Electrical and Electronic Measurements Kuestion

KuestionElectrical and Electronic Measurements www.kreatryx.com Contents Manual for Kuestion .......................................................................... 2 Type 1: Error Analysis.......................................................................... 3 Type 2: Enhancement of Instrument Range ........................................ 5 Type 3:PMMC ..................................................................................... 6 Type 4: Moving Iron ............................................................................ 8 Type 5: Bridges.................................................................................... 9 Type 6: Wattmeter ............................................................................ 12 Type 7: Energy Meter ........................................................................ 14 Type 8: Digital Meter ........................................................................ 15 Type 9: CRO....................................................................................... 16 Answer Key ....................................................................................... 20 © 2014 Kreatryx. All Rights Reserved. 1 This is the reason why Kuestion serves the purpose of being the bare minimum set of questions to be solved from each chapter during revision. 2 . All Rights Reserved. © 2014 Kreatryx. What is Kuestion? A set of 40 questions or less for each chapter covering almost every type which has been previously asked in GATE. Along with the Solved examples to refer from. How do I use Kuestion? Kuestion should be used as a tool to improve your speed and accuracy chapter wise. You should refer KNotes Theory before solving any “Type” problems from Kuestion.Manual for Kuestion Why Kuestion? It’s very overwhelming for a student to even think about finishing 100-200 questions per chapter when the clock is ticking at the last moment. It should be treated as a supplement to our K-Notes and should be attempted once you are comfortable with the understanding and basic problem solving ability of the chapter. When do I start using Kuestion? It is recommended to use Kuestion as soon as you feel confident in any particular chapter. Although it will really help a student if he/she will start making use of Kuestion in the last 2 months before GATE Exam (November end onwards). a student can try similar unsolved questions to improve his/her problem solving skills. refer to Measurement K-Notes. 20 and 50 with 100 being the full scale value for all three.7% rdg (D) 7.5% reading z d  0.0% rdg Solution: (D) is correct option xy z log   logx  log y  logz Maximum error in  Given that = d dx dy dz     x y z dx  0.5%  7% So.5%  5%  1.5% rdg (B) 5. The actual readings of the three meters are 80. %  3 .5% reading. 1% of full scale value and 1.Type 1: Error Analysis For Concept. The variables are measured with meters of accuracy 0.z as w = xy/z . The maximum uncertainty in the measurement of ‘w’ will be (A) 0.5% rdg (C) 6. Sample problem 1: A variable w is related to three other variables x.5% reading x dy  1% full scale y 1 =  100  1 100 dy 1    100  5% reading y 20 % dz  1.5% reading. 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.y. 2% Q.1 12 12 Q.z3 and z4 are 1%.1 The power in a 3.5% (C)  3% (D)  4% Q. Determine the limiting error in ohms if the above resistances are connected in parallel (A) 1. Uncertainty in the computation to total power is (A)  1.3 (B) 1.3 Consider the circuit as shown in figure. R2=100 5% and R3= 50 5%. W1 is of a accuracy class 1% and reads 100W.19 (C) 4.4 Three resistors have the following ratings R1=200 5%. The current was measured by an analog ammeter on a 25 mA range with an accuracy of 2% of full scale.phase.1 0.2% (C) 10. What is the range of error in the measurement of dissipated power? (A) 15% (B) 5% (C) 14% (D) 20% 4 .5% and reads – 50W. The overall uncertainty in the measured value of z1 is (A)  11% (B) 4% (C) 5% (D)  5% Q.wire load is measured using two 100 W full scale watt meters W1 and W2.2% (B) 5.85 Q. W2 is of accuracy class  0. Z1 is an unknown impedance and measured as z1=z2z3/z4.2% (D) 25. The uncertainties in the values of z2.5% (B)  2.6 A current of 10mA is flowing through a resistance of 820 having tolerance of 10% .2 In the circuit given on fig. 3.28 (D) 2.1mV. The mean value over a period of one year.1 (A)0 (B) (C) (D)0. 