Electrical and Electronic Measurements

CHAPTER

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ELECTRICAL AND ELECTRONIC MEASUREMENTS OPERATING FORCES Any indicating instrument requires three torques for its operation, namely deflecting force, controlling force and damping force. (a) Deflecting force l The systems that create a deflecting force are said to be deflecting systems or moving systems and their task is to convert the electrical current or potential into a mechanical force. This force causes the deflection of pointer. (b) Controlling force l The systems that produce controlling force are said to be controlling systems. These forces are incorporated in an indicating system, so that the deflection of pointer should be proportional to the signal’s magnitude. l Controlling forces are opposite to the deflecting force and the instant when the two forces becomes equal, the pointer comes to rest. Thus, due to the controlling force, the deflection of the pointer is definite for a magnitude of current. l Due to the controlling force only, the moving system comes back to the zero position when the force causing that movement is removed. l The following methods produce a controlling force in an instrument Gravity control Spring control (c) Damping force l Damping forces are necessary for a system, so that the moving system comes to its equilibrium state rapidly and without any oscillations. l Due to the damping forces, the pointer quickly comes to its final steady position without over shooting or without oscillating near its final value. l The following methods produce damping forces in an instrument: Air friction damping Eddy current damping Fluid friction damping electromagnetic damping

MOVING ELEMENT SUPPORT Power consumption of the instrument should be kept low, so that the introduction of the instrument into a circuit causes least change in the circuit condition. Therefore, in order to obtain a correct reading, frictional forces of the system should be kept minimum and hence, following types of supports may be used: Suspension Taut suspension Pivot and jewel bearings Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

2

PERMANENT MAGNET MOVING COIL (PMMC) INSTRUMENT

l l

l l l l

l l l

PMMC instruments are used only to measure DC quantities and not AC quantities. This is so, because permanent magnets are used for creating magnetic fields. Constructional diagram of a PMMC instrument is shown in the fig. 1. A PMMC type instrument consists of a moving coil, made up of enameled copper wire, which is mounted on an aluminium drum. This drum is pivoted on the spindle of a jeweled bearing. The pointer of the scale is also connected to this spindle. The end of the moving coils are connected to springs which produce a controlling force and also provide a way to lead current in and out of the coil. The moving coil is placed in the magnetic field which is produced by the two magnets. The supply to be measured is provided to the ends of the moving coil. We know that when a current carrying conductor is placed in a magnetic field, a force is induced in it. Therefore, when the current flows through the coil, a force is induced in it and since a spindle is attached to it, there is a movement in the spindle. The movement of the spindle moves the pointer. This movement is controlled by the controlling springs and thereby we get the final reading. To avoid unwanted oscillations of the pointer, a damping torque is produced by the eddy current method. Torque equation

Pointer Jewelled Bearing Control Spring Coil

Spindle Input Supply

l

S

N

Permanent Magnet

Aluminium Drum

Fig. 1: PMMC instrument

Deflecting torque, Td µ I And controlling torque, Tc = Kcq At steady position of pointer, Tc = Td And thus, I µ q since the deflection is directly proportional to the current flowing through the instrument, we get a uniform scale for the instrument. D.C. voltage and D.C. current can be measured using PMMC instruments.

MOVING IRON TYPE INSTRUMENT Such instruments are of two types – attraction type and repulsion type. Attraction Type l Constructional arrangement of attraction type moving iron type instrument is shown in the fig. 2. l Moving system of this instrument comprises a moving iron which is fixed at one end and its other end is free to move. The pointer is also attached to this moving iron. l There is a former which supports the coil and the coil is made up of enameled copper wire. l Controlling forces are provided by the gravity control method and damping forces are provided by the air friction damping torque. l Voltage or current to be measured is given as an input across the coil and when current flows through the coil, a magnetic field is produced. l Due to the magnetic field, the moving iron is attracted towards it and due to this force of attraction, the pointer deflects. This deflection of pointer is controlled by the controlling forces and thus, the pointer comes to rest and we get a reading on the scale. l Torque equation Deflecting torque, Td = KI 2

Air Damping Chamber

Scale Pointer Magnetic Coil Balanced Weight

Moving Iron Piece Control Weight Former

Fig. 2: Attraction type instrument

And controlling torque, Tc = K1 sin q At steady position of pointer, Tc = Td l

Air Damping Piston Arm

And thus, q µ I 2 Since deflection of the pointer is directly proportional to the square of the current, the scale of the attraction type moving iron type instrument is not uniform. Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

3 Repulsion Type In repulsion type instrument, two vanes of soft iron are used inside the coil. One vane is fixed and the other one is free to move. When current flows through the coil, both vanes are magnetized and therefore, there is a force of repulsion between the two and this force acts as the driving force for the instrument. Two different designs of repulsion type moving iron type instruments are common – radial type and co-axial type.

ELECTRODYNAMOM ETER TYP E INSTRU MENT l l l

Scale l

Pointer

l l

Moving Coil Input A.C/D.C

l

F2 F1 Fixed coil

l

Fig. 3: Electrodynamometer instrument l l

This type of instrument is capable of measuring AC voltage and AC current as well as DC voltage and DC current. The fig. 3 shows the arrangement of electrodynamometer type instrument. Magnetic field in this type of instrument is produced by a fixed coil which is divided into two parts: F1 and F2. The pointer of the instrument is attached to the moving coil and this moving coil is generally wounded on a non-metallic former. The moving coil is placed in series with the two fixed coils. Air friction damping system is used to provide damping force and controlling torque is provided by the control springs. When measuring input is applied to the instrument, current starts flowing through the fixed coil and moving coil. Current flowing through the fixed coil creates a magnetic field and the current carrying moving coil is placed in that magnetic field. Therefore, the moving coil experiences a force which results in the movement of pointer over the scale. This force is controlled by the control springs and thus, the pointer comes to rest and in this way, we get the reading. Torque Equation Deflecting torque, Td µ I 2 And controlling torque, Tc = K cq At steady position of pointer, Tc = Td

l

And thus, q µ I 2 Since, the deflection of the pointer is directly proportional to the square of the current, the scale of the electrodynamometer type instrument is not uniform.

INDU CTIO N TYPE INSTRUMENTS l l

Pointer Scale

Laminated Magnetic System

Fig. 4: Induction type instrument

l l

Spring l

Laminated Iron Core l

Aluminium Drum L R Supply

l

Induction type instrument works on the principle of induction and therefore, this method is used in measuring AC voltage and AC current. The maximum length of scale is possible in induction type instrument as the angle of deflection of this instrument may be 360 degrees. Ferraris type induction type instrument is shown in the fig. 4 This type of instrument consists of two coils – one coil is highly resistive and the other one is highly inductive. When we give signal to the instrument, current starts flowing through the system and this current splits up, i.e. some portion of current flows through the resistive coil and the remaining current flows through the inductive coil. These two currents produce two flux and these two flux produce the deflecting torque which rotates the laminated iron core which is mounted on the spindle. With the movement of the spindle, the pointer also deflects. This deflection of pointer is controlled by controlling forces, which is produced from the control spring.

MEAS UREMENT OF VOLTAGE Voltmeter is an instrument used for measuring the voltage between any two points in an electrical circuit. There are two types of voltmeters: Analog and Digital. In an analog voltmeter, a pointer shows deflection across the scale, which indicates the voltage Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

4 reading with respect to the applied current; whereas a digital voltmeter gives a numerical display of the voltage according to the applied current, with the help of an ADC (analogto-digital convertor). Analog Voltmeter : The fig. 5 shows the D’arsonval type moving coil galvanometer. The device consists of a pointer, a full deflection scale, magnets and a coil. The pointer is attached with the coil, which is suspended in a magnetic field, as shown in the figure 5. Under no biasing condition, the pointer is at the zero position, i.e. at the center of the scale, which also helps to notify if the voltage changes its polarity. Now, in order to measure the voltage, the galvanometer is connected in circuit with a series resistor as shown in fig. 6; the resistor is connected in series to ensure that the angular rotations of the indicator are directly proportional to the applied voltage. Normally, this device is used in case of direct current; however, we can also have an AC source by using a rectifier in the circuit. The output of this voltmeter is expressed in ‘ohms per volt’. Digital Voltmeter : A Digital voltmeter uses the analog-to-digital convertor for displaying the voltage on the numerical display. The accuracy of a Digital voltmeter is higher than that of an analog voltmeter. The main parts of Digital voltmeter are an amplifier and a numeric display as shown in fig. 7. Just like the analog voltmeter, a digital voltmeter is also connected in series with the circuit, but the value of series resistance is fixed by the manufacturer (generally about 10 mega ohms).

Deflection Scale 10 15 5 0

20

ter Poin

Magnets Coil Fig. 5: Analog Voltmeter Voltmeter Voltage to be measured

Resistor

Fig. 6: Voltmeter connection in circuit Display

MEASU REMENT OF CU RRENT Current is measured with the help of an instrument known as ‘Ammeter’. It consists of a deflection scale and a pointer. An ammeter is always connected in parallel with the circuit as shown in fig. 8, and the simplest method for determining circuit current is: Thermocouple or Resistive Heating : This technique is based on the Seebeck effect. In this method, a thermocouple made up of two dissimilar metals, is connected in parallel with an ammeter, as shown in the fig. 9. The hot junction is welded to a heater wire or conductor, while the cold junction is connected to the ammeter. Now, when current passes through the heater wire, the temperature of the junction gets raised, which leads to an increase in the thermoelectric EMF and sends more current to the ammeter. Since the deflection on the scale depends upon the heating effect and the length of the conductor, we can say that the amount of heat produced is directly proportional to the square of the current and resistance. Q µ I2R

110 Function keys

I/O port

Fig. 7: Digital Voltmeter

MEASUREMENT OF POWER Power is generally measured through Wattmeters of which there are two types – electrodynamometer type and the induction type. Electrodynamometer Type Wattmeter l Electrodynamometer type wattmeter consists of two coils as shown in the fig. 10. l Fixed coil is split up into two identical coils and are made up of thick copper wires.This fixed coils are connected in series with the load and so they carry the current in the circuit, thus, they form the current coil of the circuit. l Moving coil is mounted on the spindle and is placed in between the two fixed coil.The moving coil is connected across the voltage and therefore it forms the pressure coil of the wattmeter. l Spring control method is used for producing controlling torque and damping torque is produced by air friction damping system. l The moving coil act as a current carrying conductor placed in a magnetic field and thus force induced in it and since pointer is connected to this moving coil, it deflects on the scale. l This deflection is controlled by the controlling spring and at last pointer comes to rest showing a reading.

A

R

Fig. 8: Circuit arrangement for Current measurement using ammeter Hot junction A R

Conductor B Cold junction

Thermocouple Ammeter connection

Fig. 9: Resistive Heating

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Ammeter

5 Spindle

Moving Coil

Fixed Coil

Fig. 10: Electrodynamometer type wattmeter V

Shunt electromagnet (P.C)

(cc)

S.C. bands

Spindle f1 Al disc f2

Series electromagnet

Fig. 11: Induction type wattmeter

Induction Type Wattmeter l Induction type wattmeter consists of two electromagnets – the upper one is known as shunt electromagnet and the lower one is known as series electromagnet as shown in the fig. 11. l The coil of shunt electromagnet is made up of a thin enameled copper wire and this electromagnet also possesses some self short-circuiting bands. The shunt electromagnet acts as a pressure coil. l The coil of series electromagnet is made of thick wire and the number of turns in this coil is less compared to that of the shunt electromagnet. The series electromagnet acts as a current coil. l An aluminium disc mounted on a spindle is in between the two electromagnets. l The spindle is supported by a jeweled bearing. l Eddy current damping system is used for producing a damping force and the controlling force is provided by the spring control method. l When we apply voltage across the pressure coil, then a current starts flowing in the coil, which further produces flux, which is also of the same nature as that of the current. Let this flux be f1. l When this flux reaches the centre of shunt electromagnet, due to the flux , voltage is induced in the short circuited band and due to this voltage, current starts flowing in the band which has its own flux f2. l The difference of flux (f1 – f2) produces a rotating force due to which the aluminium disc starts rotating and since the spindle is connected to it, the pointer also deflects on the scale to give the reading.

MEASUREMENT OF ENERGY Energy is measured by an induction type energymeter, the arrangement for which is shown in the figure and its whole operation is divided into four parts which are explained below :-

Shunt Electromagnet Laminated Core PC S C Bands Braking Magnet Aluminium Disc Series Electromagnet

Fig. 12: Induction type energymeter

(a) Driving System l The driving system consists of two electromagnets – shunt electromagnet and series electromagnet. l The number of turns in the coil of the shunt electromagnet is more than that of the series electromagnet. l The coil of the series electromagnet has negligible resistance and since it carries the load current, this coil is known as the current coil. l The coil of the shunt electromagnet is connected across the supply and therefore, it carries a current which is in proportion to the voltage and hence it is known as the pressure coil. l Short circuited bands are also provided on the shunt electromagnet in order to provide some phase difference. (b) Moving system

The moving system consists of a disc which is mounted on a light alloy shaft. The disc is made up of aluminium which is not magnetic but is a conducting material. l This disc is positioned in between the air gap of the shunt and the series electromagnet. (c) Braking system l At the edge of the aluminium disc, a permanent magnet is positioned, which forms the braking system of the instrument. l It controls the speed of the aluminium disc. l The disc acts as a current carrying conductor placed in the magnetic field (of permanent magnet), and hence voltage is induced in it, due to which the current is produced. Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html l

6 This current produces a braking torque and by adjusting the position of the permanent magnet, we can change the braking torque and hence the speed of the disc. (d) Registering system l A gear arrangement is there in the instrument which is in the spindle. l It continuously records the number which is proportional to the revolutions made by the moving system. l

MEASUREMENT OF PO WER FA CTOR

1

The power factor is measured by an electrodynamic power factor meter as described below: Fixed coil q

sin q sin f = cos q cos f

Moving coil

B

A

L

R

Fig. 13: Electrodynamic type power factor meter

90°

Let the deflecting torque in A = TA = kVIM max cos f sin q And the deflecting torque in B = TB = kVIM max cos(90 - f ) sin(90 + q ) When the pointer comes to rest, both the torques are equal. kVIM max cos f sin q = kVIM max cos(90 - f )sin(90 + q )

Lag

Lead

Supply Voltage

Electrodynamic Type Power Factor Meter l The circuit arrangement is shown in the fig. 13. l It consists of a fixed coil, which is split up into two parts. This coil acts as a current coil and it carries current of the circuit under test. l Two identical coils A and B are pivoted on a spindle and are cross coupled at 90 degrees. l Pressure coil A is connected in series with a non- inductive resistance and pressure coil B is connected in series with a highly inductive choke. l Both A and B are connected across the voltage, and the value of R and L are so adjusted that the two coils have the same impedance i.e. R = w L. Thus, both of them carry the same amount of current. l The current through coil A is in the same phase with the circuit voltage, while the current in coil B lags the voltage by an angle nearly equal to 90o as shown in fig. 14. l Two torques are produced in the system, one acting on coil A and the other acting on coil B. l Since the coils are cross coupled and the torques acting on both the coils are opposite in nature, therefore it rotates and rests at that point where the two opposite torques are equal. l A pointer is attached to this moving coil and hence we get reading on the scale. l Torque Equation

IA

f

V

I

IB Fig. 14: Phasor diagram

tan q = tan f

q =f

Primary

Therefore, the deflection of the instrument is a measure of the phase angle of the circuit and the instrument can be directly calibrated in terms of power factor.

