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Solution to Previous Year Exam Questions for Basic Electrical Engineering - BEE of bput by Mitu Baral

  • Basic Electrical Engineering - BEE
  • 2018
  • PYQ Solution
  • Biju Patnaik University of Technology BPUT - BPUT
  • Electrical and Electronics Engineering
  • B.Tech
  • 160 Offline Downloads
  • Uploaded 10 months ago
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BASIC ELECTRICAL ENGINEERING –BE2102 Registration No. Mr. Mitu Baral, ECE,Dept. NIST, Berhampur : B.Tech BE2102 Second Semester Regular Examination-2015 BASIC ELECTRICAL ENGINEERING BRANCH(S): AEIE, AERO, AUTO, BIOMED, BIOTECH, CIVIL, CSE, EC, EEE, EIE, ELECTRICAL, ETC, IT, MECH, MINING, MM, PLASTIC, TEXTILE QUESTION CODE: J 367 Full Marks – 70 Time: 3 Hours Answer question No. 1 which is compulsory and any five from the rest The figures in the right – hand margin indicate marks 1. Answer the following question: 2 X10 a) Differentiate between an ideal voltage source and practical voltage source? Answer: Ideal Voltage source Practical Voltage source (i) Its internal resistance is zero. (i) It has finite value of internal (ii) Its terminal voltage is constant and independent of load. resistance. (ii) Its terminal voltage depends upon load impedance hence terminal voltage does not remain constant with change in load. b) Explain the significance of the power factor. Answer: Power factor is the ratio between active power and apparent power. So it gives the idea about how much percentage of the total power is utilized for useful work. NATIONAL INSTITUTE OF SCIENCE &TECHNOLOGY PALUR HILLS, BERHAMPUR, ORISSA – 761008, INDIA

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BASIC ELECTRICAL ENGINEERING –BE2102 Mr. Mitu Baral, ECE,Dept. NIST, Berhampur c) What do you understand by retentivity? Answer: The power of retaining residual magnetism by a magnetic specimen is called retentivity. d) Write down the equation of a sinusoidal source voltage of 60 Hz frequency having a r.m.s. value of 250V. Figure 1.c: B- H Curve Answer: V (t )  vm sin t V (t )  2  250 sin( 2    60)t  354 sin 378t e) What is the time constant of a RL circuit having R=10Ω and L= 10 H ? Answer: Time constant for a RL circuit is given by   L 10   1sec R 10 f) Two impedances of value (2+j6) Ω and (8-j12) Ω are connected in series. What would be the resulting impedance in polar form? Solution: let Z1 =2+j6, Z 2=8-j12. Then the equivalent impedance of for series connection of Z 1 and Z2 is, Z= Z1+Z2 Z= (10-j6) ohms In rectangular Z= 11.662  30.964 𝝮 g) Describe constant losses occurring in a DC Machine. Answer: Iron Losses:  Hysteresis Loss  Eddy Current Loss Mechanical Losses:  Friction Loss  Windage Loss NATIONAL INSTITUTE OF SCIENCE &TECHNOLOGY PALUR HILLS, BERHAMPUR, ORISSA – 761008, INDIA

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BASIC ELECTRICAL ENGINEERING –BE2102 Mr. Mitu Baral, ECE,Dept. NIST, Berhampur h) A 3-Phase, 50 Hz, 415V, six pole induction motor runs at 960rpm. What is the slip speed and slip? Answer: Given that, Ns  P=6, f-50Hz, Vs=415V, Nr=960rpm. 120 f 120  50   1000 rpm P 6 Slip speed = N s  Nr  1000  960  40 rpm i) What is noise and write various sources of noise? Answer: (i) Any undesirable signal interfering with measurement is known as noise. (ii) The various sources of noise in measurement are the electromagnetic fields caused by fluorescent light fixtures, video monitors, power supplies, switching circuits and high voltage or current circuits. j) What is a transducer? Answer (i) A transducer is a physical device which measures physical quantity such as forces, stresses, temperature, pressure etc. (ii) A transducer converts the change of a physical quantity (eg: humidity, pressure) to a corresponding change in an electrical quantity (eg: voltage or current). 2. (a) In an AC single phase circuit three impedances of value 530 Ω, 1060 Ω and 4-j8 Ω are connected in series with a 230 V, 50 Hz supply. i. Find the total combined impedance in rectangular form? ii. Magnitude of the current flowing in the circuit? 5 Marks Solution: Given that: Z1  530  4.33  j 2.5, Z 2  1060  5  j8.66, Z 3  4  j8 i. The total combined impedance in rectangular form Z  Z1  Z 2  Z3 Z  4.33  j 2.5  5  j8.66  4  j8  13.33  j 3.16 ii. Magnitude of the current flowing in the circuit Z  13.33  j3.16  13.7 NATIONAL INSTITUTE OF SCIENCE &TECHNOLOGY PALUR HILLS, BERHAMPUR, ORISSA – 761008, INDIA

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Mr. Mitu Baral, ECE,Dept. NIST, Berhampur BASIC ELECTRICAL ENGINEERING –BE2102 I  I V Z  230 13.33  j 3.16 230  16.79 Amps 13.7 2. (b) Using super position theorem, find the voltage across 10Ω, resistor as shown in figure 2.b.1. 5 Marks Answer: Consider 15V Source acts alone: (Means 12V source short circuited ) V10  15volts . Consider 12V Source acts alone: (Means 15V source short circuited ) V10  0volts Because, no current will flow through the 10ohm resistance. So, voltage drop across 10 ohms resistance is 0 volts. From Super position theorem: V10Ω =V10Ω when 15V source acts alone + V10Ω when12V source acts alone V10Ω=( 15)+(0) = 15 V V10  15volts 3. (a) An iron ring with circular cross section of 4cm diameter and mean circumference of 100 cm is wound with a coil of 500 turns. For an exciting current of 3 amp in the coil, the flux is found to be 2mWb. Calculate the relative permeability of NATIONAL INSTITUTE OF SCIENCE &TECHNOLOGY PALUR HILLS, BERHAMPUR, ORISSA – 761008, INDIA

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