1% and 3% respectively. The standard deviation of the set constituted by the monthly mean values will be? 0.5 The voltage of a standard cell is monitored daily over a period of one year. The mean value of the voltage for every month shows a standard deviation of 0.Unsolved Problems: Q. the limiting error in the power dissipation ’I2R’ in the resistor ‘R’ is (A) 1. then meter required the following shunt resistance? (A)0.011Ω (C)0.55KΩ (C)34.3 A DC ammeter has a resistance of 0.9k (C) 4000 (D) 4900 Q. Sample Problem 2: An ammeter has a current range of 0-5 A. If the range is to be extended to 0-500 A.0 Ω in series with the meter (C) 0. we need to add a resistance of (A) 0. external resistance is added in parallel to meter and in voltmeter additional resistance is added in series.44KΩ (B)55.025Ω 5 (D)1. (A)44. In order to change the range to 0-25 A.010Ω (B)0.Type 2: Enhancement of Instrument Range For Concept. Electro-mechanical Instruments Point to remember: In Ammeter.2 Ω. refer to Measurement K-Notes.2 What is the value of series resistance to be used to extend (0-200)V range voltmeter having 2000Ω/V sensitivity is to be extended to (0-2000)V range. The full scale range of this voltmeter can be extended to 50V by connecting an external resistance of (A) 900 (B) 499.0Ω .1 A 0-10mA DC Ammeter with internal resistance of 100 is used to design a DC voltmeter with full scale voltage of 10 V.04 Ω in parallel with the meter (D) 0.8 Ω in series with the meter (B) 1.25KΩ Q.2 R sh  0.56KΩ (D)45. and its internal resistance is 0.05 Ω in parallel with the meter Solution: (D) is correct option Given that full scale current is 5A Current in shunt I’=IR-Ifs = 25-5=20A 20  R sh  5  0.1Ω and its current range is 0-100 A.05 Unsolved Problems: Q. refer to Measurement K-Notes. 0-50 V and 0-100 V respectively. Now the same instrument is used as an ammeter by connecting a shunt resistance of 1/499Ω across it.Q. the reading of the voltmeter in Volts is? (A)4. for -ve half cycle.15 (C)2. when a resistance of 4.4 The coil of a measuring instrument has a resistance of 10Ω and the instrument reads up to 250V. ideal diode) Therefore. If Ra. diode will be forward biased (Vg = 0. Which is given by 6 . So. 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.5 A D’Arsonval movement with a full scale deflection current of 10 mA and internal resistance of 500Ω is to be converted into the different range of voltmeters.46 (B)3. then Ra:Rb:Rc is? (A)2:5:10 (B)1:1:1 (C)3:9:19 (D)5:11:21 Type 3:PMMC For Concept.999Ω is connected in series. What is the current range of the ammeter? (A)30A (B)26A (C)20A (D)24A Q. For the +ve half cycle of I/p voltage. the voltmeter will be short circuited and reads V1 = 0 volt (for +ve half cycle) Now. 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 voltmeter reads the voltage across 100 kW. diode will be reverse biased and treated as open circuit.Rb and Rc are the required series resistance for the ranges 0-20V.23 (D)0 Solution: (A) is correct option PMMC voltmeter reads average value. 707V. The value of series multiplier is (RS) and voltmeter sensitivity for ac range is (A) 9550 .3 mV (D) 25. If this voltmeter is used to measure a voltage V(t)=10 sin 314t. with a scale calibrated to read rms of a symmetrical square wave having zero mean.14.rms  V 2 Therefore.46V Vavg  1  2 2 2 2 V2  100  Unsolved Problems: Q.14 (100  1) 14 So. 11% (C) 6.1400  14.9% (B) 0.94  4.1 A PMMC has an internal resistance of 100  and requires 1 mA dc for full scale deflection.0 mV Q. The resistance of leads is 10. Diodes D1and D2 have forward resistance of 400  and infinite reverse resistance. 250 V/ Q.rms 2  14  4. The output is read by a PMMC meter having an internal resistance of 120  the output voltage indicated will be (A) 50 mV (B) 40 mV (C) 33. Its internal resistance is 50  . 