p1

s

Ltan

R

4) Energy meter

Energy = Power * Time

T

VIcos tW * kwhr

1000 3600

TW = True energy

It is based on principle of induction.

It is an integrating type instrument.

m

tW VIsin * kwhr

3600

Where mW = measured Energy

= angle between potential coil voltage & flux produced by it.

= load pf angle

Error = m TW W

Energy constant = No.ofRe voluations N

kwhr P.t

Measured Energy = m

Totalno.ofrevolutions

KW

14

True Energy =

T

VIcos t* kw.hr

1000 3600W

Error = m Tr

T

W W% *100

W

Creeping Error in energy meter

If friction is over compensated by placing shading loop nearer to PC, then disc starts rotating

slow with only PC excited without connecting any load is creeping.

Otherwise if over voltage is applied on pressure coil then also creeping may happen due to

stray magnetic fields.

To remove creeping holes are kept on either side of disc diametrically opposite & the torque

experienced by both holes is opposite & they stop creeping.

% creeping error = TotalNo.ofRew / kwhr due to creeping

*100TotalNo.ofRew / kwhr due to load

Thermal Instruments

These instruments work on the principle of heating and are called as Thermal Instruments.

These are used for high frequency measurements.

They can measure both AC & DC.

In case of AC, they measure RMS value.

Electrostatic voltmeter

Deflecting torque, 2d

1 dcT V

2 d

At Balance,

cdT T

21 dcV k

2 d

2 V

Condition for linearity

15

dc

constantd

For increasing the range, we connect another capacitor in series

To increase the range from mV to V

m

s

CC

m 1 ;

m

Vm

V

Potentiometer / Null Detector wI = working current

B

wh

VI

R l.r _____________(1)

Switch at (A)

If gI 0

s w 1V I l r

sw

1

VI

l r _____________(2)

Switch at (B)

2x wV I l r

xw

2

VI

l r ________(3)

s x

1 2

V V

l r l r

2x s

1

lV V

l

16

r = resistance of slide wire (Ω/ m)

l = Total length of slide wire (m)

1l = length at which standard cell ( sV ) is balanced

2l = length at which test voltage ( xV ) is balanced

Measuring a low resistance

R

s

VR S

V

Instrument Transformer Current transformer

Equivalent circuit

Turns Ratio = Nominal Ratio 2

1

Nn

N

1 l s

l s

X Xtan

R R

R = Actual Ratio

s

I cos I sinn

I

17

Errors in current transformer

1) Ratio Error :

Current ratio p

s

I

I is not equal to turns ratio due to no-load component of current.

r

K R% *100

R

K = n = Nominal Ratio

R = Actual Ratio

2) Phase Angel Ratio :

Ideally, Phase difference between p sI & I should be 0180 but due to no-load component of

current, it deviates from that value.

Phase angle error =

s

I cos I sin 180*

nI degrees

Phase angle between primary & secondary currents

= 180 degrees

Potential Transformer

Equivalent circuit

Turns Ratio = n = 2

1

N

N

Actual Transformation Ratio = R = P

S

V

V

SP P P P

S

I1R n R cos X sin I R I X

V n , Where

1 Xtan

R

Phases angle error

SP P P P

s

IX cos R sin I X I R

nnV

18

AC Bridges

AC Bridges

Balance condition : DI 0

41 2 3Z Z Z Z

41 2 3Z Z Z Z

41 2 3

432 3

1 24

Z ZZ

Z

Quality Factor & dissipation factor

Quality Factor (Q) Dissipation Factor

(D)

1 wLQ

R

RD

wL

2 RQ

wL

wLD

R

3 1Q

wCR

D =wcR

4 Q = wcR 1D

wCR

Measurement of Inductance

(i) Maxwell’s Inductance Bridge

Here, we try to measure 1

R & 1

L

19

2 31

4

R RR

R

2 31

4

L LL

R

(ii) Maxwell’s Inductance Capacitance Bridge

2 31

4

R RR

R

41 2 3L R R C

This bridge is only suitable for coils where 1 < Q < 10

Q = Quality Factor

(iii) Hay’s Bridge

Used for coils having high Q value

2 24 42 3

1 2

R R R CR

11

Q

42 31 2

R R CL

11

Q

4 4

1Q

R C

(iv) Anderson’s Bridge

This Bridge is used for low Q coils.

2 31 1

4

R RR r

R

341 2 2 3

4

CRL R R r R R

R

20

(v) Owen’s Bridge

431

2

R CR

C

2 41 3L R R C

Measurement Of Capacitance

De-Sauty’s Bridge

31 2 2 1

4

Rr R r R

R

42

31

RC C

R

D = dissipation factor

= 1 1

C r

1r = internal resistance of

1C

Schering Bridge

431

2

R CR

C

4 21

3

R CC

R

dissipation factor = D = 4 4C R

Measurement of frequency

Wien Bridge Oscillator

Balancing Condition

3 1 2

4 2 1

R R C

R R C

Frequency of Osculation

1 2 1 2

1f

2 R R C C

21

Measurement of Resistance

Classification of Resistance

1) Low Resistance : R ≤ 1Ω , Motor and Generator

2) Medium Resistance : 1Ω < R < 100kΩ , Electronic equipment

3) High Resistance : R > 100 kΩ, winding insulation of electrical motor

DC Bridges

Medium Resistance Measurement

1. Wheatstone Bridge

Finding Theremin Equivalent

th

ggth

VI

R R

Th

P RV V

P Q R S

Th

PQ RSR

P Q R S

For Balance Condition

gI 0

Th

V 0

PS = RQ

22

Sensitivities

1) Current sensitivity , i

g

SI

mm/mA

= deflection of Galvan meter in mm

2) Voltage sensitivity,

Th

SV

mm/V

3) Bridge Sensitivity , BS

R /R

mm

ThB

vV SS

R / R

B

vV.SS

SR 2S R

For Maximum Sensitivity

SRS R = 1

B, maxvV.S

S4

2. Carey –foster slide wire Bridge

r = slide wire resistance in m

.

for case (1).

