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2 Contents Manual for K-Notes... Error Analysis... 3 Electro-Mechanical Instruments... 6 Potentiometer / Null Detector... 5 Instrument Transformer... 6 AC Bridges... 8 Measurement of esistance... Cathode ay Oscilloscope (CO)... 5 Digital Meters... 8 Q meter / Voltage Magnifier Kreatryx. All ights eserved.

3 Error Analysis Static characteristics of measuring system ) Accuracy Degree of closeness in which a measured value approaches a true value of a quantity under measurement. When accuracy is measured in terms of error : Guaranteed accuracy error (GAE) is measured with respect to full scale deflation. Limiting error (in terms of measured value) GAE * Full scale deflection LE Measured value ) Precision Degree of closeness with which reading in produced again & again for same value of input quantity. 3) Sensitivity Change the output quantity per unit change in input quantity. qo S q i 4) esolution Smallest change in input which can be measured by an instrument 5) Threshold Minimum input required to get measurable output by an instrument 6) Zero Drift Entire calibration shifts gradually due to permanent set 3

4 7) Span Drift If there is proportional change in indication all along upward scale is called span drift. 8) Dead zone & Dead time The range of input for which there is no output this portion is called Dead zone. To respond the pointer takes a minimum time is called dead time. TYPES OF EO a) Gross Error : Error due to human negligency, i.e. due to loose connection, reading the value etc. b) Systematic error : Errors are common for all observers like instrumental errors, environmental errors and observational errors. c) andom errors : Error due to unidentified causes & may be positive or negative. Absolute Errors : A Am A A m Measured value A True value elative Errors : r r = Am AbsoluteErrors Truevalue r r A A A A T T A T m r 4

5 Composite Error : i) Sum of quantities X X X x x x ii) Difference of quantities X X X x x x So for sum & difference absolute errors are added. iii) Multiplication of quantities X X X X 3 X X X3 X X X X3 X iv) Division of quantities X X X X X X X X X So, for multiplication & division, fractional or relative errors are added. If X X X m m p X3 X X X X3 m n p X X X X3 Precision Index Indicates the precision for a distribution h 5

6 Probable Error r = r h Standard deviation of combination of quantities X X X x x x... xn X X Xn Probable Error X X X x x x xn X X Xn r r r... r Electro-Mechanical Instruments ) Permanent magnet moving Coil (PMMC) Deflecting Torque Td = niab Where n = no. of turns I = current flowing in coil A = Area of coil B = magnetic flux density Deflection G I k G = NBA & K = Spring constant Eddy current damping & spring control torque in used. For pure AC signal, the pointer vibrates around zero position. It is used to measured DC or average quantity. It can directly read only up to 50mV or 00mA. 6

7 Enhancement of PMMC i) Ammeter For using PMMC as an ammeter with wide range, we connect a small shunt resistance in parallel to meter. I I m m multiplication factor Basically, m is ratio of final range (as an ammeter) to initial range of instrument. sh m m ; m = meter resistance ii) Voltmeter A series multiples resistance of high magnitude is connected in series with the meter. M = multiplication factor m V V m s mm Sensitivity of voltmeter S v s m / V I V fsd Application of PMMC ) Half wave rectifier meter Im I Iavg 7

8 V MS s m f I avg 0.45V MS s m f ; For Ac input For DC input I avg V DC s m f avg avg AC DC I 0.45 I (Assuming V V ) DC MS (Sensitivity) AC 0.45(Sensitivity) DC ) Full wave rectifier meter I avg AC s V m MS 0.9V MS f s m f V DC avg DC s m f I I avg 0.9I AC avg DC (Assuring V MS V DC ) Sensitivity 0.9Sensitivity AC ) Moving iron meter DC Deflecting torque, T d dl I d I = current flowing throw the meter L = Inductance = deflection Under steady state 8

9 dl K I d I MI meter measures both ac & dc quantities. In case of AC, It measures MS value. T IMS i tdt T 0 If it I0 I sinwt I sinwt... IMS I0 I I... Air friction Damping is used Condition for linearity dl cons tant d MI meter cannot be used beyond 5Hz, as then eddy current error is constant. 3) Elector dynamometer dm Deflecting Torque, Td ii d For DC, i i I T d I dm d I For AC, i Im sint i I sin t m dm T I d avg I cos d Where I m I & I I 9

10 Applications of dynamometer ) Ammeter Fixed coils are connected in series. I I I 0 T d I dm d (Angel between I & I ) At balance, T T c d dm KI d I It reads both AC & DC & for AC it reads MS. ) Voltmeter s Series multiplier resistance I V I, 0 cos s T V d s dm d At balance, Td T c V K s dm d V It reads both AC & DC & for AC it reads MS. 0

11 3) Wattmeter Fixed coils carry same current as load & as called as current coils. Moving coil is connected across voltage and thus current voltage, a high non-inductive load is connected in series with MC to limit the current. T dm I I cos d d P I cos d d V dm avg dm s At balance, kt d P avg Symbol : s Two wattmeter method W V I cos V & I Y Y V I cos 30 L L W V I cos V & I BY B BY B V I cos 30 L L Where V L is line to line voltage

12 I L is line current These expression remain same for -connected load. P W W 3 3V I cos L L Q 3 W W 3 3V I sin L L 3 Q3 3 W W tan P W W 3 W W tan W W 3 W W tan W W for lag load for lead load = emember, In our case W is wattmeter connected to -phase and W is wattmeter connected to B-phase. = If one of the wattmeter indicates negative sign, then pf < 0.5 Errors in wattmeter a) Due to potential coil connection L c Ir % r * 00 P T I L = load current r = CC esistance C P T = True Power V % r * 00 P s T V = voltage across PC