INSTRUMENT TRANSFORMER The transformer used in conjunction with measuring instruments for measurement purposes is called an Instrument Transformer. To measure current, Current Transformers (CTs) are used and to measure voltage (or potential), Potential Transformers (PTs) are used. Secondary Current Transformer l Such a transformer is used to measure the current. l It is a step-up transformer whose primary (lv side) is connected to the line whose S current is to be measured and its secondary (hv side) is connected to the ammeter as shown in fig. 15. l The ammeter gives a reading and this reading when multiplied by the 0-5A transformation ratio, gives the value of line current which was flowing through A the line. Fig. 15: Current transformer l By using CT, measurement is carried out using a low rating of ammeter. Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

7

V

Fig. 16: Potential transformer b

BRIDGES Q

P G

a

Wheatstone Bridge This bridge is used to measure the medium resistance and its arrangement is shown in fig. 17. Let P and Q = known resistances known as ratio arms S = known variable resistance R = unknown resistance Kb and Kg= Battery and Galvanometer key E = supply voltage P The value of the unknown resistance when the bridge is in balanced state is R = S Q

c

KG S

R d

Kb

E

Fig. 17: Wheatstone bridge

b G KG

P

p

R

a I

q

f

Q

d S

e E

c

KB

Fig. 18: Kelvin’s bridge

I1 b E1

Z1 R1

a

L1

E2

Z2 R2

I1

D I2

I2

R4

R3 Z E3 3 L3

Potential Transformer l Such a transformer is used to measure voltage. l It is a step-down transformer whose primary (hv side) is connected to the line whose voltage is to be measured and its secondary (lv side) is connected to a voltmeter as shown in fig. 16. l The voltmeter gives a reading and this reading when multiplied by the transformation ratio, gives the value of the line voltage of the line. l By using PT, measurement is carried out using a low rating of the voltmeter and insulation required for measuring such a high voltage is minimized.

c

Z4

E4 d

Kelvin’s Double Bridge Kelvin’s double bridge is used to measure low resistance and its arrangement is shown in the fig. 18. p and q = first set of ratio arms P and Q = second set of ratio arms S = known variable resistance R = unknown resistance r = lead resistance E and KG= Battery and Galvanometer key p P Also, = q Q The value of unknown resistance when the bridge is in a balanced state is P R= S Q Maxwell’s Induction Bridge This bridge is used to measure self inductance is shown in fig. 19. R1 = resistance of the coil (unknown) L1 = inductance of the coil (unknown) R2, R4 = known non-inductive resistances R3 = variable resistance L3 = known inductance r3 = internal resistance of L3 At balanced condition, R R1 = 2 ( R3 + r3 ) R4 L1 = L3

E

Fig. 19: Maxwell’s Induction Bridge

R2 R4

Q – factor =

w L3 R3 + r3

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8 Maxwell’s Inductance-Capacitance Bridge This bridge is also used to measure self inductance. Its circuit diagram and vector b diagram is shown in fig. 20 and 21 respectively I1 L 1 R1 = resistance of the coil (unknown) Z1 R2 Z2 R1 L1 = inductance of the coil (unknown) I1 c R2, R3, R4 = known non-inductive resistances D c a C = known variable capacitor I2 IC At balanced condition, R I R3 IR 4 I Z3 Z4 R2 R3 I2 d R1 = R4 E L1 = CR2 R3 Fig. 20: Maxwell’s inductance Q – factor = wCR4 capacitance bridge Desaultay Bridge This bridge is used to measure the value of capacitance circuit diagram is shown in fig. 22. b C1 R2 I1 a c D I1 I2 I R4 C3 I I2 d

I 1 jwL1

I2 I1R1

I1,I2

I1R2=E2

Fig. 21: Vector diagram

E4 = I2R4 E2 = I1R2

Fig. 22: Desaultay Bridge C1 = unknown capacitor C3 = known variable capacitor R2,R4 = known standard resistance I I E1 = 1 E3 = 1 jwC1 jwC3 At balanced condition, C1 =

E

E1

I1, I2

E

R4 C3 R2

Fig. 23: Vector diagram

Anderson Bridge This bridge is used to measure self inductance. Its circuit diagram and vector diagram are shown in fig. 24 and 25 respectively Bridge arrangement b r1 R1L1

R2 e

a I

D

I1

R3

IC

r I2

c

I1 c

R4

Fig. 24: Anderson bridge R1 =

R2 R3 - r1 R4

L1 = C

I

E

I2R3

d IR = I2 – IC E

At balanced condition

E1 I 1 jwL1

I2 IC ICr

IRR4 IR

I1(R1+r1 ) E2=I1R2 I = C jwc Fig. 25: Vector diagram

R2 [ R3 R4 + r ( R3 + R4 )] R4 Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

I1

9

POT ENTI OM ETER Pin No-1

A potentiometer is a three terminal device shown in fig. 26 which is used as a variable resistor. The outer terminals are fixed, while the middle terminal can vary; the middle terminal is either in the screw-shape or in the form of a control shaft along with a wiper. The screw moves over the resistive element and shows continuous variation in the resistance of the element, which is connected between the outside and the middle terminal of the potentiometer. A potentiometer is basically used to control the voltage of the circuit. Although there is one more application of the potentiometer: it can be used as a rheostat by connecting its middle terminal with an outside terminal. (A rheostat is mainly used to control the circuit current.)

Top view

Side view

Fig. 26: Potentiometer I/P Voltage

Ranking & attenuator

DIGITAL VOLTMETERS

Input Comparator

Start Pulse Clock Osc.

Gate

Counter

Read out 0000

Stop Pulse Ramp Gen.

Ground Comp.

Sample rate MV

Fig. 27: Ramp type voltmeter Ramp voltage +12 V Unknown Voltage 0

Time

-12 V Gating time interval

t

Clock Pulses

Fig. 28: Timing diagram Error Amplifier

Potentiometer Read out adjustment 0000 device

Error Signal +

-

R

Ref. Voltage Feedback source Sliding voltage contact

Comparator

On Known Voltage

Fig. 29: Potentiometric digital voltmeter Display

110 I

V

I/O port -

Ramp Type Voltmeter l The fig. 27 shows the arrangement of a ramp type voltmeter. l Input comparator receives two voltages – one from the input and the other from the ramp generator. l the instant when both the voltages are equal, the comparator generates a pulse which opens the gate. l The ramp voltage continuous to decrease till it reaches the ground level. At this instant, another comparator generates a pulse and closes the gate. l The time elapsed between this opening and closing of gate is as indicated in the timing diagram as shown in fig. 28. l During the time interval, pulses from a clock pulse generator pass through the gate and are counted by the counter and the result is displayed at the read out. l The sample rate multivibrator sends a pulse to the counters which sets the reading of the display again back to zero. Potentiometric Digital Voltmeter l Arrangement of this type of voltmeter is shown in the given fig. 29. l The input voltage to be measured is applied to a comparator which is also known as the error detector. l The value of feedback voltage depends upon the position of the sliding contact. l The unknown voltage and feedback voltage are compared in the comparator and the output of the comparator is the difference between the voltages which is known as the error signal. l This signal is amplified and is fed to the potentiometer adjustment device which moves the sliding contact of the potentiometer. l The sliding contact moves to such a place where the unknown voltage equalizes feedback voltage and when this happens, the sliding contact comes to rest. l The position of the potentiometer adjustment device at this point is indicated in numeric form on the display.

MULTIMETER

R Transistor testing

Digital voltmeters are the instruments which measure AC or DC voltages and display the results in a numeric form. This method is more advantageous than the analog method as observational errors are eliminated. There are two types of digital voltmeters – ramp type voltmeter and potentiometric digital voltmeter.

+

A Multimeter is a measuring instrument which is used to measure several functions like current, voltage and resistance. Multimeters are of two types: analog and digital. In an analog multimeter, a micro ammeter is used which consists of a pointer and a deflection scale; whereas in digital Multimeter, the measured value is displayed in numerals on a digital screen, the display can either be in the seven segment format or in the liquid style display of an LCD. Nowadays, digital multimeters are preferred over analog for all types of measurements. But analog multimeters are still used when we have to monitor the values which rapidly vary over a wide range.

Fig. 30: Multimeter Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

10 Main applications of multimeter: (a) Measuring of AC and DC (b) Measuring voltage and current (c) Measuring resistance (d) Testing of continuity in circuit

PHASE AND FREQUENCY MEASUREMENT Phase Measurement In phase measurement, phase of the signal which is fed into the system is compared with the phase of the signal responded by the system. The difference between the phases of both the signals is recorded as phase difference, and it is due to the electrical properties of the system. The fig. 31 shows the phase difference or phase shift between the applied and responded signal: In time and frequency domain, phase shift is defined in terms of time instead of angle. During a complete cycle, time taken by a signal to constitute a 3600 phase shift is given as — T=

Phase difference

Fig. 31: Phase difference

1 f 360

Or 1 f 360 There is a device named Phase Analyzer, which is used to display the phase-frequency curve over +1800 to –1800 ranges. There are different types of phase measurement processes – (1) Complex impedance (2) Group delay (3) Deviation from linear phase (4) Amplitude modulation –to– phase modulation conversion

T=

Frequency Measurement Frequency is defined as the number of oscillations or repetitions of a signal per unit of time. Three methods are used for frequency calculations: (1) Counting – Since we know that frequency is inversely proportional to time, i.e. F=

(2)

(3)

1 T

So, by calculating the number of oscillations of a signal or events which occur in a specific time period, we can calculate the frequency of that signal. Frequency counter – A frequency counter is an electronic device which is used to display the frequency of a signal in Hertz. In a frequency counter, with the help of transducer, the processes which are not electrical in nature are converted into electronic signals. These processes are known as cyclic processes, and they include mechanical vibrations, rate of rotating shaft, etc. These electronic signals are then applied to the frequency counter, where digital logic counts the number of cycles of a specific time interval. The results obtained are displayed on the digital display. This device is used to determine a frequency up to 100 GHz. Heterodyne or frequency conversion method – In this method, the signal whose frequency is to be measured is nonlinearly mixed up with a reference signal of known frequency. The difference of frequency between these two signals is small enough to be measured by the frequency counter. The rest of the procedure for frequency calculation is followed by the frequency counter. This method is suitable for high frequency calculation. Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

11

Q-ME TER Q-meter is a device which is used to determine the quality factor of a circuit. Generally, this device is used in radio frequency circuits, where it is desirable to know how much amount of energy is dissipated from the system in a non-ideal reactive form. Q-factor is given asPeak Energy Stored Q = 2p ´ Energy dissipated per Cycle

OSCILLOSCOPES Input Signal

Vertical amplifier

CRT

Delay Line

To CRT HV Supply LV Supply Electron To all Circuit Gun Trigger Time base Horizontal amplifier generator Circuit

Electron beam

l l

l Screen

l

Time/division

Fig. 32: Oscilloscopes

l l

Block diagram of the cathode ray oscilloscope is shown in the fig. 32. The cathode ray tube (CRT) generates the electron beam, accelerates it to a high velocity, focuses it and deflects the beam to produce the image on a phosphor screen. The power supply block provides the input required by the CRT to generate the beam. Horizontal and vertical amplifiers are required to strengthen the weak signal and also to focus it correctly on the screen. The time base generator provides the potential of variable frequency to be applied to the x-plate. The function of the vertical delay line is to allow the operator to observe the leading edge of the signal waveform by delaying the signal drive.

POTENTIOMETRIC RECORDER A Potentiometric recorder is a device which is used to record and monitor the voltage which a voltmeter measures. The most important application of Potentiometric recorder is that it enables us to monitor even minute unattended voltages during a measurement, which varies over a wide range of 0 – 500 V. This device is highly sensitive in nature and produces accurate results. The gain adjustment panel of recorder helps in determining the sample voltage over full-scale recorder deflection. The differential voltage determined by the recorder is not impaired in any way due to its high leakage resistance, which is approximately 500,000 Mega Ohms.

ERROR ANALYSIS There are three basic types of errors which are obtained while taking measurements in lab with any electrical system and these are – (1) Random Error (2) Systematic Error (3) Gross Error (1) Random Error : Those uncontrolled or uncertain fluctuations which randomly affect the results of experiments are called random errors. Examples include the change in temperature due to sunlight around temperature sensors, air fluctuations caused by opening and closing of doors, etc. This error type is difficult to remove but can be solved by calculating the estimated standard deviation of collected data. (2) Systematic Error : Instrumental mistakes, methodological and personal mistakes fall under this category. Instrumental errors can be caused in many ways, for e.g. an improper placement of device. Methodological errors are caused because of the selection of wrong alternatives for experiment, and personal mistakes are caused by observers and performers like noting down incorrect readings, etc. These errors can be easily eliminated by careful observation and correct operation of instruments. (3) Gross Error : This error is caused either by an instrument failure or the carelessness of the experimenter. In order to minimize this error, a set of precision measurements must be taken by the experimenter.

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12

SOME SOLVED EXAMPLES 1.

A DC ammeter has a resistance of 0.1 ohm and its current range is 0 – 100 A. If this range is to be extended to 0 – 500 A, then the meter requires the following shunt resistance: (a) 0.010 ohms (b) 0.011 ohms (c) 0.025 ohms (d) 1.00 ohms Solution : Maximum current which can flow through the ammeter is 100 A. So, the value of the shunt resistance which is required to measure the 500 A current must be chosen in such a way that 400 A current flows through the shunt. From current division formula, ( I - I a ) Rshunt = I a Ra Ia 100 ´ Ra = ´ 0.1 Therefore, Rshunt = (I - I a ) (500 - 100) 1 Thus, Rshunt = ´ 0.1 = 0.025W 4 Hence, option (c) is the correct answer. 2. The setup in the fig. 1 is used to measure resistance R. The ammeter and voltmeter resistances are 0.1 ohm and 2000 ohm, respectively. Their readings are 2 A and 180 V respectively, yielding a measured resistance of 90 ohm. The percentage error in the measurement is (a) 2.25% (b) 2.35% (c) 4.5% (d) 4.71% Solution : From Ohm’s law, the current flowing through voltmeter, 180 V IV = = = 0.09 A R 2000 Therefore, actual ammeter current should be 2 - 0.09 = 1.91A 180 Therefore, actual value of resistance should be = = 94.24 ohm 1.91 94.24 - 90 ´ 100 = 4.71% The percent error = 90 Hence, option (d) is the correct answer. 3. Two wattmeters, which are connected to measure the total power on a three – phase system supplying a balanced load, read 10.5 kW and – 2.5 kW respectively. The total power and the power factor, respectively, are (a) 13.0 kW, 0.334 (b) 13.0 kW, 0.684 (c) 8.0 kW, 0.52 (d) 8.0 kW, 0.334 Solution : We know that the total power of the circuit when measured by the twowattmeter method is the algebric sum of the readings of the two wattmeters.

I2

R A 1000 W V

Fig 1:

Therefore, P = W1 + W2 = 10.5 + ( -2.5) = 8 W To find the power factor, firstly we have to calculate the value of f Therefore, 3 ´ (W1 - W2 ) 3 ´ (10.5 - ( -2.5) 3 ´13 f = tan -1 = tan -1 = tan -1 W1 + W2 10.5 + ( -2.5) 8

f = tan -1 (2.81458) = 70.440

4.

Power factor = cos f = cos 70.440 = 0.334 0.023W 20A Hence, option (d) is the correct answer. The circuit in fig. 2 is used to measure the power consumed by the load. The 1000W current coil and the voltage coil of the wattmeter have 0.02 ohm and 1000 ohm 200V resistances respectively. The measured power compared to the load power will be (a) 0.4 % less (b) 0.2 % less Fig 2: (c) 0.2 % more (d) 0.4 % more Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

Unity PF

13 Solution : We know that the reading which we get from a wattmeter comprises the power which is consumed by the load and the power which is lost across the current coil of the galvanometer. Therefore, Power consumed by the load, P = V ´ I Or, P = 200 ´ 20 = 4000W 2 And power loss across current coil, p = I R 2 Therefore, p = 20 (0.02) = 8W Thus, ideally the wattmeter should show 4000W as the final reading, but due to the loss, the reading which is shown by the wattmeter is 4000W + 8W = 4008W

So, the measured power is

I2

100 W

1000 W I

4008 - 4000 ´ 100 = 0.2% more than the actual 4000

load power. Hence, option (c) is the correct answer. 5. A galvanometer with a full scale current of 10 mA has a resistance of 1000 ohms shown in fig. 3. The multiplying power (the ratio of measured current to galvanometer current) of a 100 ohm shunt with this galvanometer is (a) 110 (b) 100 (c) 11 (d) 10 Solution : We know that a shunt is always connected in parallel with a galvanometer. In the question, it is given that the internal resistance of galvanometer is 1000 ohms and the value of shunt resistance is 100 ohms. From the current division formula, Let measured current = I

I1

Current through galvanometer, I1 = Fig 3:

100 I 1100

I I1 On substituting the values, we get

Multiplying factor power =

I = 11 1 100 I 1100 Hence, option (c) is the correct answer. A moving coil of a meter has 10 turns, and a length and depth of 100 mm and 20 mm respectively. It is positioned in a uniform radial flux density of 200 mT. The coil carries a current of 50 mA. The torque of the coil is (a) 200 mNm (b) 100 mNm (c) 2 mNm (d) 1 mNm

Multiplying factor =

6.

Solution : Given, uniform radial flux density, B = 200 ´ 10 -3 T length, l = 100 ´ 10-3 m depth, d = 20 ´10-3 m current, I = 50 ´10-3 A number of turns, N = 10 As we know that torque, T = BldIN m Nm 3 3 3 3 Thus, T = (200 ´10- ) ´ (100 ´10- ) ´ (20 ´10- ) ´ (50 ´10- ) ´10 Therefore, T = 0.0002 Nm = 200 m Nm Hence, option (a) is the correct answer.

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14 7.

