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Cornelius Pope
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1 UNIVERSITY OF PRETORIA / UNIVERSITEIT VAN PRETORIA Dept. of Electrical, Electronic & Computer Eng. / Dept. Elektriese, Elektroniese & Rekenaar-Ing. EIR NOVEMBER 2012 EXAM PART A1 / EKSAMEN DEEL A1 12 Questions, 12 Marks & 20 Minutes / 12 Vrae, 12 Punte & 20 Minute Initials & Surname / Voorletters & Van : Student No. / Studentenommer:.. Signature / Handtekening :.... Mark the best answer to every question on the answer sheet / Merk die beste antwoord vir elke vraag op die antwoordvel Marks awarded: Correct answer = 1, no answer or wrong answer = 0 Punttoedeling: Regte antwoord = 1, geen antwoord of verkeerde antwoord =0 HAND IN THE QUESTION PAPER AND THE ANSWER SHEET: ANSWER SHEET WILL NOT BE MARKED IF THERE IS NO QUESTION PAPER WITH IT HANDIG DIE VRAESTEL EN DIE ANTWOORDSTEL IN: ANTWOORDSTEL SAL NIE GEMERK WORD AS DAAR NIE 'N VRAESTEL SAAM MET DIE ANTWOORDSTEL IS NIE Question 1: Which is the right equation for calculating the energy stored in the magnetic field of an inductor when a steady current I is flowing through the inductor?: A E = ½P t 2 B E = ½P t 2 C E = ½L V 2 D E = ½C I 2 E E = ½L I 2 Vraag 1: Wat is die korrekte vergelyking vir die berekening van die energie gestoor in die magneetveld van n inductor wanneer n konstante stroom I deur die inductor vloei?: A E = ½P t 2 B E = ½P t 2 C E = ½L V 2 D E = ½C I 2 E E = ½L I 2 Question 2: Give the units of average electrical power drawn: A Joule B Watt C Watt hrs D Watt / hr E kw hrs Vraag 2: Gee die eenhede van gemiddelde elektriese drywing voorsien: A Joule B Watt C Watt uur D Watt / uur E kw uur Question 3: Give the units of a series R, L and C circuit impedance: A Henry B Farad C Ohm D Siemens E Farad Vraag 3: Gee die eenhede van n series R, L en C strombaan-impedansie: A Henry B Farad C Ohm D Siemens E Farad Question 4: Which is the equation for calculating the input torque of a separately excited rotating DC generator?: A T = Bli B E = Blu C T = P ω D T = kφi E T = kφω Vraag 4: Welke is die vergelyking vir die berekening van die inset-draaimoment van n apart-opgewekte GS generator?: A T = Bli B E = Blu 1 of 5

2 C D E T = P ω T = kφi T = kφω Question 5: For an ideal power transformer the number of windings on the primary side of the steel core, divided by the number of winding on the secondary of the steel core, is equivalent to: A V Primary / V Secondary B V Secondary / V Primary C I Primary / I Secondary D P Primary / P Secondary E Z Primary / Z Secondary Vraag 5: Vir n ideale krag- (drywings-) transformator is die aantal windinge op die primêkant van die staal kern, gedeel deur die aantal winding op die sêkondêre kant van die staal kern, is gelyk aan:: A V Primêr / V Sêkondêr B V SêKondêr / V Primêr C I Primêr / I Sêkondêr D P Primêr / P Sêkondêr E Z Primêr / Z Sêkondêr Question 6: For a balanced three phase load with the RMS Line voltage = V Line and the RMS Phase current = RMS Line current = I Phase, the total Active Power delivered to the load is: A P Total = 3 x V Phase x I Line x cosθ B P Total = 3 x V Phase x I Phase x cosθ C P Total = 3 x V Line x I Line x cosθ D P Total = 3 x V Line x I Phase x cosθ E All of the above Vraag 6: Vir n gebalanseerde drie-fase las met WGK Lynspanning = V Lyn en WGK Fasestroom = WGK Lynstroom = I Fase, is die totale Aktiewe Drywing gelewer die volgende: A P Total = 3 x V Phase x I Line x cosθ B P Total = 3 x V Phase x I Phase x cosθ C P Total = 3 x V Line x I Line x cosθ D P Total = 3 x V Line x I Phase x cosθ E Al bogenoemde Question 7: For power factor improvement (closer to unity) of a three phase squirrel cage induction motor, the following can be done : A Increase the speed of the induction motor B Put capacitors in parallel with the motor C Increase the current of the induction motor D Put inductors in series with the motor E None of the above Vraag 7: Vir arbeidsfaktor- (drywingsfaktor-) verbetering (nader aan een) vir n driefase kourotor-induksiemotor, die volgende kan gedoen word: A Verhoog die spoed van die induksiemotor B Verbind kapasitore in parallel met die motor C Verhoog die stroom van die induksiemotor D Verbind induktore in serie met die motor E Geen van bogenoemde Question 8: Why would you prefer a Series DC motor to a Squirrel Cage Induction motor?: A cheaper B smaller C more energy efficient D higher starting torque Vraag 8: Waarom sal jy n Serie GS motor verkies bo n Kourotor Induksie-motor?: A goedkoper B kleiner C meer energie-effektief D hoër aansit-draaimoment Question 9: Why would you prefer a Squirrel Cage Induction motor to a Shunt Wound DC motor?: A more reliable B easier speed control C operate from battery D higher starting torque 2 of 5