11. then The reading of the voltmeter and magnitude of percentage error in reading respectively are? (A) 7.3 An Electronic AC voltmeter is constructed using a full wave bridge Rectifier.7 K .1V. V2. 500 /V (D) 1800 .7 K (C) 83 K 7 (D) 91. the average voltage for the whole time period is obtained as 14 0 V  V2.36V. Resistance that must be put in the series with the coil to give one volt per division is? (A) 50 K (B) 63. 0% Q. 9. 225 /V (C) 5000 .2wb/m2.4 The coil of moving coil voltmeter is 50 mm long and 40 mm wide and has 120 turns on it. 250 /V (B) 4550 . flux density of the magnetic field in the air gap is 1.1% (D) 11. When the deflection is 120 divisions on full scale. Shunting resistor Rsh placed across the movement has a value of 100.m.07V. Neglect the resistance of the coil.2 A Thermocouple produces a voltage of 50 mv. The control spring exerts a torque of 180 x 10-6 N. For 10 V ac range. Flex density in the air gap is 0.6 Two 100V full scale PMMC type DC voltmeters having a figure of merits of 10 k/V and 20K/V are connected in series.74 (D)none Solution: (C) is correct option From the graph. so Vrms T 1 2  v (t)dt T 0 1 = 20  10 3 2010 3 25  106 = 20  103  t3     3 0  (5  103  t)2 dt 0 20103  57.08 wb/m 2. dimension of coil 30mm x 30mm. The deflecting torque produced by a voltage of 200v is (A) 60Nm (B) 144N – m (C) 78. number of turns on coil 100.74 A 8 .C voltage of (A) 100V (B) 300 V (C) 150 V (D) 200 V Type 4: Moving Iron For Concept.Q. refer to Measurement K-Notes. we write mathematical expression of voltage v(t) 100 v(t)  t  5  103 t volts 3 20  10 A moving iron voltmeter reads rms value of voltage. Its reading would be close to __________ (A)48.56 (C)57. Sample Problem 4: The saw-tooth voltage waveform shown in the figure is fed to a moving iron voltmeter.5 The following date refers to a moving coil voltmeter resistance 10K.41 (B)66. Electro-mechanical Instruments Point to remember: Moving Iron Instruments measures the rms value of output.6N – m (D) 178N – m Q. The series combination can be used to measure maximum D. Total change in mutual inductance is (A) 18x10-3 H/rad (C) 26. the balance condition is Z1Z4 = Z2Z3 9 (D)200V .2 A 50 V range spring controlled.25H. electrodynamic voltmeter having a square law scale response takes 0. If the moving iron type instrument reaches 5A.5 A (C) 3.7x10-3 H/rad Q.540 (C)1.580 Q. if the spring constant is 10  10 6 N-m/rad. If it draws a current of 0. The control constant is 0. (A) Zero (B) 2. the meter reading is? (A)25V (B)550V (C)100V Type 5: Bridges For Concept.5 x 10-6 N-m/degree and the initial mutual inductance of the instrument is 0.3x10-3 H/rad (D) 13.5x10-3 H/rad (B) 28.Unsolved Problems: Q. What will be the deflection of the instrument.5 A spring controlled moving iron voltmeter draws a current of value 100V.1 A permanent magnet moving coil type ammeter and a moving iron type ammeter are connected in series in a resistive circuit fed form output of a half wave rectifier voltage source.18 A (D) 5 A Q.5mA.560 (D)1.250 (B)1.A current is 18 x 10-6N. (A)1. the permanent magnet moving coil type instrument is likely to read. refer to Measurement K-Notes.H.3 For certain dynamometer ammeter the mutual inductance M varies with deflection  (expressed in degrees) as m=-5Cos(+30)M. Determine the   deflection in degree for a current of 8A. AC Bridges Point to remember: For any bridge.4 The inductance of a moving iron ammeter is given by the expression L  20  10  32 H where ϴ is the angle of deflection in radians. if the deflection torque produced by 60m.05 A on dc for full scale deflection of 900.m (A) 60 (B) 90 (C) 30 (D) 40 Q. RR (A) R  2 3 .M and the rheostat is adjusted so that balance is obtained when the sliding contact is at 101. R  C4R 2R3 R4 (C) R  R4 1 . L R 2R 3 C4R2R3 (D) L  R4 1 . R R 2R 3 C4R 2R3 Solution: (A) is correct option At balance condition (R  jL)(R 4 || R4 j C 4 j )  R 2R 3  (R  jL)   R 2R 3 C 4  j  R4   C 4   jR R jR R  jRR 4 LR 4  jRR 4 LR 4   R 2R3R 4  2 3    R 2R3R 4  2 3 C4 C4 C4 C4 C4 C4 By comparing real and imaginary parts RR RR 4 R 2R 3   R  2 3 and  C 4 C 4 R4 LR 4  R 2R 3R 4  L  R 2R 3R 4 C4 Unsolved Problems: Q. He resistance of slide wire is 100  and it’s length 200 cm. L  C4R 2R3 R4 (B) L  R 2R3 .8 cm.Sample Problem 5: The Maxwell’s bridge shown in the figure is at balance.018 V is used for standardizing P.1 A slide wire potentiometer has a battery of 4 V and negligible internal resistance. The parameters of the inductive coil are. Find the working current in slide wire (A) 10 mA (B) 20 mA (C) 30 mA 10 (D) 40 mA . A standard cell of 1. 