« Explain the following electrical faults: © Open circuit; * Short circuit. * State five measures that can keep an electric blender in good working condition. . Observation (a) (i) Open circuit: This fault occurs as a result of failure of one or more conductors in an electrical circuit leading to very high resistance value and zero current. (ii) Short circuit: This fault occurs when contact is made between two wires of different polarity (live and neutral) thereby leading to the flow of excessive current in the circuit. (b) Measures that can keep electric blender in good working condition * Regular clean up with the use of vacuum cleaners to get rid of dirts. * Periodic lubrication of electric motors that are not sealed. ¢ Appliance should be kept away from sources of heat (oven, fire, etc.) * Ensure compliance with manufacturer's instructions. * Always start the blender from low to high variable speed. * Do not submerge the motorized blender base in water. « Rinse jar thoroughly with warm water after use and allow to dry before storage. * Unplug the blender when cleaning and when not in use. * Avoid overloading the jar with content(s) to be blended. The few candidates who attempted this question did poorly. =BOOPPPAORSA 26 waeconline.org.ng/e-le + Weakness/Remedies Candidate's Strength Question 6 * (a) State three advantages of a hydroelectric power station over thermal power station. * (b)Explain how a gas turbine is used to generate electricity. Observation The expected answers were; (a) Advantages of hydroelectric over thermal station: i)It incurs no losses when idle. i)It can start and take up load quickly (iii)Low running cost (iv)Low maintenance cost (b) The mixture of air and fuel is ignited in the combustion chamber. The resulting hot gases in the confined chamber develop a high pressure. The high pressure is directed against the turbine blades to turn the turbine which in turn, turns the generator or alternator through the shaft. Many candidates who attempted this question could not do it well. Es k=Peete e an Teer eran Ceres Pes Subject Home a Ps 3 cs General Comments Weakness/Remedies Candidate's Strength Question 5 * (a)State three advantages of a ring distribution system of electrical supply. * (b)State the two operating voltage levels for a transmission line Observation The expected answers were; 1. (a) Advantages of a ring distribution system of electrical supply (i) | Canuse smaller supply cables than a radial distribution system. (ii) Offers greater security of supply than a radial distribution. (iii) more reliable (iv) more efficient (b) Operating Voltages (i) 132 kV (ii) 330kV Candidates did not perform well in this question. eo osQuestion 4 A shunt motor has the following features: Back e.m.f (Eb) = 220V; Field resistance (rf) = 150W; Field current (If) = 1.54; Supply current (I) = 31.5A. Calculate the: (i) terminal voltage; (ii) armature resistance; (ii) power output at the motor shaft. Observation The expected answers were; (i) Terminal voltage (Vt) = rfif = 1.5 x 150 = 225V (i) Armature Resistance: lara +Eb = Vt Ih= I - Ie 31.5 - 1.5 225 —220R=5a H=628n X= 63-650 240V7soHz Fig. 1 Figure 1 is a series RLC circuit. Calculate the: (i) impedance of the circuit; (ii) total current; (iii) phase angle. Observation The expected answers were; (0) Impedance: Z=5+(-0.857 Z? =25+0.7225 Z= 425+ 0.7225oo BETO) Menu General Comments Question 3 (a)Define a fuse. (b)List two types of fuse. + (c)State three reasons for earthing in an electrical installation. Observation The expected answers were; 1. (a) A fuse is a protective device for opening a circuit by means of a conductor designed to melt when an excessive current flows through it. (b) (i) Rewirable fuse (ii) Cartridge fuse (iii) High breaking capacity (HBC) fuse (iv)High rupturing capacity (HRC) fuse (c)(i) To protect equipment / appliances. (ii) To prevent electric shock. (iii)To prevent fire outbreak. Copyright © 2018. The West African Examinations Council. All rights reserved.Question 1 1. Sketch and label the following: (a)a bar magnet; (b) an electromagnet. 