A dc A-h meter is rated for 15 A, 250 V. The meter constant is 14.4 A-second/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 : Given, Meter constant = 14.4 A - sec/ rev 1 hr and 1sec = 3600 14.4 A - hr / rev Therefore, the meter constant can be written as 3600 A - hr W - hr ´V = also, rev rev 14.4 Ah / rev thus, ´ 250 = 1Wh / rev 3600 1 now, 1Wh = kWh 1000 1 so, 1Wh / rev = kWh / rev = 1000rev / kWh 1000 Hence, option (c) is the correct answer. 8. A moving iron ammeter produces a full scale torque of 240 m Nm with a deflection of 120 degrees at a current of 10 A. The rate of change of self inductance ( m H / radian ) of the instrument at full scale is (a) 2.0 mH/radian (b) 4.8 mH/radian (c) 12.0 mH/radian (d) 114.6 mH/radian Solution : Full scale torque in a moving coil instrument, 1 dl T = I2 2 dq On substituting the value of current and torque in the above equation, we get dl 1 240 ´ 10 -6 = (10) 2 2 dq dl -6 Thus, 240 ´ 10 = 50 dq -6 Therefore, 240 ´ 10 = dl dq 50

= 4.8 ´ 10 -6 H/radian = 4.8 μH/radian Hence, option (b) is the correct answer.

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15

1.

2.

The Q-meter works on the principle of [2005, 1 Mark] (a) mutual inductance (b) self inductance (c) series resonance (d) parallel resonance A PMMC voltmeter is connected across a series combination of a DC voltage source V1 = 2V and an AC voltage source V2(t) = 3 sin (4t). The voltmeter reads [2005, 1 Mark] (a) 2 V (b) 5 V (c)

3.

æ 3ö çç 2 + ÷ 2 ÷ø V è

(d)

æ 17 ö çç ÷÷ V è 2 ø

± 0.5 % rdg (b) ± 5.5% rdg (c) ± 6.7% rdg (d) ± 7.0 rdg An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWH) in 1 hour. The heater is removed from the supply and now connected to a 400V peak to peak square wave source of 150 Hz. The power in kW dissipated by the heater will be [2006, 2 Marks] (a) 3.478 (b) 1.739 (c) 1.540 (d) 0.870 Consider the following statements with reference to the

5.

equation

dp dt

[2006, 2 Marks]

(1) This is a point form of the continuity equation (2) Divergence of current density is equal to the decrease of charge per unit volume per unit at every point (3) This is Maxwell's divergence equation (4) This represents the conservation of charge Select the correct answer. (a) Only 2 and 4 are true (b) 1, 2 and 3 are true (c) 2, 3 and 4 are true (d) 1,2 and 4 are true

R1 and R4 are the opposite arms of a Wheatstone bridge as are R3 and R2. The source voltage is applied across R1 and R3. Under balanced conditions which one of the following is true? [2006, 2 Marks] R 2R 3 R4

(a)

R1 =

R 3R 4 R2

(b)

(c)

R1 =

R2R4 R3

(d) R1 = R2 + R3 + R4

R1 =

200 Current Transformer (CT) is wound with 200 1 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be –0.5% and 30 min respectively. If the number of secondary turns is reduced by 1, the new ratio error (%) and phase error (min) will be respectively [2006, 2 Marks] (a) 0.0, 30 (b) –0.5, 35 (c) –1.0, 30 (d) –1.0, 25

7.

A

8.

A current of -8 + 6 2(sin wt + 30°) A is passed through

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 [2006, 2 Marks] (a)

4.

6.

9.

three meters. They are at centre zero PMMC meter, true rms meter and a moving iron instrument. The respective readings (in ampere) will be [2006, 2 Marks] (a) 8, 6, 10 (b) 8, 6, 8 (c) –8, 10, 10 (d) –8, 2, 2 The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 kHz, 5 V p-p square wave calibration pulse to channel 1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel 2 (lower trace) of the scope. If the time/div and V/div on both channels are the same, the amplitude (pp) and period of the unknown signal are respectively. [2006, 1 Mark]

(a) 5 V, 1 ms (c) 7.5 V, 2 ms

(b) 5V, 2 ms (d) 10 V, 1 ms

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16 10. A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5A (p-p) as shown in the figure. The period is 25 ms. The reading in watt will be [2006, 1 Mark]

B

A

Ch1

R Vin

GND12

L Ch2 C

(a) A, B, C, A (b) A, B, C, B (c) C, B, A, B (d) B, A, B, C 13. The ac bridge shown in the figure is used to measure the impedance Z. If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be [2008, 2 Marks]

0

0

B 0.

0W 50

H 30

D

0W

R4

0W

Z

m

R2

R3

30

C

1

R1

mH

D

A .9

jX1

~

Oscillator

8

15

11.

(a) zero (b) 25 W (c) 50 W (d) 100 W A bridge circuit is shown in the figure below. Which one of the sequences given below is most suitable for balancing the bridge? [2007, 2 Marks]

39

(a) (260 + j0) W (b) (0 + j200) W (c) (260 – j200) W (d) (260 + j200) W 14. The sinusoidal signals p(w1t) = A sin w1t and q (w2t) are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen is shown below Y

–jX4

(a) First adjust R4 and then adjust R1 (b) First adjust R2 and then adjust R3 (c) First adjust R2 and then adjust R4 (d) First adjust R4 and then adjust R2 12. The probes of a non-isolated, two-channel oscilloscope are clipped to points A, B and C in the circuit of the adjacent figure. Vin is a square wave of a suitable low frequency. The display on Ch1 and Ch2 are as shown on the right. Then the "Signal" and "Ground" probes S1, G1 and S2, G2 of Ch1 and Ch2 respectively are connected to points. [2007, 1 Mark]

The signal q (w2t) will be represented as [2008, 2 Marks] (a) q(w2t) = A sin w2t, w2 = 2w1 (b) q(w2t) = A sin w2t, w2 =

w1 2

(c) q(w2t) = A cos w2t, w2 = 2w1 (d) q(w2t) = A cos w2t, w2 =

w1 2

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17 15. Two 8-bit ADCs, one of single slope integrating type and other of successive approximate type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5 V, the approximate time taken two ADCs will respectively, be [2008, 1 Mark] (a) TA, TB (c)

(b)

TA , TB

(d)

2

TA , TB 2

TA , TB 2

2

16. The figure shows a three-phase delta connected load supplied from a 400V, 50Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensurer a positive deflection. The wattmeter reading will be [2009, 2 Marks]

J0 + 1

Z

CC

z1

=

(1 00

R3

0)W

400 volts 50 Hz.

R + jw L

+J

Z2

)W

0 (10

3-Phase Balanced supply

z2=

a

19. The pressure coil of a dynamometer type wattmeter is [2009, 1 Mark] (a) highly inductive (b) highly resistive (c) purely resistive (d) purely inductive 20. The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. Which of the following inference can be made from this? [2009, 1 Mark] (a) The signals are not sinusoidal (b) The amplitudes of the signals are very close but not equal (c) The signals are sinusoidal with their frequencies very close but not equal (d) There is a constant but small phase difference between the signals 21. The Maxwell's bridge shown in the figure is at balance. The parameters of the inductive coil are [2010, 2 Marks]

b

PC

C

R4 R2

(a) 0 (b) 1600 Watt (c) 800 Watt (d) 400 Watt 17. An average reading digital multimeter reads 10 V when fed with a triangular wave, symmetric about the time axis. For the same input, an rms reading meter will read [2009, 2 Marks] (a)

20

(b)

3

–j/(w C4)

~ (a)

(b)

1 G2 and . The relative small errors associated with G3 each respective subsystem G1, G2 and G3 are e1, e2 and e3. The error associated with the output is [2009, 1 Mark]

1 G3

Input

G1

(a)

e1 + e 2 +

1 e3

(b)

e1 .e 2 e3

(c)

e1 + e 2 - e3

(d)

e1 + e 2 + e3

G2

Output

R = C4 R 2 R 3 L=

R 2R 3 R4

(c)

R=

R4 , L = C4 R 2 R 3 R 2R 3

(d)

L=

R4 , R = C4 R 2 R 3 R 2R3

(c)

18.

(d) 10 3 20 3 The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as G1,

R 2R3 R4

L = C4 R 2 R 3

10 3

R=

22. An ammeter has a current range of 0–5 A, and its internal resistance is 0.2 W. In order to change the range to 0 – 25A, we need to add a resistance of [2010, 1 Mark] (a) 0.8 W in series with the meter (b) 1.0 W in series with the meter (c) 0.04 W in parallel with the meter (d) 0.05 W in parallel with the meter

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18 23. A wattmeter is connected as shown in the figure. The wattmeter reads [2010, 1 Mark]

27. The bridge circuit shown in the figure below is used for the measurement of an unknown element Zx. The bridge circuit is best suited when Zx is a [2011, 1 Mark] C1

Current coil

~

Z1

Vs ~

Wattmeter

(d) power consumed by Z2 1 24. A 4 digit DMM has the error specification as : 0.2% 2 or 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 [2011, 2 Marks] (a) (c)

± 0.1% ± 0.3%

(b) (d)

Zx

(a) low resistance (b) high resistance (c) low Q inductor (d) lossy capacitor 28. An analog voltmeter uses external multiplier setting. With a multiplier setting of 20 kW, it reads 440 V and with a multiplier setting of 80 kW, it reads 352 V. For a multiplier setting of 40 kW, the voltmeter reads [2012, 2 Marks] (a) 371 V (b) 383 V (c) 394 V (d) 406 V 29. For the circuit shown in the figure, the voltage and current expressions are [2012, 1 Mark] v (t ) = E1 sin(wt ) + E 3 sin(3wt ) and

± 0.2% ± 0.4%

i (t ) = I1 sin(wt - f1 ) + I3 sin(3 wt - f3 ) + I5 sin(5 wt ) The average power measured by the Wattmeter is

25. Consider the following statements : (i)

D

R4

Z2

(a) zero always (b) total power consumed by Z1 and Z2 (c) power consumed by Z1

R1

+

The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil.

i(t) + v(t) –

(ii) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit. [2011, 1 Mark] (a) (i) is true but (ii) is false

Wattmeter

1 E1 I1 cos f1 2 1 [ E1 I1 cos f1 + E1 I3 cos f3 + E1 I5 ] (b) 2 1 [ E1I1 cos f1 + E3 I3 cos f3 ] (c) 2 1 [ E1I1 cos f1 + E3 I1 cos f1 ] (d) 2 30. The bridge method commonly used for finding mutual inductance is [2012, 1 Mark] (a) Heavyside Campbell bridge (b) Schering bridge (c) De Sauty bridge (d) Wien bridge

(a)

(b) (i) is false but (ii) is true (c) both (i) and (ii) are true (d) both (i) and (ii) are false 26. A dual trace oscilloscope is set to operate in the alternate mode. The control input of the multiplexer used in the ycircuit is fed with a signal having a frequency equal to [2011, 1 Mark] (a) the highest frequency that the multiplexer can operate properly (b) twice the frequency of the time base (sweep) oscillator (c) the frequency of the time base (sweep) oscillator

Load

Potential coil

R2

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19 31. A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent Magnet Moving Coil (PMMC) meter connected across the same load reads [2012, 1 Mark]

the strain gauge resistance is increased by 1% over the nominal value, the output voltage V0 in mV is [2013, 2 marks] R5

V(t) 10V Vi

5V 0 – 5V

10

(a) 4 V (c) 8 V 32.

12

20

R1

+ –

–

(b) 5 V (d) 10 V

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 [2013, 1 mark]

R2

(a) 56.02 (b) 40.83 (c) 29.85 (d) 10.02 35. In an oscilloscope screen, linear sweep is applied at the (a) vertical axis [2014, Set-1, 1 Mark] (b) horizontal axis (c) origin (d) both horizontal and vertical axis 36. The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is _________. R = 0.5W

14.14 sin (314t) V

– 100 kW +

1jW

Voltmeter

1/jW

100 sin(wt)

V 1jW

1/jW

(a) 4.46 (b) 3.15 (c) 2.23 (d) 0 Three moving iron type voltmeters are connected as shown below. Voltmeter readings are V, V1 and V2 as indicated. The correct relation among the voltmeter readings is [2013, 1 mark]

–j 1W

V1

–j 2W

V2

V

(a)

V=

V1

+

2 (c) V = V1V2 34.

+

R3

Time (ms)

1 kW

33.

V0

V2 2

(b) V = V1 + V2

[2014, Set-1, 2 Marks] 37. The dc current flowing in a circuit is measured by two ammeters, one PMMC and anothrr electrodynamometer type, connected in series. The PMMC meter contains 100 turns in the coil, the flux density in the air gap is 0.2 Wb/m2, and the area of the coil is 80 mm 2. The electrodynamometer ammeter has a change in mutual inductance with respect to deflection of 0.5 mH/deg. The spring constants of both the metres are equal. The value of current, at which the deflections of the two meters are same, is _________. [2014, Set-1, 2 Marks] 38. The total power dissipated in the circuit, shown in the figure, is 1 kW. 10 A 2 A 1 W

(d) V = V2 – V1

A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as R5 = 300 W. Other bridge resistances are R1 = R2 = R3 = 300 W. The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by maximum permissible voltage and

A.C. Source

XL

XC1

XC2

V

R

Load 200V

The voltmeter, across the load, reads 200 V. The value of XL is _________. [2014, Set-2, 2 marks]

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20 39. Two ammeters X and Y have resistances of 1.2 W and 1.5 W respectively and they give full-scale deflection with 150 mA and 250 mA respectively. The ranges have been extended by connecting shunts so as to give full scale deflection with 15A. The ammeters along with shunts are connected in parallel and then placed in a circuit in which the total current flowing is 15A. The current in amperes indicates in ammeter X is _________. [2014, Set-2, 2 marks] 40. An LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150 V and 75 V, and two current settings 5 A and 10 A. The full scale reading is 150. If the wattmeter is used with 150 V voltage setting and 10 A current setting, the multiplying factor of the wattmeter is _________. [2014, Set-3, 1 Mark] 41. The two signals S1 and S2, shown in figure, are applied to Y and X deflection plates of an oscilloscope.

v

1

Y 1

X

-1

(d)

Y 1

S1 X

T

v

(c)

1

2T

t

-1

S2

42.

T

2T

A periodic waveform observed across a load is represented by 0 £ wt < 6 p ì 1 + sin wt V(t) = í î-1 + sin wt 6 p £ wt < 12p

t

The measured value, using moving iron voltmeter connected across the load, is [2014, Set-3, 2 Marks] The waveform displayed on the screen is [2014, Set-3, 1 Mark] (a)

(a)

Y 1

3 2

(b)

2 3

2 3 (d) 3 2 In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance C 1 in microfarad is _________. [2014, Set-3, 2 Marks]

(c)

X

43.

-1

(b)

Y 1

X -1

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0.1 F

21

1.

2.

What is the correct sequence of the following types of ammeters and voltmeters with increasing accuracy? 1. Moving-iron 2. Moving-coil permanent magnet 3. Induction Select the correct answer using the codes given below (a) 1, 3, 2 (b) 1, 2, 3 (c) 3, 1, 2 (d) 2, 1, 3 The total current I = I1 + I2 in a circuit is measured as I1 = 150 ±1 A, I2 = 250 ± 2A , where the limits of error are given as standard deviations. I is measured as (a)

3.

4.

5.

6.

7.

( 400 ± 3) A ( 400 ± 1/ 5 ) A

(b)

( 400 ± 2.24) A ( 400 ± 1) A

(d) (c) A C.R.O. is operated with X and Y settings of 0.5 mV/cm and 100 mV/cm. The screen of the C.R.O. is 10 cm × 8 cm (X and Y). A sine wave of frequency 200 Hz and r.m.s. amplitude of 300 mV is applied to the 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 cycle of the sine wave with clipped amplitude A Wien-bridge is used to measure the frequency of the input signal. However, the input signal has 10% third harmonic distortion. Specifically signal is 2 sin 400 pt + 0.2 sin 1200 pt (with t in sec.). With this input the balance will (a) Lead to a null indication and setting will correspond to a frequency of 200 Hz. (b) Lead to a null indication and setting will correspond to 260 Hz. (c) Lead to a null indication and setting will correspond to 400 Hz. (d) Not lead to null indication Measurement of an unknown voltage with d.c. potentiometer loses its advantage of open circuit measurement when (a) primary circuit battery is changed. (b) standardisation has to be done again to compensate for drifts. (c) Voltages larger than the range of the potentiometer are measured. (d) range reduction by a factor of 10 is employed in the potentiometer to improve resolution. A set of independent current measurements taken by four observers was recorded as: 117.02 mA, 117.11 mA, 117.08 mA and 117.03 mA. What is the range of error? (a) ± 0.045 (b) ± 0.054

8.

9.

10.

11.

12.

13.