13 s = Series multiplier resistance P T = True Power b) Due to self inductance of PC If PC has finite inductance Z jwlp p p s p s Zp s jwlp % r tantan *00 = load pf angle L p tan s 4) Energy meter Energy = Power * Time VIcos t W T * kwhr W T = True energy It is based on principle of induction. It is an integrating type instrument. t W VIsin * kwhr 3600 m Where W = measured Energy m = angle between potential coil voltage & flux produced by it. Error = W W m Energy constant = = load pf angle T No.ofe voluations N kwhr P.t W Measured Energy = m Totalno.ofrevolutions K 3

14 VI cos t True Energy = W T * kw.hr Wm WT Error = % r * 00 W Creeping Error in energy meter T If friction is over compensated by placing shading loop nearer to PC, then disc starts rotating slow with only PC excited without connecting any load is creeping. Otherwise if over voltage is applied on pressure coil then also creeping may happen due to stray magnetic fields. To remove creeping holes are kept on either side of disc diametrically opposite & the torque experienced by both holes is opposite & they stop creeping. % creeping error = TotalNo.ofe w / kwhr due to creeping * 00 TotalNo.ofe w / kwhr due to load Thermal Instruments These instruments work on the principle of heating and are called as Thermal Instruments. These are used for high frequency measurements. They can measure both AC & DC. In case of AC, they measure MS value. Electrostatic voltmeter Deflecting torque, At Balance, T d dc V d Td T c dc V k d V Condition for linearity 4

15 dc cons tant d For increasing the range, we connect another capacitor in series To increase the range from V m to V C s Cm m ; V m V m Potentiometer / Null Detector I w = working current I w V B l.r () h Switch at (A) If I g 0 V I l r s w I w Vs () lr Switch at (B) V I l r x w I w Vx (3) lr Vs Vx l r l r V x l Vs l 5

16 r = resistance of slide wire (Ω/ m) l = Total length of slide wire (m) l = length at which standard cell ( l = length at which test voltage ( V s ) is balanced V x ) is balanced Measuring a low resistance V V s S Instrument Transformer Current transformer Equivalent circuit Turns atio = Nominal atio N n N X X l s tan l s = Actual atio Icos Isin n I s 6

17 Errors in current transformer ) atio Error : Current ratio K % r * 00 K = n = Nominal atio = Actual atio I I p s is not equal to turns ratio due to no-load component of current. ) Phase Angel atio : Ideally, Phase difference between I & I s p s should be 0 80 current, it deviates from that value. Icos Isin 80 Phase angle error = * degrees ni Phase angle between primary & secondary currents = 80 degrees but due to no-load component of Potential Transformer Equivalent circuit Turns atio = n = N N Actual Transformation atio = = IS n cos X sin I IX V n S IS Phases angle error n P P P P V V P S, Where X cos sin I X I P P P P nvs X tan 7

18 AC Bridges AC Bridges Balance condition : Z Z Z Z 4 3 Z Z Z Z 4 3 ID 0 Z 4 3 Z Z 3 Z4 3 4 Quality Factor & dissipation factor Quality Factor (Q) wl Q Dissipation Factor (D) D wl Q wl D wl 3 Q wc D =wc 4 Q = wc D wc Measurement of Inductance (i) Maxwell s Inductance Bridge Here, we try to measure & L 8

19 L 3 4 LL 3 4 (ii) Maxwell s Inductance Capacitance Bridge L 3 4 C 3 This bridge is only suitable for coils where < Q < 0 Q = Quality Factor 4 (iii) Hay s Bridge Used for coils having high Q value 3 4 C4 Q C L Q Q C (iv) Anderson s Bridge This Bridge is used for low Q coils. 3 r 4 C L r

20 (v) Owen s Bridge L C 3 C 4 C 3 4 Measurement Of Capacitance De-Sauty s Bridge 3 r r 4 C 4 3 C D = dissipation factor = Cr r = internal resistance of C Schering Bridge C 3 4 C C C 4 3 dissipation factor = D = C 4 4 Measurement of frequency Wien Bridge Oscillator Balancing Condition 3 C C 4 Frequency of Osculation f C C 0

21 Measurement of esistance Classification of esistance ) Low esistance : Ω, Motor and Generator ) Medium esistance : Ω < < 00kΩ, Electronic equipment 3) High esistance : > 00 kω, winding insulation of electrical motor DC Bridges Medium esistance Measurement. Wheatstone Bridge Finding Theremin Equivalent I g th V th VTh V P Q S Th g P PQ S P Q S For Balance Condition I 0 g Th V 0 PS = Q

22 Sensitivities ) Current sensitivity, ) Voltage sensitivity, 3) Bridge Sensitivity, S S B S i I g mm/ma = deflection of Galvan meter in mm V Th / mm/v mm S B VThS v / S B V.S v S S For Maximum Sensitivity S S = S B, max V.S v 4. Carey foster slide wire Bridge m r = slide wire resistance in. for case (). At balance P r.() Q S L r For case () & S is reversed P S r..() Q L r From () & () r S r S L r L r

23 3. Voltmeter Ammeter Method a) Ammeter near the load Vv I m X A A V v = voltage across voltmeter I A = Ammeter current X = Test resistance, % error = m T A 00 00% T A = Animator resistance x b) Voltmeter near the load % error = m m Vv VX I I I m A X X v I I X V V X v v X X X X v 00% If X a v, voltmeter is connected near the load, ammeter is connected near the load X a v 4. Ohmmeter a) Series Type when X 0 I I = Full scale deflection m when FSD X Im 0 = zero deflection 3