2 State two differences each between a bar magnet and an electromagnet Observation The expected answers were; (a) @ (b)(i) A Bar magnet is a material which attracts ferrous metals (ii)Electromagnet needs external supply to function while Bar magnet does not. (iii)Bar magnet has permanent magnetism while that of electromagnet depends on the current in the coil. (iv)The polarities of bar magnet are permanent while those of electromagnet can be changed. Copyright © 2018. The West African Examinations Council. Il tights reserved.Question 7 (a) Explain the quantities of lamp rated 60W, 240V. (b) (i) Sketch the circuit of a 12V,12W parallel lamps connected in series with a battery of e.m.f 12V. (ii) Calculate the value of the rated current in each 12W lamp. Observation The expected answers were; (a) A lamp rated 60W, 240V is one that dissipates or consumes a maximum power of 60W when connected to an a.c voltage flow of 240V. (b) a eacrcay = | z. T | 12V (c) The candidates were expected to explain lamp ratings, sketch parallel connection of lamps in series with R across a battery and to calculate the rated current in each lamp. The candidate's performance was reported to be averagely good.Question 6 (a) List four parts of a d.c. machine. (b) State the function of any three components mentioned in 6(a) Observation The expected answers were; (a) Substation components (i) Transformer (ii) Isolators (iii) Fuses (iv) Circuit breaker (v) Feeder (vi) Contactor (b) Explanation of any three listed above (i) Transformer; it is used to step up or step down voltage. (ii) Fuse; it is used to break circuit on short circuit and overload. (iii) Isolator; it is used to isolate the transformer from the system in case of a fault. (iv) Feeder; feed the input and the output voltage of the transformer. (v) Contactor; it is used for monitoring of electrical qualities i.e V or | in the system. The candidates were expected to list four substation components and state the function of three of the components. It was reported that many candidates who attempted this question could not do well.BOOPPPAORSA 26 waeconline.org.ng/e-le + Candidate's Strength Question 5 (a) Define a generator. (b) List the three parts of a de machine. (c) State ; (i) the part of a generator that enables it to produce a d.c. output; (ii) three methods of connecting field winding in a d.c. machine. Observation The expected answers were; (a) A generator is a machine that converts mechanical energy into electrical energy. (b) The three parts of d.c. machine; (i) Field windings (ii) Armature (iii) Commutator and its associated brush gear (c) (i) Commutator; (ii) (a) Series (b) Shunt (c) Compound (d) Separately excited The candidates were expected to define a generator, list three parts of a d.c. machine, state the part of a generator which enables d.c. output and state three methods of field-winding connection. Candidates were reported to have performed very good in this question.BOOPPPAORSA 26 waeconline.org.ng/e-le + Question 4 (a) Define coulomb. (b) State the units of the following quantities; (i) Resistance; (ii) Potential difference. (c) A de motor supplied from 230V supply for 1hour consumes an energy of 36mJ. Calculate; (i) Power rating of the motor; (ii) Current taken from the supply. Observation The expected answers were; (a) Coulomb (Q) =Ixt Is the quantity of a steady current of one ampere in one second. (b) {Quantity Unit Resistance Ohm () Potential difference Volt (V) (c) (i) Power rating of the motor = (Electrical energy)/time = (36x (10) *(-3))/(1x60x60) =0.01x {10} C3) OR =0.01mwW (ii) Current taken from the supply = Energy/VI = (36x (10) 4(-3))/(230x60x60) =0.04348x [10) *6 =0.043/A OR I=P/V =(1x (10] *(-3))/230 =4.3x (10) 4¢-3) A =0.043LA The candidates were to define Coulomb, state unit of measurement of resistance and potential difference. They were to calculate the power rating and current taken from supply by an excited de motor. The performance of the candidates was reported to have been very good in this question.DOOPPPAOPS ae ri 26 waeconline.org.ng/e-le + General Comments Weakness/Remedies | Candidate's Strength Question 3 A balanced three-phase delta connected system has a line voltage of 415V and a line current of 20A. If the total load on the system is 6000W, calculate the; (i) Load power