Kuestion

Electrical and Electronic

Measurements

www.kreatryx.com

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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.

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Manual for Kuestion

Why Kuestion?

It’s very overwhelming for a student to even think about finishing 100-200 questions per

chapter when the clock is ticking at the last moment. This is the reason why Kuestion serves

the purpose of being the bare minimum set of questions to be solved from each chapter

during revision.

What is Kuestion?

A set of 40 questions or less for each chapter covering almost every type which has been

previously asked in GATE. Along with the Solved examples to refer from, a student can try

similar unsolved questions to improve his/her problem solving skills.

When do I start using Kuestion?

It is recommended to use Kuestion as soon as you feel confident in any particular chapter.

Although it will really help a student if he/she will start making use of Kuestion in the last 2

months before GATE Exam (November end onwards).

How do I use Kuestion?

Kuestion should be used as a tool to improve your speed and accuracy chapter wise. It should

be treated as a supplement to our K-Notes and should be attempted once you are comfortable

with the understanding and basic problem solving ability of the chapter. You should refer K-

Notes Theory before solving any “Type” problems from Kuestion.

© 2014 Kreatryx. All Rights Reserved.

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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%

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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

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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

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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

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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

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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

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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

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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

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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

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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