At balance

1

1

R rP

Q S L r

………….(1)

For case (2)

R & S is reversed

2

2

S rP

Q R L r

………..(2)

From (1) & (2)

1 2

1 2

R r S r

S L r R L r

23

3. Voltmeter Ammeter Method

a) Ammeter near the load

vm X A

A

VR R R

I

vV = voltage across voltmeter

A

I = Ammeter current

XR = Test resistance, A

R = Animator resistance

% error = m T A

xT

R R R100 100%

R R

b) Voltmeter near the load

X

v Xm

vA

VVR

I I I

vXm

X v vX

X X

R R1R

I I R R

V V

% error = m X

X

R R100%

R

If a vXR R R , voltmeter is connected near the load

a vXR R R , ammeter is connected near the load

4. Ohmmeter

a) Series Type

when XR 0

m FSDI I = Full scale deflection

when XR

mI 0 = zero deflection

24

for Half scale deflection

mshseX h

msh

R .RR R R

R R

b) Shunt Type

S

R = current limiting resistor

If X

R 0

mI 0 = zero deflection

If xR

m FSDI I = Full scale deflection

For Half scale Deflection

m S

x hm S

R RR R

R R

Measurement of Low Resistance

Kelvin’s Double Bridge Method

Unknown resistance

qr pP P

R SQ p q r Q q

P, Q = outer ratio arms

p, q = inner ratio arms

S = standard resistance

r = lead resistance

R = Test resistance

High Resistance Measurement

Loss of charge Method

tRc

eCV t V

10

C

0.4343tR

VC log

V

25

t = time in (seconds)

V = source voltage

CV = Capacitor voltage

Cathode Ray Oscilloscope (CRO)

The velocity of e is changed by changing the pre-accelerating & accelerating anode

potential

KE =PE

2

a1

mv qV2

a2qV

m

Deflection sensitivity

D = deflection height on screen

d = distance between plates

d = length of vertical deflecting plates

L = distance between centre of plate & screen

aV = anode potential

yV = Vertical plate Potential

yd

a

L VVD

mm2dV

deflection sensitivity

d

y a

LD VSmmV 2dV

26

Lissajous Pattern

If both horizontal & vertical deflection plates of CRT is applied with the sinusoidal signal,

the wave form pattern appearing on screen is called Lissajous Pattern.

Case – 1: Both signals have same frequency

x m xV V sin w t

y m yV V sin w t

x y mV V V

x yw w w

= variable

S.No Lissayous Pattern

1

0 or 360

2

0 90

Or

270 360

3

90 or 270

4

90 270

Or

180 270

5

180

27

Finding

1) Lissajous Pattern in Ist & IIIrd Quadrant

1 11 1

2 2

X Ysin sin

X Y

for anti-clockwise orientation phase difference = (360 - )

for clockwise orientation, phase difference =

2) Lissajous Pattern in IInd & IVth Quadrant

2

1 1X

180 sinX

2

1 1Y

180 sinY

for clockwise orientation, phase difference =

for anti-clockwise orientation = 360

Case – 2

x yw w

x m xV V sinw t

y m yV V sinw t

y y

x x

w f Number of horizental tangencies

w f Number of vertical tangencies

y

x

f 42

f 2

28

Digital Meters

Type of converter Maximum Conversion Time

1) Dual slope ADC n 12 Clocks

2) Successive Approximation Register (ADC) n Clocks

3) Counter ADC n2 Clocks

4) Flash ADC 1 Clock

Dual slope A/D Converter

aV = analog input

RV = Reference input

Ra 2 1

1

VV T T

T

1n

CLKT 2 T

Maximum conversion time = n 1

CLK2 T

Successive Approximation Register

Suppose = REF aV 1 V

and aV = 12V

3D

2D

1D

0D

10 5 2.5 1.25

1

T 1 0 0 0 10V < 12V

2T 1 1 0 0 15V > 12 V

3T 1 0 1 0 12.5 > 12 V

4T 1 0 0 1 11.25 < 12 V

In first clock cycle, MSB is set to get voltage corresponding to the digital o/p

If 0

V < aV , then in next cycle next bit is set else,

If 0

V > aV , MSB is reset & next bit is set

We continue the same process till we reach LSB.

29

Specifications of Digital Voltmeter

1) Resolution

The smallest value of input that can be measured by digital meter is called resolution.

n

1R

10

n = No. of full Digits (0, 1,….., 9)

2) Sensitivity

S = Resolution x Range

3) Over – Ranging

The extra 12

digit is called over-ranging

If n = 3, we can measure from 0 – 999

Resolution , 3

1R 0.001

10

if 1n 32

digit, 12

digit can be 0 & 1.

we can measure from 0 – 1999

Resolution, 1

R 0.0052000

if 34

digit is there than MSB can be 0 – 3.

4) Total Error

Error = (% error in reading) x reading + (NO. of counts) Full Scale

Range of meter

30

Q–meter / Voltage Magnifier

If works on the principal of series resonance.

At series resonance

L C

X X

V

IR

CCV IX

LCX X

V VR R

C

V = V. Q

C

V Q

Practical Q-meter

Also includes series resistance of source (oscillator)

True T

wLQ

R

Measured Q,

Tm

sh shsh

QwL wLQ

R RR RR 1 1

R R

msh

T

RQ Q 1

R

31

Measurement of unknown capacitance

Test capacitance is connected at 43T & T .