factor; (ii) Current in each phase; (iii) Power delivered by each phase. Observation The expected answers were; (i) The load power factor; Power (P) = V3 VLIL COS@ Pf=P/(v3 EL IL) = 6000/(1.7x415x20) = 0.418 (ii) The current in each phase; |_P= |_L/V3 = 20/1.73 = 11.6A (iii) Power per phase; = (Total Power)/3 = 6000/3 =2000W OR P= V_L |_Ph COS@ =415x11.6 x0.412 =2012.3W The candidates were to calculate load power factor, current in each phase and power per phase. Few candidates were reported to have attempted this question. Use of incorrect formula was responsible for their weakness. v3 was found missing in the fomula the candidates used in the calculation.figure 2 In figure 2 calculate the; (i) Equivalent capacitor; (ii) Total energy stored in the circuit; (iii) Charge in the 6uF capacitor; (iv) Potential difference across the 6uF capacitor. Observation The expected answers were; (i) Equivalent capacitor (Ceq) = (6x12)/(6+12 ) UF = 72/18 = 4x (10) *(-6) F = 4uF (ii) Total energy stored = [CV] *2/2=(4x (10) *(-6 )x [30] *2)/2=1.8x [10] *(3) joules = 1.8 m joules (iii) Charge in the 6uF capacitor (Q) =CV Q = 4x 110] \(-6)x30 =1201C (iv) Potential difference across the 6,1F Capacitor V = Q/C = (120x [103 *(-6))/(6x [10] *-6)) = 20V Candidates were expected to calculate the equivalent capacitance of a series- parallel connection of capacitors and the total energy stored in the circuit, the charge and potential difference across the series (6uF) capacitor. The candidates were reported to have performed very good in this question. Copyright © 2018. The West African Examinationsfigure 1 (a) Sketch the magnetic field pattern of the current- carrying coil indicating its polarity. (b) State one rule for the determination of the polarity of the field in 1 (a).PPPAD 23 waeconline.org.ng/e-le Question 7 (a) Define the capacitance of a capacitor. (b) () List three types of capacitor. (li) State two applications of a capacitor. (c) Write the equivalent capacitance of two capacitors connected in: (i) series; (ii) parallel. ‘The expected answers were: (a) Capacitance of a capacitor is the ability of a capacitor to store electric charges. OR Capacitance of a capacitor isthe ratio of the quantity of charge to the potential difference across the terminals of the capacitor, OR (C= where Q is the quantity of charge and V is the potential difference across the terminals of the capacitor. (b) (i) Any three types of capacitor Paper Ceramic Electrolytic Mica Air capacitor Polyester (i) Any two applications of eapacitor Power factar correction ‘Starting of single phase motor Blocking of direct current Decoupling ‘Stores charges ‘Smoothening Filtering Tuning cireuit ©W= Oe ‘The question was on Magnetic Field and Electromagnetism. The question Was not well attempted by candidates. The few candidates who attempted this question could not define capacitance, list types of capacitor and state applications of a capacitor.a | 23 waeconline.org.ng/e-le Question 7 (a) Define the capacitance of a capacitor. (b) () List three types of capacitor. (li) State two applications of a capacitor. (c) Write the equivalent capacitance of two capacitors connected in: (i) series; (ii) parallel. ‘The expected answers were: (a) Capacitance of a capacitor is the ability of a capacitor to store electric charges. OR Capacitance of a capacitor isthe ratio of the quantity of charge to the potential difference across the terminals of the capacitor, OR (C= where Q is the quantity of charge and V is the potential difference across the terminals of the capacitor. (b) (i) Any three types of capacitor Paper Ceramic Electrolytic Mica Air capacitor Polyester (i) Any two applications of eapacitor Power factar correction ‘Starting of single phase motor Blocking of direct current Decoupling ‘Stores charges ‘Smoothening Filtering Tuning cireuit ©W= Oe ‘The question was on Magnetic Field and Electromagnetism. The question Was not well attempted by candidates. The few candidates who attempted this question could not define capacitance, list types of capacitor and state applications of a capacitor.ae aon] 23 waeconline.org.ng/e-le edn Pop 3 284 Question 5 State: iny three precautions to be taken to avoid accident at an