SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 2 Measurements of Basic Electrical Quantities 1 (Current Voltage, Resistance)

Transcription

1 SAMPLE OF THE STUDY MATERIAL PART OF CHAPTER 2 Measurements of Basic Electrical Quantities 1 (Current Voltage, Resistance) 2.1 Indicating Instruments Analog Instruments: An analog device is one in which the output or display is a continuous function of time and bears a constant relation to its input. Analog instruments are classified in one way as a) Indicating b) Recording c) Integrating Instruments. (A) Indicating: Indicating instruments are those instruments which indicate the magnitude of a quantity being measured. Eg: Analog Ammeter, Voltmeter. (B) Recording: Instruments give a continuous record of the quantity being measured over a specified period. The variations of the quantity being measured by a pen on a sheet of paper fixed. Eg: Plotter, ECG Recorder (C) Integrating: Integrating instruments totalized events over a period of time. The summation which they give is the product of time and an electrical quantity. Eg: Ampere hour meter, energy meter etc. Indicating instruments: These instruments make use of a dial and pointer for showing or indicating magnitude of unknown quantity. The examples are ammeters, voltmeter etc. Recording instruments: These instruments give a continuous record of the given electrical quantity which is being measured over a specific period. The examples are various types of recorders. In such recording instruments, the readings are recorded by drawing the graph. The pointer of such instruments is provided with a marker i.e. pen or pencil, which moves on graph paper as per the reading. The X Y plotter is the best example of such an instrument. Page : 1

2 Integrating instruments: These instruments measure the total quantity of electricity delivered over period of time. For example a household energymeter registers number of revolutions made by the disc to give the total energy delivered, with the help of counting mechanism consisting of dials and pointers. Essential Requirements of an Instrument In case of measuring instruments, the effect of unknown quantity is converted into a mechanical force which is transmitted to the pointer which moves over a calibrated scale. The moving system of such instrument is mounted on a pivoted spindle. For satisfactory operation of any indicating instrument, following systems must be present in an instrument. 1. Deflecting system producing deflecting toque 2. Controlling system producing controlling torque 3. Damping system producing damping torque. Deflecting System In most of the indicating instruments the mechanical force proportional to the quantity to be measured is generated. This force or torque deflects the pointer. The system which produces such a deflecting torque is called deflecting system and the torque is denoted as. The deflecting torque overcomes, 1. The inertia of the moving system 2. The controlling torque provided by controlling system 3. The damping torque provided by damping system. The deflecting system uses one of the following effects produced by current or voltage, to produce deflecting torque. 1. Magnetic Effect: When a current carrying conductor is placed in uniform magnetic field, it experiences a force which causes to move it. This effect is mostly used in many instruments like moving iron attraction and repulsion type, permanent magnet moving coil instruments etc. 2. Thermal Effect: The current to be measured is passed through a small element which heats it to cause rise in temperature which is converted to an e.m.f. by a thermocouple attached to it. When two dissimilar metals are connected end to end, to form a closed loop and the two junctions formed are maintained at different temperatures, then e.m.f. is induced which causes the flow of current through the closed circuit which is called a thermocouple. Page : 2

3 3. Electrostatic Effect: When two plates are charged, there is a force exerted between them, which moves one of the plates. This effect is used in electrostatic instruments which are normally voltmeters. 4. Induction Effect: When a non-magnetic conducting disc is placed in a magnetic field produced by electromagnets which are excited by alternating currents, an e.m.f. is induced in it. If a closed path is provided, there is a flow of current in the disc. The interaction between induced currents and the alternating magnetic fields exerts a force on the disc which causes to move it. This interaction is called an induction effect. This principle is mainly used in energymeters. 5. Hall Effect: If a bar of semiconducting material is placed in uniform magnetic field and if the bar carries current, then an e.m.f. is produced between two edges of conductor. The magnitude of this e.m.f. depends on flux density of magnetic field, current passing through the conducing bar and hall effect co-efficient which is constant for a given semiconductor. This effect is mainly used in flux-meters. Thus the deflecting system provides the deflecting torque or operating torque for movement of pointer from its zero position. It acts as the prime mover for the deflection of pointer. Controlling System This system should provide a force so that current or any other electrical quantity will produce deflection of the pointer proportional to its magnitude. The important functions of this system are, 1. It produces a force equal and opposite to the deflecting force in order to make the deflection of pointer at a definite magnitude. If this system is absent, then the pointer will swing beyond its final steady position for the given magnitude and deflection will become indefinite. 2. It brings the moving system back to zero position when the force which causes the movement of the moving system is removed. It will never come back to its zero position in the absence of controlling system. Controlling torque is generally provided by springs. Sometimes gravity control is also used. Spring control is provided by using spiral spring. It can be used in any electromechanical instrument. Gravity control can be used only in vertical mounted or panel mounted instruments because the gravity cannot act along the horizontal direction. In the electromechanical system the deflection torque and the spring torque acts opposite to each other. At the output point, the deflection torque is balanced by the control torque. Page : 3