Circuit is resonated at C = 1

C

fr=

T1

1

2 2 C C ………(1)

TC = Test Capacitance

TC is removed & circuit is resonated at C =

2C

fr =

2

1

2 LC ………(2)

from (1) & (2)

T 2 1C C C

Measurement of self-capacitance

Resonance is achieved at C = 1

C

1

1 d

1f

2 L C C

At C = 2

C , resonance is achieved at 2

fr

2

2 d

1f

2 L C C

= n f1,

21 2

d 2

C n CC

n 1

Kuestion

Electrical and Electronic

Measurements

www.kreatryx.com

1

Contents

Manual for Kuestion .......................................................................... 2

Type 1: Error Analysis .......................................................................... 3

Type 2: Enhancement of Instrument Range ........................................ 5

Type 3:PMMC ..................................................................................... 6

Type 4: Moving Iron ............................................................................ 8

Type 5: Bridges .................................................................................... 9

Type 6: Wattmeter ............................................................................ 12

Type 7: Energy Meter ........................................................................ 14

Type 8: Digital Meter ........................................................................ 15

Type 9: CRO ....................................................................................... 16

Answer Key ....................................................................................... 20

© 2014 Kreatryx. All Rights Reserved.

3

Type 1: Error Analysis

For Concept, refer to Measurement K-Notes, Error Analysis

Point to remember:

While using the limiting error concept in division we need to remember that all variables

must be independent of each other and hence this rule does not hold for parallel

combination of resistance.

Sample problem 1:

A variable w is related to three other variables x,y,z as w = xy/z . The variables are measured with meters of accuracy 0.5% reading, 1% of full scale value and 1.5% reading. The actual readings of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum uncertainty in the measurement of ‘w’ will be (A) 0.5% rdg (B) 5.5% rdg (C) 6.7% rdg (D) 7.0% rdg

Solution: (D) is correct option

xyGiven that =

z

log logx logy logz

Maximum error in

d dx dy dz%

x y z

dx0.5% reading

x

dy1% full scale

y

1 = 100 1

100

dy 1100 5% reading

y 20

dz1.5% reading

z

So, d

% 0.5% 5% 1.5% 7%

4

Unsolved Problems:

Q.1 The power in a 3- phase, 3- wire load is measured using two 100 W full scale watt meters

W1 and W2. W1 is of a accuracy class 1% and reads 100W. W2 is of accuracy class 0.5% and reads – 50W. Uncertainty in the computation to total power is

(A) 1.5% (B) 2.5% (C) 3% (D) 4%

Q.2 In the circuit given on fig, the limiting error in the power dissipation ’I2R’ in the resistor ‘R’ is

(A) 1.2%

(B) 5.2%

(C) 10.2%

(D) 25.2%

Q.3 Consider the circuit as shown in figure. Z1 is an unknown impedance and measured as

z1=z2z3/z4. The uncertainties in the values of z2,z3 and z4 are 1%, 1% and 3% respectively. The overall uncertainty in the measured value of z1 is

(A) 11%

(B) 4%

(C) 5%

(D) 5%

Q.4 Three resistors have the following ratings R1=200 5%, R2=100 5% and R3= 50 5%. Determine the limiting error in ohms if the above resistances are connected in parallel

(A) 1.3 (B) 1.19 (C) 4.28 (D) 2.85

Q.5 The voltage of a standard cell is monitored daily over a period of one year. The mean value over a period of one year. The mean value of the voltage for every month shows a standard deviation of 0.1mV. The standard deviation of the set constituted by the monthly mean values will be?

(A)0 (B) 0.1

12 (C)

0.1

12 (D)0.1

Q.6 A current of 10mA is flowing through a resistance of 820 having tolerance of 10% . The current was measured by an analog ammeter on a 25 mA range with an accuracy of 2% of full scale. What is the range of error in the measurement of dissipated power? (A) 15% (B) 5% (C) 14% (D) 20%

5

Type 2: Enhancement of Instrument Range

For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments

Point to remember:

In Ammeter, external resistance is added in parallel to meter and in voltmeter additional

resistance is added in series.

Sample Problem 2:

An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 Ω. In order to change the range to 0-25 A, we need to add a resistance of (A) 0.8 Ω in series with the meter (B) 1.0 Ω in series with the meter (C) 0.04 Ω in parallel with the meter (D) 0.05 Ω in parallel with the meter

Solution: (D) is correct option Given that full scale current is 5A Current in shunt I’=IR-Ifs

= 25-5=20A

sh

sh

20 R 5 0.2

R 0.05

Unsolved Problems:

Q.1 A 0-10mA DC Ammeter with internal resistance of 100 is used to design a DC voltmeter with full scale voltage of 10 V. The full scale range of this voltmeter can be extended to 50V by connecting an external resistance of

(A) 900 (B) 499.9k (C) 4000 (D) 4900

Q.2 What is the value of series resistance to be used to extend (0-200)V range voltmeter having 2000Ω/V sensitivity is to be extended to (0-2000)V range.

(A)44.44KΩ (B)55.55KΩ (C)34.56KΩ (D)45.25KΩ

Q.3 A DC ammeter has a resistance of 0.1Ω and its current range is 0-100 A. If the range is to be extended to 0-500 A, then meter required the following shunt resistance?

(A)0.010Ω (B)0.011Ω (C)0.025Ω (D)1.0Ω

6

Q.4 The coil of a measuring instrument has a resistance of 10Ω and the instrument reads up to 250V, when a resistance of 4.999Ω is connected in series. Now the same instrument is used as an ammeter by connecting a shunt resistance of 1/499Ω across it. What is the current range of the ammeter?