4 A schering bridge is used for measuring the power loss in dielectrics.3.17K (B) 4. 75 mH (D) 75 . 150 mH (C) 37.3 . Ratio of sensitivity of galvanometer ‘A’ to galvanometer ‘B’ is (A) 1 (B) 1. The values of RX and LX will be respectively be (A) 375 . the values of resistance RX and inductance LX of a coil are to be calculated after balancing the bridge. Determine the value of the variable resistor required. Galvanometer ‘A’ has a resistance of 50  and a sensitivity of 200 mm/A and galvanometer ‘B’ has values of 600  and 500 mm/A. The component values are shown in the figure at balance.05% then the value of Vout will be (A) 50 mV (B) 5 mV (C) 0.54K Q. (A) 3. In the Maxwell bridge as shown below.73K (D) 4.3cm thick having a dielectric constant of 2. Two galvanometer are available.75 (D) 2 Q. 75 mH Q.26K (C) 3.Q.25 (C) 1.5 In the Wheatstone Bridge shown in the given figure. if the resistance in each arm is increased by 0.2 A Wheat stone bridge has ratio arms of P-1000  and Q – 100  and is being used to measure an unknown resistance of R as 25 as shown.1 V (D) zero 11 . 75 mH (B) 75 . The specimen are in the form of discs 0. The fixed resistor of the network has a value of 100 and the fixed capacitance is 50pF.8 . The area of each electrode is 314cm2 and the loss angle is known to be 9 for a frequency of 50Hz. The wattmeter will read? (A) −795 W (B) −597 W (C) +597 W (D) +795 W Solution: (B) is correct option In the figure VRY = 415300 VBN= 415 1200 3 Current in current coil  power factor=0. symmetrical.67% high (C) 0.8 lagging.87  VRY 415300 Ic    4.8   0 Cos=0.870  994.8  =36.1 The resistance of two coils of a Watt meter are 0.67% low (D) 0.01 and 1000  respectively and both are non-inductive.60 =-597 W Unsolved Problems : Q.870 3 Reading of wattmeter P=994.870 0 Z 10036. 4-wire system with phase sequence RYB.870 ) =994.87 Power  VI *  415 1200  4. 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. Find the error in the readings for two methods of connection (A) 0.15% high. refer to Measurement K-Notes. 0. These all quantities can calculated from the phasor diagrams.156.15% low. 0. 0.Type 6: Wattmeter For Concept.67% low (B) 0.67% low 12 .3126.3  cos(126.15  6.15% high.15% low. The power factor of the load is 0.3  -0. 3-phase. Sample Problem 6: A single-phase load is connected between R and Y terminals of a 415 V. The load current is 20 A and voltage applied to the load is 30 V. 0. A wattmeter is connected in the system as shown in figure. 4 The power in a 3- circuit is measured with the help of 2 wattmeter. It can be concluded that the power factor of the circuit is (A) Unity (B) zero (lagging) (C) 0.8 The line to line voltage to the 3-phase.866 and the torque produced is 229. Assuming that the phase sequence is RYB the wattmeter readings would be? (A)W1=500W.05 A and the power factor is 0.5x10-6 N. Reading of the wattmeter is (A) 282.7 A certain circuit takes 10 A at 200 V the power absorbed is 1000 W .5 (lagging) Q.57 W (D) None Q. W2=500W 13 .2 The current coil of dynamometer wattmeter is connected to 30 V DC source in series with a 8 resistor. The current in moving coil is 0.m (A) 0 (B) 70 (C) 80 (D) 90 Q.2 Pressure coil resistance 2 K Load: power factor =1 The reading of wattmeter is (A) 980W (B) 1030W (C) 1005W (D) 1010W Q.3 A voltage: 100 sin t + 40 cos (3t . The Error due to resistance of two coil of the Wattmeter. W2=1000W (C)W1=1000W.Q. if the flux density produced by field coil is 15x10-3 wb/m2. The magnitude of readings is different. Find the angle between the axis of the field and moving coil. The readings