1. Maxwell’s bridge 2. Schering bridge 3. Wein-bridge 4. Hay’s bridge 5. Wheatstone bridge Select the correct answer using the codes given below (a) 1 and 2 (b) 2 and 3 (c) 3, 4 and 5 (d) 1 and 4 The secondary winding of a current transformer is open when current is flowing in the primary then, (a) there will be high current in primary. (b) there will be very high secondary voltage. (c) the transformer will burn out immediately. (d) the meter will burn out. Which one of the following digital voltmeters is most suitable to eliminate the effect of period noise? (a) Ramp type digital voltmeter (b) Integrating type digital voltmeter (c) Successive approximation type digital voltmeter (d) Servo type digital voltmeter Match List - I (Instrument) and List - II (Error) and select the correct answer using the code given below the lists: List - I List - II A. PMMC voltmeter 1. Eddy current error B. AC ammeter 2. Phase angle error C. Current transformer 3. Braking system error D. Energy meter 4. Temperature error Codes: A B C D (a) 2 3 4 1 (b) 4 1 2 3 (c) 2 1 4 3 (d) 4 3 2 1 Which of the following can be used/modified for measurement of angular speed? 1. LVDT 2. Magnetic pick-up 3. Tacho-generator 4. Strain gauge Select the correct answer using the code given below (a) Only 1 and 2 (b) Only 2 and 3 (c) Only 3 (d) Only 2, 3 and 4 A meter has full scale deflection at 90° at a current of 1 A, the response of meter is square law. Assuming spring control, the current for a deflection of 45° will be (a) 0.25 A (b) 0.50 A (c) 0.67 A (d) 0.707 A The sensitivity of voltmeter using 0 to 5 mA meter movement is (a) 50 W/volt (b) 100 W/volt (c) 200 W/volt (d) 500 W/volt

(c) ± 0.065 (d) ± 0.056 Which of the following bridges can be used for inductance measurement? Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

22 14. The full scale deflection current of a meter is 1 mA and its internal resistance is 100 W. This meter is to have full deflection when 100 V is measured. What is the value of series resistor to be used? (a) 99.99 kW (b) 100 kW (c) 99.99 W (d) 100 W 15. Calculate the maximum velocity of the beam of electrons in a CRT having a cathode anode voltage of 1000 V. Assume the electrons to leave the cathode with zero velocity. Charge of electron = 1.6 × 10–19 C and mass of electron = 9.1 × 10–31 kg. (a) 0.1875 × 106 m/s (b) 0.1875 × 107 m/s (c) 0.1875 × 108 m/s (d) 0.1875 × 109 m/s 16. In the Maxwell bridge as shown in the figure 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 respectively be LX RX

2000 W R4 0.05 mF

750 W

4000 W

(a) 375 W, 75 mH (b) 75 W, 150 mH (c) 37.5 W, 75 mH (d) 75 W, 75 mH 17. A simple dc potentiometer is to be standardised by keeping the slider wire setting at 1.0183 V. If by mistake, the setting is at 1.0138 V and the standardisation is made to obtain a source voltage of 1.0138 V, then the reading of the potentiometer will be (a) 1.0138 V (b) 1.0183 V (c)

(1.0138)2

V

(d) (1.0138)2 V

1.0183 18. Two-wattmeter method is employed to measure power in a 3-phase balanced system with the current coils connected in the A and C lines. The phase sequence is ABC. If the wattmeter with its current coil in A-phase line reads zero, then the power factor of the 3-phase load will be (a) zero lagging (b) zero leading (c) 0.5 lagging (d) 0.5 leading 19. A digital voltmeter uses a 10 MHz clock and has a voltage controlled generator which provides a width of 10 m sec per volt of unit signal. 10 volt of input signal would correspond to a pulse count of (a) 500 (b) 750 (c) 1000 (d) 1500 20. In a digital voltmeter, the oscillator frequency is 400 kHz and the ramp voltage fills from 8 V to 0 V in 20 m sec. The number of pulses countered by the counter is (a) 800 (b) 2000 (c) 4000 (d) 8000 21. A high frequency a.c. signal is applied to a PMMC instrument. If the rms value of the a.c. signal is 2 V, then the reading of the instrument will be

(a) zero

(b) 2 V

(c)

(d) 4 2V 2 V 22. A current i = (10 + 10 sin t) amperes is passed through an ideal moving iron type ammeter. Its reading will be (a) zero (b) 10 A (c)

(d) 10 2A 150A 23. In PMMC instrument, the central spring stiffness and the strength of the magnet decrease by 0.04% and 0.02% respectively due to a rise in temperature by 1°C. With a rise in temperature of 10°C, the instrument reading will (a) increase by 0.2% (b) decrease by 0.2% (c) increase by 0.6% (b) decrease by 0.6% 24. Standard cell (a) Will have precise and accurate constant voltage when current drawn from it is few microamperes only. (b) Will have precise and accurate constant voltage when few milliamperes are drawn from it. (c) Will continue to have constant voltage irrespective of loading conditions. (d) Can supply voltages up to 10 V. 25. If the reading of the two wattmeters are equal and positive in two-wattmeter method, the load pf in a balanced 3-phase 3-wire circuit will be (a) zero (b) 0.5 (c) 0.866 (d) unity 26. A voltage of {200 2 sin 314t + 6 2 sin ( 942t + 30° ) + 8 2 cos (1570t + 30° )}V is given to a harmonic distortion

meter. The meter will indicate a total harmonic distortion of approximately (a) 4.55% (b) 6.5% (c) 7.5% (d) 8.5% 27. In the measurement of power on balanced load by twoWattmeter method in a 3-phase circuit, the readings of the Wattmeters are 3 kW and 1 kW respectively, the latter being obtained after reversing the connections to the current coil. The power factor of the load is (a) 0.277 (b) 0.554 (c) 0.625 (d) 0.866 28. A CRO screen has ten divisions on the horizontal scale. If a voltage signal 5 sin (314 t + 45°) is examined with a line base setting of 5 m sec/div, the number of cycles of signal displayed on the screen will be (a) 0.5 cycles (b) 2.5 cycles (c) 5 cycles (d) 10 cycles 29. A 0-10 mA PMMC ammeter reads 4 mA in a circuit. Its bottom control spring snaps suddenly. The meter will now read nearly (a) 10 mA (b) 8 mA (c) 2 mA (d) Zero

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23 30. A Lissajous pattern, as shown in figure below, is observed on the screen of a CRO when voltages of frequencies fx and fy are applied to the x and y plates respectively. fx : fy is then equal to

31.

32.

33.

34.

35.

36.

37.

(a) 3 : 2 (b) 1 : 2 (c) 2 : 3 (d) 2 : 1 Two voltmeters have the same range 400 V. The internal impedances are 30,000 ohms and 20,000 ohms. If they are connected in series and 600 V be applied across them, the readings are (a) 360 V and 240 V (b) 300 V and 300 V (c) 400 V and 200 V (d) None of these A meter has a full-scale angle of 90° at a current of 1 A. This meter has perfect square-law response. What is the current when the deflection angle is 45°? (a) 0.5 A (b) 0.65 A (c) 0.707 A (d) 0.87 A Which of the following electronic instruments (or equipment) can be used to measure correctly the fundamental frequency component of a waveform and its higher harmonics? 1. Cathode ray oscilloscope 2. Vacuum tube voltmeter 3. Spectrum analyzer 4. Distortion factor meter Select the correct answer using the codes given below (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 (d) 1 and 4 In d.c. potentiometer measurements, a second reading is often taken after reversing the polarities of the d.c. supply and the unknown voltage, and the average of the two reading is taken. This is with a view to eliminate the effects of (a) ripples in the d.c. supply (b) stray magnetic fields (c) stray thermal emf’s (d) erroneous standardisation Electrostatic instruments are normally used for (a) low current measurements (b) high current measurements (c) low voltage mesurements (d) high voltage measurements A certain oscilloscope with 4 cm × 4 cm screen has its own sweep output fed to its input. If the x and y sensitivities are same, the oscilloscope will display a (a) triangular wave (b) diagonal line (c) sine wave (d) circle Which instrument has necessarily the ‘square law’ type scale? (a) Permanent magnet moving coil. (b) Hot wire instruments (c) Moving iron repulsion (d) None of the above

38. In a single-phase power factor meter, the controlling torque is (a) provided by spring control (b) provided by gravity control (c) provided by stiffness of suspension (d) not required 39. In electrodynamometer type wattmeter, the inductance of pressure coil produces error. The error is (a) constant irrespective of the power factor of the load (b) higher at higher power factor loads (c) higher at lower power factor loads (d) highest at unity power factor loads 40. If an induction type energy meter runs fast, it can be slowed down by (a) lag adjustment (b) light load adjustment (c) adjusting the position of braking magnet and moving it closer from the centre of the disc (d) adjusting the position of braking magnet and moving it away from the centre of the disc. 41. The reflecting mirror mounted on the moving coil of a vibration galvanometer is replaced by a bigger size mirror. This will result in (a) lower frequency of resonance & lower amplitude of vibration (b) lower frequency of resonance but the amplitude of vibration is unchanged (c) higher frequency of resonance & lower amplitude of vibration (d) higher frequency of resonance but the amplitude of vibration is unchanged 42. In calibration of a dynamometer Wattmeter by potentiometer, phantom loading arrangement is used because (a) the arrangement gives accurate results. (b) the power consumed in calibration work is minimum. (c) the method gives quick results. (d) the onsite calibration is possible. 43. The accuracy of Kelvin’s double bridge for the measurement of low resistance is high because the bridge (a) use two pairs of resistance arms (b) has medium value resistance in the ratio arms. (c) uses a low resistance link between standard and test resistances (d) uses a null indicating galvanometer. 44. In a Q meter measurement to determine the self-capacitance of a coil, the first resonance occurred at f1 with C1 = 300 pF. The second resonance occurred at f2 = 2f1, with C2 = 60 pF. The self-capacitance of coil works out to be (a) 240 pF (b) 60 pF (c) 360 pF (d) 20 pF 45. Which one of the following statements is correct? The deflection of hot wire instrument depends on (a) r.m.s. value of the a.c. current (b) r.m.s. value of the a.c. voltage (c) average value of the a.c. current (d) average value of the a.c. voltage

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24 46. Which one of the following decides the time of response of an indicating instrument? (a) Deflecting system (b) Controlling system (c) Damping system (d) Pivot and Jewel bearing 47. Chopper stabilized d.c. amplifier type electronic voltmeter overcomes the effect of (a) Amplifier CMRR (b) Amplifier sensitivity (c) Amplifier drift (d) Electromagnetic interference 48. Which one of the following defects is responsible for creeping in an induction type energy meter? (a) Imperfect lag compensation (b) Over friction compensation (c) Imperfect overload compensation (d) Misalignment of brake magnet 49. What is the range for a 3½ digital meter? (a) 0 to 1999 (b) 0 to 1500 (c) 0 to 999 (d) 0 to 19999 50. Piezoelectric crystal is generally employed for the measurement of which one of the following? (a) Flow (b) Velocity (c) Acceleration (d) Temperature 51. The deflection expression q µ V2 =

(a) Permanent magnet moving coil voltmeter (b) Vacuum Tube voltmeter (VTVM) (c) Transistor voltmeter (TVM) (d) Phase Sensitive Detector (PSD). 55. An Ayrton shunt is used to make a d’ Arsonval galvanometer into a (a) single range voltmeter (b) single range ammeter (c) multirange voltmeter (d) multirange ammeter 56. In the ac circuit shown in the given figure, when the ammeter reads 10A, the readings on a voltmeter placed across the entire circuit and then across each element are given below. Match List - I (Position of the voltmeter) with List-II (readings on the voltmeter) and select the correct answer using the codes given below the lists:

3

M

A.C. Source

DISPLACEMENT

–j6

A

dC corresponds to dq

(a) moving iron type instrument (b) electrodynamic type instruments (c) electrostatic type instrument (d) induction type instruments 52. A sinusoidal waveform, when observed on an oscilloscope, has peak-to-peak amplitude of 6 cm. If the vertical sensitivity setting is 5 V/cm, the rms value of the voltage will be (a) 10.6 V (b) 11.1 V (c) 12.6 V (d) 15 V 53. In the two-wattmeter method of measuring power in a balanced three-phase circuit, one wattmeter shows zero and the other positive maximum. The load power factor is (a) zero (b) 0.5 (c) 0.866 (d) 1.0 54. Which one of the following instruments should be used at M in the given figure for the measurement of magnitude as well as direction of the displacement?

j2

VT List - I (Position of the voltmeter A. VT B. VR C. VL D. VC

List - II (readings on the voltmeter 1. 60 2. 20 3. 30 4. 50 5. 110

Codes A B C D (a) 5 4 2 1 (b) 5 3 1 2 (c) 4 5 1 2 (d) 4 3 2 1 57. A 10 V full-scale voltmeter having 100 k-ohm / V sensitivity is used to measure the output of a photovoltaic cell having an internal resistance of 1 M-ohm. The voltmeter reads 5 V. The voltage generated by the photovoltaic cell is (a) 5 V (b) 10 V (c) greater than 5 V but less than 10 V (d) greater than 10 V 58. Differential transformer transducer is used for measurement of (a) speech and music (b) displacement (c) liquid level (d) temperature

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25 59. A shunt type ohmmeter is shown in the figure given below. With Rx disconnected, the meter reads full scale ‘S’ represents the meter current as a fraction of full scale current with Rx connected such that S =

62. In the multimeter circuit shown in the figure for a.c. voltage measurement, the function of diode D1 is to D2

R

Rx R x + Rp M

D1

Rp

A Im

E Rx

Rm

B

The value of RP is given by (a) Rm (c)

(b) R1 + Rm

R1R x R ( 1 + Rm )

(d)

(a) Provide half wave rectification (b) Make the rectifier D2 perform full-wave rectification (c) By-pass reverses leakage current of D2 in the negative cycle of the input. (d) Short-circuit over range voltages. 63. Two sine waves of same frequency are impressed on the X and Y plates of a CRO and the Lissajous figure seen are shown in the given diagram. The phase difference between the signals is

R1.R m R ( 1 + Rm )

Y2

60. The power in a resistor R is estimated by measuring the voltage and current using the voltmeter-ammeter method. Two different arrangements can be used as shown in circuits I and II. Less erroneous results are obtained by adopting +

Rsh

A +

R

V

Circuit - I

+

+ V

A

R

Circuit - II

(a) circuit I for low values of R (b) circuit II for low values of R (c) circuit I for high values of R (d) circuit II for low and high values of R. 61. A simple dc potentiometer is to be standardised by keeping the sidewire setting at 1.0813 V. If by mistake, the setting is at 1.0138 V and the standardisation is made to obtain a source voltage of 1.0318 V, then the reading of the potentiometer will be (a) 1.0138 V (b) 1.0813 V (c) (1.0138)2 / 1.0813 V (d) (1.0138)2 V

Y1 = 0.5 Y2

Y1

(a) 30° or 330° or 150° or 210° (b) 30° or 330° or 150° (c) 30° or 330° (d) 30° 64. When the horizontal deflecting plates of a CRO are kept -at the ground potential and 30 V dc is applied to the vertical deflecting plates, the bright spot moves 1 cm away from the centre. If with the same setting, a 30 V ac is applied to the vertical deflecting plates, then the picture observed on the screen would be (a) a spot approximately 3 cm away from the center (b) a vertical line 2 cm long (c) a vertical line approximately 3 cm long (d) two spots 2 cm vertical above each other 65. A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type a.c. voltmeter connected in parallel. If the meter readings are V1 and V2 respectively and the meters are free from calibration errors, then the form factor of the ac voltage may be estimated as (a) V1/V2 (b) 1.11 V1 / V2 (c)

2V1 / V2

(d) p V1 / 2V2

66. The resistances of two coils of a wattmeters are 0.01 ohm and 1000 ohms respectively and both are non-inductive. The load current is 20 A and the voltage across the load is 30 V. In one of the two ways of connecting the voltage coil, the error in the reading would be (a) 0.1% too high (b) 0.2% too high (c) 0.15% too high (d) zero

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26 67. In the statement “the wattmeter commonly used for power measurement at commercial frequencies is of the X type. This meter consists of two coil systems, the fixed system being the Y coil and moving system being Z coil”. X, Y and Z stand respectively for (a) dynamometer, voltage and current (b) dynamometer, current and voltage (c) induction, voltage and current (d) induction, current and voltage 68. A galvanometer is tested shown in figure, in the circuit where E = 1.5 V, R1 = 1.0 ohm, R2 = 2500 ohm and R3 is variable. With R3 set at 450 W, the galvanometer deflection is 140 mm and with R3 set at 950 ohm, the galvanometer deflection is 70 mm. The resistance of the galvanometer is R3

RG

E 15 V

1.0 W

G

R1

2500 kW R2

(a) 99 ohm (b) 49 ohm (c) 25 ohm (d) 10 ohm 69. In the particular form of frequency meter, a 1 mF capacitor is connected across a symmetrical square wave signal of IV peak value. If the average value of the current taken by the capacitor, after full wave rectification is measured as 2 mA, then the frequency of the signal will be (a) 1000 / p Hz (b) 500 Hz (c) 1000 Hz (d) 1000 p Hz 70. Consider the following statements about LVDT as a transducer: 1. The relationship between input displacement and output voltage is almost linear. 2. The range of displacement that can be measured is wide 3. It does not form a loading on the mechanical system. Of these statements: (a) 1, 2 and 3 are correct (b) 2 and 3 are correct (c) 1 and 2 are correct (d) 1 and 3 are correct 71. In the Maxwell bridge as shown in the figure 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 respectively be: LX