(A)30A (B)26A (C)20A (D)24A

Q.5 A D’Arsonval movement with a full scale deflection current of 10 mA and internal resistance of 500Ω is to be converted into the different range of voltmeters. If Ra,Rb and Rc are the required series resistance for the ranges 0-20V, 0-50 V and 0-100 V respectively, then Ra:Rb:Rc is?

(A)2:5:10 (B)1:1:1 (C)3:9:19 (D)5:11:21

Type 3:PMMC

For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments

Point to remember:

PMMC always measures the average value of the output and the pointer vibrates around the

zero position for pure AC input.

Sample Problem 3:

The input impedance of the permanent magnet moving coil (PMMC) voltmeter is infinite. Assuming that the diode shown in the figure below is ideal, the reading of the voltmeter in Volts is? (A)4.46

(B)3.15

(C)2.23

(D)0

Solution: (A) is correct option

PMMC voltmeter reads average value. For the +ve half cycle of I/p voltage, diode will be forward biased (Vg = 0, ideal diode) Therefore, the voltmeter will be short circuited and reads V1 = 0 volt (for +ve half cycle) Now, for -ve half cycle, diode will be reverse biased and treated as open circuit. So, the voltmeter reads the voltage across 100 kW. Which is given by

7

0

2

2,rms

14.14 0V 100 14.14

(100 1)

14So, V V

2

Therefore, the average voltage for the whole time period is obtained as

1 2,rms

avg

140

V V 142V 4.94 4.46V2 2 2 2

Unsolved Problems:

Q.1 A PMMC has an internal resistance of 100 and requires 1 mA dc for full scale deflection.

Shunting resistor Rsh placed across the movement has a value of 100. Diodes D1and D2 have

forward resistance of 400 and infinite reverse resistance. For 10 V ac range, The value of series multiplier is (RS) and voltmeter sensitivity for ac range is

(A) 9550 , 250 /V

(B) 4550 , 225 /V

(C) 5000 , 500 /V

(D) 1800 , 250 V/

Q.2 A Thermocouple produces a voltage of 50 mv. Its internal resistance is 50 . The

resistance of leads is 10. The output is read by a PMMC meter having an internal resistance of 120 the output voltage indicated will be

(A) 50 mV (B) 40 mV (C) 33.3 mV (D) 25.0 mV

Q.3 An Electronic AC voltmeter is constructed using a full wave bridge Rectifier, with a scale calibrated to read rms of a symmetrical square wave having zero mean. If this voltmeter is used to measure a voltage V(t)=10 sin 314t, then The reading of the voltmeter and magnitude of percentage error in reading respectively are?

(A) 7.07V, 11% (B) 0.707V, 11.1% (C) 6.36V, 9.9% (D) 11.1V, 0%

Q.4 The coil of moving coil voltmeter is 50 mm long and 40 mm wide and has 120 turns on it. The control spring exerts a torque of 180 x 10-6 N.m. When the deflection is 120 divisions on full scale, flux density of the magnetic field in the air gap is 1.2wb/m2. Neglect the resistance of the coil. Resistance that must be put in the series with the coil to give one volt per division is?

(A) 50 K (B) 63.7 K (C) 83 K (D) 91.7 K

8

Q.5 The following date refers to a moving coil voltmeter resistance 10K; dimension of coil 30mm x 30mm; number of turns on coil 100, Flex density in the air gap is 0.08 wb/m2. The deflecting torque produced by a voltage of 200v is

(A) 60Nm (B) 144N – m (C) 78.6N – m (D) 178N – m

Q.6 Two 100V full scale PMMC type DC voltmeters having a figure of merits of 10 k/V and

20K/V are connected in series. The series combination can be used to measure maximum D.C voltage of

(A) 100V (B) 300 V (C) 150 V (D) 200 V

Type 4: Moving Iron

For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments

Point to remember:

Moving Iron Instruments measures the rms value of output.

Sample Problem 4:

The saw-tooth voltage waveform shown in the figure is fed to a moving iron voltmeter. Its reading would be close to __________

(A)48.41

(B)66.56

(C)57.74

(D)none

Solution: (C) is correct option

From the graph, we write mathematical expression of voltage v(t)

3

3

100v(t) t 5 10 t volts

20 10

A moving iron voltmeter reads rms value of voltage, so

3

T

2

rms

0

20 10

3 2

3

0

1V v (t)dt

T

1 = (5 10 t) dt

20 10

320 106 3

3

0

25 10 t = 57.74 A

320 10

9

Unsolved Problems: Q.1 A permanent magnet moving coil type ammeter and a moving iron type ammeter are connected in series in a resistive circuit fed form output of a half wave rectifier voltage source. If the moving iron type instrument reaches 5A, the permanent magnet moving coil type instrument is likely to read.

(A) Zero (B) 2.5 A (C) 3.18 A (D) 5 A

Q.2 A 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on dc for full scale deflection of 900. The control constant is 0.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 0.25H. Total change in mutual inductance is

(A) 18x10-3 H/rad (B) 28.3x10-3 H/rad (C) 26.5x10-3 H/rad (D) 13.7x10-3 H/rad

Q.3 For certain dynamometer ammeter the mutual inductance M varies with deflection

(expressed in degrees) as m=-5Cos(+30)M.H. What will be the deflection of the instrument, if the deflection torque produced by 60m.A current is 18 x 10-6N.m

(A) 60 (B) 90 (C) 30 (D) 40

Q.4 The inductance of a moving iron ammeter is given by the expression

2L 20 10 3 H where ϴ is the angle of deflection in radians. Determine the

deflection in degree for a current of 8A, if the spring constant is 610 10 N-m/rad.

(A)1.250 (B)1.540 (C)1.560 (D)1.580

Q.5 A spring controlled moving iron voltmeter draws a current of value 100V. If it draws a current of 0.5mA, the meter reading is?