of wattmeter is? (A) 939 W (B) 539 W (C) 439 W (D) 1039 W Q. 50Hz AC circuit shown in figure is 100V rms.84 W (B) 405 W (C) 202. The potential circuit is connected through an ideal rectifier in series with a 50 Hz source of 120 V. W2=0W (D) W1=1000W.5 In a dynamometer wattmeter the moving coil has 500 turns of mean diameter 30mm.120) amps. if the pressure coil of the meter connected on the load side (A) 15 W (B) 8 W (C) 11 W (D) 13 W Q.15  and its pressure coil a resistance of 5000  and an inductance of 0.2 reading 5A Voltmeter: Resistance 2K reading 200V Wattmeter: Current coil resistance 0.6 Consider the following data for the circuit shown below Ammeter: Resistance 0.If the wattmeter’s current coil has a resistance of 0. The readings of one of wattmeter is positive and that of other is negative.30) + 50 sin (5t + 45) volts is applied to the pressure circuit of a wattmeter and through the current coil is passed a current of 8 sint + 6 cos(5t . W2=1000W (B) W1=0W.3 H. The inductance of pressure coil circuit and current coil resistance are negligible.5 (lagging) (D) less than 0. lagging? (A) 4.4 rpm (C) 4.F.4  250 1 1000  3600 K   1000 rev/kWh 14.1 rpm (D) 4.5 P.5 rpm Q.4 A-sec/rev.2 Single phase wattmeter operating on 230 V and 5 A for 5 hour makes 1940 Revolutions. 250 V. The speed of the disc at full load and unity power factor is 40 rpm. 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. 1 K  VI 15 1 14. The meter constant at rated voltage may be expressed as (A) 3750 rev/kWh (C) 1000 rev/kWh (B) 3600 rev/kWh (D) 960 rev/kWh Solution: (C) is correct option Meter constant (A-sec/rev) is given by 1 14.mechanical Instruments. Electro.8 (C) 0.4  Q.4  speed 1 14. Using these two values we can compute error in energy meter. The meter constant is 14.Type 7: Energy Meter For Concept.4 250 3600  3600 1000 14. but there is an angular departure of 30 from quadrature between voltage and shunt magnet flux.3 rpm (B) 4. Meter constant in revolutions is 400.1 The current and flux produced by series magnet of an induction watt-hour energy Meter are in phase.4  K  K  15  250 14.4  K  Power Where ‘K’ is the meter constant in rev/kWh. The power factor of the load will be (A) 1 (B) 0. calculate it’s speed at 1 /4 full load and 0. Sample Problem 7: A dc A-h meter is rated for 15 A. Assuming the meter to read correctly under this condition. refer to Measurement K-Notes.6 .7 14 (D) 0. -85.2% (C)  0. the maximum error is e = (100  0.10 V 2 For 100 V.Q.4 The voltage-flux adjustment of a certain 1-phase 220V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it 85 0(instead of 900). the maximum error that can be expected in the reading is? A4 (A)  0.2% of reading + 10 counts. If a dc voltage 2 of 100 V is read on its 200 V full scale.187 fast (C) 2.5 lagging are respectively.1% (B)  0. (A)3.8mW. Digital Meters.1)  0. The meter is found to make so revolutions in 138 seconds.002  0.8% slow (D) 7. The percentage error at half load is (A) 1. 50Hz single phase house service meter has a meter constant of 360 rev/KWhr. The meter takes 50 sec for making 51 revolutions of the disc when connected to a 10KW unity power factor load.1 W Q.3% (D)  0. Sample Problem 8: 1 digit DMM has the error specification as: 0.72% fast (B) 0.3% of reading 15 .0% Type 8: Digital Meter For Concept. The error introduced in the reading of this meter when the current is 5A at power factors of unity and 0.00 to 199.3 The meter constant of a 230v. 85. 77.5% (C)-2.4mW (D)4. Point to remember: The fractional error in a digital meter is the most significant digit. 10A watt hour meter is 1800 rev/kwh. The meter is tested at half load.3  100 Percentage error   % 100  0.0% (D)+2. rated voltage and unity power factor.3V 0.8mW.1 W (B)-3.2 W.4mW (C)-4. refer to Measurement K-Notes.99 So error of 10 counts is equal to=  0. -77.5 A 230V.4% Solution: (C) is correct option 1 4 digit display will read from000. The error in the reading of the meter is? (A)0% (B)+0.7% fast Q.2mW. 5A. 000 V (D) 19. (A)  0. 