24 for Half scale deflection. sh m X h se sh m b) Shunt Type S If If = current limiting resistor X 0 Im 0 x I m I FSD For Half scale Deflection ms x h = zero deflection = Full scale deflection m S Measurement of Low esistance Kelvin s Double Bridge Method Unknown resistance P qr P p S Q p q r Q q P, Q = outer ratio arms p, q = inner ratio arms S = standard resistance r = lead resistance = Test resistance High esistance Measurement Loss of charge Method t V e c C t V t V C log0 VC 4

25 t = time in (seconds) V = source voltage V C = Capacitor voltage Cathode ay Oscilloscope (CO) The velocity of e is changed by changing the pre-accelerating & accelerating anode potential KE =PE mv qva qv a m Deflection sensitivity D = deflection height on screen d = distance between plates d = length of vertical deflecting plates L = distance between centre of plate & screen V a = anode potential V y = Vertical plate Potential L D dv dvy V a mm deflection sensitivity D L d S V dv y a V mm 5

26 Lissajous Pattern If both horizontal & vertical deflection plates of CT is applied with the sinusoidal signal, the wave form pattern appearing on screen is called Lissajous Pattern. Case : Both signals have same frequency V V sin w t S.No V Vx Vy Vm x m x V sin w t y m y w w w x y = variable Lissayous Pattern 0 or Or or Or

27 Finding ) Lissajous Pattern in I st & III rd Quadrant sin X Y sin X Y for anti-clockwise orientation phase difference = (360 - ) for clockwise orientation, phase difference = ) Lissajous Pattern in II nd & IV th Quadrant X 80 sin X Y 80 sin Y for clockwise orientation, phase difference = for anti-clockwise orientation = 360 Case w w V V x y V sinw t x m x V sinw t y m y wy fy Number of horizental tangencies w f Number of vertical tangencies x x fy 4 f x 7

28 Digital Meters Type of converter Maximum Conversion Time ) Dual slope ADC Clocks ) Successive Approximation egister (ADC) n Clocks 3) Counter ADC Clocks 4) Flash ADC Clock n n Dual slope A/D Converter V a = analog input V = eference input V V T T a T n T T CLK Maximum conversion time = n T CLK Successive Approximation egister Suppose = VEF a V and V a = V D 3 D D D 0 T V < V T 0 0 T V > V.5 > V T < V In first clock cycle, MSB is set to get voltage corresponding to the digital o/p If V 0 < V a, then in next cycle next bit is set else, If V 0 > V a, MSB is reset & next bit is set We continue the same process till we reach LSB. 8

29 Specifications of Digital Voltmeter ) esolution The smallest value of input that can be measured by digital meter is called resolution. 0 n = No. of full Digits (0,,.., 9) n ) Sensitivity S = esolution x ange 3) Over anging The extra digit is called over-ranging If n = 3, we can measure from esolution, if n 3 digit, digit can be 0 &. if 3 4 we can measure from esolution, digit is there than MSB can be ) Total Error Error = (% error in reading) x reading + (NO. of counts) Full Scale ange of meter 9

30 Q meter / Voltage Magnifier If works on the principal of series resonance. At series resonance X L VC V C X V I VC C IX C XC XL V V = V. Q Q Practical Q-meter Also includes series resistance of source (oscillator) True Q T Measured Q, Q wl wl wl QT sh sh sh m QT Qm sh 30

31 Measurement of unknown capacitance Test capacitance is connected at Circuit is resonated at C = C T & T 3 4. C T f r = C T C C T = Test Capacitance is removed & circuit is resonated at C = fr = LC from () & () C C C T C () () Measurement of self-capacitance esonance is achieved at C = At C = C f C f L C C d C, resonance is achieved at L C C d C n C n d = n f, fr 3

32 Kuestion Electrical and Electronic Measurements

33 Contents Manual for Kuestion... Type : Error Analysis... 3 Type : Enhancement of Instrument ange... 5 Type 3:PMMC... 6 Type 4: Moving Iron... 8 Type 5: Bridges... 9 Type 6: Wattmeter... Type 7: Energy Meter... 4 Type 8: Digital Meter... 5 Type 9: CO... 6 Answer Key Kreatryx. All ights eserved.

34 Type : Error Analysis For Concept, refer to Measurement K-Notes, Error Analysis Point to remember: While using the limiting error concept in division we need to remember that all variables must be independent of each other and hence this rule does not hold for parallel combination of resistance. Sample problem : A variable w is related to three other variables x,y,z as w = xy/z. The variables are measured with meters of accuracy actual readings of the three meters are 80, 0 and 50 with 00 being the full scale value for all three. The maximum uncertainty in the measurement of w will be (A) rdg (B) rdg (C) rdg (D) 0.5% Solution: (D) is correct option 0.5% reading, 5.5% xy Given that = z log logx logy logz Maximum error in d dx dy dz % x y z dx 0.5% reading x dy % full scale y = dy 00 5% reading y 0 dz.5% reading z d So, % 0.5% 5%.5% 7% % of full scale value and 6.7%.5% reading. The 7.0% rdg 3