electrical installation work site; ‘+ two types of substation; + three differences between overhead and underground distribution systems. (a) Any three precautions Adhere to safety precautions Do not misuse or interfere with equipment provided for health and safety Dress appropriately Behave appropriately and with care Avoid over enthusiasm and foolishness Stay alert and avoid fatigue Do not use aleohol or drugs at work site ‘Work within your level of competence Attend safety courses and read safety literature ‘Take a positive decision to act and work safer Use the right/correct tool for each job (b) Any two substations listed below Main substations istribution substations ‘Outdoor substations. Indoor substations ‘Transmission substations Generator substations ©), ‘Overhead distribution system Underground distribution system Itis easy to repair Cannot be easily repaired Fault can be detected easily Fault cannot be easily detected Has lower installation cost Costly installation Jointing is simple Jointing is difficult Maintenance not costly Maintenance costly Air insulation Liquid and/or solid insulation ‘The question was on Electricity Transmission and Distribution. The question 15 not well attempted by candidates. The few candidates who ‘attempted this question did not do well.BOPPPERS i Question 5 State: ‘any three precautions to be taken to avoid accident at an electrical installation work site; + two types of substation; ‘three differences between overhead and underground distribution systems. [Saco (a) Any three precautions Adhere to cafoty procautione Do not misuse or interfere with equipment provided for health and safety Dress appropriately Behave appropriately and with care ‘Avoid over enthusiasm and foolishness Stay alert and avoid fat Do not use aleohol or drugs at work site Work within your level of competence Attend safety courses and read safety literature ‘Take a positive decision to act and work safer Use the right/correct tool for each job (b) Any two substations listed below ‘Main substations Distribution substations ‘Outdoor substations. Indoor substations ‘Transmission substations Generator substations ©) ‘Overhead distribution system Underground distribution system Itis easy to repair Cannot be easily repaired Fault can be detected easily Fault cannot be. Has lower installation cost Costly installation Jointing is simple Jointing is difficutt ‘Maintenance not costly Maintenance costly Air insulation Liquid and/or solid insulation ‘The question was on Electricity Transmission and Distribution. The question ‘was not well attempted by candidates. The few candidates who attempted this question did not do well.ae aan) 23 waeconline.org.ng/e-le | Question 4 1. List: () twoessential devices of an analogue electrical measuring instrument; (i)_ two basic types of scale found in the instrument in 4(a)(i). Fig. 1 is a schematic diagram of a moving-coil instrument. (b) Calculate the value of Rs. ‘The expected answers were: (2) i) Any two essential devices Deflecting (or operating) device Controlling device Damping device (i) Linear scale Non-linear scale (b)V= lara =IsRs Is=1-la=40- Re== 030 = 39.97 A Ro= 15,01 m@=15.01x 10-30 ‘The question was on Measuring Instruments. t was very popular with the candidates. It showed that many candidates that attempted thie did very well but missed part (a)ae aan) 23 waeconline.org.ng/e-le Question 3 + Explain briefly the basic principle of mutual induction in a transformer. ‘List two factors that determine the induced e.m.f. in a transformer winding. ‘= Calculate the volts per turn for a transformer with primary turns of 1,100 when supplied with an alternating input of 110 V. The expected answers were: (a) An alternating flux set up by the circult in the primary coll links the secondary Coil of the transformer thereby inducing an e.m4. in the secondary coil. The primary and secondary coils are electrically separated but magnetically coupled. The e.m{. induced can be expressed as e=-m where m = mutual inductance = flux from the primary coil inking the secondary coil tstime (b) Any two factors listed below ‘Number of turns in the winding ‘Alternating current in the winding or coil Alternating magnetic flux