ASSOCIATE DEGREE IN ENGINEERING RESIT EXAMINATIONS SEMESTER 1. "Electrical Eng Science"


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1 ASSOCIATE DEGREE IN ENGINEERING RESIT EXAMINATIONS SEMESTER 1 COURSE NAME: "Electrical Eng Science" CODE: GROUP: "[ADET 2]" DATE: December 2010 TIME: DURATION: 9:00 am "Two hours" INSTRUCTIONS: 1. This paper consists of SIX questions. 2. Candidates must attempt ANY FOUR questions on this paper. 3. All working MUST be CLEARLY shown. 4. Keep all parts of the same question together. 5. The use of nonprogrammable calculators is permitted. DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO 1
2 Instructions: Answer any FOUR (4) questions. Question 1 a. In the circuit below calculate the following: i. The total resistance in the circuit [4 marks] ii. The total current [1 mark] iii. The current through R 2 [2 marks] iv. The power dissipated through R 3 [2 marks] R1 2.2k R2 6.2k V 38 V R3 1.0k R4 2.0k R5 510 b. Using superposition theorem, find the current through R 3 R1 2.0 R2 10 R3 3.0 V1 15 V V2 20 V [6 marks] c. If the value of R 2 in part(b) is changed to 13Ω find the current that would flow through by applying Kirchhoff s voltage law. [6 marks] d. How is a mesh difference from a loop in circuit analysis? [2 marks] e. A coil of copper wire has a resistance of 200Ω, at temperature coefficient of when its temperature is 0 o C. Calculate the resistance of the coil when its mean temperature is 80 o C. [2 marks] 2
3 Question 2 (a) (i) Define the following: Farad, electric flux density, relative permittivity [ 6 marks ] (ii) Three capacitors are connected to a 15V dc supply. Calculate the total capacitance, capacitor voltages, and total energy stored when the capacitors are connected in (a) series, (b) in parallel. The capacitor values are: C1 = 0.1 F, C2 = 0.3F, and C 3 = 0.5 F [ 6 marks ] (b) A 600 ml! inductor is connected to a 25V supply via a l5 K resistor and a switch. Calculate the inductor voltage 40 s. after the supply is switched on. Also, determine a suitable value for a resistor connected directly in parallel with the inductor to limit the counter emf to 50V [ 7 marks ] (c ) V 2 20 kω 1.0 µf (i) For the circuit shown above the capacitor is initially uncharged and the switch is thrown to position 1. Determine 1. the time constant, [2 marks] 2. the current after 40 ms, [2 marks] 3. the time taken for the capacitor to be fully charged. [2 marks] 3
4 Question 3 Question 4 a) What is meant by the term resonance of a series RLC circuit? b) Briefly draw the relationship of the, versus the frequency of a series RLC circuit. c) If most appliances in Jamaica are designed to operate from a 110V, 50Hz supply; Determine:  i. The peaktopeak voltage ii. The effective voltage iii. The average voltage iv. The supply frequency v. The periodic time vi. The equation that describes the voltage, with an assumed phase angle of [25 Marks] a. Define the following terms: i. Permeability ii. Magnetic field strength iii. Magnetic flux density b. [3 marks] i. Determine the inductance of a coil 0.01m long, 4 turns, crosssectional area of 0.1m 2 with a permeability of the core material is 0.25x103 H/m. [4 marks] ii. If a current of 3A flow for 5ms what is the induce voltage. [2 marks] c. The air gap in a certain machine is 0.5m 2 in crosssection and is 5mm long. The magnetizing force is provided by a coil of 1000 turns. Calculate the current which must flow in the coil to produce a total flux of 0.25Wb in the gap. [7 marks] d. Calculate the flux density produced by a magnetizing force of 2000 At/m in a steel of relative permeability 400. [4 marks] 4
5 Question 5 a) Give the name of the instrument shown in the figure above b) Label and explain the principle of operation of the instrument in figure. 2 e b a a c d [ 8 Mks] c) An ammeter has an internal resistance of Ώ and take a full scale deflecting current of 30 μ A. Determine: (i) (ii) (iii) The resistance necessary to convert this instrument to measure 10 A dc The resistance necessary to convert this instrument to measure 250V dc Draw the circuit diagram for each case above. [17 Mks] 5
6 Question 6 a. Describe the difference between ntype and ptype materials. (4mks) b. Show the input and output waveforms of a PN diode halfwave and fullwave rectifier in figure a and b respectively. [5marks] [a] [b] c. Draw and explain how the bridge rectifier works. d. Explain what is meant by the terms Mutual and self inductance. [10 marks] ( 6 mks) 6