35 Unsolved Problems: Q. The power in a 3- phase, 3- wire load is measured using two 00 W full scale watt meters W and W. W is of a accuracy class % and reads 00W. W is of accuracy class 0.5% and reads 50W. Uncertainty in the computation to total power is (A).5% (B).5% (C) 3% (D) 4% Q. In the circuit given on fig, the limiting error in the power dissipation I in the resistor is (A).% (B) 5.% (C) 0.% (D) 5.% Q.3 Consider the circuit as shown in figure. Z is an unknown impedance and measured as z=zz3/z4. The uncertainties in the values of z,z3 and z4 are %, % and 3% respectively. The overall uncertainty in the measured value of z is (A) % (B) 4% (C) 5% (D) 5% Q.4 Three resistors have the following ratings =00 5%, =00 5% and 3= 50 5%. Determine the limiting error in ohms if the above resistances are connected in parallel (A).3 (B).9 (C) 4.8 (D).85 Q.5 The voltage of a standard cell is monitored daily over a period of one year. The mean value over a period of one year. The mean value of the voltage for every month shows a standard deviation of 0.mV. The standard deviation of the set constituted by the monthly mean values will be? (A)0 (B) 0. (C) 0. (D)0. Q.6 A current of 0mA is flowing through a resistance of 80 having tolerance of 0%. The current was measured by an analog ammeter on a 5 ma range with an accuracy of % of full scale. What is the range of error in the measurement of dissipated power? (A) 5% (B) 5% (C) 4% (D) 0% 4

36 Type : Enhancement of Instrument ange For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments Point to remember: In Ammeter, external resistance is added in parallel to meter and in voltmeter additional resistance is added in series. Sample Problem : An ammeter has a current range of 0-5 A, and its internal resistance is 0. Ω. In order to change the range to 0-5 A, we need to add a resistance of (A) 0.8 Ω in series with the meter (B).0 Ω in series with the meter (C) 0.04 Ω in parallel with the meter (D) 0.05 Ω in parallel with the meter Solution: (D) is correct option Given that full scale current is 5A Current in shunt I =I-Ifs = 5-5=0A sh 0.05 sh Unsolved Problems: Q. A 0-0mA DC Ammeter with internal resistance of 00 is used to design a DC voltmeter with full scale voltage of 0 V. The full scale range of this voltmeter can be extended to 50V by connecting an external resistance of (A) 900 (B) 499.9k (C) 4000 (D) 4900 Q. What is the value of series resistance to be used to extend (0-00)V range voltmeter having 000Ω/V sensitivity is to be extended to (0-000)V range. (A)44.44KΩ (B)55.55KΩ (C)34.56KΩ (D)45.5KΩ Q.3 A DC ammeter has a resistance of 0.Ω and its current range is 0-00 A. If the range is to be extended to A, then meter required the following shunt resistance? (A)0.00Ω (B)0.0Ω (C)0.05Ω (D).0Ω 5

37 Q.4 The coil of a measuring instrument has a resistance of 0Ω and the instrument reads up to 50V, when a resistance of 4.999Ω is connected in series. Now the same instrument is used as an ammeter by connecting a shunt resistance of /499Ω across it. What is the current range of the ammeter? (A)30A (B)6A (C)0A (D)4A Q.5 A D Arsonval movement with a full scale deflection current of 0 ma and internal resistance of 500Ω is to be converted into the different range of voltmeters. If a,b and c are the required series resistance for the ranges 0-0V, 0-50 V and 0-00 V respectively, then a:b:c is? (A):5:0 (B):: (C)3:9:9 (D)5:: Type 3:PMMC For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments Point to remember: PMMC always measures the average value of the output and the pointer vibrates around the zero position for pure AC input. Sample Problem 3: The input impedance of the permanent magnet moving coil (PMMC) voltmeter is infinite. Assuming that the diode shown in the figure below is ideal, the reading of the voltmeter in Volts is? (A)4.46 (B)3.5 (C).3 (D)0 Solution: (A) is correct option PMMC voltmeter reads average value. For the +ve half cycle of I/p voltage, diode will be forward biased (Vg = 0, ideal diode) Therefore, the voltmeter will be short circuited and reads V = 0 volt (for +ve half cycle) Now, for -ve half cycle, diode will be reverse biased and treated as open circuit. So, the voltmeter reads the voltage across 00 kw. Which is given by 6

38 V (00 ) 4 So, V V,rms Therefore, the average voltage for the whole time period is obtained as 4 0 V V,rms 4 V V avg Unsolved Problems: Q. A PMMC has an internal resistance of 00 and requires ma dc for full scale deflection. Shunting resistor sh placed across the movement has a value of 00. Diodes Dand D have forward resistance of 400 and infinite reverse resistance. For 0 V ac range, The value of series multiplier is (S) and voltmeter sensitivity for ac range is (A) 9550, 50 /V (B) 4550, 5 /V (C) 5000, 500 /V (D) 800, 50 V/ Q. A Thermocouple produces a voltage of 50 mv. Its internal resistance is 50. The resistance of leads is 0. The output is read by a PMMC meter having an internal resistance of 0 the output voltage indicated will be (A) 50 mv (B) 40 mv (C) 33.3 mv (D) 5.0 mv Q.3 An Electronic AC voltmeter is constructed using a full wave bridge ectifier, with a scale calibrated to read rms of a symmetrical square wave having zero mean. If this voltmeter is used to measure a voltage V(t)=0 sin 34t, then The reading of the voltmeter and magnitude of percentage error in reading respectively are? (A) 7.07V, % (B) 0.707V,.% (C) 6.36V, 9.9% (D).V, 0% Q.4 The coil of moving coil voltmeter is 50 mm long and 40 mm wide and has 0 turns on it. The control spring exerts a torque of 80 x 0-6 N.m. When the deflection is 0 divisions on full scale, flux density of the magnetic field in the air gap is.wb/m. Neglect the resistance of the coil. esistance that must be put in the series with the coil to give one volt per division is? (A) 50 K (B) 63.7 K (C) 83 K (D) 9.7 K 7