(A)25V (B)550V (C)100V (D)200V

Type 5: Bridges

For Concept, refer to Measurement K-Notes, AC Bridges

Point to remember:

For any bridge, the balance condition is Z1Z4 = Z2Z3

10

Sample Problem 5:

The Maxwell’s bridge shown in the figure is at balance. The parameters of the inductive coil are.

(A) 2 3

4 2 3

4

R RR , L C R R

R

(B) 2 3

4 2 3

4

R RL , R C R R

R

(C) 4

2 3 4 2 3

RR , L

R R C

1

R R

(D) 4

2 3 4 2 3

RL , R

R R C

1

R R

Solution: (A) is correct option

At balance condition

4

4

4 2 3 2 3

4

4

4

jR

Cj(R j L)(R || ) R R (R j L) R R

C jR

C

2 3 2 34 4 4 4

2 3 4 2 3 4

4 4 4 4 4 4

jR R jR RjRR LR jRR LRR R R R R R

C C C C C C

By comparing real and imaginary parts

2 3 2 34

4 4 4

4

2 3 4 2 3 4

4

R R R RRRR and

C C R

LRR R R L R R R

C

Unsolved Problems:

Q.1 A slide wire potentiometer has a battery of 4 V and negligible internal resistance. He

resistance of slide wire is 100 and it’s length 200 cm. A standard cell of 1.018 V is used for standardizing P.M and the rheostat is adjusted so that balance is obtained when the sliding contact is at 101.8 cm. Find the working current in slide wire

(A) 10 mA (B) 20 mA (C) 30 mA (D) 40 mA

11

Q.2 A Wheat stone bridge has ratio arms of P-1000 and Q – 100 and is being used to

measure an unknown resistance of R as 25 as shown. Two galvanometer are available.

Galvanometer ‘A’ has a resistance of 50 and a sensitivity of 200 mm/A and galvanometer

‘B’ has values of 600 and 500 mm/A. Ratio of sensitivity of galvanometer ‘A’ to galvanometer ‘B’ is (A) 1

(B) 1.25

(C) 1.75

(D) 2

Q.3 . In the Maxwell bridge as shown below, the values of resistance RX and inductance LX of a coil are to be calculated after balancing the bridge. The component values are shown in the figure at balance. The values of RX and LX will be respectively be

(A) 375 , 75 mH

(B) 75 , 150 mH

(C) 37.8 , 75 mH

(D) 75 , 75 mH

Q.4 A schering bridge is used for measuring the power loss in dielectrics. The specimen are in the form of discs 0.3cm thick having a dielectric constant of 2.3. The area of each electrode

is 314cm2 and the loss angle is known to be 9 for a frequency of 50Hz. The fixed resistor of

the network has a value of 100 and the fixed capacitance is 50pF. Determine the value of the variable resistor required.

(A) 3.17K (B) 4.26K (C) 3.73K (D) 4.54K

Q.5 In the Wheatstone Bridge shown in the given figure, if the resistance in each arm is

increased by 0.05% then the value of Vout will be

(A) 50 mV

(B) 5 mV

(C) 0.1 V

(D) zero

12

Type 6: Wattmeter

For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments

Point to remember:

The power reading of a wattmeter is equal to product of voltage across Potential Coil and

Current through the Current Coil and the cosine of angle between them. These all quantities

can calculated from the phasor diagrams.

Sample Problem 6:

A single-phase load is connected between R and Y terminals of a 415 V, symmetrical, 3-phase, 4-wire system with phase sequence RYB. A wattmeter is connected in the system as shown in figure. The power factor of the load is 0.8 lagging. The wattmeter will read? (A) −795 W (B) −597 W (C) +597 W (D) +795 W

Solution: (B) is correct option

In the figure 0

RY

0

BN

V = 415 30

415V = 120

3

Current in current coil 0

0RY

c 0 0

power factor=0.8V 415 30I 4.15 6.87

Z 100 36.87 Cos =0.8 =36.87

* 0 0 0

0

415Power VI 120 4.15 6.87 994.3 126.87

3

Reading of wattmeter P=994.3 cos(126.87 )

=994.3 -0.60

=-597 W

Unsolved Problems :

Q.1 The resistance of two coils of a Watt meter are 0.01 and 1000 respectively and both are non-inductive. The load current is 20 A and voltage applied to the load is 30 V. Find the error in the readings for two methods of connection

(A) 0.15% high, 0.67% high (B) 0.15% low, 0.67% low (C) 0.15% high, 0.67% low (D) 0.15% low, 0.67% low

13

Q.2 The current coil of dynamometer wattmeter is connected to 30 V DC source in series with

a 8 resistor. The potential circuit is connected through an ideal rectifier in series with a 50 Hz source of 120 V. The inductance of pressure coil circuit and current coil resistance are negligible. Reading of the wattmeter is

(A) 282.84 W (B) 405 W (C) 202.57 W (D) None

Q.3 A voltage: 100 sin t + 40 cos (3t - 30) + 50 sin (5t + 45) volts is applied to the pressure circuit of a wattmeter and through the current coil is passed a current of

8 sint + 6 cos(5t - 120) amps. The readings of wattmeter is?