in the winding or coll Winding or coil inductance Alternating flux in the winding or coil ‘Change of current in the winding or coil ‘Change of flux in the winding or coll (©) Volts per turn = =0.1 Vfturn The question was on Transformers. It was unpopular with the candidates. The candidates that attempted this question did well only in parts (b) and (c). The candidates’ performance was fair.ae aan) 23 waeconline.org.ng/e-le = Question 2 (a) Define the following terms as applied to magnetic circul (i) magnetic field; (i) magnetic flux. (b) Aconductor of length 0.4 m moves with a velocity of 2 m/s at an angle to a magnetic field of flux density 0.9 T. If an e.m.f. of 0.5 V was induced in it, calculate the angle of movement of the conductor relative to the magnetic field. Soar The expected answers were: (a) () Magnetic field is a rogion of space where magnatic effectis experionced. (Accept any other reasonable definition) (i) Magnetic fuxis the total number of lines of flux threading through a magnetic field. (Accept any other reasonable definition) ()e= Blvsin volts sin = sin = = 0.6944 = 43.980 ‘The question was on Magnetic field and Electromagnetism. The question was attempted by many candidates who could not define the terms nor get the calculation right. The candidates’ performance was not encouraging.Ol =o) —— 4 ‘ep eset Paper, vibe. 2014 a } Question 1 + State Ohms law. + Calculate the resistance of a conductor that has a conductance of 20 pS. (©) An electric heater consumes 2.16 MJ when connected to a 240 V supply for 30 minutes. (i) power rating of the heater; (ii) current taken from the supply. The expected answers were: (0) Ohm's law states that the current |, flowing through a metallic conductor is directly proportional to the potential difference across its ends provided temperature and all other physical factors at kept constant. oR Valorlav V= ki where V=voltage, I= current and kis the constant of proportionality (b) Conductance, in Siemens, G = ORR= ing of heater, P = = 1200J/s=1.2kW The question was on Direct Current Circuit Theory. The question was popular with the candidates The performance was reperted good.Oa] 23 waeconline.org.ng/e-le Question 6 a. State two methods of connecting 13 A socket outlets. b. Sketch a wiring diagram of a two-way system controlling a lamp in OFF position ©. The expected answers were. @) Radial connection Ring connection (b) The question was on Electrical Wiring. This question was unpopular with the candidates. It v ‘eported that the performance of the very few that attempted it was very poor.Sa 23 waeconline.org.ng/e-le | Question 4 2. List two types of electrical measuring instrument. b. The coil of a moving-coll meter has a resistance of 10 0 and gives a full-scale deflection when a current of 20 mA passes through it. Calculate the resistance required to enable the Instrument to read up to: 1.104; i10v, The expected answers were: 1. Any.two ofthe instruments stated below i. Moving-coil i. Moving-iron Multimeter iv. Insulation Resistance Meter v. Ohmmeter vi, Watimeter Vil. Watt-hour meter 1.) 20x10-3x10= 9.98% Reh Reh = 0.0200 i) 20x 10-3(10+ R) R =500-10=4900 ‘The question was on Measuring Instruments. It was very popular with the candidates. Many candidates that attempted this question performed well.PPPHOPSE 6° 23 waeconline.org.ng/e-le ‘a, Draw the symbol used to represent: i. energy metre; li, lamp; iil. main switch; iv. fuse. b. State one function of each of the following in a substation: i. transmission transformer; li. distribution transformer; li, isolate iv. fuse. ‘The expected answers wer: @) (©) @)__A transmission transformer (always step-up) receives electric power from a nearby generating facility and uses a large power transformer to Increase the voltage for transmission to distant locations, Gi) A distribution transformer steps down transmission or sub- transmission voltage to distribution voltage. (lll) Isolator is used to open or close a circult to enable maintenance (iv) A fuse is used to protect circuit (transformer) and distributio tion was on Electrical Energy Supply. It was unpopular with the candidates. The repair work to be done, tes’ performance was reported to be poor.Ol a aoa) svonsione Question 2 mem a, State Faraday’ law of electromagnetism, . A conductor of active length 0.4m moving ata velocity of 10 m/s carries a current of 10 Aat right angles to.a magnetic field of strength 0.5 T. Caloulate the «. force exerted; dem. induced in it; ‘© power dissipated in watts ‘The expected answers were: 1.Faraday’s Law of electromagnetism states that whenever magnetic hx linked with a circuit ‘changes, an emf is always Induced In the cieult OR there is a relative motion between a conductor and a magnetic field, an emis induced in the ‘conductor. 