4 Direction of control Direction of the deflection Torque Pointer balanced at the position where Deflection = Control Damping System Fig. 2.1 The deflecting torque provides some deflection and controlling torque acts in the opposite direction to that of deflecting torque. So before coming to the rest, pointer always oscillates due to inertia, about the equilibrium position. Unless pointer rests, final reading can not be obtained. So to bring the pointer to rest within short time, damping system is required. The system should provide a damping torque only when the moving system is in motion. Damping torque is proportional to velocity of the moving system but it does not depend on operating current. It must not affect the controlling torque or increase the friction. The quickness with which the moving system settles to the final steady position depends on relative damping. If the moving system reaches to its final position rapidly but smoothly without oscillations, the instrument is said to be critically damped. If the instrument is under damped, the moving system will oscillate about the final steady position with a decreasing amplitude and will take sometime to come to rest. While the instrument is said to be over damped if the moving system moves slowly to its final steady position. In over damped case the response of the system is very slow and sluggish. In practice slightly under damped system are preferred. The time response of damping system for various types of damping conditions is shown in the Fig. 2.2 Page : 4

5 Deflection Under damped Over damped Steady final position Critically damped 0 Time Fig. 2.2 Effect of damping on deflection The following methods are used to produce damping torque. 1. Air friction damping 2. Fluid friction damping 3. Eddy current damping. (1) and (2) are Used when operating magnetic field of the instrument is weak (in case or electromechanical instruments) 1. Galvanometers: A galvanometer is an instrument used for detecting presence of small currents or voltages in a circuit or for measuring their magnitudes. Galvanometers find their application in bridge and potentiometer measurements. 2. D' Arsonval Galvanometer: This instrument is very commonly used in various resistance measurement methods and also in d.c. potentiometer work. It mainly consists of (A) permanent magnet to produce required magnetic flux (B) Moving coil which is placed in between magnetic poles and carries the current to be measured. (C) A metal former in which required damping torque is produced by induction of eddy currents. (D) A spring to produce useful control torque. 3. Ballistic Galvanometer: A ballistic galvanometer is used for measurement of quantity of electricity (charge) passed through it. It is very much useful in magnetic measurements. Page : 5

6 4. Vibration Galvanometer: In these galvanometer, alternating current is passed through moving coil due to which an alternating deflecting torque is produced. It makes the coil to vibrate with frequency equal to alternating current frequency, on account of inertia on moving parts, the amplitude vibrations is small. The galvanometer finds useful application as a detector in AC bridges. 2.2 Measurement of Voltage and Current Analog Ammeter and Voltmeters: Analog ammeters and voltmeters are classified together as there are no fundamental differences in their operating principles. The action of all ammeter and voltmeters with the exception of electrostatic type of instruments depends upon a deflecting torque produced by electric current. Ammeter is connected in series in the circuit whose current is to be measured. The power loss in an ammeter is I 2 R a where I is the current to be measured and R a is the resistance of ammeter. Therefore, ammeters should have low resistance so that they cause a small voltage drop and consequently absorb small power. Voltmeters are connected in parallel with the circuit whose voltage is to be measured. The power loss in voltmeters is V 2 /R V where v is the voltage to be measured and R v is the resistance of voltmeter. Therefore, voltmeters should have a high electrical resistance in order that the current drawn by them is small and consequently the power consumed is small. The Instruments used for measurement of voltage and current can be classified as: a) Moving coil instruments: (i) Permanent magnet type (ii) dynamometer type b) Moving iron instruments: c) Electrostatic Instruments: d) Rectifier instruments: e) Induction instruments: f) Thermal instruments: (i) Hot - wire type (ii) Thermocouple type. a) Moving Coil Instruments: (i) Permanent Magnet Type: It works on the principle of magnetic effect. Here coil placed in between the pole pieces of a permanent magnet deflects due to the current passed through the coil. Page : 6