39 Q.5 The following date refers to a moving coil voltmeter resistance 0K; dimension of coil 30mm x 30mm; number of turns on coil 00, Flex density in the air gap is 0.08 wb/m. The deflecting torque produced by a voltage of 00v is (A) 60Nm (B) 44N m (C) 78.6N m (D) 78N m Q.6 Two 00V full scale PMMC type DC voltmeters having a figure of merits of 0 k/v and 0K/V are connected in series. The series combination can be used to measure maximum D.C voltage of (A) 00V (B) 300 V (C) 50 V (D) 00 V Type 4: Moving Iron For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments Point to remember: Moving Iron Instruments measures the rms value of output. Sample Problem 4: The saw-tooth voltage waveform shown in the figure is fed to a moving iron voltmeter. Its reading would be close to (A)48.4 (B)66.56 (C)57.74 (D)none Solution: (C) is correct option From the graph, we write mathematical expression of voltage v(t) 00 3 v(t) t 50 t volts 3 00 A moving iron voltmeter reads rms value of voltage, so V rms T T v (t)dt = (5 0 t) dt t = A

40 Unsolved Problems: Q. A permanent magnet moving coil type ammeter and a moving iron type ammeter are connected in series in a resistive circuit fed form output of a half wave rectifier voltage source. If the moving iron type instrument reaches 5A, the permanent magnet moving coil type instrument is likely to read. (A) Zero (B).5 A (C) 3.8 A (D) 5 A Q. A 50 V range spring controlled, electrodynamic voltmeter having a square law scale response takes 0.05 A on dc for full scale deflection of The control constant is 0.5 x 0-6 N-m/degree and the initial mutual inductance of the instrument is 0.5H. Total change in mutual inductance is (A) 8x0-3 H/rad (C) 6.5x0-3 H/rad (B) 8.3x0-3 H/rad (D) 3.7x0-3 H/rad Q.3 For certain dynamometer ammeter the mutual inductance M varies with deflection (expressed in degrees) as m=-5cos(+30)m.h. What will be the deflection of the instrument, if the deflection torque produced by 60m.A current is 8 x 0-6 N.m (A) 60 (B) 90 (C) 30 (D) 40 Q.4 The inductance of a moving iron ammeter is given by the expression L H where ϴ is the angle of deflection in radians. Determine the deflection in degree for a current of 8A, if the spring constant is N-m/rad. (A).5 0 (B).54 0 (C).56 0 (D).58 0 Q.5 A spring controlled moving iron voltmeter draws a current of value 00V. If it draws a current of 0.5mA, the meter reading is? (A)5V (B)550V (C)00V (D)00V Type 5: Bridges For Concept, refer to Measurement K-Notes, AC Bridges Point to remember: For any bridge, the balance condition is ZZ4 = ZZ3 9

41 Sample Problem 5: The Maxwell s bridge shown in the figure is at balance. The parameters of the inductive coil are. 3 (A), L C (B) L, C (C), L C (D) L, C Solution: (A) is correct option At balance condition j C ( jl)( ) ( jl) 4 j C 4 j 4 C4 j L j j L j C C C C C C By comparing real and imaginary parts and C C L C L Unsolved Problems: Q. A slide wire potentiometer has a battery of 4 V and negligible internal resistance. He resistance of slide wire is 00 and it s length 00 cm. A standard cell of.08 V is used for standardizing P.M and the rheostat is adjusted so that balance is obtained when the sliding contact is at 0.8 cm. Find the working current in slide wire (A) 0 ma (B) 0 ma (C) 30 ma (D) 40 ma 0

42 Q. A Wheat stone bridge has ratio arms of P-000 and Q 00 and is being used to measure an unknown resistance of as 5 as shown. Two galvanometer are available. Galvanometer A has a resistance of 50 and a sensitivity of 00 mm/a and galvanometer B has values of 600 and 500 mm/a. atio of sensitivity of galvanometer A to galvanometer B is (A) (B).5 (C).75 (D) Q.3. In the Maxwell bridge as shown below, the values of resistance X 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 X and LX will be respectively be (A) 375, 75 mh (B) 75, 50 mh (C) 37.8, 75 mh (D) 75, 75 mh Q.4 A schering bridge is used for measuring the power loss in dielectrics. The specimen are in the form of discs 0.3cm thick having a dielectric constant of.3. The area of each electrode is 34cm and the loss angle is known to be 9 for a frequency of 50Hz. The fixed resistor of the network has a value of 00 and the fixed capacitance is 50pF. Determine the value of the variable resistor required. (A) 3.7K (B) 4.6K (C) 3.73K (D) 4.54K Q.5 In the Wheatstone Bridge shown in the given figure, if the resistance in each arm is increased by 0.05% then the value of Vout will be (A) 50 mv (B) 5 mv (C) 0. V (D) zero

43 Type 6: Wattmeter For Concept, refer to Measurement K-Notes, Electro-mechanical Instruments Point to remember: The power reading of a wattmeter is equal to product of voltage across Potential Coil and Current through the Current Coil and the cosine of angle between them. These all quantities can calculated from the phasor diagrams. Sample Problem 6: A single-phase load is connected between and Y terminals of a 45 V, symmetrical, 3-phase, 4-wire system with phase sequence YB. A wattmeter is connected in the system as shown in figure. The power factor of the load is 0.8 lagging. The wattmeter will read? (A) 795 W (B) 597 W (C) +597 W (D) +795 W Solution: (B) is correct option In the figure V = 4530 Y V = 0 BN 3 Current in current coil 0 VY power factor=0.8 I c 0 0 Z Cos =0.8 =36.87 * Power VI eading of wattmeter P=994.3cos(6.87 ) Unsolved Problems : = =-597 W Q. The resistance of two coils of a Watt meter are 0.0 and 000 respectively and both are non-inductive. The load current is 0 A and voltage applied to the load is 30 V. Find the error in the readings for two methods of connection (A) 0.5% high, 0.67% high (C) 0.5% high, 0.67% low (B) 0.5% low, 0.67% low (D) 0.5% low, 0.67% low