2000 W

RX 0.05 mF 750 W

4000 W

(a) 375 ohm, 75 mH (c) 37.5 ohm, 75 mH

(b) 75 ohm, 150 mH (d) 75 ohm, 75 mH

72. Consider the following statements regarding the sources of error in a Q-meter: 1. If a coil with resistance R is connected in the direction measurement mode and if the residual resistance of the Q meter is 0.1 R, then the measured Q of the coil would be 1.1 times the actual Q. 2. If the inductance to be measured i.e. less than 0.1 mH, the error due to the presence of residual inductance cannot be neglected 3. The presence of the distributed capacitance in a coil modifies the effective Q of the coil. Of these statements: (a) 1, 2 and 3 are correct (b) 1 and 2 are correct (c) 2 and 3 are correct (d) 1 and 3 are correct 73. Which of the following are the advantages of a balanced bridge vacuum tube voltmeter in comparison with a conventional VTVM? 1. Higher input impedance 2. Meter zero has less tendency to shift 3. Effects of changes due to variation of value characteristics are minimized. 4. Power supply fluctuations have a smaller effect on the measuring circuit. Select the correct answer using the codes given below: Codes (a) 3 and 4 (b) 1 and 2 (c) 1, 2 and 3 (d) 1, 2, 3 and 4 74. A permanent magnet moving coil type ammeter and a moving iron type ammeter are connected in series in a resistive circuit fed from the output of a half wave rectifier voltage source. If the moving iron type instrument reads 5A, the permanent magnet moving coil type instrument is likely to read. (a) zero (b) 2.5 A (c) 3.18 (d) 5 A 75. The coil is having self-inductance of 10 mH and 15 mH have an effective inductance of 40 mH, when connected in series aiding. What will be the equivalent inductance if we connect them in series-opposing? (a) 20 mH (b) 10 mH (c) 5 mH (d) zero 76. In a digital voltmeter, the oscillator frequency is 400 kHz and the ramp voltage falls from 8V to 0V in 20 m sec. The number of pulses counted by the counter is (a) 800 (b) 2000 (c) 4000 (d) 8000 77. Pulses of a frequency of 1 MHz are applied to the time base selector of a digital frequency meter, which consists of 6 frequency dividers, each dividing the incoming frequency by a factor of 10. The time-base setting at the output of 4th frequency divider starting from the input is (a) 1 ms (b) 10 ms (c) 100 ms (a) 1 s

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27 78. Which of the following error is likely to occur in bridge method of measurement? I. Residual error II. Frequency and waveform error III. Leakage and eddy current error (a) I only (b) II only (c) I and II only (d) I, II and IV 79. An ideal meter movement has a full scale deflection of 1 mA. This meter is to be used in series with a vary-non-ideal voltage source of 10 V to produce an ohm meter. It is noted that the ohm-meter shows a full scale deflection when the terminals are shorted (zero resistance). When the terminals are open circuited, the meter movement deflection is zero (infinite resistance). If 5 k ohm is placed across the terminals, the meter movement will have a deflection of (a) 1/5 (b) 1/2 (c) 1/3 (d) 2/3 80. A high - Q quartz crystal exhibits series resonance at the frequency ws and parallel resonance at the frequency wp. Then (a) ws is very close to, but less than wp (b) ws > wp 81. An unshielded moving iron voltmeter is used to measure the voltage in an a.c. circuit. If a stray d.c. magnetic field having a component along the axis of the meter coil appears, the meter reading would be (a) unaffected (b) decreased (c) increased (d) either decreased or increased depending on the direction of the d.c. field. 82. A resistance is measured by the voltmeter ammeter method employing d.c. excitation and a voltmeter of very high resistance connected directly across the unknown resistance. If the voltmeter and ammeter readings are subject to maximum possible errors of ± 2.4% and ± 1.0% respectively, then the magnitude of the maximum possible percentage error in the value of resistance deduced form the measurement is nearly (a) 1.4% (b) 1.7% (c) 2.4% (d) 3.4% 83. In d.c. potentiometer measurements, a second reading is often taken after reversing the polarities of the d.c. supply and the unknown voltage, and the average of the two readings is taken. This is with a view of eliminate the effects of (a) ripples in the d.c. supply (a) stray magnetic field (c) stray thermal emf’s (d) erroneous standardization. 84. In a dual slope integrating type digital voltmeter the first integration is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V, the total conversion time for an input of 1 V is (a) 0.01 sec (b) 0.05 sec (c) 0.1 sec (d) 1 sec

85. A CRO screen has ten divisions on the horizontal scale. If a voltage signal 5 sin (314 t + 45°) is examined with a line base setting of 5 m sec/ div, the number of cycles of signal displayed on the screen will be (a) 0.5 cycles (b) 2.5 cycles (c) 5 cycles (d) 10 cycles 86. The “accuracy” of a measuring instrument is determined by (a) closeness of the value indicated by it to the correct value of the measured (b) repeatability of the measured value (c) speed with which the instrument’s reading approaches the final value (d) least change in the value of the measured that could be detected by the instrument 87. Given Za = 100 50° ; Z b = 300 -90° ; Zc = 200 0° The value of Zd of the bridge shown in the figure to be balanced is (a)

600 -40° Zb

Za

(b)

600 140°

(c)

600 -140°

CRO Zc

Zd

(d) 150 40° 88. For the ac bridge circuit shown in the figure, at balance, the value of Rd, Ld and Qd will be (a)

(b)

Ra Rb Ra Rc : : wCb R b Rc Cb

Cb

RbRc Ra Rc : : wC b R b Ra Cb

Rc Rb

D Ld

Ra

(c)

RbRc : R a R c C b : wCb R c Ra

(d)

R a Rc : R a R c Cb : wCb R b Rb

Rd

89. Two in-phase 50 Hz sinusoidal waveforms of unit amplitude are fed into channel 1 and channel 2 respectively of an oscilloscope. Assuming that the voltage scale, time scale and other settings are exactly the same for both the channels, what would be observed if the oscilloscope is operated in X-Y mode? (a) A circle of unit radius (b) An ellipse (c) A parabola (d) A straight line inclined at 45° with respect to the x-axis.

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28 90. The Wein bridge circuit shown in the below figure can be used as a frequency-measuring device, provided (a)

R2 =2 R4

(b)

R4 =2 R2

(c)

(d)

z2

z1

R1

Wattmeter

R2

C1

z3

D

z4

R3

R2 =4 R4

C3

R4 VS

ZS

+ –

R2 =3 R4

91. A 0-150 V voltmeter has an accuracy of 1% of full-scale reading. The voltage measured by this instrument is 75 V. The limiting error is (a) 1% (b) 2% (c) 2.5% (d) 3% 92. The reading of high impedance voltmeter V in the bridge circuit shown in the given figure is 10 W

20 W v

20 W

10 W +

– 10 V

(a) zero (b) 3.33 V (c) 4.20 V (d) 6.66 V 93. Consider the following data for the circuit shown below: W

A

LOAD

V 200

The overall uncertainty in the measured value of Z1 is (a) (c)

(b) ± 4%

11 %

± 5%

(d)

95. Match List - I (Instrument) with List - II (Frequency) and select the correct answer using the codes given below the lists: List - I List - II A. Vibration Galvanometer 1. 100 Hz B. Head phone 2. Zero Hz C. D’Arsonval 3. 1 kHz Galvanometer D. CRO 4. Large frequency range Codes: A B C D (a) 2 3 1 4 (b) 1 4 2 3 (c) 2 4 1 3 (d) 1 3 2 4 96. For the Owen-bridge circuit shown below, when balanced, the values of L and R are R

C1

Ammeter: Resistance 0.2 W; Reading 5.0 A Voltmeter: Resistance 2 kW; Reading 200 V Wattmeter: Current coil resistance 0.2 W, Pressure coil resistance 2 kW Load: Power factor = 1 The reading of the Wattmeter is (a) 980 W (b) 1000 W (c) 1005 W (d) 1010 W 94. Consider the circuit as shown below. Z1 is an unknown impedan ce and measured as Z 1 = (Z 2Z 3 )/Z 4 . The uncertainties in the value of Z2, Z3 and Z4 are ± 1% , ± 1% and ± 3% respectively.

5%

R1

L D

Cs

Rs

(a)

L = Cs R s / R1, R = R s Cs / C1

(b)

L = Cs R s R1, R = R s Cs / C1

(c)

L = C1R s R1 , R = R s Cs C1

(d)

L = Cs / C1R s R1, R = R1Cs / C1

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29 97. In the bridge circuit shown below, at balance condition, the value of Cs = 0.5 mF and Rs = 1000 W. LX

1000 W

RX Cs 1000 W

Rs

The values of inductance Lx and resistance Rx are (a)

L x = 0.5 H, R x = 1000 W

(b)

L x = 0.25 H, R x = 2000 W

(c)

L x = 0.5 H, R x = 3000 W

(d)

L x = 0.25 H, R x = 500 W

98. The X and Y inputs to a CRO are respectively 10 cos (100t + f) and 10 sin (100 t + f). The resulting Lissajous pattern is (a) A straight line inclined at an angle f (b) A horizontal line (c) An ellipse with axis making an angle f (d) A circle 99. Damping torque in the disc of an a.c. energy meter is provided by which one of the following? (a) Electrostatic effect (b) Magnetostatic effect (c) Eddy current effect (d) Chemical effect 100. A voltmeter has a range 0-20 V and manufacturer rates its accuracy as ± 1% fsd. Match List - I (Voltage values) with List - II (Error as Percentage of True Value) and select the correct answer using the codes given below the lists: List - I List - II A. 2 V 1. 4% B. 5 V 2. 10% C. 10 V 3. 2% D. 20 V 4. 1% Codes: A B C D (a) 3 1 2 4 (b) 2 1 3 4 (c) 2 4 3 1 (d) 3 4 2 1 101. Which one of the following statements is not correct? (a) The power requirements of digital instruments are considerably lower than that of analog instruments (b) In analog instruments, the resolution limit is one part in several hundreds, whereas digital instruments can be made with a resolution of one part in several thousands (c) Digital instruments are extremely portable and usually do not require an outside source of supply for measurements (d) Digital instruments indicate the readings directly decimal numbers and therefore errors due to parallax and approximation, etc., are eliminated

102. How can the resolution of a wire-wound potentiometer be improved? (a) By increasing the applied voltage (b) By decreasing the applied voltage (c) By reducing the diameter of the resistance wire (d) By increasing the diameter of the resistance wire 103. The voltage across an impedance is measured by a voltmeter having input impedance comparable with the impedance causing an error in the reading. What is this error called? (a) Random error (b) Gross error (c) Systematic error (d) Loading effect error 104. The current passing through a 10 W resistance is given by I = 3 + 4 2 sin 314t. This current is measured by a PMMC meter. What is the measured value? (a) 3A (b) 4A (c) 5A (d) 4 2 A 105. For measurement of modulation factor by a CRO, a trapezoidal pattern is obtained on the CRO screen, as shown in figurg given above. What is the value of the modulation factor ‘M’?

Modulated Wave

Modulated Wave

(a) 0 (b) 0.5 (c) 0.75 (d) 1 106. Voltages VY = 100 sin 1000 t and VX =50 sin 1000 t are connected to Y and X terminals of a CRO, respectively. What is the shape of the figure seen on the CRO? (a) A circle (b) A straight line (c) An ellipse (d) A parabola 107. Which one of the following measuring devices has minimum loading effect on the quantity under measurement? (a) PMMC (b) CRO (c) Hot wire (d) Electrodynamometer 108. The figure shows ‘Owen Bridge’ arranged to measure incremental inductance of the unknown inductance LX, RX. At balance, what are the values of LX and RX? RC

Cb

CC

D LX

Ra RX

(b)

R a Rc R C ;Rx = a b Cb Cc L x = R a R cCb ; R x = R x Cb Cc

(c)

Lx =

(d)

Lx = R a R cCb ; R x =

(a)

Lx =

R a Rc R C ;Rx = a c Cb Cb R a Cb Cc

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30 109. Which one of the following statements is not correct? (a) Data loggers are usually of digital types (b) A digital voltmeter is essentially an A-D converter (c) A servo-type Potentiometric recorder has a frequency response better than that of a galvanometric recorder (d) In digital transducers, there are ergonomic advantages in presenting digital data 110. The Wheatstone bridge method of measuring resistance is ideally suited for the measurement of resistance values in the range of frequencies of (a) 0.001 to 1 W (b) 0.1 to 100 W (c) 100 W to 10 kW (d) 100 kW to 10 MW 111. Which of the following meters requires an external power source for its operation? (a) PMMC meter (b) Hot wire meter (c) Electronic voltmeter (d) Electrodynamometer 112. In a digital voltmeter the oscillator frequency is 400 kHz and the ramp voltage falls from 8 V to 0 V in 20 ms. What is the number of pulses counted by the counter? (a) 800 (b) 2000 (c) 4000 (d) 8000 113. Which of the following devices is used at the first stage of an electronic voltmeter? (a) BJT (b) SCR (c) MOSFET (d) UJT 114. A single strain gauge of resistance 120 W is mounted along the axial direction of an axially loaded specimen of steel (E = 200 GPa). The percentage change in length of the rod due to loading is 3% and the corresponding change in resistivity of strain gauge material is 0.3%. For a Poisson’s ratio of 0.3, the value of the gauge factor is (a) 1.3 (b) 1.5 (c) 1.7 (d) 2.0 115. A voltmeter with an internal resistance of 200 kW when connected across an unknown resistance reads 250 V. The milliammeter internal resistance » 0 connected in series with the above combination reads 10 mA. The actual value of the unknown resistance is (a) 25 kW (b) 200 kW (c) 28.56 kW (d) 20 kW 116. When a sinusoidal signal of 220 V, 50 Hz produces on CRO a vertical deflection of 2 cm at a particular setting of the vertical gain control, what would be the value of the voltage to be applied to produce a deflection of 3 cm for the same vertical gain? (a) 330 V (b) 110 V (c) 220 V (d) 55 V 117. Maximum value of voltage can be measured by 3 digit digital voltmeter is (a) 199 V (b) 999 V (c) 1999 V (d) 900 V

118. A galvanometer is tested as shown in Fig., in the circuit where E = 1.5 V, R1 = 1 W, R2 = 2.5 kW, and R3 is variable. With R3 set at 450 W, the galvanometer deflection is 140 mm and with R3 set at 950 W, the galvanometer deflection is 70 mm. The resistance of the galvanometer is R3

E R1

15 V

G RG

10 W

2.5 kW R2

(a) 99 W (b) 40 W (c) 28 W (d) 10 W 119. The arrangement shown in the given fig. represents an RC potentiometer for measuring ac voltage. What should be the value of C so that Vo/Vin is independent of the frequency of the input signal? 1 mF Input Signal

10 kW

Vin C

1kW V To meter

(a) 10 mF (b) 11 mF (c) 0.01 mF (d) 0.09 mF 120. A scale of a 0-500 V voltmeter is divided into ten large divisions representing 50 V each and each large division a further subdivided into 10 small divisions, each representing 5 V. It is used for measurement of output voltage of a potentiometer which can be varried form 0 to 500 V. It is observed that when the sliding contact moved form its zero position, there is no perceptible movement of pointer of the voltmeter till the sliding contact reaches a position where the output voltage should be 5 V. Therefore, it can be concluded that (a) threshold of the voltmeter is 5 V (b) resolution of the voltmeter is 5 V (c) sensitivity of the voltmeter is 5 V (d) none of these 121. A slide wire potentiometer has 20 wires at 1 m each. With the help of a standard voltage source of 1.018 V it is standardised by keeping the jockey at 101.8 cm. If the resistance of the potentiometer wires is 1000 ohm, then the value of the working current will be (a) 0.1 mA (b) 0.5 mA (c) 1 mA (d) 10 mA

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31 122. A 35 V d.c. supply is connected across a combined resistance of 600 ohms and an unknown resistance of R ohms in series. A voltmeter having a resistance of 1.2 kW is connected across 600 ohm resistor and reads 5 V. The resistance R will be 1200 W 5V

600 W 35 V R

(a) 120 ohms (b) 500 ohms (c) 1.7 k ohms (d) 2.4 k ohms 123. A basic d’ Arsonval movement with internal resistance 100 W and full scale current 1 mA is to be converted into a multirange dc voltmeter with voltage ranges 0 – 10 V, 0 – 30 V, 0 – 100 V and 0 – 300 V. The circuit arrangement is shown in the figure. The value of R1 will be R4

R3

125.

126.

127.

V

124.