7 Torque equation: The deflecting torque is given by = NBAI Where deflecting torque in N m B Flux density in air gap. N Numbers of turns of the coil A effective coil area ) length; b breadth of the coil I current in the moving coil, amperes = GI Where G = NBA = Constant The controlling torque is provided by springs and is proportional to angular deflection of the pointer Where = controlling torque K = spring constant, = Angular deflection. For the final steady state position, = GI = = or I = Thus we get a linear relation between current and deflection angle. Damping used in this type of instrument is eddy current damping. Key Point: Whenever the operating magnetic field is strong, eddy current damping can be used. If the operating magnetic field is weak as in the case of instruments using electromagnets (like Electrodynamometer type), Eddy current damping cannot be used. This is because the field due to eddy current damping will distort the operating filed. In canse of Permanent magnet Moving Coil Type Instrument, the operating field is strong and the error encountered due to this is negligible. Page : 7

8 Advantages: i) Uniform scale ii) Highest accuracy iii) Low power consumption because driving power is small iv) Lower power consumption of the order 25 W to 200 W. v) No hysteresis loss as the former is of copper or aluminum vi) Very effective and reliable eddy current damping vii) High torque / weight ratio viii) No effect of stray magnetic field as intense polarized or unidirectional field is employed. ix) Range can be extended with shunts or multipliers. x) High Sensitivity. Disadvantages: i) These cannot be used for a.c. measurements. i,e suitable only for D.C. measurements. ii) These are costlier in comparison with moving iron instruments. iii) Friction and temperature might introduce errors as in the case of other instruments. iv) Ageing of control springs and of the permanent magnets might cause errors. Example: A permanent magnet moving coil instrument has a coil of dimensions 15 mm 12 mm. The flux density in the air gap is 1.8 Wb/ and the spring constant is 0.14 Nm/rad. Determine the number of turns required to produce an angular deflection of 90 degrees when a current of 5 ma is flowing through the coil. Solution: Deflection rad. At equilibrium = or NB d I = K Number of turns N = = 136. Example: The coil of a moving coil voltmeter is 40 mm long and 30 mm wide and has 100 turns on it. The control spring exerts a torque of 240 N m when the deflection is 100 divisions on full scale. If the flux density of the magnetic field in the air gap is 1.0 Wb/, estimate the resistance that must be put in series with the coil to give one volt per division. The resistance of the voltmeter coil may be neglected. Page : 8

9 Solution: Controlling torque at full scale deflection = 240 N-m. Deflecting torque at full scale deflection = NB di = I = 120 I N-m At final steady position, or 120 I = 240 Current at full scale deflection I 2 A = 2 ma Let the resistance of the voltmeter circuit be R Voltage across the instrument = 2 R. This produces a deflection of 100 divisions Volts per division = 2 R/100 This value should be equal to 1 in order to get 1 bolt per division. 2 R.100 = 1 or R = 50,000 Ω = 50 kω. (ii) Dynamometer Type: Unlike PMMC instruments, where the magnetic field is produced by permanent magnets, dynamometer instrument uses the current under measurement to produce the magnetic field. The filed is produced by fixed coils which are divided into two sections to give a more uniform field near the center. The fixed coil (or coils) is called the current coil and the movable coil is known as potential coil. SCALE MOVABLE COIL FIXED COILS Fig. 2.3 Page : 9

10 These instruments can be used for both d.c as well as a.c Operating principle: When the instrument is connected in the circuit, the fixed coil carries load current and movable coil carries current proportional to load voltage. Due to currents in the coil, mechanical force exists between them The result is that movable coil moves the pointes over the scale Pointer comes to rest at a position where deflecting torque is equal to controlling Torque. Here as the same current flows through fixed as well as movable coils, a unidirectional torque is produced for a.c currents Controlling Torque: It is provided by two control springs. These springs act as leads to the instrument. Damping Torque: Air friction damping is employed and is provided by a pair of aluminum vanes, attached to the spindle at the bottom Eddy current damping cannot be used as the operating filed is very weak and introduction of permanent magnet may distort the operating magnetic field. Deflecting Torque: Torque equation: T = Where instantaneous value of current in fixed coils; A = Instantaneous value of current in moving coil; A M = Mutual inductance between fixed and moving coil; H Operation with D.C: Deflecting torque deflects the moving coil to such a position where the controlling torque of spring is equal to deflecting torque. = K: spring constant; = = = Operation with A.C: = are instantaneous values of current in moving and fixed coils. Page : 10

11 Average deflection over a complete cycle (T) is = = T = time period for one complete cycle. Case: If are sinusoidal and are displaced by a phase angle i.e sin t and sin ( t ) Average deflecting torque = sin ( t ) d( t) = ( ). Where are rms values of current flowing in coils Electrodynamometer Ammeters: Above fig (2.3) shows the arrangement as ammeter where fixed and movable coils are connected in series and therefore carry the same current i,e = = I and = 0 = = Range: upto 100 ma. Electrodynamometer Voltmeter: Below Fig. (2.4) is used as voltmeter where a high non-inductive resistance is connected in series with moving and fixed coils. Page : 11