3 Vraag 9: Waarom sal jy n Kourotor Induksie-motor verkies bo n Apart-opgewekte GS motor?:?: A meer betroubaar B makliker spoedbeheer C werk vanaf battery D hoër aansit-draaimoment Question 10: ESKOM s electricity problem (which results in load-shedding ), is a result of insufficient?: A water B coal C nuclear fuel D ozone E generating capacity Vraag 10: ESKOM se elektrisiteits probleme (wat las-skedulering veroorsaak) is n gevolg van onvoldoende?: A water B steenkool C kernbrandsof D osoon E generator-kapasiteit Question 11: Which of the followingis the most cost effective and environmentally friendly source of renewable energy for conversion into electrical energy for South Africa?: A coal B nuclear C hydro D solar E wind Vraag 11: Welke van die volgende is die mees koste-effektiewe en omgewingsvriendelike bron van hernubare energie vir die omvorming na elektriese energie in Suid-Afrika?: A steenkool B kernkrag C hidro D sonkrag E windkrag Question 12: In a squirrel cage induction motor the field winding is: A separately excited B in series with the armature winding C the squirrel cage rotor D on the stator Vraag 12: In n kourotor induksie-motor is die veldwinding: A apart-opgewek B in serie met die anker C die kourotor D op die stator 3 of 5

4 EXAM PART A2 / EKSAMEN DEEL A2 3 Questions & 28 Marks / 3 Vrae & 28 Punte Complete all the questions on this questionnaire / Beantwoord al die vrae op hierdie vraestel Final score / Finale punt Initials & Surname / Voorletters & Van :.... Student No. / Studentenommer:.. Signature / Handtekening :.... Answer all questions on the question paper / Beantwoord al die vrae op die vraestel Question 1: DC machines / Vraag 1: GS masjiene a. For a linear DC motor complete the two equations / Vir n GS lineêre motor voltooi die volgende twee vergelykings: 1 1 f = e = b. Draw a labelled equivalent circuit diagram of the above linear machine / Teken n betitelde ekwivalente stroombaan-diagram vir bogenoemde lineêre masjien 3 [5] Question 2: For an RL circuit the following is known: R = 20Ω and L = 500mH. The components are connected in series with a DC current source of 5A through a switch that has been closed for a long time. At t = 0 the current source is short circuited by another switch across the source (without any other changes to the circuit). Calculate a.) the inductor current and b.) the voltage at t = 25ms. Vir n RL stroombaan is die volgende bekend: R = 20Ω en L = 500mH. Die komponente is in serie verbind met n GS stroombron van 5A deur n skakelaar wat vir n lang tyd gesluit is. By t = 0 word die stroombron gekortsluit deur n ander skakelaar oor die stroombron (sonder enige ander wysigings aan die stroombaan). Bereken a.) die induktorstroom en b.) die induktorspanning by t = 25ms. a. b. 5 2 [7] 4 of 5

5 Question 3: For a 132kV transmission line consisting of four conductors in parallel (each conductor with a resistance of 0.11Ω/km & current ampacity of 750A) for each phase is the inductive reactance of the line 0.147Ω/km and the capacitive reactance is Ω km. The line is km long and the load voltage is kept at 132kV (line voltage). The load at a specific time is Q Total = 41.49MVAr (capacitive) at a power factor of 0. Vir n 132kV transmissielyn wat bestaan uit vier geleiers in parallel (elk met weerstand van 0.11 & stroomvermoë van 750A per geleier) vir elke fase is die iduktiewe reaktansie van die lyn 0.147Ω/km en die kapasitiewe reaktansie Ω km. Die lyn is km lank en die lasspanning is 132kV (lynspanning). Die las op n spesifieke oomblik is Q Total = 41.49MVAr (kapasitief) by n arbeidsfaktor van 0. a. Calculate the load current at 41.49MVAr and PF = 0. Will the line be capable of carrying this current?/ Bereken die lasstroom by 41.49MVAr en AF = 0. Sal die lyn die stroom kan dra? 2 b. Draw the per phase equivalent circuit for the transmission line and load and add the quantities for the passive components /Teken die per fase ekwivalente stroombaan van die transmissielyn en las en voeg die groothede van die passiewe komponente by. 4 c. Calculate the supply voltage (line voltage) / Bereken die toevoerspanning (lynspanning). 5 d. Plot the per phase phasor diagrams for voltage and current to confirm Kirchhoff s laws. / Plot die per fase fasordiagramme van spanning en stroom om Kirchhoff se wette te bevestig. 5 a. b. 2 c. 4 5 d. 5 [16] 5 of 5

Examination Copyright reserved. Eksamen Kopiereg voorbehou. Module EBN122 Elektrisiteit en Elektronika 13 November 2009

Examination Copyright reserved. Eksamen Kopiereg voorbehou. Module EBN122 Elektrisiteit en Elektronika 13 November 2009 Departement Elektriese, Elektroniese en Rekenaar-Ingenieurswese Eksamen Kopiereg voorbehou Module EBN Elektrisiteit en Elektronika 3 November 009 Department of Electrical, Electronic and Computer Engineering