0

+ –

R1

30 V

10 30 0

R2

Detector

V

10 V

128. Current in the RF range is measured by (a) simple ammeter (b) ammeter using thermocouples (c) multirange ammeters (d) aryton shunt 129. An analogue voltage signal whose highest significant frequency is 1 kHz is to be coded with a resolution of 0.01 percent for a voltage range of 0–1V. The minimum sampling frequency and the minimum number of bits should respectively be (a) 1 kHz and 12 (b) 1 kHz and 14 (c) 2 kHz and 12 (d) 2 kHz and 14 130. A ring core current transformer having ratio of 1000/5 A is operating at full primary current with a secondary burden of non-inductive resistance of 1.1 W. Its exciting current is 1 A at a power factor of 0.45. What is the ratio error at full load? (a) – 0.089% (b) 0.099% (c) 0.027% (d) – 0.045% 131. A current transformer has a rating of 1000/5 A. Its magnetizing current and loss component of exciting current are 10 A and 6 A respectively. The phase angle between secondary winding induced voltage and current is 30°. The phase angle error of the transformer at rated current is (a) 0.65° (b) 0.305° (c) 0.496° (d) none of these 132. The wein’s bridge shown in the figure is balanced at a frequency of

(a) 0.0 W (b) 100 W 4.7 kW (c) 9.9 kW (d) 10 kW 20 kW 5 nF A single-phase energymeter is operating on 230 V, 50 Hz supply with a load of 20 A for two hours at upf. The meter D makes 1380 revolutions in that period. The meter constant 10 kW is 100 kW (a) 695 rev/kWh (b) 150 rev/kWh (c) 0.15 rev/kWh (d) 1/150 rev/kWh 10 nF In the case of power measurement by two wattmeter method (a) 32.8 kHz (b) 15.9 kHz in a balanced 3-phase system with a pure inductive load (c) 3.28 kHz (d) 1.59 kHz (a) both the wattmeters will indicates the same value but 133. Two sinusoidal voltage signals of equal frequency are applied to the vertical and horizontal deflection plates of a with opposite signs CRO. The output observed on the screen is shown in the (b) both the wattmeters will indicate zero following figure. (c) both the wattmeters will indicate the same value with same signs Y (d) one wattmeter will indicate zero and the other will indicate some non-zero value A 230 V, 10 A, single-phase energymeter makes 90 revolutions in 3 minutes at half load rated voltage and unity 5.0 3.0 pf. If the meter constant is 1800 revolutions / kWh, then its X X' error at half load will be (a) 13.04% slow (b) 13.04% fast (c) 15% slow (d) 15% fast The expected value of the voltage across a resistor is 80 V, and the measured value is 79 V. What is the percentage Y' error and relative accuracy of the measurement? The phase difference between the applied signals is (a) 1. 265%, 98.735% (b) 1%, 99% (a) 31° (b) 53.13° (c) 2.625%, 97.375% (d) 1.25%, 98.75% (c) 143.1° (d) 59° Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

32 134. The shunt type ohmmeter is NOT suitable for high resistance measurements because (a) a very low resistance of the meter would short the high unknown resistance (b) scale is highly cramped for high resistance values (c) full-scale value of the meter may be exceeded (d) battery cannot supply the necessary current for proper meter detection 135. A CRO probe has an impedance of 500 kW in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in figure. The measured voltage will be 100 kW

10 V rms 100 kHz

136.

137.

138.

139.

To CRO through probe

100 kW

(a) 3.53 V (b) 4.37 V (c) 4.54 V (d) 5.00 V Which of the following are disadvantages of using multipliers with voltmeters at high voltages? (a) The power consumption of multipliers becomes large average voltages. (b) The multipliers at high voltages have to be shielded in order to prevent capacitive currents. (c) The metering circuit is not electrically from the power circuit (d) All of these. When testing a coil having a resistance of 10 ohms, resonance occurred when the oscillator frequency was 10 MHz and the rotating capacitor was set at 500 /2 p pF. The effective value of the Q of the coil is (a) 200 (b) 254 (c) 314 (d) 542 A dc circuit can be represented by an internal voltage source of 50 V with an output resistance of 100 kW. In order to achieve accuracy better than 99% for voltage measurement across its terminals, the voltage measuring device should have a resistance of atleast (a) 10 MW (b) 1 MW (c) 10 kW (d) 1 kW In an oscilloscope, two Lissajous figure (X) and (Y) are observed. This indicates that ratio of vertical input signal frequency to that of horizontal input frequency are

(X)

(Y)

(a)

5 3 for X and for Y 3 2

(b)

5 3 for X and for Y 3 2

5 5 3 3 for X and for Y (d) for X and for Y 3 3 2 2 140. Three d.c. voltmeters are connected in series across a 120 V d.c. supply. The voltmeters are specified as follows: Voltmeter A : 100 V, 5 mA Voltmeter B : 100 V, 250 ohms/V Voltmeter C : 10 mA, 15,000 ohms The voltages read by the meters A, B and C are, respectively (a) 40, 50 and 30 V (b) 40, 40 and 40 V (c) 60, 30 and 30 V (d) 30, 60 and 30 V 141. Consider the signal Vm sin 100 t + 2 Vm sin 200 t to be sampled and stored in a data acquisition system. The same is to be extracted off-line later on. In order to extract the signal effectively, the original sampling frequency has to be (a) 100 rad/s (b) 200 rad/s

(c)

142.

143.

144.

145.

(c) 210 rad/s (d) 100 2 + 2002 rad / s The maximum temperature rise of a transformer is 50°C. It 1 attains a temperature of 31.6° hour. What is its thermal 2 time constant? 1 (a) 2 hours (b) hour 2 1 (c) 1 hour (d) hour 4 A moving-coil instrument gives full-scale deflection for 1 mA and has a resistance of 5 W. If a resistance of 0.55 W is connected in parallel to the instrument, what is the maximum value of current it can measure? (a) 5 mA (b) 10 mA (c) 50 mA (d) 100 mA A (0 – 25) Amp ammeter has a guaranteed accuracy of 1 per cent of full scale reading. The current measured by this ammeter is 10 Amp. The limiting error in percentage for this instrument is (a) 2.5% (b) 0.5% (c) 0.25% (d) 0.025% A moving coil ammeter having a resistance of 1ohm gives full scale deflection when a current of 10 mA is passed through it. The instrument can be used for the measurement of voltage up to 10 V by (a) connecting a resistance of 999 ohm in series with the instrument (b) connecting a resistance of 999 ohm parallel to the ammeter (c) connecting a resistance of 999 ohm parallel to the load (d) connecting a resistance of 1000 ohm in series with the load

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33 NUMERICAL TYPE QUESTIONS 146. A dc circuit can be represented by an internal voltage source of 50 V with an output resistance of 100 kW. In order to achieve accuracy better than 99% for voltage measurement across terminals, the voltage measuring device should have a resistance of at least _____ MW. 147. In the measurement of power on balanced load by two Wattmeter method in a 3-phase circuit, the readings of the Wattmeters are 3 kW and 1 kW respectively, the latter being obtained after reversing the connections to the current coil. The power factor of the load is _____. 148. A 1 cm piezoelectric transducer having a g-coefficient of 58 V/kg/m2 is subjected to a constant pressure of 10–3 kg/m2 for about 15 minutes. The Piezo electric voltage developed by the transducer will be mV _____. 149. A moving-coil instrument gives full-scale deflection for 1 mA and has a resistance of 5 W. If a resistance of 0.55 W is connected in parallel to the instrument, what is the maximum value of current it can measure? 150. A single side wire is used for the measurement of current in a circuit. The voltage drop across a standard resistance of 1.0 W is balanced at 70 cm. What is the magnitude of the current, if the standard cell having an e.m.f. of 1.45 volts is balanced at 50 cm? 151. Pulses from the clock of frequency 100 kHz pass through the counter of digital multimeter during a gate period 5.75 m sec. The number of pulses counted by the counter will be 152. In a CRT, 3 × 1017 electrons are accelerated through a potential difference of 10,000 V over a distance of 40 mm per minute. Calculate the average power supplied to the beam of electrons. 153. A zero to 300 V voltmeter has a guaranteed accuracy of 1% full scale reading. The voltage measured by the instrument is 83 V. The percent limiting error is 154. In a dual slope integrating type digital voltmeter the first integration is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V the total conversion time for an input of 1 V is

155. When testing a coil having a resistance of 10 W, resonance occured when the oscillator frequency was 10 MHz and the rotating capacitor was set at 500/2p pF. The effective value of the Q of the coil is 156. A Wheatstone bridge requires a change of 6 ohms in the unknown arm of the bridge to produce a change in deflection of 3 mm of the galvanometer. The sensitivity of the instrument is Statement for Linked Answer Questions 157 and 158 Consider a R–L–C circuit shown in figure.

R = 10 W

L = 1 mH

C = 10 µF ei

e0

157. For a step-input ei, the overshoot in the output e0 will be (a) 0, since the system is not under-damped (b) 5% (c) 16% (d) 48% 158. If the above step response is to be observed on a nonstorage CRO, then it would be best to have the ei as a (a) step function (b) square wave frequency 50 Hz (c) square wave frequency 300 Hz (d) square wave frequency 2.0 kHz

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34

PAST GATE QUESTIONS EXERCISE 1. 2.

3.

(c) (a) The permanent magnet moving coil measures the DC value only. The torque will be reversed if the current is reversed. If the instrument is connected to AC, the deflection responds to the mean torque which is zero. Hence, voltmeter reads 2 V.

i1 R2 R4 = = i2 R1 R3

Þ R1 = 7.

xy z Taking log, we get, log w = log x + log y + log z By considering only positive error, differentiating with respect to w, we get, for maximum uncertainly

(d) Given, w =

4.

kn - R R –0.5 R1 = 100 kn – 100 R1 (kn = 200) R1 =201

l0 cos a =1 lg

8.

V 2 ´ t 230 ´ 230 = ´1 R R Þ R = 23.W \ Power dissipated in

9.

(d) (b) At balance condition,

2

= 10A

Moving iron also reads rms value, so its reading will also be 10 A, hence ultimately it will be (–8, 10, 10). (c) p-p division of upper trace voltage = 2 and value of p-p voltage = 5V Voltage = 2.5 divisions Now it will same for unknown voltage, p-p division of unknown voltage = 3 So p-p voltage = 3 × 2.5 = 7.5 voltage Frequency of upper trace = 1 kHz.

R3

i1

C

A R4

R2

æ6 2 ö 8 +ç ç 2 ÷÷ è ø 2

\

B

i2

\ n becomes 199. % error in ratio = 0 (c) PMMC instrument leads only DC value and since, it is centre zero type so, it will give –8 value. RMS meter will read rms value of the current So, RMS =

V 2 200 ´ 200 = kW = R R 40 ´103 = 1.739 kW.. = 23

R1

l0 cos(d + a) = 200 + l0 cos a lg

But % error ratio = –0.5 =

2.3 ´103 =

5. 6.

(a) Here, we assume that (d = 0) nls = lp R=n+

é dx dy dz ù dw = ±ê + + ú ´ 100 w y zû ë x

é 0.4 1 0.75 ù + + ´100 = ±ê ë 80 20 50 úû = 7% of reading (b) Reading of Energy Meter = 2.3 Units (kWH)

R2 R3 R4

1 = 10-3 = 1 ms 103 and division on x-axis = 4

\ Time Period =

Time 1 ms = Division 4 and division of Lower trace voltage on x-axis = 8

Thus,

D

~ V VB = VD V – i1R1 = V – i2R2 and V – i1R3 = V – i2R4

Thus,

Time 1 ms = Division 4

æ Time ö Period of unknown signal, ç × 8 = 2 ms è Division ÷ø

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35 Z1 Z4 = Z2 Z3 Þ R1 (R4 + jX4) = (R2 + jX2) (R3 – jX3) Þ 500 (R4 + jX4) = (300 + j200) (300 – j200) Þ 500 (R4 + jX4) = 90000 + 40000

10. (a) From waveform it is clear that for half of time period the power is of opposite sign of equal amplitude to another half of the time period. The total power in one period is zero. Hence, the reading of wattmeter is zero. 11.

Þ R4 + jX4 =

1 X1 = wL1 , X 4 = , Z1 = R1 + jX1 , wC4

(c)

Z4 = R4 –jX4 At balance,

14. (d)

Z1 Z3 = Z2 Z4

æ 1 ö Þ (R1 + jwL1 ) ç R 4 - j wL ÷ = R 2 R 3 4ø è Equating real and Imaginary terms, we get,

L1 =

1 + (w C 4 R 4 ) 2

Quality factor, Q =

, R1 =

w2 R 2 R 3R 4C42 1 + ( wC 4 R 4 ) 2

wL1 1 = wC 4 R 4 R1

B 0W 50

Z1 A X2

X3 D

R4

Z3

0.398 mF R3 300W C

D

\q(w2t ) = A sin(w2t + 90°) Þ q(w2t ) = A cos(w2t ) 15. (a) The time taken by both ADCs will remain TA, TB 16. (c) V Ð - 240°

1

2p´ 2000 ´ 0.398 ´10-6 Under balanced condition

= 200 W

120°

b 120° V Ð - 0°

c

V Ð - 120°

Assume Vab as reference phasor, we have Vab = VÐ0° Vbc = 400 Ж120° Vca = 400 Ж240° Current through current coil, V 400Ð - 240° I cc = ca = Z2 100 + j 0 = 4 Ж240°A and, voltage across pressure coil, Vpc = Vbc = 400 Ж120° Hence, wattmeter reading, P = Vpc × Icc cos f f is the angle between Vpc and Icc = 400 × 4 × cos 120° = – 800 watts 17. (d) For triangular wave, Vm Average value = 3 rms =

1 XC = 2pfC

=

w2 1 w = Þ w2 = 1 w1 2 2 and q(w2t ) will lead p(w1t ) by 90° Þ

2 4

Then, L1 =

~

Number of times tangent touches top or bottom Number of times tangent touches either side

a

w R2 R3 R4 C R2 R3C4 , R1 = 1 1 1+ 2 1+ 2 Q Q For Q > 10, L1 » R2R3C4, R1 = w2R2R3C42R4 As R2 is common in both the expression of L1 and R1, then R2 would have to be adjusted first and then R4 should be adjusted. 12. (b) Here, B is a common point S1 ® A G1 ® B S2 ® C G2 ® B Hence answer is A, B, C, B 13. (a) XL = 2p fL = 2p × 2000 × 15.91 × 10–3 = 200W 2

Oscillator

w2 w y = w1 wx =

Þ Z1Z4 = Z3Z2

R 2 R 3C 4

\ Z = 260 + j0

130000 = 260 500

Vm 3

Vm = 10V Þ Vm = 30 3

rms =

30 3

= 10 3

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36 18. (d) The relative error of product or division of different quantities is equal to the sum of relative errors of individual quantities. 19. (b) Pressure coil of a dynamometer type wattmeter is made to have very high resistance which includes small unavoidable inductance. 20. (b) In X-Y mode, if figure shows ellipse that indicates the amplitude is not equal but it changes to circle means amplitudes are equal. The inference can be made from the amplitudes of the signal are very close but not equal.

Z3

R3

R

+

Z1

jw L

21. (a)

22. (d)

I

A1

0–5A

0.2W

For 25A, Let 5A through ammeter A1. So, remaining current through Rsh, I = 25 – 5 = 20 A \ 20 × Rsh = 0.2 × 5 1 = 0.05 W 20 23. (c) The fixed coil or field coil is connected in series with the load and so carries the current in the circuit. The fixed coil therefore forms the current coil. The moving coil is connected across the voltage and therefore carries a current proportional to voltage. 24. (c) Number of full digits = 4

Þ Rsh =

1 digits display = 19999 2 Full-scale reading = 200 V

C

4)

Maximum count with 4

200V 19999 Then, error corresponding to 10 counts

–j / (w

R2

R4

(0 – 25A) Rsh

Z4 Z2

A

1 Count =

200 = 0.1 V 19999 and, error corresponding to 0.2% of reading

= 10 ´

~

100 = 0.2 V 100 Hence total error, E = E1 + E2 = 0.3V

= 0.2 ´

At balance, Z1 Z3 = Z2 Z4

E ´ 100% = ± 0.3% 100 (b) The compensating coil is made nearly identical with the current coil and is also coincident with the current coil and opposes the field produce by current coil. The additional conductor is an internal connections, corresponding to the lead from P to Q of Fig. which carries the voltage-coil current in a reverse direction through the winding.

% percentage error =

æ 1 ö R + j wL = R3 ç + j w C4 ÷ R2 è R4 ø R wL R 3 +j = + jw R 3 C 4 R2 R 2 R4

Equating real and imaginary parts,

25.

I = I P + IC M

R R = 3 R2 R4

Supply V1

R 2R 3 R= R4

Q

IC

P IP M.C OR P.C

R

V

I

Compensated wattmeter

wL = w R 3 C4 R2

L = R2R3C4

Thus any extra torque due to the voltage-coil current in the current coil itself is neutralized by the torque due to the voltage-coil current in the additional winding.

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37 26. (c) Oscilloscope with two vertical inputs, referred to as dual trace oscilloscope. Using a single-beam CRT, they multiplex the inputs, usually switching between them fast enough to display two traces apparently at once. Switching channels can be asynchronous, i.e., free running with trace blanking while switching or after each horizontal sweep is complete. Asynchronous switching is usually designated chopped while sweep-synchronised is designated att [esnate]. 27. (c) The circuit shown in the figure is the Maxwell's inductance-capacitance bridge. In this bridge, an inductance is measured by comparison with a standard capacitance. This bridge is limited to measurement of low Q coil. 28. (d) Let resistance of voltmeter be RkW.