12 Moving coil (V) SUPPLY Fixed coils High noninductive resistance = = = = Fig. 2.4 And = Example: In an electrodynamometer instrument the total resistance of the voltage coil circuit is 8.00 Ω and the mutual inductance changes uniformly from 173 H at zero deflection to H at full scale, the angle of full scale being 95. If a potential difference of 100 V is applied across the voltage circuit, and a current of 3 A at a power factor of 0.75 is passed through the current coil, what will be the deflection, if the spring control constant is 4.63 N-m/rad? Solution: Change in mutual inductance = (-173) = 348 H. Deflection = 95 = 1.66 rad. Rate of change of mutual inductance dm/d = 348/1.66 = H/rad. This problem relates to a dynamometer type instrument where the two coils carry different values of current and also there is a phase difference between them. Current through fixed coil = 3 A. Current through moving coil = 100/8200 = 12.2 A and cos = 0.75 Spring constant K = 4.63 N-m/rad = N-m/degree. From Eqn deflection = = = Page : 12

13 Example: For a certain dynamometer ammeter inductance M varies with deflection (expressed in degrees) as M = 6 cos ( + 30 ) mh. Find the deflecting torque produced by a direct current of 50 ma corresponding to a deflection of 60 Solution: Rate of change of mutual inductance with deflection = [ -6 cos ( + 30 )] = 6 sin ( + 30 ) mh at a deflection of 60 is, = 6 sin ( ) mh = 6 H/degree. Deflecting torque = = Nm = 15 Nm. Example: A 50 volt range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on d.c. for full scale deflection of 90. The control constant is 0.5 N- m/degree and the initial mutual inductance of the instrument is 0.25 H. Find the true potential difference across the instrument when it reads 50 V at 50 Hz Solution: The expression for deflection for a square law response voltmeter is: = For d.c., impedance Z = resistance R and, therefore, = But V/R = 0.05 A, K = 0.5 and = 90. Thus we have, 90 = or = 18 H/rad. Angle of deflection = 90 = /2 rad. Total change in mutual inductance = 18 /2 = 28.3 H. Mutual inductance at is : = initial mutual inductance + change in mutual inductance = = H Reactance of instrument at 50 Hz = = Ω Resistance of instrument = 50/0.05 = 1000 Ω. Impedance of instrument at 50 Hz = = Ω. Current through the instrument when reading 50 V is 0.05 A. Actual voltage across the instrument = = V. Page : 13

14 Advantages: i) Such instruments can be used on both ac as well as d.c. systems. ii) These instruments are free from hysteresis and eddy current errors because of the absence of iron in the working parts of the instruments. Disadvantages: i) Since the coils are air cored, magnetic field produced is of small strength thus requiring large number of ampere turns to create necessary deflecting torque. The current is also limited because two springs are used as leads. This results in heavy moving system giving friction losses somewhat larger than with other type. ii) Since the torque / weight ratio is small friction errors tend to be serious. iii) The scale is not uniform because θ α i 2. [Non linear scale] iv) Adequate screening of the movements against stray magnetic field is essential. v) These instruments are more expensive than other types of instruments so these are only used as voltmeters and ammeters for precision measurements. vi) Though MI instruments can be used with both ac & dc, they have to be separately calibrated for ac separation & dc operation. Moving Iron Instruments: Moving-Iron instruments depend for their indication upon the movement of a piece of soft iron in the field of a coil produced by the current to be measured. Moving-iron instruments are divided into two general groups, namely attraction and repulsion types. Sometimes combined attraction and repulsion type is also used. Operating principle: In attraction type instruments, a flat sector of soft iron is drawn into the narrow opening of the 2 coil. The deflecting torque is given by T d = (1/2)I. Where I is the current through the coil and L is the inductance. The angular deflection is proportional to the square of the current and the instrument has a square-law response. In repulsion type instrument, there are two rods of iron inside the coil such that one is fixed and the other is movable. When current flows through the coil, repulsion of moving iron from the fixed one takes place. The equation for torque developed is same as that of attraction type instrument. Advantages: i) Moving iron instruments can be used both for a.c. and d.c. measurements and are cheap in cost and robust in construction. Page : 14