(A) 939 W (B) 539 W (C) 439 W (D) 1039 W

Q.4 The power in a 3- circuit is measured with the help of 2 wattmeter. The readings of one of wattmeter is positive and that of other is negative. The magnitude of readings is different. It can be concluded that the power factor of the circuit is (A) Unity (B) zero (lagging) (C) 0.5 (lagging) (D) less than 0.5 (lagging)

Q.5 In a dynamometer wattmeter the moving coil has 500 turns of mean diameter 30mm. Find the angle between the axis of the field and moving coil, if the flux density produced by field coil is 15x10-3 wb/m2. The current in moving coil is 0.05 A and the power factor is 0.866 and the torque produced is 229.5x10-6 N.m

(A) 0 (B) 70 (C) 80 (D) 90

Q.6 Consider the following data for the circuit shown below

Ammeter: Resistance 0.2 reading 5A

Voltmeter: Resistance 2K reading 200V

Wattmeter: Current coil resistance 0.2

Pressure coil resistance 2 K Load: power factor =1 The reading of wattmeter is (A) 980W (B) 1030W (C) 1005W (D) 1010W

Q.7 A certain circuit takes 10 A at 200 V the power absorbed is 1000 W .If the wattmeter’s

current coil has a resistance of 0.15 and its pressure coil a resistance of 5000 and an inductance of 0.3 H. The Error due to resistance of two coil of the Wattmeter, if the pressure coil of the meter connected on the load side

(A) 15 W (B) 8 W (C) 11 W (D) 13 W

Q.8 The line to line voltage to the 3-phase, 50Hz AC circuit shown in figure is 100V rms. Assuming that the phase sequence is RYB the wattmeter readings would be?

(A)W1=500W, W2=1000W (B) W1=0W, W2=1000W

(C)W1=1000W, W2=0W (D) W1=1000W, W2=500W

14

Type 7: Energy Meter

For Concept, refer to Measurement K-Notes, Electro- mechanical Instruments.

Point to remember:

The measured value of energy in an energy meter is calculated in terms of meter constant

and number of revolutions and true value of energy is derived from Power and time. Using

these two values we can compute error in energy meter.

Sample Problem 7:

A dc A-h meter is rated for 15 A, 250 V. The meter constant is 14.4 A-sec/rev. The meter constant at rated voltage may be expressed as

(A) 3750 rev/kWh (B) 3600 rev/kWh (C) 1000 rev/kWh (D) 960 rev/kWh

Solution: (C) is correct option

Meter constant (A-sec/rev) is given by

114.4

speed

114.4

K Power

Where ‘K’ is the meter constant in rev/kWh.

114.4

K VI

15 114.4 K

K 15 250 14.4 250

1 1000 3600K 1000 rev/kWh

14.4 250 3600

3600 1000

Q.1 The current and flux produced by series magnet of an induction watt-hour energy Meter are in phase, but there is an angular departure of 30 from quadrature between voltage and shunt magnet flux. The speed of the disc at full load and unity power factor is 40 rpm. Assuming the meter to read correctly under this condition, calculate it’s speed at 1 /4 full load and 0.5 P.F. lagging?

(A) 4.3 rpm (B) 4.4 rpm (C) 4.1 rpm (D) 4.5 rpm

Q.2 Single phase wattmeter operating on 230 V and 5 A for 5 hour makes 1940 Revolutions. Meter constant in revolutions is 400. The power factor of the load will be

(A) 1 (B) 0.8 (C) 0.7 (D) 0.6

15

Q.3 The meter constant of a 230v, 10A watt hour meter is 1800 rev/kwh. The meter is tested at half load, rated voltage and unity power factor. The meter is found to make so revolutions in 138 seconds. The percentage error at half load is

(A) 1.72% fast (B) 0.187 fast (C) 2.8% slow (D) 7.7% fast

Q.4 The voltage-flux adjustment of a certain 1-phase 220V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it 850(instead of 900). The error introduced in the reading of this meter when the current is 5A at power factors of unity and 0.5 lagging are respectively.

(A)3.8mW, 77.4mW (B)-3.8mW, -77.4mW (C)-4.2mW, -85.1 W (D)4.2 W, 85.1 W

Q.5 A 230V, 5A, 50Hz single phase house service meter has a meter constant of 360 rev/KWhr. The meter takes 50 sec for making 51 revolutions of the disc when connected to a 10KW unity power factor load. The error in the reading of the meter is?

(A)0% (B)+0.5% (C)-2.0% (D)+2.0%

Type 8: Digital Meter

For Concept, refer to Measurement K-Notes, Digital Meters.

Point to remember:

The fractional error in a digital meter is the most significant digit.

Sample Problem 8:

A 1

42

digit DMM has the error specification as: 0.2% of reading + 10 counts. If a dc voltage

of 100 V is read on its 200 V full scale, the maximum error that can be expected in the

reading is?

(A) 0.1% (B) 0.2% (C) 0.3% (D) 0.4%

Solution: (C) is correct option

14

2digit display will read from000.00 to 199.99 So error of 10 counts is equal to= 0.10 V

For 100 V, the maximum error is e = 100 0.002 0.( 0. V1) 3

0.3 100%

100

Per

0.3% of readin

centag

g

e error

16

Unsolved Problems:

Q.1 A 1

32

DVM has an accuracy specification of 0.5% of reading 1 digit. What is the

possible error in volt’s when reading 0.1V on 10 V range and also percentage error.

(A) 0.0105 V, 10.5% (B) 0.0015V, 1.5%

(C) 0.015V, 15% (D) None of the above

Q.2 A 010V, 1

42

digit dual slope integrating type DVM can read up to

(A) 99.999 V (B) 199.99 V (C) 20.000 V (D) 19.999 V

Q.3 In a dual slope integrating type digital voltmeter the first integration is carried out for 10 periods of the supply frequencies of 50 HZ. If the reference voltage used is 2 V, the total conversion time for an input of 1 V is

(A) 0.02 Sec (B) 0.05 Sec (C) 0.2 Sec (D) 0.1 Sec

Q.4 A 4-digit DVM (digital Volt-meter) with a 100mV lowest full-scale range would have a sensitivity of how much value while resolution of this DVM is 0.0001?