200 kW + R kW

–

+

440 V V –

R

–

æ 352 ö V = 352 + ç ÷ 80 è R ø

...(ii)

+ 220

–

VL –

480 ´ 220 = 406 V 40 + 220 (c) Average power,

VL =

1 + 2p

2p

v (t ).i (t ).d (wt )

[ E1 sin wt + E3 sin 3wt ].[ I1 sin(wt - f1 )].d wt

2p 0

ò

E1 I3 [cos(f3 - 2wt ) + cos(4wt - f3 )].d wt

0

2p

ò

E3 I1[cos(2wt + f1 ) - cos(4wt - f1 )].d wt

0

1 4p

2p

ò

E3 I3 (cos f3 - cos 4wt - f3 ).d wt

0

[ E1 sin wt + E3 sin 3wt ].I 3 sin(3wt - f3 ).d wt

2p

1 4p

ò

E1 I5 (cos 3wt - cos 2wt ).d wt

0

2p

ò

E3 I5 (cos wt - cos 4wt ).d wt

0

E1I1 cos f1 1 + .E3 I3 cos f3 2 2

é ù ê ú ê All other terms are zero.ú ê 2p ú ê ú ê as cos(2wt - f1 ).d wt ú ê 0 ú ê ú ê sin(2wt - f1 ) 2 p ú =0 ú ê= 2 êë úû 0

ò

30. (a) 31. (a) PMMC type of instrument measure average value

0

0

ò

1 4p

2p

1 + 4p

=

2p

ò

E1 I1[cos f1 - cos(2wt - f1 )].d wt

+

< P >=

40 kW

1 = 2p

0

+

+

ò

[ E1 sin wt + E3 sin 3wt ].I 5 sin 5wt.d wt

0

– ...(i)

480

ò

352 V

æ 440 ö V= ç ÷ 20 + 440 è R ø

1 = 2p

ò

1 + 4p

+

2p

2p

+

Solving, V = 480 V R = 220 W

29.

1 = 4p

1 2p

80 kW

+ V

+

Area of graph Total time

1 ´10 ´ 10 - 5 ´ 2 + 8 ´ 5 = 2 20

=

80 = 4V 20

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38 when Rs is increased by 1% new value of

1 kW

R 'e = 303W 32.

(a)

v0 = v0 - v1 = v mn - v

–

D

100 kW

Voltmeter

During the half cycle, the diode is forward biased (anode is connected to positive, Cathode is connected to negative). The voltage across forward biased ideal diode is zero. During –ve half cycle the diode is reversed biased. Cathode is connected to negative, cathode is connected to positive). The average voltage across the voltmeter will be 14.14 ´

100 ´ 102 3

100 ´10 + 1´10

3

´

300 ´ 12 = ( 6 - 5.97 ) v = 0.03 v = 30 mv 603 (b) horizontal axis 140 to 142 3.0 to 3.4 17.34 W Total power dissipated in the circuit is 1kW. P = 1kW 1000 = I2.1+ I2.R. 1000 = (2)2 . 1 + (10)2 . R.

= 6-

+

35. 36. 37. 38.

Þ R = 9.96W

1 = 4.46V 2

|Z| =

Input Voltage

V 200 Þ = 20 1 10 R 2 + XL2

|Z| =

Þ X 2L = (Z)2 - R 2

14.14

Þ X 2L = (20)2 – (9.96)2 20 ms 10 ms

wt, t

Þ X L = 17.34W 39.

10.157 X and Y ammeters are connected in parallel Shunt Registration of X and Y meters: 15A

Voltage across the voltmeter

1.2W 10 ms

20 ms

Rshx

t

Imx= 150mA

1.5W Rshy Imy = 250mA

14.14

33.

34.

(d) V2 > V1 and phase difference is 180 and hence MI/ voltmeter will read V = V2 – V1 (c) Maximum permissible voltage = (300 + 300) × 20 × 10–3 = 12 V R1

R2

b + –

Rs d

Rshx =

R3

æ 15 ´ 103 ö ç 150 - 1÷ è ø

Rshx = 0.01212W Rshy =

Q

V0

1.2

1.5 æ 15 ´ 103 ö ç 250 - 1÷ è ø

Rshy = 0.02542W Current through X ammeter is

0.02542 ´ 15 (0.1212 + 0.2542) = 10.157 amperes

=

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39 40.

2 Power at full scale reading = 150 Current range used = 10A Voltage range used = 150V Power factor = 0.2 Multiplying factor =

PRACTICE EXERCISE

Current range used ´ voltage range ´ p.f. Power at full scale reading

1.

(c) Induction principle is more generally used for Watthour meters than for ammeters and voltmeters owing to their comparatively high cost, and inaccuracy of induction instruments of the latter types.

2.

(b)

2

10 ´ 150 ´ 0.2 15 =2

=

41.

(a)

2

æ ¶I ö 2 æ ¶I ö 2 s1 = ç s + s è ¶I1 ÷ø I1 çè ¶I2 ÷ø I2 ¶I ¶I = =1 ¶I1 ¶I2

(1) 2 (1)2 + (1) 2 ( 2) 2 I = ( 4.00 ± 2.24 ) A

\ s1 =

Points Y(S1 )

X (S2 )

x 2 + y2

f = 100 -1

1

0

A

0

1

B

1

1

C

0

0

D

-1

-1

45°

2 0

3.

(a)

6.

(a) Average current =

0 225°

2

y 1

–1

\

1

x

4. (d)

5. (c)

117.02 + 117.08 + 117.11 + 117.03 4 = 117.06 mA as Imax = 117.11 mA & Imin = 117.02 mA

( Imax - Iav ) + ( Iav - Imin ) \ range of error = ± 2 0.05 + 0.04 ± = ± 0.045mA 2 7. (d) 8. (b) 9. (d) Refer potentiometric type digital voltmeter 10. (b) 11. (b) LVDT & strain gauge measure linear displacement. 12. (d)

Td = Cq, Td = KI 2 2 \I =

–1

= 2.24A

C q = C 'q K

I2 C ' p / 4 = 1 C'p / 2 2 1 Þ I = ; I = 0.707A 2 1 = 200 W / V 13. (c) Sensitivity (I) = 5mA

Þ

42.

(a)

2 æ 1 ö RMS value = 1 + ç è 2 ÷ø

= 1+

= 43.

1 æ ö kW÷ = 99.99 kW 14. (a) Series resistor = ç100 kW - 1 ´ è ø 100 15. (c) Work done by the electric field = eV

1 2

3 2

Kinetic energy =

0.3 z1z4 = z2z3

or v =

1 1 35kx jw 0.1 = 105k jwc 1

c1 = 0.3 mF

2

2eV = m

1 mv 2 = eV 2

2 ´1.6 ´ 10 -19 ´1000 9.1´ 1031

= 0.1875 × 108 m/s R 750 16. (a) R1 = R 2 3 = 2000 ´ = 375W R4 4000 L1 = R2R3C4 = 2000 × 750 × 0.05 × 10–5 = 75 mH

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40 17. 19. 21. 23. 24.

25. 27. 29. 31. 33.

34. 36. 38.

39. 40. 41.

(a) 18. (c) (c) 20. (d) (a) 22. (c) (a) (a) The makers specify the maximum value as 100 mA. This means that the current drawn from the cell should be less than 100 mA and this current should flow momentarily. (d) 26. (a) (a) 28. (b) (b) 30. (c) (a) 32. (c) (c) Both spectrum analyzer and distortion factor meter are signal analysers. A spectrum analyser sweeps the signal frequency band and displays a plot of amplitude versus frequency. Distortion factor meter tunes out the fundamental signal and gives an indication of the harmonics. (c) 35. (d) (b) 37. (b) (d) There will be two deflecting torques, one acting on coil A and the other on coil B. The coil windings are so arranged that the torques due to the two coils are opposite in direction. So, there is no controlling torque. (c) The error caused by pressure coil inductance is VI sin f tan b. With low power factor, the value of f is large and therefore, the error is correspondingly large. (d) Braking torque TD ¥ R2 where R is the radial position of braking magnet with respect to the centre of the disc. (a) Bigger size mirror results in increased inertia constant J.

GIm 1 K &A= 2 2p J (Dw ) + (K - Jw 2 ) 2 as J increases both fn & A decrease. 42. (b) Actual loading arrangements would involve a considerable waste of power. In order to avoid this ‘Phantom’ or ‘Fictitious’ loading is done. 43. (a) Kelvin bridge overcomes the difficulties that arise in a Wheatstone bridge on account of the resistance of the leads and the contact resistances while measuring low valued resistors. The Kelvin double bridge incorporates the idea of a second set of ratio arms, hence the name double bridge and the use of four terminal resistors for the low resistance arms. The second set of ratio arms eliminate the effect of connecting lead resistance. fn =

C - 4C2 44. (d) Self-capacitance Cd = 1 3 45. (a) The expansion (deflection) is proportional to the heating effect of the current and hence to the square of the rms value of the current. Therefore, the meter may be calibrated to read the rms value of the current. 46. (c) 47. (c) Simple a.c. amplifiers may be used to amplify a dc input through use of an additional circuit component known as chopper. The approach is used to build a dc amplifier

where in the dc is first converted to an equivalent a.c. signal that is amplified by a standard a.c. amplifier. The a.c. signal is finally converted back to a dc signal. Chopper stabilized amplifier eliminates the effects of dc offset currents & the drift of other dc parameters by using an a.c. coupled amplifier for necessary gain. 48. (b) In some meters a slow but continuous rotation is observed even when there is no current flowing through the current coil and only pressure coil is energized. This is called creeping. The major cause for creeping is over-compensation for friction. 49. (a) A three digit display on a digital voltmeter for 0-1 V range will be able to indicate values from zero to 999 mV, with smallest increment or resolution of 1 mV. In practice a fourth digit usually capable of indicating either 0 or 1 only is placed to the left of active digits. This permits going above 999 to 1999 to give overlap between ranges for convenience. This is called over ranging. This type of display is known as a 3½ digit display. 50. (c) Refer Piezoelectric type accelerometer. 51. (c) 6´5 52. (a) Amplitude of the signal is = 15 V 2 15 \ rms value is = 10.6 V 2 53. (b) The wattmeters read One µ cos (30 + f) The other µ cos (30 – f) Thus whenever load p.f. angle f becomes 60° one of the wattmeters would read zero. \ For the given condition, the load p.f. is cos 60° = 0.5 54. (d) M should be able to detect phase reversal if not strictly phase sensitive. Phase sensitive detector can well serve the purpose. 55. (d) Strictly speaking, an Ayrton shunt is used to protect the galvanometer during initial stages of balancing by reducing the galvanometer sensitivity. However, it can also be used to convert a galvanometer into a multistage ammeter 5

4

R

I

3 I

2

G

1

r1 + r2 I' = I r1 + r2 + r3 + r4 + r5

56. (d) It is simple to see VR = 30 V, VL = 20 V, VC = 60 V and VT = 50 V.

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41 = 30°, therefore, the phase difference is 30° or – 30°, i.e., 30° or 330°. Y1 is the deflection on the y-plates when x-plates deflection is zero, i.e., when Vx(t) = 0 and Vy(t) = Vmy sinf and Y2 is the deflection due to Vmy. 64. (c) The total length observed will be proportional to

57. (b) I = 5 mA E 1 MW 5V 1 MW

Voltmeter resistance = 100 × 10 = 1000 k-ohm = 1 M-ohm Photovoltaic cell internal resistance = 1 M-ohm The voltmeter reads 5 V. Therefore, the internal voltage is 10 V. 58. (b) Differential transformer transducer is used for measurement of displacement. E 59. (d) Ifs = R1 + R m With R connected, Rx E I= R x + R m R1 + ( R m R x ) / ( R m + R x ) ER x ER x = R1 ( R m + R x ) + R m R x R x ( R 1 + R m ) + R1R m Rx Rx E = Ifs = R + R R é ù ( 1 x ) ëR x + ( R1R m ) / ( R1 + R m )û x +Rp

=

R1R m R1 + R m 60. (a) It is an elementary piece of knowledge. For low resistance measurements, if circuit II is used, the ammeter resistance, which may be comparable with the low resistance being measured, is added to the unknown and thereby causing larger error. 61. (c) The setting of the potentiometer is at 1.0138 instead of 1.0813. Therefore, the working current in the slide wise is more than what it would have been if the setting was at 1.0813, i.e. the correct value for applied voltage of 1.0813 V,

where, R p =

I 1.0813 = Ic 1.0138 If the setting is at 1.0138 and the applied voltage is also 1.0138, the standardisation is then correct. (Statement of the problem appears ambiguous) 62. (c) If D1 were not there in the negative half cycle of the applied voltage, there would be some small reverse current through D2 (in the reverse direction) and the meter, thereby giving erroneous mean value and hence of the applied voltage. With D1 as in circuit, the current in the negative half cycle flows through Rsh and D1 and there is no voltage across D2 and hence no current through D2. 63. (c) When major area of the Lissajous figure lies in the 1st and 3rd quadrant, then the angle is within 0 and ± 90°. If the loop lies in 2nd and 4th quadrant, then the angle is 180° – q where q is within 0 and ± 90°. In the present case the angle is given by q = sin–1 (Y1/Y2) = sin–1 0.5

2Vm = 2 2 × 30. 30 V dc causes 1 cm deflection from the centre point. Therefore, the length observed is 2 2 cm » 3 cm. RMS Value 65. (b) Form factor = of the wave M.I. Instrument Mean Value will show RMS value. Rectifier voltmeter is calibrated to read rms value of sinusoidal voltage, i.e., with form factor of 1.11. Therefore, mean value of the applied V2 voltage is . 1.11 V1 1.11V1 = Hence, form factor = V2 / 1.11 V2 66. (c) rcc = 0.01 ohm, rvc = 1000 ohm. The current coil and voltage coil can be connected in two ways. The wattmeter reads high by an amount equal to power loss in current coil in one case and in voltage coil in the other. The two possible power losses are (i) Ic2 rcc = 202 ´ 0.01 = 4W V2 302 = = 0.9W rve 1000 The two percentage errors are 4 ´ 100 2 (i) = 600 3 0.9 ´ 100 = 0.15 (ii) 600 67. (b) In the induction type of instrument there is no moving coil. Therefore, the wattmeter is question is of dynamometer type. The fixed coil is the current coil and moving coil the pressure or voltage coil. 68. (b) It is simple resistive circuit calculation,

(ii)

Ic =

1 = 1+ R3 + RG

1 1( R 3 + R G ) 2500 + 1 + ( R3 + R G )

From the given data, 1 1 + 450 + R G 1 1 + 950 + R G

1 450 + R G 2500 + 1 + 450 + R G

2E 950 + R G 2500 + 1 + 950 + R G 2500 (451 + RG) + 450 + RG = 2500 (951 + RG) + 50 Þ 2501 RG = 2500 (951 – 902) + 50 = 2500 × 49 + 50 = 2501 × 49 + 1 Hence, RG = 49 + 1 / 2500 » 49 W. Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

42 69. (a)

2

f=

=

1000 Hz p

2p 10 -6 70. (d) Only 1 and 3 statements are correct. Measurement range is not wide. 71. (a) Applying the usual balance condition relation Z1Z4 = Z2Z3, we have

79. (d) For 10 V to produce 1 mA requires a 10 k ohm resistor. Thus, the meter will appear as: 10 kW M

+ 10 V –

LX

R2

RX S

C4 R3

When 5 kilo-ohms is placed across the terminals, the current will be 10 V / (10 + 5) k. Thus, 10 V/15 K = 2/3 mA and thus 2/3 deflection.

R4

80. (a) 81. (d) The meter reading will either decrease or increase depending upon the direction of flux. Fm

1/ R 4 jC 4 w ( R1 + jL1w ) R + 1/ jC 4w 4

Or, ( R1 + jL1w)

( R 4 / jC4w )

R 4 + 1/ jC4 w

= R 2R3

Þ R1R 4 + jL1wR 4 = R 2 + R 3 + jR 2 R 3R 4C4w \ R1 = R 2

R3 ; R4

L1 = R 2 R 3C4

750 = 375 ohm. And, 4000 L1 = 2000 × 750 × 0.05 × 10–6 = 75 mH. (c) Q = Lw / R. With Q-meter resistance considered, the measured or indicated Q is 1/1.11 times the actual Q. Thus, statement 1 is not correct. Statements 2 and 3 are correct. (d) (c) For a half radiated wave

\ R1 = 2000 ×

72.

73. 74.