44 Q. The current coil of dynamometer wattmeter is connected to 30 V DC source in series with a 8 resistor. The potential circuit is connected through an ideal rectifier in series with a 50 Hz source of 0 V. The inductance of pressure coil circuit and current coil resistance are negligible. eading of the wattmeter is (A) 8.84 W (B) 405 W (C) 0.57 W (D) None Q.3 A voltage: 00 sin t + 40 cos (3t - 30) + 50 sin (5t + 45) volts is applied to the pressure circuit of a wattmeter and through the current coil is passed a current of 8 sint + 6 cos(5t - 0) amps. The readings of wattmeter is? (A) 939 W (B) 539 W (C) 439 W (D) 039 W Q.4 The power in a 3- circuit is measured with the help of wattmeter. The readings of one of wattmeter is positive and that of other is negative. The magnitude of readings is different. It can be concluded that the power factor of the circuit is (A) Unity (B) zero (lagging) (C) 0.5 (lagging) (D) less than 0.5 (lagging) Q.5 In a dynamometer wattmeter the moving coil has 500 turns of mean diameter 30mm. Find the angle between the axis of the field and moving coil, if the flux density produced by field coil is 5x0-3 wb/m. The current in moving coil is 0.05 A and the power factor is and the torque produced is 9.5x0-6 N.m (A) 0 (B) 70 (C) 80 (D) 90 Q.6 Consider the following data for the circuit shown below Ammeter: esistance 0. reading 5A Voltmeter: esistance K reading 00V Wattmeter: Current coil resistance 0. Pressure coil resistance K Load: power factor = The reading of wattmeter is (A) 980W (B) 030W (C) 005W (D) 00W Q.7 A certain circuit takes 0 A at 00 V the power absorbed is 000 W.If the wattmeter s current coil has a resistance of 0.5 and its pressure coil a resistance of 5000 and an inductance of 0.3 H. The Error due to resistance of two coil of the Wattmeter, if the pressure coil of the meter connected on the load side (A) 5 W (B) 8 W (C) W (D) 3 W Q.8 The line to line voltage to the 3-phase, 50Hz AC circuit shown in figure is 00V rms. Assuming that the phase sequence is YB the wattmeter readings would be? (A)W=500W, W=000W (C)W=000W, W=0W (B) W=0W, W=000W (D) W=000W, W=500W 3

45 Type 7: Energy Meter For Concept, refer to Measurement K-Notes, Electro- mechanical Instruments. Point to remember: The measured value of energy in an energy meter is calculated in terms of meter constant and number of revolutions and true value of energy is derived from Power and time. Using these two values we can compute error in energy meter. Sample Problem 7: A dc A-h meter is rated for 5 A, 50 V. The meter constant is 4.4 A-sec/rev. The meter constant at rated voltage may be expressed as (A) 3750 rev/kwh (C) 000 rev/kwh Solution: (C) is correct option Meter constant (A-sec/rev) is given by 4.4 speed 4.4 K Power Where K is the meter constant in rev/kwh. 4.4 K VI K K K 000 rev/kwh (B) 3600 rev/kwh (D) 960 rev/kwh Q. The current and flux produced by series magnet of an induction watt-hour energy Meter are in phase, but there is an angular departure of 3 0 from quadrature between voltage and shunt magnet flux. The speed of the disc at full load and unity power factor is 40 rpm. Assuming the meter to read correctly under this condition, calculate it s speed at /4 full load and 0.5 P.F. lagging? (A) 4.3 rpm (B) 4.4 rpm (C) 4. rpm (D) 4.5 rpm Q. Single phase wattmeter operating on 30 V and 5 A for 5 hour makes 940 evolutions. Meter constant in revolutions is 400. The power factor of the load will be (A) (B) 0.8 (C) 0.7 (D) 0.6 4

46 Q.3 The meter constant of a 30v, 0A watt hour meter is 800 rev/kwh. The meter is tested at half load, rated voltage and unity power factor. The meter is found to make so revolutions in 38 seconds. The percentage error at half load is (A).7% fast (B) 0.87 fast (C).8% slow (D) 7.7% fast Q.4 The voltage-flux adjustment of a certain -phase 0V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it 85 0 (instead of 90 0 ). The error introduced in the reading of this meter when the current is 5A at power factors of unity and 0.5 lagging are respectively. (A)3.8mW, 77.4mW (C)-4.mW, -85. W (B)-3.8mW, -77.4mW (D)4. W, 85. W Q.5 A 30V, 5A, 50Hz single phase house service meter has a meter constant of 360 rev/kwhr. The meter takes 50 sec for making 5 revolutions of the disc when connected to a 0KW unity power factor load. The error in the reading of the meter is? (A)0% (B)+0.5% (C)-.0% (D)+.0% Type 8: Digital Meter For Concept, refer to Measurement K-Notes, Digital Meters. Point to remember: The fractional error in a digital meter is the most significant digit. Sample Problem 8: A 4 digit DMM has the error specification as: 0.% of reading + 0 counts. If a dc voltage of 00 V is read on its 00 V full scale, the maximum error that can be expected in the reading is? (A) 0.% (B) 0.% (C) 0.3% (D) 0.4% Solution: (C) is correct option 4 digit display will read from to So error of 0 counts is equal to= 0.0 V For 00 V, the maximum error is e = ( ) 0. 3V Percentage error % % of reading 5