15 Disadvantages: i) The possible sources of errors in moving-iron instruments are friction, temperature change, hysteresis, stray magnetic field, frequency variation, and wave form changes. It is possible to design a moving-iron instrument which has a linear scale from the upper limit down to about 10% of full scale, by suitably adjusting the variation of with θ. Linearization of Scale: θ.(d L /d θ ) = K. Compensation towards frequency errors can be done by connecting a capacitor across a part of series resistance in MI voltmeter, C = 0.41 x (L/R 2 ) Example: A ma moving iron ammeter is converted to a V, 50 Hz voltmeter by adding a series resistance with the coil. The coil has negligible resistance and an inductance L = ( )/4 henry, where is the deflection in radian. The total angular span of the meter is 100. Compute (a) the spring constant of the meter and (b) the series resistance required. Solution: Now, L = H/rad (a) Spring constant K = = 45.6 Nm/rad (b) Impedance required to convert the meter to 500 V voltmeter Z = 500/(100 Inductance of meter coil full scale deflection L = = Reactance of meter coil at 50 Hz = 2 = 8.98 Ω Series resistance required = 5000 Ω = 5000 Ω Example: The coil of a 300 V moving iron voltmeter has a resistance of 500 Ω and an inductance of 0.8 H. The instrument reads correctly at 50 Hz a.c. supply and takes 100 ma at full scale deflection. What is the percentage error in the instrument reading when it is connected to 200 V d.c. supply. Solution: Reactance of coil at 50 Hz, X = = Ω Current taken by meter at 50 Hz and 300 V, = 100 ma = 0.1 A. Impedance of meter at 50 Hz, Z = 300/0.1 = 3000 Ω Resistance of meter circuit = = = 2989 Ω. Current taken by meter at 200 V, 50 Hz supply = 200/3000 = A. Page : 15

16 Current taken by instrument with 200 volt d.c. = 200/2989 = A The deflection is proportional to the operating current. Also the instrument reads correctly with 50 Hz a.c. Hence reading of instrument with 200 V d.c. = = V. % Error = Electrostatic Instruments: All electrostatic instruments are essentially voltage measuring devices. The deflecting torque is produced from the action of an electric field on charged conductors. Electrostatic instruments may be classified into three types: Repulsion, symmetrical, and attraction. For linear motion: F = (1/2) V 2 For angular motion: T d = (1/2) V 2 While all other measuring instruments are operated by current, electrostatic instruments depend on voltage for their operation. Their deflection is directly proportional to the square of the r.m.s voltage applied across the terminals. When used for d.c. measurements the steady state Current drawn is practically zero. On a.c. measurements, the current drawn by the instrument depends on the frequency of the supply. Electrostatic voltmeter is devoid of most of the errors occurring in other measuring instruments. Example: An electrostatic voltmeter reading upto 2000 V is controlled by a spring with a torsion constant of 5 Nm/ rad has a full scale deflection of 90. The capacitance at zero voltage is 15 pf. What is the capacitance when the pointer indicates 2000 V? Solution: Deflection = 90 = /2 rad. From Eqn. 9.74, final steady deflection or rate of change of capacitance = 3.93 F/rad= 3.93 pf/rad. Change in capacitance when reading from 0 to 2000 V is 3.93 Capacitance when reading 2000 V, C = = pf. Example: = 6.17 pf. The movable vane of a quadrant electrometer turned through 25 scale divisions when Page : 16

17 idiostatically connected to a potential difference of 60 V. When used heterostatically with the quadrants connected to a small voltage V and the needle connected to 900 V supply, the deflection was 10 scale divisions. Determine V. Solution: From Eqn deflection for idiostatic connection and from Eqn deflection for heterostatic connection where V = voltage being measured, and = Voltage of needle. Substituting the data given, we have : V Voltage V = 0.8 volt Example: An absolute electrometer has a movable circular plate 80 mm in diameter. If the distance between the plates during a measurement is 4 mm, find the potential difference when the force of attraction is 2 N. The dielectric is air, having a permittivity of 8.85 F/m. Solution: Distance between plates d = 4 mm = m. Area of plates A = ( /4) = From Eqn. 9.84, Voltage V = = 1200 V. Rectifier Instruments: Rectifier instruments consist of a full wave rectifier having four diodes arranged in a bridge fashion and a permanent magnet instrument as shown in fig: 2.5. The indication of PMMC instruments depends on the mean value of the current flowing through it, but the scale is calibrated to give r.m.s values (by multiplying with 1.11). Hence, for non sinusoidal waveforms, these instruments give erroneous readings. Page : 17

18 + Full wave Rectifier Bridge Input (Multiplier resistance) _ DIODE + _ PMMC Instrument Fig. 2.5 For d.c. measurements, the indication of rectifier instrument will be 11% high Half wave Rectifier type Instruments V = In half wave rectifier type instrument, the sensitivity of a.c. is 0.45 times that of d.c. Full wave rectifier type instruments: In full wave rectifier type instrument, the sensitivity of a.c. is 0.9 times that of d.c. Advantages: Rectifier instruments have high sensitivity, are very rugged, and generally withstand overload conditions upto several hundred percent without permanent damage. They are specifically designed for measurements where very little operating power is available and where moderate accuracy is acceptable. Disadvantages: The current voltage characteristic of the rectifier is not linear even in forward direction, particularly at low voltage. This means that the resistance of the rectifier is not constant for all Page : 18