Idc =

I peak p

and Irms =

Ipeak 2

\ PMMC reading would be Idc = 75. (b)

2´5 = 3.18 A p

L = L1 + L 2 ± 2M

When ordinary, L = L1 + L2 + 2M 40 = 10 + 15 + 2M. \ 2M = 15 mH When opposing, L = L1 + L2 – 2M = 10 + 15 – 15 = 10 mH 76. (d) Assuming that gate open when the ramp is 8V and closes when the ramp reached zero, then the number of pulses counted by counter would be : 400 × 103 × 20 × 10–3 = 8000 77. (b) f = 1 × 106 / 1 × 104 = 100 1 = 10 ms 100 (d) All the three errors may occur in bridge method of measurement.

\ T=

78.

F Fs As shown in fig. let Fm is the flux due to meter coil and Fs is due to stray magnetic field, then resultant will be F. Similar but opposite will happen when Fs is in opposite direction. Thus, the reading will either increase or decrease. 82. (d) Ammeter error, DI = ± 4% Voltmeter error, DV = ± 24% Resistance to be measured, R = V/I In case, error in resistance measurement is ± DR , then the maximum possible error will be R + DR = V + DV / I - DI or RI + IDR - RDIDR = V + DV Neglecting DR being small and substituting V = RI we have, V + IDR - RDI = V + DV Þ IDR - RDI = DV Dividing by IR, DR / R - DI / I = DV / IR = DV / V Þ D R / R = DV / V + D I / I Therefore, maximum percentage error = 2.4% + 1% = 3.4% 83. (c) Error in the potentiometer is due to thermo electric emf’s set up at junctions of dissimilar metals and also by the heat from the operator’s hand during adjustment of the working parts of the potentiometer. Reversing of polarity and taking reading will give error in the negative direction. Suppose, the reading with error due to one direction is D + DR and in the other direction is D – DR, then the mean value is (D + DR + D – DR) / 2 = D; which is the actual value.

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43 84. (c) In a dual slope integrating type digital voltmeter,

Vin = Vref ( t 2 / t1 )

Then, V = V1 – V2 =

Where, t1 is the first integration time. t1 = 10 × 1/50 = 0.2 s Vm = 1 V Vref = 2 t2 = VinT1 / Vref = 1 × 0.2 / 2 = 0.1 sec 85. (b) The lines base setting = 5 msec/div. 1

Input frequency =

-3

=

87. (c)

20 10 - = 3.33V 3 3

(100 50°)Zd = (300 - 90°) (200 0°)

Rb 1 + j wCb R b

1.5 = 0.02 75 voltage between the limits,

Relative error, e r =

75 (1 ± 0.02) = 75 ± 1.5V 1.5 ´ 100 = 2% % limiting error = 75 (b) As voltmeter has high impedance, therefore, current through it is zero or negligible. 1

v

+

–

I=0

20 W

10 W 2

LOAD

V = (0.2 + 0.2) × 5 + 200 = 202 V; I=5A Wattmeter reading, P = V I cos f = 202 × 5 × 1 = 1010 W. 94. (a) The overall uncertainty;

=

(1)2 + (1)2 + (3)2

= 11%

95. (d) A – 1, B – 3, C – 2, D – 4 Instruments A.Vibration Galvanometer B. Head phone C. D ' Arsonval Galvanometer D. CRO

Frequency 100 Hz 1 kHz 0 Hz L arg e frequency range

96. (b)

R

C1 R1 V

L D

Cs

Rs

In balanced condition;

wC1 ( R + jwL ) Rs R + jwL = Þ = jwCs R s 1 1 wC1R1 - j R1 - j wC1 jwCs Þ C1 ( R + jwL) = jCs R s ´ ( wC1R1 - j) Equating real terms

10 W

20 W

V

V

Wx = w 2x + w x2 + w 2x 1 2 3

Þ (1 + jwCb R b ) .R a R c = R b ( R d + jwLd ) Þ RbRd = RaRc Þ Rd = RaRc / Rb And, wCbRbRaRc = wRbLd Þ Ld = RaRcCb 89. (d) As both the signals are of same frequency, same phase and of same amplitude hence these are applied at the X-Y input of the CRO, hence on the CRO’s screen, these will produce a straight line with 45° with respect to x-axis. 90. (a) 91. (b) Magnitude of the limiting error, ¶A = e r .As = 0.01 × 150 = 1.5 V

92.

IR

A IV

Þ Zd = 600 -140° 88. (d) From balance condition R a R c = ( R d + jwLd )

I

93. (d)

1000 = 200 Hz 50

5 ´ 10 Frequency of input signal = 314 = w, or, 2pf = 314 or f = 314 / 2p = 50 Hz Number of cycles of signal displayed = 50 × 10 / 200 = 2.5 86. (a) The “accuracy” of a measuring instrument is determined by the closeness of the value indicated by it to the correct value of the measurement.

10 10 .20 - .10 30 30

C R C1R = Cs R s Þ R = s s C1

Equating imaginary terms: wLC1 = wCs R s C1R1

\ L = Cs R s R1

10 V

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44 97. (a)

LX

1000 W

RX V Cs 1000 W

Rs

In balance condition: Rs 1000 = R x + jwL x ( jwR s Cs + 1) ´ 1000

Þ ( jwR s Cs + 1) ´106 = R s ( R x + jwL x )

103. (d) Loading effect error: When the voltage across impedance is measured by a voltmeter, the voltmeter acts as a load and due to its internal impedance an error in reading occurs. This error is called loading effect error. 104. (a) The PMMC is used for measurement of DC. Therefore, the current i = 3 + 4 2 sin 314t = 3A . 105. (d) In a trapezoidal pattern, modulation factor (M) M=

Large line – Smaller line Large line + Smaller line

Equating real terms: R s R x = 106 éGiven ù 106 Þ Rx = = 103 = 1000 W ê ú R 1000 = W Rs ë s û

Imaginary terms; 106 jwR s Cs = jwR s L x Þ 106 Cs = L x

Þ L x = 106 ´ 0.5 ´ 10 -6 = 0.5H [Given Cs = 0.5 mF = 0.5 × 10–6 F] 98. (d) Inputs to the CRO are: Vx = 10 cos (100t + f); and Vy = 10 sin (100t + f) = 10 cos (100t + f + 90°). The resulting Lissajous pattern is a circle as the magnitudes of the two signal are same and phase difference is 90°. Note: A circle can be formed in Lissajous pattern only when the magnitudes of the two signals are equal and the phase difference between them is 90° or 270°. 99. (c) The damping torque in the disc of an ac energy meter is produced by Eddy current effect (by interaction of the eddy current and permanent magnet’s field). 100. (b) Error as percentage of true value; =

Full scale voltage ´ Accuracy Voltage value

Given, V.M. range = 0 to 20 v and Accuracy = ± 1% Hence, for voltage values 2V, 5 V, 10 V and 20 V respectively; errors are: A.

20 ´1 = 10% 2

20 ´ 1 = 2% C. 10

B.

20 ´ 1 = 4% 5

20 ´ 1 = 1% D. 20

Smaller line Larger line

So, in the trapezoidal pattern smaller line = 0. Large line – 0 Hence, M = Large line + 0 = 1

106. (b) The shape of the figure seen will be a straight line as both the voltages are equal and in phase with each other (the shape will be a straight line even if the two voltages are out of phase by 180°). 107. (b) CRO has minimum loading effect on the quantity under measurement. 108. (d) Owen’s bridge is used for measurement of inductance in terms of capacitance. At balance condition; æ æ 1 ö 1 ö çè R c + jwC ÷ø R a = ( R x + jwL x ) çè jwC ÷ø c b

By equating real terms; Lx = RaCbRc R C Equating imaginary terms; R x = a b Cc

109. (c) Ser vo type Potentiometric r ecorder has better frequency response than galvanometric recorder. 110. (c) The Wheatstone bridge is ideally suited for measuring resistance in the range of 100 W to 10 kW. 111. (c) Electronic voltmeter requires external source for its operation. Electronic voltmeter consumes low power. 112. (d) Given; Oscillator frequency = 400 kHz = 400 × 103 Hz. Time period of oscillation; T =

1

400 ´103 Number of pulses counted by the counter (in 20 ms); -3

-3

20 ´10 20 ´10 101. (c) = = 200 ´ 400 = 8000 1 102. (c) The resolution of a wire wound potentiometer can be T . improved by reducing the diameter of the resistance 400 ´103 wire. Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

45 113. (c) In electronic voltmeter MOSFET is used at the first stage. 114. (c) Percentage change in length = 3%; Change in resistivity = 0.3%; hence, Gauge factor G f = 1 + 2v +

¶r / r [where, v = poisson’s ratio] ¶l / l

= 1 + 2 ´ 0.3 +

0.3 = 1.7 3

115. (c) Given; Internal resistance of voltmeter (Rv) = 200 kW. Voltmeter Reading = 250 V Ammeter reading = 10 mA Internal resistance of Ammeter » 0 Value of unknown resistance (say RU) = ? Now, effective resistance, R e =

250

10 ´ 10-3 To calculate unknown resistance (RU): R e = R U || R v Þ 25 =

= 25 kW

R U ´ 200 R U + 200

Þ 25 (RU + 200) = 200 RU Þ RU = 5000/175 = 28.571 kW = 28.56 kW 116. (a) Given: 220 V produces deflection of 2 cm. So, for 3 cm deflection: 220 ´ 3 = 330V 2

117. (d) 118. (b) 119. (a) As given VO/Vin is independent of frequency then; C = 1 × 10/1 = 10 mF. 120. (a) 121. (c) Total length of the slide wire = 1000 cm Total resistance of the slide wire = 1000 ohm \ Resistance per cm = 1 ohm Resistance of 101.8 cm segment of the wire = 101.8 ohm This corresponds to a voltage of 1.018 V 1.018 \ Current = = 1 mA 101.8 122. (d) Parallel sum of 600 ohm and 1200 ohm 600 ´ 1200 = = 400 ohm 1800

R 400 = \ 35.5 5 or R = 2400 ohms = 2.4 k ohms

123. (c)

10 –3 R1 + 100 = 10

or or

230 ´ 20 ´ 1 ´ 2 = 9. 2 kWh 1000 Revolution made = 1380

124. (b) Energy consumed =

1380 = 150 rev/kWh 9.2 125. (a) Two wattmeter readings are W1 = VLIL cos (30 – f) = VLIL cos (30 – 90°) = VLIL cos 60° W2 = VLIL cos (30 + f) = VLIL cos 120° = – VLIL cos 60° 126. (c) Energy consumed as indicate by energy meter

\ Meter constant =

90 = 0.05 k Wh 1800 Actual energy consumed

=

æ 10 ö æ 3 ö = 230 × ç ÷ × 1 × ç ÷ ×10–3 = 0.0575 kWh è 2ø è 60 ø \ Error = 0.0500 – 0.0575 = – 0.0075 kWh Hence meter is slow by 127. (d) Absolute error =

0.00075 ´ 100 = 15% 0.0500

exp ected value - measured value exp ected value

=

80 - 79 ´ 100 = 1.25% 80

measured value Relative Accuracy = exp ected value ´ 100

or Relative Accuracy = 100 – % absolute error = 100 % – 1.25 % = 98.75 % 128. (b) Thermocouple instruments are suitable for very high frequencies upto 50 Mhz. 129. (d) By nyquist theorem minimum sampling frequency should be twice of highest signal frequency. Here sampling frequency is 2 kHz. Resolution = or

1 2n - 1

= 0.01 ´ 100

n = 14

130. (d) Nominal ratio Secondary burden

Knom =

1000 = 200 5

Re = 1.1W

Since the burden of secondary winding is purely resistive therefore, secondary winding power factor is unity or d = 0. The power factor of exciting current is 0.45

\ cos (90° – a) = 0.45 R1 + 100 = 10000 or a = 90° – cos–1 0.45 = 26.74° R1 = 9900 ohm = 9.9 kW Buy books : http://www.dishapublication.com/entrance-exams-books/gate-exams.html

46 Since there is no turn compensation, the turns ratio is equal to nominal ratio or Kt = Knom = 200. When the primary winding carries rated current of 1000 A, the secondary winding carries a current of 5 A. Rated secondary winding current, Is = 5A Actual transformation ratio, I0 sin ( d + a ) Kact = Kt + Is

1 1 reactance Xc = = jwC 2p ´ 100 ´ 103 ´ 10 ´ 10 -12

writing node equation at P

VP - 10 1 j ö æ 1 + VP ç + =0 è 100 500 159 ÷ø 100

\ I0 = 1A

10 – VP = VP (1.2 – j 0.628)

1 = 200 + sin ( 0 + 26.74°) = 200.09 5

\

Ratio error =

Nominal ratio - Actual ratio ´ 100 Actual ratio 200 - 200.09 ´ 100 = - 0.045% 200.09

=

131. (d) 132. (c) The frequency is obtained by following equation f= =

10 = (2.2 – j 0.628) VP

136. (d) 137. (a)

Q=

1 =

2p C1R1R 3C3

Lw 0 1 = R Cw0 R 1

( 500 / 2) ´ 10 -12 ´ 2p ´ 107 ´ 10

=

104 = 200 500

1 2p 5 ´ 10

-9

´ 4.7 ´ 103 ´ 10 ´ 10 -9 ´ 10 ´ 103

1

= 3.283 kHz 2p 5 ´ 10-10 ´ 4.7 133. (c) Pattern observed on the screen is an ellipse. So, phase angle

=

æ 3ö f = sin –1 ç ÷ = 36.9° or 143.1° è 5ø We can see from the figure that ellipse is in second and fourth quadrants so the valid value of phase angle is 143.1°. 134. (b) Meter current, Im = Ifs

138. (a) 139. (d) 140. (a)

= 20000 W

141. (b) fs = 400 rad/s For reconstruction in SH circuit for signal we keep frequency fs/2 to have a linear phase response. \ Sampling rate of signal = 142. (b)

(

31.6 = 50 1 - e t/T

)

400 = 200 rad/s 2

Þ 1 - et/T = 0.632

\ T = 1/2 hrs 5W

143. (b)

100 kW

XC

5 ´ 10-3

Voltage developed across voltmeter will be in proportional to their resistance.

Rx Rx + Rp

R1R m (quite a low value) R1 + R m Variation in the value of Rx when Rx is in the high value range, cause hardly any variation in Im and thus the scale is highly cramped for high resistance values. 135. (b) In the following configuration

100 kW

100

R3 = 15000 W

and R p =

10 V rms

R1 =

R2 = 250 × 100 = 25000 W

where, Rx is unknown

100 kHz

10 = 4.38 V 2.28

VP

I

Ife

500 kW

0.55

Ife = I ´

0.55 1 Þ 1 = I´ Þ I = 10 mA 5 + 0.55 10

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47 144. (a)

154. In a dual slope integrating type digital voltmeter. \ Vin = Vref (t2/t1) where t1 is the first integration line t1 = 10 × 1/50 = 0.2 sec Vm = 1 V Vref = 2

X 1 ´ 25 d x = ±d l . FS = ± = ± 2.5% IA 10

R

145. (a)

Rint = 1 W 10 mA M 10 V

10 Þ R + 1 = 1000 R +1 Þ R = 999 ohm (in series with first resistance) 10 ´10 -3 =

146.

E L 49.5 = = E0 50

t1 0.2 = 1× = 0.1 sec Vref 2

t2 = Vin ×

From figure

155. Q =

1 Þ ZL » 10 MW 100 kW 1+ ZL

=

147. When p.f. < 0.5 one of the wattmeter reads negative power; which is read after reversing the connections to the current coil or pressure coil. Also by formula: æ P -P ö cos f = cos ç tan -1 3 1 2 ÷ = 0.277 P1 + P2 ø è

=

Lw 0 1 = R Cw 0 R 1

( 500 / 2) ´ 10 -12 ´ 2p ´ 107 ´ 10 104 10000 = = 20 500 500

156. 0.5 mm/ohm. R = 10 W L = 1 mH

157. (c)

where, P1 = 3 kW and P2 = – 1 kW. ei

V Þ 58 ´ 10 -3 = 58 mV P 149. v = 5 × 1 = 5 mV

148. g =

New resistance =

s2 +

5 ´ 111 » 10 mA 55 150. Voltage drop per unit length = 1.45/50 = 0.029 V/cm Voltage drop across 70 cm. length = 0.029 × 70 = 2.03 V

\ Imax =

R 1 s+ =0 L LC

s 2 + 2xwn s + wn2 = 0

2xwn =

2.03 \ Current through resistor = 1 W = 2.03A

wn =

151. No. of pulses = 5.75 × 100 = 575 pulses. 152. Total energy supplied by the source in one minute. W = 3 × 1017 × 1.6 × 10–19 × 10000 = 4.8 × 102 J = 480 J \ Average power supplied to the beam = 480/60 = 8 W 300 ´ 1 100

= 3 may be present in any reading. Since the deflection is 83 V the present limiting error is

e0

C = 10 m F

The characteristics equation

5 ´ .55 55 = W 5.55 111

153. 1% accuracy means that a maximum possible error of

~

x=

R L

1 LC

R LC ´ = 0.5 L 2 e-px

Peak overshoot =

1 - x2

´ 100%

= 16% 158. (a)

3 ´ 100 = 3.62 83

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