4 digit dual slope integrating type DVM can read up to 2 (A) 99.3 In a dual slope integrating type digital voltmeter the first integration is carried out for 10 periods of the supply frequencies of 50 HZ. refer to Measurement K-Notes. Choose the correct value of Φ from Group-I to match with the corresponding figure of Group-II.2 A 010V.02 Sec (B) 0.999 V Q.0001? (A)0.5% (D) None of the above 1 Q.2 Sec (D) 0.999 V (B) 199. an unknown signal voltage is integrated over 100 cycles of the clock.015V. the maximum clock frequency can be? (A)50 Hz (B)5 KHz (C)10 KHz (D)50 KHz Type 9: CRO For Concept.1 Sec Q.5% (C)  0. Sample Problem 9: Group-II represents the figures obtained on a CRO screen when the voltage signals Vx = Vxmsinωt and Vy = Vymsin(ωt + Φ) are given to its X and Y plates respectively and Φ is changed.0105 V.1 A 3 DVM has an accuracy specification of  0. CRO Point to remember: The best method to draw Lissajous figure is to plot the points on x-y plane at various time instants.5 In a dual slope type digital voltmeter.05 Sec (C) 0. If the signal has a 50 Hz pick up.Unsolved Problems: 1 Q. 1.4 A 4-digit DVM (digital Volt-meter) with a 100mV lowest full-scale range would have a sensitivity of how much value while resolution of this DVM is 0.0015V.01mV (C)1. If the reference voltage used is 2 V.5% of reading  1 digit. the total conversion time for an input of 1 V is (A) 0.1mV (B)0. 10. What is the 2 possible error in volt’s when reading 0.1V on 10 V range and also percentage error. 15% (B)  0.99 V (C) 20. 16 .0mV (D)10MV Q. S=4 Solution: (A) is correct option We can obtain the Lissaju pattern (in X-Y mode) by following method.Group-I P. R=6. π < Φ < 3π/2 S. Vx = Vxmsinωt Vy = Vymsin(ωt + 00) = sinωt Draw Vx and Vy as shown below 17 . S=4 (B) P=2. Φ = 3π/2 Codes: (A) P=1. R=5. Q=5. R=6. S=5 (D) P=1. Φ = 0 Q. Φ = π/2 R. Q=3. S=5 (C) P=2. For φ = 00. R=4. Q=3. Q=6. Similarly for φ = 900 Vx = Vxmsinωt Vy = Vymsin(ωt + 900) Similarly for 2 3  2 we can also obtain for 0    3 2 18 .Divide both Vy and Vx equal parts and match the corresponding points on the screen. 5. the number of cycles of signal observed on the screen will be (A) 8 cycles (B) 2 cycles (C) 2.2 A voltage signal 10sin(314t+450) is examined using an analog single channel cathode ray oscilloscope with a time base setting of 10 msec per division.5 cycles (D) 4 cycles Q.3 A lissajous pattern.4 Voltage E1 is applied to the horizontal input and E2 to the vertical input of CRO. the length of defalcating plates is 4. The slope of major axis is negative. The maximum vertical value is 3 divisions and the point where the ellipse crosses the vertical axis is 2. 1 19 . 2. 4 (B) 3. The phase angle difference then is (A) 2100 (B) 1400 (C) 2400 (D) 1300 Q. The CRO screen has 8 divisions on the horizontal scale. E1 and E2 have same frequency. 1. Then.1 In a cathode ray tube the distance between the deflecting plates is 1. The ellipse is symmetrical about a horizontal and vertical axis.5 mm/V (B) 4. 2. Consider the following patterns obtained in the CRO The correct sequence of these patterns in increasing order of the values of  is (A) 3. If the accelerating plate’s voltage is 300 V.0cm. 5. 3.5 cm/V (D) 2. 4. 4.6 divisions. 3.5 Horizontal deflection in a CRO in due to E sint while vertical deflection is due to E sin(t + ) with a positive . 5.Unsolved Problems: Q.5cm and the distance of the screen from centre of the deflecting plates is 33 cm.5 mm/V (C) 3. 5. The trace on the screen is an ellipse. 1 (C) 2. 1 (D) 2. then the deflection sensitivity of the tube is (A) 3.5 mm / V Q. 4. as shown in figure is observed on the screen of a CRO when voltage of frequencies fx and fy are applied to the x and y plates respectively fx : fy is then equal to (A) 3:2 (B) 1:2 (C) 2:3 (D) 2:1 Q. Q. 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 1 Type 2 Type 3 Type 4 Type 5 Type 6 Type 7 Type 8 Type 9 1 C C B C B A D A D 2 C A C B C C B D D 3 B C C A A C B D B 4 A D C D B D C C A 5 D C B B D D D B D 20 6 D 7 8 B B C D C .5 ms/cm and 100mV/cm.6 A CRO is operated with X and Y setting of 0. The screen of the CRO is 10cm  8cm (X and Y). A sine wave of frequency 200 Hz and rms amplitude of 300 mV is applied to Y-input.