47 Unsolved Problems: Q. A 3 DVM has an accuracy specification of 0.5% of reading digit. What is the possible error in volt s when reading 0.V on 0 V range and also percentage error. (A) V, 0.5% (B) 0.005V,.5% (C) 0.05V, 5% (D) None of the above Q. A 00V, 4 digit dual slope integrating type DVM can read up to (A) V (B) V (C) V (D) V Q.3 In a dual slope integrating type digital voltmeter the first integration is carried out for 0 periods of the supply frequencies of 50 HZ. If the reference voltage used is V, the total conversion time for an input of V is (A) 0.0 Sec (B) 0.05 Sec (C) 0. Sec (D) 0. Sec Q.4 A 4-digit DVM (digital Volt-meter) with a 00mV lowest full-scale range would have a sensitivity of how much value while resolution of this DVM is 0.000? (A)0.mV (B)0.0mV (C).0mV (D)0MV Q.5 In a dual slope type digital voltmeter, an unknown signal voltage is integrated over 00 cycles of the clock. If the signal has a 50 Hz pick up, the maximum clock frequency can be? (A)50 Hz (B)5 KHz (C)0 KHz (D)50 KHz Type 9: CO For Concept, refer to Measurement K-Notes, CO Point to remember: The best method to draw Lissajous figure is to plot the points on x-y plane at various time instants. Sample Problem 9: Group-II represents the figures obtained on a CO screen when the voltage signals Vx = Vxmsinωt and Vy = Vymsin(ωt + Φ) are given to its X and Y plates respectively and Φ is changed. Choose the correct value of Φ from Group-I to match with the corresponding figure of Group-II. 6

48 Group-I P. Φ = 0 Q. Φ = π/. π < Φ < 3π/ S. Φ = 3π/ Codes: (A) P=, Q=3, =6, S=5 (B) P=, Q=6, =4, S=5 (C) P=, Q=3, =5, S=4 (D) P=, Q=5, =6, S=4 Solution: (A) is correct option We can obtain the Lissaju pattern (in X-Y mode) by following method. For φ = 0 0, Vx = Vxmsinωt Vy = Vymsin(ωt ) = sinωt Draw Vx and Vy as shown below 7

49 Divide both Vy and Vx equal parts and match the corresponding points on the screen. Similarly for φ = 90 0 Vx = Vxmsinωt Vy = Vymsin(ωt ) Similarly for 3 we can also obtain for 3 0 8

50 Unsolved Problems: Q. In a cathode ray tube the distance between the deflecting plates is.0cm, the length of defalcating plates is 4.5cm and the distance of the screen from centre of the deflecting plates is 33 cm. If the accelerating plate s voltage is 300 V, then the deflection sensitivity of the tube is (A) 3.5 mm/v (B) 4.5 mm/v (C) 3.5 cm/v (D).5 mm / V Q. A voltage signal 0sin(34t+45 0 ) is examined using an analog single channel cathode ray oscilloscope with a time base setting of 0 msec per division. The CO screen has 8 divisions on the horizontal scale. Then, the number of cycles of signal observed on the screen will be (A) 8 cycles (B) cycles (C).5 cycles (D) 4 cycles Q.3 A lissajous pattern, as shown in figure is observed on the screen of a CO when voltage of frequencies fx and fy are applied to the x and y plates respectively fx : fy is then equal to (A) 3: (B) : (C) :3 (D) : Q.4 Voltage E is applied to the horizontal input and E to the vertical input of CO. E and E have same frequency. The trace on the screen is an ellipse. The slope of major axis is negative. The maximum vertical value is 3 divisions and the point where the ellipse crosses the vertical axis is.6 divisions. The ellipse is symmetrical about a horizontal and vertical axis. The phase angle difference then is (A) 0 0 (B) 40 0 (C) 40 0 (D) 30 0 Q.5 Horizontal deflection in a CO in due to E sint while vertical deflection is due to E sin(t + ) with a positive. Consider the following patterns obtained in the CO The correct sequence of these patterns in increasing order of the values of is (A) 3,, 5,, 4 (B) 3,, 4, 5, (C), 3, 4, 5, (D), 3, 5, 4, 9

51 Q.6 A CO is operated with X and Y setting of 0.5 ms/cm and 00mV/cm. The screen of the CO is 0cm 8cm (X and Y). A sine wave of frequency 00 Hz and rms amplitude of 300 mv is applied to Y-input. The screen will show? (A) One cycle of the undistorted sine wave (B) Two cycle of the undistorted sine wave (C) One cycle of the sine wave with clipped amplitude (D) Two cycles of the sine wave with clipped amplitude Answer Key Type C C B A D D Type C A C D C Type 3 B C C C B C Type 4 C B A D B Type 5 B C A B D Type 6 A C C D D D B B Type 7 D B B C D Type 8 A D D C B Type 9 D D B A D C 0

52 Kreatryx Subject Test Electrical and Electronic Measurements

53 KST- General Instructions during Examination. Total Duration of KST is 60 minutes.. The question paper consists of parts. Questions -0 carry one mark each and Question - 0 carry marks each. 3. The question paper may consist of questions of Multiple Choice Type (MCQ) and Numerical Answer Type. 4. Multiple choice type questions will have four choices against A, B, C, D, out of which only ONE is the correct answer. 5. All questions that are not attempted will result in zero marks. However, wrong answers for multiple choice type questions (MCQ) will result in NEGATIVE marks. For all MCQ questions a wrong answer will result in deduction of /3 marks for a -mark question and /3 marks for a -mark question. 6. There is NO NEGATIVE MAKING for questions of NUMEICAL ANSWE TYPE. 7. Non-programmable type Calculator is allowed. Charts, graph sheets, and mathematical tables are NOT allowed during the Examination.