19 values of current through it. It is essential to use a large series resistance to reduce the error due to variation in rectifier resistance. Extension of range of rectifier type voltmeters: Diode Multiplier PMMC Fig. 2.6 v = Voltage drop across PMMC instrument V = Applied voltage For DC Operation Applied voltage V = (By KVL) for half wave rectification for full wave rectification. For AC Operation V = = 0.45 (for half wave rectification) = 0.9 (for full wave rectification) Example: Compute the value of multiplier resistor for a 10 V rms sinusoidal a.c. range of the voltmeter shown in Fig The forward resistance of the diode is zero and the reverse resistance is infinite. Page : 19

20 Multiplier resistance Diode Input sin t PMMC Fig. 2.7 Solution: The sensitivity of meter movement for d.c. Ω/V For half wave rectifier, the a.c. sensitivity = = 450 Ω/V Total resistance of circuit for a.c. operation = = V = = 4500 Ω. Resistance of multiplier = V - - = = 4200 Ω. Example: Compute the value of the multiplier resistor for a 10 V r.m.s. sinusoidal a.c. range of the voltmeter shown in Fig The forward resistance of the diode is zero and the reverse resistance is infinite. Solution: The d.c. sensitivity is: For full wave rectifier circuit, a.c. sensitivity = = 900 Ω/V Ω/V Page : 20

21 Input Multiplier 1O Rectifier Bridge PMMC = 1 ma = 500 Ω Resistance of multiplier = V - 2 = = 8500 Ω. Example: A rectifier type instrument uses a bridge rectifier and has its scale calibrated in terms of rms values of a sine wave. It indicates a voltage of 2.22 V when measuring a voltage of triangular waveshape. Estimate the peak and r.m.s. values of applied voltage. Also calculate the error. Solution: Fig. 2.8 The meter uses a full wave rectifier circuit and it indicates a value of 2.22 V. The form factor for full wave rectified sinusoidal waveform is Average value of voltage = 2.22/1.11 = 2 V For a triangular waveshape, peak value of voltage = 2 = 4 V. Rms value of voltage V = / 3 = 4/ 3 = 2.31 V Error = = -3.9 %. Thermal Instruments: (i) Hot wire type: Thermal instruments make use of heat developed by the passage of current through a conductor, for the measurement. The heating action of current is employed in many ways for indication. In a hot wire instrument, the deflection of the pointer is produced by the linear expansion of a wire heated by the current. Page : 21

22 Generally, the change in length is not measured, but the sag caused in the hot-wire due to current is magnified and measured. As the sag is proportional to square root of the current, the scale obtained is very nearly a square scale. (ii) Thermo couple instruments: A fine resistance wire is heated by the current to be measured or the current proportional to the voltage to be measured. The temperature rise of the wire is measured by means of a thermocouple. The thermo e.m.f generated is dependent on the square of the current. The scale is a square-law scale and the PMMC instrument can be calibrated in terms of heater current or voltage. The e.m.f generated in thermocouple is given by Thermocouple V PMMC Instrument Fig. 2.9 E = a( T) + B( T) 2 Where T is the temperature difference Between hot and cold junctions a is a constant (40 to 50 μv/ 0 C) b is a constant (1 to 2 μv/ 0 C) Advantages: There is no error due to stray magnetic field. Free from frequency errors. High sensitivity. They correctly indicate rms value of voltage or current irrespective of the waveform. Disadvantages: The speed of response is poor due to the time taken for the wire to heat up. The power consumption is quite high and the overload capacity is also limited. The skin effect at higher frequencies limit the size of the conductors which are usually made of very fine wire. Shunts and Multipliers: Shunts and multipliers are the resistance connected in shunt or series with ammeter and Page : 22