1. 120.4; 10m/s; 12104; B=05T F=Bil=0.5x04x10=2N (W) induced emt, E = 81 0Sx04x 10-20 (ii) Power dissipated, P= BWV x 0.50.4 10x 10-20 “The question was on Electromagnetic Feld. The question was unpopular with the candidates. The ‘candidates were reported to show ahallow knowledge of tis topic O O‘A moving-coil instrument has a coil resistance of 50 0 and gives a full-scale deflection (fed) for a current of 250 mA. 1. Sketch the clreuit diagram, 2. Calculate the: 3. shunt resistance required for the meter to read an fsd of § A; 4. resistance required for the meter to read 25 V. ruts IO The question was on Measuring Instruments, The candidates who attempted this ‘question were reported to do better in part (b) than in part (a) Question 2 1. Sketch the phasor dlagram of the RL circuit shown in Fig. 7 1. Acurront flowing through a elrcult is represented by the expression = 10 sin 628t amperes ‘Calculate th 1am 8, value; 2. average value The expected answers were: fo @ vsorz Vlora VRorR 1 <9 50.707 x maximum value =0.707%10= 7.07 (W average value = 0.637 x maximum value 0.697 410 = 6.27 A The question was on Alternating Current Ciroult Theory. The question was ‘unpopular with the candidates, Many candidates who attempted the question could not draw the phasor diagram. They did not know that the cms. and average values should be computed from the peak value of the instantaneous 0 kms 1. Acurrent flowing through a elcult is represented by the expression 0 sin 628t Amperes ‘Caleulate the The expected answers were:ROR a aay a+] 26 waeconline.org.ng/e-le The expected answers wer (2) @__Faraday’s law of electromagnetic induction states that the magnitude of the induced e.m.f. is proportional to the rate at which the conductor cuts or is cut by the magnetic fux OR Faraday’ law states that the magnitude of the e.m/. induced in a circuit is proportional to the rate of change of magnetic flux through the circuit. (i) Electric bells, buzzers, telephone receivers, loudspeakers and relays. () — 2poles give 1 cycl 10 poles give 1x = 5 cycles (ii) T(revolutions per second) = = = 20 mS. 1 revolution ‘question was on Electromagnetic Fields/A.C. Machines. The question was well attempted by t! ndidates. However, the (b) part was not well done by many; attributed to level of understanding and exposureQDOPPHOPS waeconline.org.ng/e-le Question 4 fren The expected answers were: £ pesmi nmeensne mire 1 () Ammeter application Current measurement (i) Ohmmeter application Impedance measurement Resistance measurement Insulation resistance measurement The question was on Measuring Instruments. It was very popular with the candidates. Many candidates that attempted this question could not put complete/right answer in each cell. They ‘could not state the applications of ammeter and ohmmeter.PPHOPE 26 waeconline.org.ng/e-le + 1. State one difference between the voltages generated by a single-phase and a three-phase synchronous generator, ©). List four components of a high voltage overhead transmission system, (@) Slate one advantage each of overhead and underground transmission systems. The expected answers wer Single-phase synehonous generator ‘Thvee-phase synchronous generator Constant Varies 250 caries connection Between 400 and 230 V Any four of the items listed below Steel towers Glass/porcelain insulators Aluminium conductors High voltage transformers ‘Overhead advantage ‘Underground advantage Low invactmentcoste Long fe expectancy ‘Les expensive maintenance Reduced aintenarae osts Service uninterrsted by storms ‘Concave valuable and ‘Reduced fe-fighting Nazarde Provertion of arsenite The question was on