(A)0.1mV (B)0.01mV (C)1.0mV (D)10MV

Q.5 In a dual slope type digital voltmeter, an unknown signal voltage is integrated over 100 cycles of the clock. If the signal has a 50 Hz pick up, the maximum clock frequency can be? (A)50 Hz (B)5 KHz (C)10 KHz (D)50 KHz

Type 9: CRO

For Concept, refer to Measurement K-Notes, CRO

Point to remember:

The best method to draw Lissajous figure is to plot the points on x-y plane at various time

instants.

Sample Problem 9:

Group-II represents the figures obtained on a CRO screen when the voltage signals Vx = Vxmsinωt and Vy = Vymsin(ωt + Φ) are given to its X and Y plates respectively and Φ is changed. Choose the correct value of Φ from Group-I to match with the corresponding figure of Group-II.

17

Group-I P. Φ = 0

Q. Φ = π/2

R. π < Φ < 3π/2

S. Φ = 3π/2

Codes: (A) P=1, Q=3, R=6, S=5 (B) P=2, Q=6, R=4, S=5 (C) P=2, Q=3, R=5, S=4 (D) P=1, Q=5, R=6, S=4

Solution: (A) is correct option

We can obtain the Lissaju pattern (in X-Y mode) by following method. For φ = 00, Vx = Vxmsinωt Vy = Vymsin(ωt + 00) = sinωt Draw Vx and Vy as shown below

18

Divide both Vy and Vx equal parts and match the corresponding points on the screen. Similarly for φ = 900 Vx = Vxmsinωt Vy = Vymsin(ωt + 900)

Similarly for 2

3

2

we can also obtain for 3

02

19

Unsolved Problems: Q.1 In a cathode ray tube the distance between the deflecting plates is 1.0cm, the length of defalcating plates is 4.5cm and the distance of the screen from centre of the deflecting plates is 33 cm. If the accelerating plate’s voltage is 300 V, then the deflection sensitivity of the tube is

(A) 3.5 mm/V (B) 4.5 mm/V (C) 3.5 cm/V (D) 2.5 mm / V

Q.2 A voltage signal 10sin(314t+450) is examined using an analog single channel cathode ray oscilloscope with a time base setting of 10 msec per division. The CRO screen has 8 divisions on the horizontal scale. Then, the number of cycles of signal observed on the screen will be

(A) 8 cycles (B) 2 cycles (C) 2.5 cycles (D) 4 cycles

Q.3 A lissajous pattern, as shown in figure is observed on the screen of a CRO when voltage of frequencies fx and fy are applied to the x and y plates respectively fx : fy is then equal to

(A) 3:2

(B) 1:2

(C) 2:3

(D) 2:1

Q.4 Voltage E1 is applied to the horizontal input and E2 to the vertical input of CRO. E1 and E2 have same frequency. The trace on the screen is an ellipse. The slope of major axis is negative. The maximum vertical value is 3 divisions and the point where the ellipse crosses the vertical axis is 2.6 divisions. The ellipse is symmetrical about a horizontal and vertical axis. The phase angle difference then is

(A) 2100 (B) 1400 (C) 2400 (D) 1300

Q.5 Horizontal deflection in a CRO in due to E sint while vertical deflection is due to

E sin(t + ) with a positive . Consider the following patterns obtained in the CRO

The correct sequence of these patterns in increasing order of the values of is (A) 3, 2, 5, 1, 4 (B) 3, 2, 4, 5, 1 (C) 2, 3, 4, 5, 1 (D) 2, 3, 5, 4, 1

20

Q.6 A CRO is operated with X and Y setting of 0.5 ms/cm and 100mV/cm. The screen of the CRO is 10cm 8cm (X and Y). A sine wave of frequency 200 Hz and rms amplitude of 300 mV is applied to Y-input. The screen will show?

(A) One cycle of the undistorted sine wave (B) Two cycle of the undistorted sine wave (C) One cycle of the sine wave with clipped amplitude (D) Two cycles of the sine wave with clipped amplitude

Answer Key

1 2 3 4 5 6 7 8

Type 1 C C B A D D

Type 2 C A C D C

Type 3 B C C C B C

Type 4 C B A D B

Type 5 B C A B D

Type 6 A C C D D D B B

Type 7 D B B C D

Type 8 A D D C B

Type 9 D D B A D C

Kreatryx

Subject Test

Electrical and Electronic

Measurements

www.kreatryx.com

1

KST- General Instructions during Examination

1. Total Duration of KST is 60 minutes.

2. The question paper consists of 2 parts. Questions 1-10 carry one mark each and Question 11-

20 carry 2 marks each.

3. The question paper may consist of questions of Multiple Choice Type (MCQ) and

Numerical Answer Type.

4. Multiple choice type questions will have four choices against A, B, C, D, out of which

only ONE is the correct answer.

5. All questions that are not attempted will result in zero marks. However, wrong

answers for multiple choice type questions (MCQ) will result in NEGATIVE marks.

For all MCQ questions a wrong answer will result in deduction of 𝟏/𝟑 marks for a 1-mark

question and 𝟐/𝟑 marks for a 2-mark question.

6. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE.

7. Non-programmable type Calculator is allowed. Charts, graph sheets, and

mathematical tables are NOT allowed during the Examination.

2

Q.1 The total current I=I1+I2 in a circuit is measured whereas 1 2I 150 1A , I 1 50 2A the

limits of error are given as standard deviations. I is measured as

(A) (300 ± 1.24)A (B) (300 ± 1.73)A

(C) (300 ± 2)A (D) (300 ± 2.24)A

Q.2 An ac bridge shown below has the following specifications 1

C 0.5 F and 1

R 1k ,

3

C 0.5 F . If the supply frequency is 1 kHz, determine the dissipation factor.

