Industrial Electronics 1 Exam for Electrical Engineering Students, Exams of Industrial Engineering

Download Industrial Electronics 1 Exam for Electrical Engineering Students and more Exams Industrial Engineering in PDF only on Docsity! 1 CORK INSTITUTE OF TECHNOLOGY INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ Semester 1 Examinations 2012/13 Module Title: Industrial Electronics 1 Module Code: ELEC 6024 School: Mechanical, Electrical & Process Engineering Programme Title: Bachelor of Engineering in Electrical Engineering – Year 1 Bachelor of Engineering (Hons) in Electrical Power Systems – Year 1 Programme Code: EELEC_7_Y1 EEPSY_8_Y1 External Examiner(s): Ms Mary Desmond, Mr Colm Murray Internal Examiner(s): Dr Pascal O’Connor Instructions: Answer THREE questions. All formula for calculations must be shown. Duration: 2 Hours Sitting: Winter 2012/13 Requirements for this examination: Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have received the correct examination paper. If in doubt please contact an Invigilator. 2 Q1. (a) Name and explain in detail how the circuit in Fig.Q1(a) operates. Voltage Probes (1, 2 and 3) are shown at specific positions on the circuit. Draw exact waveforms for each probe position. (Input = ±40V sinewave at 50Hz, the TR ratio is 4:1) (10 Marks) (b) Now modify the circuit in Fig.Q1(a), using only two diodes, which must give the same Probe waveforms as in part (a). (4 Marks) (c) When a capacitor (C=20µF) is placed in parallel with R1, what effect if any, does this have on either circuit? (6 Marks) (d) With regard to the three circuits in Fig.Q1(d), sketch the output waveforms using suitable scales if the input voltage in each case is ±5V sinewave at 50Hz. (9 Marks) (e) For the circuit shown in Fig. Q1(e), determine the output waveform if the input signal is a ±5V sinewave at 50Hz. (4 Marks) Q2. (a) Explain, using a circuit diagram your understanding of a voltage divider circuit with 2 resistors (i.e. R1 and R2) and derive its equation. (9 Marks) (b) Assuming a source voltage VS = 15V, R1 = 3k and R2 = 12k, calculate the voltage drop across each resistor (VR1 and VR2) and the circuit current. (6 Marks) (c) Three load resistors, RL = 60k, 12k and 3k are now connected in parallel across the output resistor R2 of the voltage divider, one at a time. Analyse the loading consequences for the voltage divider by calculating the following values for each load and completing the table below. RL VR1 VR2 VRL IR1 IR2 IRL 60k 12k 3k No RL Comments/remarks on the completed table are required. (18 Marks) 5 Fig.Q1(d)(ii) Fig.Q1(d)(iii) V2 VSINE VA=5 FREQ=50 R2 1k D2 DIODE Input-2 Output-2 D3 DIODE-ZEN BV=2V V3 VSINE VA=5 FREQ=50 D4 1N4372A BV=4V Input-3 Output-3 R3 1k R4 1k 6 Fig.Q1(e) Fig.Q3(a)(i) Fig.Q3(a)(ii) Input-1 V1 VSINE VA=5 FREQ=50 Output-1 R1 3k R2 2k R1 1.5k Probe-1 Z1 1N4734A BV=3.5V Z2 1N4734A BV=2.8V Z3 1N4734A BV=5.2V D1 1N914 B1 30V Probe-2 Probe-4 I- c u rr e n t Probe-3 R2 2.2k Probe-1 Z4 1N4734A BV=4.5V Z5 1N4734A BV=2.5V Z6 1N4734A BV=4.2V D2 1N914 B2 30V Probe-2 Probe-4 I- c u rr e n t Probe-3 7 Fig.Q3(b) Fig.Q3(c) R1 250 Z1 1N4734A BV=7V B1 12V R2 1k R1 280 B1 18V R2 450 D1 LED-RED VF=2V