Alternating Current Machines/Electrical Supply. candidates. The candidates that attempted the part (a) could not state th: generated by the two systems. They were not at home with the comp transmission system but rather stated overhead distribution sys [Prev ][ next oe eee) Saoirsea) eNO RCL PP) EECMaTIET-TeXo) a1 TNT KO ]K@ Rule) -o (a a pected answers were: 1. When a conductor (e.g. a wite) is moved through a magnetic field in such a way that it cuts the ‘magnetic lines of force (ie. magnetic flux), voltages induced in the conductor, An electric current will iow through the conductor if the two ends of the conductor are connected by a wire to form a circuit. Suppose the conductor has a length L motres and moves perpendicularly to the lines of flux in a uniform magnetic feld of B tesla ata velocity of V metre per sec. then, in ‘one second, the flux cut, E=Bivvolts wet ‘rmutual induetance) = 5.22 mH ‘The question was on Elactromagnati Field. The question was unpopular with tho candidates. The candidates were reported to show litle knowledge ofthis topic.-Aeplid Eleticity Paper 2, NowDec. 2011 Question 1 The expected answers were: 1. Impedance is the total opposition to the flow of an alternating current. The question was on Alternating Current Circuit Theory. It was quite popular with the can Many of the candidates who attempted the question got the definition right. However, the importance of resonance was not well understood due to faulty circuit <—Yy—> «<—y—> ==40 The question was on Direct Current Circuit Theory. It was well attempted by candidates. 1 candidates’ performance was reported good.i Question 5 (a) Define the following in terms of alternating current: (i) peak value; (ii) instantaneous value. (b) A sinusoidal signal generated from an a.c. mains is represented by the expression v=5sin 3741 volts. Calculate its: () amplitude; (i) em.s. value; (iil) peak-to-peak. The aipected enone Wer (2)(i) Peak value - This is the highest or maximum value attained by an alternating quantity over half nape {i) Instantaneous vale - This i the vahe of an aerating quantity a any instant of ine (owen Seinatat vos Recall fmsinwvt volts: () Amplitude yn) * 5 vols 707 xv (li) Peak-to-peak value = 2 x Amplitude (2x5) volts = 10 volts ‘The question was on Alternating Current Circuit Theory. The question was popular. The performance was average. Many of the candidates answered the recall question well. However, those who did not understand the23 waeconline.org.ng/e-le [= Question 4 (a) State one application of each of the following instruments: () ohmmeter, Li} watt-hour meter. (b) State the function of a choke in a fluorescent fitting. (©) The coil of a loudspeaker has 20 tums and a diameter of 15 mm. If the coil is placed in a magnetic field of flux density 0.8 T, calculate the force produced when a current of 0.1 A flows through the coll The expected answers were: (a)( Ohmmoter is used for measuring resistance / continuity test (ii) Watthour meter is used for measuring energy consumed (b) The function of a choke in a fluorescent fitting is to limit the current flowing In an a.c.elroult. () Active length of the coil, L = 20 Xx 15, 1000 = 0.4m LI (Newtons) 8x 0.943 x 0.1 = 0.07SN The question was on Measuring Instruments and Electromagnetic Induction. It was quite unpopular. The performance was poor. The recall question, part (a), was got by most of the candidates, while the part (b) was not attempted by many. Most of the candidates did not attempt part(c) and this was attributed to lack of exposure and experience as pointed out by the Chief Examiner.Ol Scie] waeconline.org.ng/e-le (@) A coil of 300 turns is wound uniformly over a wooden ring having a mean circumference of 900 mm and a uniform cross-sectional area of 400 mrrr. If the current through the coil of 6 A, calculate the: (i) magnetic field strength; (H) flux density; (Hi) total flux. [Take p1, of wood = 1; [lg = 47x10-7H/m] The expected answers were: Number of turns, n = 300 Mean length, 1= 900mm = 0.9m Cross-sectional area, A = 400mm7? = 400 x 10-8m? Current through coil, = 6A (@)The magnetic field strength H= NII = 300x 6/a = 2000A/m (b)The flux density, B = HottgH = AIX 10-7x 1x 2000 = 2.513mT/0.00251T/2.513 