7 ASSOCIATE DEGREE IN ENGINEERING SOLUTIONS SEMESTER DECEMBER COURSE NAME: CODE: [COURSE NAME] [8 CHARACTER COURSE CODE] GROUP: "[ADENG 1 OR 2]" DATE: TIME: DURATION: "[EXAM DATE]" "[TIME OF PAPER]" "[LENGTH OF PAPER]" Question 1 a. i. R p1 = = R s1= R 2 +R p1 =6.2k =6606.4Ω R p2 = = R T =R 1+ R p2 = 2200Ω Ω = Ω ii A iii. V 3 =V T V 1 =38V(2.2kΩ x 0.012A )=11.6V I 2 =I T I 3 =0.012A =0.012A =0.012A A=0.0004A iv. =0.135W b. with V 1 alone in the circuit R T = =4.3 7
8 =3.49A =0.802A with V 1 alone in the circuit R T = =4.3 =3.49A =0.802A with V 2 alone in the circuit R T = =4.67 =4.28A =0.713A I 3 =I 21 +I 22 =0.802A A=1.515A c. Loop1 V 1 =R 1 I 1 +R 2 (I 1 + I 2 ) =2I 1 +13I I 2 15V =15I I 2 Loop2 V 2 =R 3 I 2 +R 2 (I 1 + I 2 ) 20V=3I 2 +13I I 2 20V =13I I 2 Multiply 20V =13I I 2 x 15 15V =15I I 2 x V =195I I 2 15V =15I I 2 Subtracting give 105V=71I 2 =1.48A 20V =13I I 2 20V =13I x V=13I V I 1 = =0.28A I R2 =I 1 +I 2 =1.48A0.28A=1.198A d. A loop is a closed path in a circuit in which no element or node is encountered more than once. A mesh is a loop that contains no other loop within it. e. R 1 =R o (1+α o θ)=200( x80)=268.5ω 8
9 Question 4 a. i. Permeability is the measure of the ability of a material to support the formation of a magnetic field within itself. ii. Magnetic field strength, also called magnetic intensity, is an expression of the force that a magnetic field exerts on a theoretical unit magnetic pole in free space. iii. The amount of magnetic flux through a unit area taken perpendicular to the direction of the magnetic flux. b. i. c. ii. E= d. 9
10 Question 5 a) Give the name of the instrument shown in figure  below The instrument is a D Arsonval /Moving coil Meter b) Lable and explain the principle of operation of the instrument in figure. 2 c) An ammeter has an internal resistance of Ώ and take a full scale deflecting current of 30 μ A. (i) The resistance necessary to convert this instrument to measure 10 A dc Since the meter can only carry 30 μ A and we require the circuit to measure 10 A the we must connect a shunt resistance to carry 10A  30 μ A. which is A as shown in figure 6.4 below 10A I shunt = A meter 30 μ A Ώ A Figure 6.4 R shunt 10
11 By current divider rule I shunt R me te r I T R me te r R shunt which implies that ( 0.025) ( 10) R shunt from which R shunt x10 8 or R I fsd R me te r I shunt 30x 10 6 ( 0.025) x10 8 (ii) The resistance necessary to convert this instrument to measure 250V dc Since the meter can only carry 30 μ A and we require the max current at 250 V is 30 μ A. then we must connect a resistance in series with the Ώ of the meter such that I = V/ (R add +R meter ) = 30 μ A as shown in figure 6.5 below V meter R add 30 μ A Ώ V d = V V m = 75 μ V 250 V Figure 6.5 I fsd V R add R meter R add V I fsd R meter I fsd x 10 6 x x0 6 R add 8.33x M 11
12 (Question 6) a. An ntype semiconductor material has an excess of electrons for conduction established by doping an intrinsic material with donor atoms having more valence electrons then needed to establish the covalent bonding. The majority carrier is the electron while the minority carrier is the hole. A ptype semiconductor material is formed by doping an intrinsic material with acceptor atoms having an insufficient number of electrons in the valence shell to complete the covalent bonding thereby creating a hole in the covalent structure. The majority carrier is the hole while the minority carrier is the electron. (b) Half wave rectification input waveform output waveform Full wave rectification [2] input waveform output waveform (c) In the diagrams below, when the input connected to the left corner of the diamond is positive, and the input connected to the right corner is negative, current flows from the upper supply terminal to the right along the red (positive) path to the output, and returns to the lower supply terminal via the blue (negative) path. 12
13 When the input connected to the left corner is negative, and the input connected to the right corner is positive, current flows from the lower supply terminal to the right along the red (positive) path to the output, and returns to the upper supply terminal via the blue (negative) path. b. In each case, the upper right output remains positive and lower right output negative. 13
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