(A) 2.142

(B) 3.142

(C) 4.142

(D) 5.142

Q.3 A length of cable is tested for insulation resistance by loss of charges method. An

electrostatic voltmeter of infinite resistance is connected between the cable conductor and

earth, forming there with a joint capacitance of 500pF. If is observed that after charging it,

the voltage falls from 250 V to 92 V in 1 minute. The insulation resistance of the cable is

10(A) 2 10 (B) 100,000M Ω 11(C) 1.2 10 7(D) 86.8 10

Q.4 To check the distributed capacitance of a coil, it is resonated at 10 MHz with 120pF and

then is resonated at 15 MHz with 40 pF. What is its equivalent distributed capacitance?

(A) 12pF (B) 18pF (C) 24 pF (D) 30pF

Q.5 A

14

2 digit voltmeter is used to measure the voltage value of 0.3861 V on a 1 range. It

would be display

(A) 0 .3 8 6 1 (B) 0 0 .3 8 6

(C) 0 0 0 .3 8 (D) .3 8 6 1 0

Q.6 A voltmeter reading 80 V on its 100V range and an ammeter reading 80 mA on its 150 mA

range, are used to determine the power dissipated in a resistor. Both these instruments are

guaranteed to be accurate within ±2%, at full scale deflection. The limiting error in the

measurement of the power dissipation is

(A) 3.25% (B) 4.25% (C) 5.25% (D) 6.25%

3

Q.7 A set of independent current measurement taken by four observers was recorded as :

115.02 mA, 115.11,A, 115.08mA and 115.04 mA. What is the average range of error?

(A) 0. 045 mA (B) 0.04 mA (C) 0.0425 mA (D) 0.5 mA

Q.8 In a thermocouple based instrument, if 20% harmonic is present then error in the current

reading is

(A) 0% (B) 2% (C) 4% (D) 5%

Q.9 Calculate the value of the multiplier resistor for a 10 V rms range on the voltmeter as

shown in the figure given below.

(A) 2.3 k Ω

(B) 3.3 k Ω

(C) 4.3 k Ω

(D) 5.3 k Ω

Q.10 A voltmeter, having a sensitivity of 2k Ω/V, connected across an unknown resistance in

series with a milliammeter, reads 100 V on 150 V scale. If the milliammeter reads 10 mA, the

error due to loading effect of voltmeter would be – y% , then the value of y is __________.

Q.11 A coil with a resistance of 3 Ω is connected to the terminals of a Q meter. Resonance

occurs at an oscillator frequency of 5 MHz with a capacitance of 100 pF. If –x% is the error

introduced by the insertion resistance Rsh,=0.1 Ω then the value of x is ___________.

Q.12 Calculate the constants of a shunt to extend the range of 0 – 5 A moving iron ammeter to

0 – 50 A. The instrument constants are R = 0.09 Ω and L = 90µH If the shunt is made non-

inductive and the combination is correct on d.c. find the full scale error at 50 Hz.

(A) 3.8% high (B)2.8%low (C) 2.8% high (d)3.8%low

Q.13 The coil of a 300 V moving iron voltmeter has a resistance of 500 W and an inductance of

0.8H.The instrument reads correctly at 50Hz. A.C. supply and takes 100 mA at full scale

deflection. What is the percentage error in the instrument reading, when it is connected to 200V

d.c.supply.

(A)200.6 V (B)205.6V (C) 210 V (D)212.2V

4

Q.14 A single-phase load is connected between R and Y terminals of 415 V, symmetrical 3-

phase, 4 wire systems with phase sequence RYB. A wattmeter is connected in the system as

shown in figure below. The power factor of the load is 0.8 lagging. The wattmeter will read

(A)–197.23 watt

(B)–248.58watt

(C)–295.29 watt

(D)–298.23 watt

Q.15 An energy meter rated as 5A, 230 V makes 500 revolutions per kWh. If in a test at full load

unity power factor, it makes 5 revolutions in 30 seconds. Then which of the following statement

is true?

(A) The meter runs faster and error is 4.16%

(B) The meter runs faster and error is 4.35%

(C) The meter runs slower and error is 4.16%

(D) The meter runs slower and error is 4.35%

Q.16 Figure shown below is the circuit of a capacitance comparison bridge. If R1 = 20 Ω

R2=30 Ω R3 = 25 and C3 = 10 pF than the value of Rx and Cx respectively are.

(a) 27.5 , 26/3 pF

(b) 30.5 , 28/3 pF

(c) 35.5 , 22/3 pF

(d) 37.5 , 20/3 pF

Q.17 Figure gives parameter values of an AC bridge at balance when supply frequency is

1500Hz. Find Zx assuming it to be a resistance Rx in series with an inductance or capacitance

and choose the correct option

(A) 210 F

(B) 210

F

(C) 210

H

(D) 210

5

Q.18 A CRO is used to measure the voltage as shown below in the diagram. CRO probe has the

impedance of 600 k with a capacitance in parallel of a value of 20 pF. The reading of CRO will

be _________ V.

Q.19 For an electrodynamometer ammeter. The mutual inductance M varies with deflection ϴ

(ϴ is in degree) as M = -6cos(ϴ +45o) mH. For a direct current of 50 mA corresponding to a

deflection of 600, the deflecting torque will be _______µNm.

Q.20 A current of i 0.5 0.3 sin t 0.2 sin 2 t A is passed through the circuit shown in

figure given below. _______ ampere is the sum of readings of each instrument. 6( = 10 rad/s)

6

Power Systems KST

Answer Key

1 2 3 4 5 6 7 8 9 10

C C A B 4 C B B C D

11 12 13 14 15 16 17 18 19 20

D C A C C D D 16.8-17.7 B C