54 Q. The total current I=I +I in a circuit is measured whereas I 50 A, I 50 A the limits of error are given as standard deviations. I is measured as (A) (300 ±.4)A (C) (300 ± )A (B) (300 ±.73)A (D) (300 ±.4)A Q. An ac bridge shown below has the following specifications C 0.5F and C3 0.5F. If the supply frequency is khz, determine the dissipation factor. k, (A).4 (B) 3.4 (C) 4.4 (D) 5.4 Q.3 A length of cable is tested for insulation resistance by loss of charges method. An electrostatic voltmeter of infinite resistance is connected between the cable conductor and earth, forming there with a joint capacitance of 500pF. If is observed that after charging it, the voltage falls from 50 V to 9 V in minute. The insulation resistance of the cable is 0 (A) 0 (B) 00,000M Ω (C).0 (D) Q.4 To check the distributed capacitance of a coil, it is resonated at 0 MHz with 0pF and then is resonated at 5 MHz with 40 pf. What is its equivalent distributed capacitance? (A) pf (B) 8pF (C) 4 pf (D) 30pF 4 Q.5 A digit voltmeter is used to measure the voltage value of V on a range. It would be display (A) (B) (C) (D) Q.6 A voltmeter reading 80 V on its 00V range and an ammeter reading 80 ma on its 50 ma range, are used to determine the power dissipated in a resistor. Both these instruments are guaranteed to be accurate within ±%, at full scale deflection. The limiting error in the measurement of the power dissipation is (A) 3.5% (B) 4.5% (C) 5.5% (D) 6.5% 7

55 Q.7 A set of independent current measurement taken by four observers was recorded as : 5.0 ma, 5.,A, 5.08mA and 5.04 ma. What is the average range of error? (A) ma (B) 0.04 ma (C) ma (D) 0.5 ma Q.8 In a thermocouple based instrument, if 0% harmonic is present then error in the current reading is (A) 0% (B) % (C) 4% (D) 5% Q.9 Calculate the value of the multiplier resistor for a 0 V rms range on the voltmeter as shown in the figure given below. (A).3 k Ω (B) 3.3 k Ω (C) 4.3 k Ω (D) 5.3 k Ω Q.0 A voltmeter, having a sensitivity of k Ω/V, connected across an unknown resistance in series with a milliammeter, reads 00 V on 50 V scale. If the milliammeter reads 0 ma, the error due to loading effect of voltmeter would be y%, then the value of y is. Q. A coil with a resistance of 3 Ω is connected to the terminals of a Q meter. esonance occurs at an oscillator frequency of 5 MHz with a capacitance of 00 pf. If x% is the error introduced by the insertion resistance sh,=0. Ω then the value of x is. Q. Calculate the constants of a shunt to extend the range of 0 5 A moving iron ammeter to 0 50 A. The instrument constants are = 0.09 Ω and L = 90µH If the shunt is made noninductive and the combination is correct on d.c. find the full scale error at 50 Hz. (A) 3.8% high (B).8%low (C).8% high (d)3.8%low Q.3 The coil of a 300 V moving iron voltmeter has a resistance of 500 W and an inductance of 0.8H.The instrument reads correctly at 50Hz. A.C. supply and takes 00 ma at full scale deflection. What is the percentage error in the instrument reading, when it is connected to 00V d.c.supply. (A)00.6 V (B)05.6V (C) 0 V (D).V 3

56 Q.4 A single-phase load is connected between and Y terminals of 45 V, symmetrical 3- phase, 4 wire systems with phase sequence YB. A wattmeter is connected in the system as shown in figure below. The power factor of the load is 0.8 lagging. The wattmeter will read (A) 97.3 watt (B) 48.58watt (C) 95.9 watt (D) 98.3 watt Q.5 An energy meter rated as 5A, 30 V makes 500 revolutions per kwh. If in a test at full load unity power factor, it makes 5 revolutions in 30 seconds. Then which of the following statement is true? (A) The meter runs faster and error is 4.6% (B) The meter runs faster and error is 4.35% (C) The meter runs slower and error is 4.6% (D) The meter runs slower and error is 4.35% Q.6 Figure shown below is the circuit of a capacitance comparison bridge. If = 0 Ω =30 Ω 3 = 5 and C 3 = 0 pf than the value of x and C x respectively are. (a) 7.5, 6/3 pf (b) 30.5, 8/3 pf (c) 35.5, /3 pf (d) 37.5, 0/3 pf Q.7 Figure gives parameter values of an AC bridge at balance when supply frequency is 500Hz. Find Z x assuming it to be a resistance x in series with an inductance or capacitance and choose the correct option (A) (B) (C) (D) 0 F 0 F 0 H 0 4

57 Q.8 A CO is used to measure the voltage as shown below in the diagram. CO probe has the impedance of 600 k with a capacitance in parallel of a value of 0 pf. The reading of CO will be V. Q.9 For an electrodynamometer ammeter. The mutual inductance M varies with deflection ϴ (ϴ is in degree) as M = -6cos(ϴ +45 o ) mh. For a direct current of 50 ma corresponding to a deflection of 60 0, the deflecting torque will be µnm. Q.0 A current of i sint 0. sin t A is passed through the circuit shown in 6 figure given below. ampere is the sum of readings of each instrument. ( = 0 rad/s) 5

58 Power Systems KST Answer Key C C A B 4 C B B C D D C A C C D D B C 6