23 voltmeters to enhance their measuring capacity. Shunts: A resistance placed in parallel to an instrument (or galvanometer) to control the current passing through it, when placed in a circuit carrying a fairly large current is called a shunt. The shunt resistance used with a basic instrument may consist of a length of constant temperature resistance wire within the box of the instrument. Alternatively, there may be an external (manganin or constantan) shunt having very low resistance. General requirements of a shunt are as follows: 1) The temperature coefficient of the shunt and instrument should be low and nearly identical. 2) The resistance of the shunt should not vary with time. 3) It should carry the current without excessive temperature rise. 4) It should have a low thermal emf. Manganin is usually used as a shunt for d.c. instruments since it gives a low value of thermal emf with copper. Constantan is a useful material for a.c. circuits since its comparatively high thermal emf being unidirectional, is ineffective on these circuits. Shunts for low current are enclosed in the meter casing but for currents above 200A, they are mounted separately. Shunt with ammeter: (Fig: 2.10) S I S A I I m Fig shunt with an Ammeter Low resistance shunt is connected across the coil of ammeter. Let I be the circuit current to be measured. Let = Ammeter resistance S = Shunt resistance = Full scale deflection current of ammeter (f.s.d) = Shunt current I = or = I Page : 23

24 Since the voltage across shunt and ammeter is same, = (I ) [ ] or or I = i.e. Circuit current = F.s.d current x Instrument Constant: The ratio of current to be measured to full scale deflection current is called instrument constant. Instrument constant, m = It is clear that with different shunts, the same instruments will have different instrument constants. Multi Range Shunt: (i) Individual Multi range shunt: Meter Switch Fig Where ; is the rated current through the meter and are the different ranges to which the ammeter has to be extended. (ii) Universal (or) Ayrton Shunt: Page : 24

25 1 2 By KVL, (a) Position (2), Fig Similarly, where Example: Design a multi-range dc-milti ammeter which has a meter resistance of Ω 50 & a ful scale current of 1mA. The ranges required are, ma, ma Design the shunt using (i) Individual shunt (ii) Ayrton shunt Solution: (i) For 10 ma here = 10 ma = 1 ma = 10 = 5.55 Ω for ma for (ii) Ayrton shunt ; ma Page : 25

26 ; Multipliers: A high resistance in series with a galvanometer is connected, to limit the current flowing through the meter so that it does not exceed the value for full scale deflection and thus prevents the instrument from being damaged. Such a resistance it called multiplier. Multiplier with Voltmeter: (Fig: 2.13) High resistance is connected in series with voltmeter coil. R V Supply A B Load Let = voltmenter resistance R = high resistance in series with voltmeter coil = full scale deflection current for voltmeter Since voltmeter is connected in parallel with circuit AB, Voltage across AB = Voltage across voltmeter or, V = (R + ) or, R + = R = Multiplier constant, m = = Fig Multiplier with a voltmeter Page : 26

27 = 1 + i.e. Required high resistance, R = voltmeter resistance Example: A moving coil instrument gives a full scale deflection with a current of 40 A, while the internal resistance of the meter is 500 Ω. It is to be used as a voltmeter to measure a voltage range of 0 10 V. Calculate the multiplier resistance needed. Solution: Given values are, Now This is the required multiplier resistance. Example: = kω. A moving coil instrument gives a full scale deflection for a current of 20 ma with a potential difference of 200 mv across it. Calculate : i) shunt required to use it as an ammeter to get a range of A ii) Multiplier required to use it as a voltmeter of range V. Solution: The meter current ma Drop across meter, = 200 mv Now = mv = (20 ma) = 10 Ω i) For using it as an ammeter, I = 200 A = Ω This is the required shunt. ii) For using it as a voltmeter V = 500 V = = k Ω This is the required multiplier. Page : 27

28 Shunt for a.c. instruments: In order to maintain the current division between the shunt and instrument, constant for all frequencies, the ratio of impedances of the instrument and leads to that of the shunt must remain constant., If and are inductance of the instrument and shunt respectively in fig: 2.14 then the ratio should remain constant for all frequencies, which is possible only when time constants of the shunt and instrument are same. Instrument I I V Of shunt Fig i.e. Multiplying factor, N = = = = ] Multipliers for Moving Iron Instruments: Page : 28

29 (Multiplier) L R (R, L) resistance and inductance of the instrument A large resistance is connected in series to extend the range of the instrument as a voltmeter, as shown in fig Voltage drop across the meter ( Let V be the voltage to be measured. Total resistance = R + Total Inductance = L Total Impedance = and current in the meter ( ) = (or) v = Fig voltage multiplying factor (m) = Clearly m is dependent on frequency and to make it independent of frequency, series resistance is shunted by a capacitor as shown below in fig R L Of the meter Fig C = Page : 29