x 10-8T (©)Total flux, = BA = 2.513 X 10-3X 400 X 0-6 = 1.005yb The question was on Electromagnetic fields. The question was unpopular few candidates attempted this question with fair performance. the c .HOPPHYTRA [OR BELL Abst 26 waeconline.org.ng/e-le + ‘Sma Question 2 = (a) State two losses that occur in a d.c.machine. From fig.8, calculate the: (load current, I (armature current, Ja (ii)armature current, Ig (jii)generated e.m-f. ‘The expected answers were: 1. (a)(Any_two ofthe following () Copper loss (H) Frictional windage loss (ii) Brush contact loss (iv) Iron loss (b)Terminal voltage, V = 230V Power to the load, P = 1150 Armature resistance, R, ‘The question was on Direct Current Machine. Many of the candidates who att gave the losses correctly but was very difficult to do any calculation due to ince .‘The expected answers were: (0) Inductive reactance is dined as te opposton offered to the fow of curent nan a, erat y an inductor (i) Capacitive reactance is the opposition offered to the flow of current in an ac. circuit by @ ‘capacitor (ii) Impedance is defined as the total opposition offered to the flow of current in an a.. eireuit ‘The question was on Alternating Current Circuit Theory, It was quite popular with the candidates, ‘Many of the candidates who attempted the question got the definition right. However, the importance of resonance was not well understood due to faulty cireult diagram.8 e+) 23 waeconline.org.ng/e-le Question 6 1. @) Define trunking, (List three types of trunking. 1, State one purpose of polarity test in an instal 2. List two types of commonly used metal condui ‘The expected anewers were: (@)() Trunking is @ fabricated tobe lai .ctangular casing with a removable lid which allowe the conductors is a system in which a bunch of cables are placed in a rectanguler long run boxes with cover lids. (Types of trunking Tap-on trunking Floor trunking Busbar trunking Conerete trunking Plain-section trunking ‘Skirting trunking Bench trunking Compactmented trunking 1. Purpose is to ensure that: 2.All ole switches are connected to the live conductor. 3. Live conductor to live terminal of socket, neutral terminal and earth to earth terminal of socket outlet. 4. Fuses protection are elways on the live conductor. 1. Types of commonly used eondult Light gauge steel conduit Heavy gauge steel conduit ‘Aluminium conduit Copper conduit Flexible conduit ‘The question was on Electrical Wiring. The question was unpopular. The performance was faitly ‘good. Many candidates could not define trucking. Only a few listed trucking types and the purpose of Polarity test in an installation. Majority of the candidates could not list two types of commonly used ‘metal conduit.SS | opted lett Papr 2, vibes. 2008 Question 5 Explain the cause of each of the following power losses in a transformer: hysteresis loss; ‘copper loss. ‘State one application each of the following transformer: single-phase; three-phase. A transformer is connected to a 240 V,50 Fiz a.c. supply. If the primary winding has 1800 turns and the secondary winding provides an output of 12 V, calculate the number of turns in the secondary winding. ‘The expected answers were: () Cause of Hysteresis loss: This is the energy expended in causing the magnetisation in a core in the form of heat. (Cause of Copper loss: It is the heating of the conductors due to the fect that they posses resistance. (b)(V) Application of single-phase transformer Constant voltage application; impedance matching; protection; inverters; communication system, domestic appliances. (i) Application of three-» Power distribution system, 3-phase inverter communication system Power transmission Power distribution (Ep = NE Es Ns Ns = 1800x12 =90 tums 240 This question was on Alternating Current Machine. It was quite popular. The performance was satisfactory. A handful of the candidates attempted this question. However, not many of the candidates that ‘attempted this question could explain the cause of hysteresis and copper losses nor give any application. Almost all of them who attempted this got part(c), but report show that causes of losses in transformer was not understood by them probably they were not taught.