30 2.3 Measurement of Resistance From the point of measurement, resistances are classified as (i) low, (ii) medium and (iii) high resistances. i. Low Resistances: Resistances of the order of ohms and less are included in this class. Measurements of low resistances are required for determination of resistances of armatures and series field windings of large machines, ammeter shunts, cable length, contacts etc. ii. Medium Resistances: Resistances ranging from about 1Ω to 100 kω are included in this class. Most of the electrical apparatus employed are of medium resistances. iii. High Resistances: Resistances of 100 kilo ohms and above are usually termed as high resistances. Measurements of high resistances are required for determination of insulation resistance of components and built resistance of high resistance circuit elements. Measurement of Low Resistance: Four wires are used for representation of low resistance. Kelvin's Double Bridge is used for the measurement of low resistance as shown in fig: 2.17 R = b P p d q Q Measurement of Medium Resistance: R r S a m n c E Fig Two wires are required to represent a medium resistance: This can be measured by: Page : 30

31 a) Ammeter voltmeter method b) Substitution method c) Wheatstone bridge method d) Ohm meter a) Voltmeter - Ammeter method: From fig: 2.18 R a A I R R V V V a R V R Fig Measured value of resistance, Where R is the true value of the resistance. Error = R a % Error = (R a /R) In this method, always the measured value of resistance is greater than true value of resistance. This method is suitable for measurement of high resistance, among the range. Ammeter - Voltmeter Method: From fig: 2.19 Measured value of resistance, Where I = + = Which Error = % error = I A R V V I v Fig In this method, the measured value of resistance is always less than the true value of R I R V R Page : 31

32 resistance. This method is suitable for measurement of low resistance among the range. The resistance where both the methods give same error is obtained by equating the two errors. R = b) Substitution Method: 1 S resistance R h A 2 E Fig R R unknown S known variable resistance c) Wheat stone Bridge: From fig: 2.21 b P Q a R d S c E Fig balanced condition = Sensitivity of the galvanometer, Where = deflection of the galvanometer Page : 32

33 e = emf across galvanometer e = Sensitivity of galvanometer, Sensitivity of the Bridge or = = = when Bridge sensitivity is useful for (A) Selecting the galvanometer with which the given unbalance can be observed. (B) Determining the deflection to be expected for a given unbalance. Measurement of High Resistance: G Fig Three wires are required to represent a high resistance as shown in fig: The third, guard terminal, G, is utilized to reduce the errors because of leakage currents caused by insulation. The methods used for the measurement of high resistance are (A) Loss of charge method (B) Direct deflection method (C) Mega ohm Bridge Resistance method (D) Megger (A) Loss of charge method: From fig: 2.23 at t, voltage across capacitor Page : 33

34 V S 1 + V - S 2 C R V Fig v = ln R = R = R = (B) Direct deflection method: This method is suitable for the measurement of volume resistivity, surface resistivity of any insulating material available in sheet form. (C) Mega ohm Bridge Resistance Method: A P R B G Q S By connecting the guard terminal to one of the terminals of galvanometer. We can make the insulation resistance not effect the value of high resistance to be measured. AG, BG are the insulation resistances of the order of 1MΩ. (D) Megger: i) This meter principle is based upon ratio meter, ohm meter principle E Fig Page : 34

35 ii) It consists of self driven generator or pre charged capacitor to develop the required voltage for the operation of the instrument. iii) Operating voltages of-the Megger will be high to make considerable amount of current to flow in the circuit. Example: In the measurement of a resistance, R, by the Ammeter-voltmeter method, connections as in Fig and 2.19 are used. The resistance of ammeter is 0.01 Ω and that of voltmeter, 2000 Ω. In case of (2.19) the current measured is 2 A and the voltage 180 V. Find the percentage error in calculating resistance R as a quotient of the readings and the true value of R. Also find the reading of the voltmeter in case of (fig: 2.18.) if the current indicated by the ammeter is 2 A. Solution: Case (b) see fig Measured value of resistance = V/I = 180/2 = 90 Ω. Current through the voltmeter = 180/2000 = 0.09 A. Current through the resistance = ( ) A. True value of resistance R = = = Ω Percentage error = = = -4.5%. Case (a) See Fig Reading of voltmeter V = + = I ( + R) = 2 ( ) = V. Example: A resistance R is measured using the connections of Fig The current measured is 10 A on range 10 A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform, the total scale divisions of ammeter are 100 and that of voltmeter are 150. The scales of these instruments are such that 1/10 of a scale division can be distinguished. The constructional error of the ammeter is 0.3% and that of voltmeter 0.4%. The resistance of the ammeter is 0.25 Ω. Calculate the value of R and the limits of possible error in the results. Solution: Reading error of ammeter = Reading error of voltmeter = Total error of ammeter I = 0.4%. Total error of voltmeter V = 0.467%. Page : 35

36 Now, resistance R = V/I and therefore total systematic error in measurement = Measured value of resistance = 125/10 = 12.5 Ω. True value of resistance R = = 12.5 Ω. Therefore the value of R is specified as : Ω % = Ω. Page : 36