Intermediate 2 Electronic and Electrical Fundamentals Specimen Question Paper NATIONAL QUALIFICATIONS [C025/SQP068] Time: 2 hours 30 minutes

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1 [C05/SQP068] Intermediate Electronic and Electrical Fundamentals Specimen Question Paper Time: hours 0 minutes NATIONAL QUALIFICATIONS 00 marks are allocated to this paper. Attempt all questions in Section A (50 marks). Attempt any two questions from Section B (50 marks). [C05/SQP068]

2 Section A Attempt ALL questions in this Section (50 marks).. The system illustrated below in Figure Q is set up for a test with voltmeters and ammeters. In the condition shown, the transistor is fully saturated with a base current of 500 µa. (a) State the expected reading on voltmeter V. (b) Determine the expected readings on voltmeters V and V and ammeters A and A. 4 (5) V V A kω V h fe = Ω V 0V A Figure Q [C05/SQP068] Page two

3 . The circuit shown below in Figure Q represents a zener diode controlling the operation of a small cassette recorder from the cigar lighter socket on a car. The recorder requires 7. 5 V dc. The car battery voltage is nominally V, but when travelling at speed the alternator is charging and the terminal voltage could be as high as 4. 6 V. When stopped at traffic lights, the terminal voltage could be as low as V. Determine: (a) the maximum and minimum current; (b) the required power rating of the zener diode (assume load disconnected); (c) the required power rating of R S. (6) R S Ω CR = Cassette Recorder V S 7. 5 V CR Figure Q [C05/SQP068] Page three

4 . The electric motor shown below in Figure Q has an output power of 00 kw and operates at an efficiency of 80%. Determine: (a) (b) the electrical energy in joules delivered in 5 minutes; the electrical energy in kwh delivered in 7 hours. (5) P in Motor P out L O A D Figure Q [C05/SQP068] 4 Page four

5 4. An electronic security lock is shown in Figure Q4 below, with 5 volts providing a logic and 0 volts a logic 0. The lock opens on receiving a logic signal. It makes use of switches A, B, C and D. The lock opens only when switches A and C are pressed but should not open when switches B or D are pressed. 5V 0V A B C D Logic to open lock Z Figure Q4 (a) Describe the operation of the lock in terms of the switches A, B, C and D; (b) Determine the output Z in terms of A, B, C and D. (5) [C05/SQP068] 5 Page five

6 5. In the network shown below in Figure Q5, determine: (a) (b) (c) the total network resistance; the supply current; the potential difference across the 0 Ω resistor. (7) 40 Ω 0 Ω V 0 Ω 60 Ω 0 Ω 50 Ω Figure Q5 [C05/SQP068] 6 Page six

7 6. Access to a compound containing dangerous high voltage equipment can be obtained by a maintenance electrician under the following conditions: (i) the high voltage is off (Logic 0); (ii) a keyswitch on the control panel, a distance away, is turned off (Logic 0); (iii) a keyswitch on the gate of the compound is turned on (Logic ). Under all other conditions the gate cannot be physically opened. Given the following: A = High voltage switch B = Control panel switch C = Gate switch For this logic requirement: (a) draw up a truth table in terms of A, B, C and the conditions for entry Z; (b) derive a boolean expression from part (a); (c) design a combinational logic network that meets the requirement. (7) [C05/SQP068] 7 Page seven

8 7. Convert the following numbers. (a) Binary to decimal 0 (b) Hexademical to decimal F 6 (c) Binary to hexadecimal 0000 (5) [C05/SQP068] 8 Page eight

9 8. The system shown below in Figure Q8 controls the operation of a filament lamp via a light sensitive device termed a light dependent resistor (LDR). The resistance of the LDR is approximately 0 MΩ in total darkness and approximately 50 Ω in bright light. 00 kω kω 9V 6V Filament Lamp. kω T T LDR 680 Ω 0V Figure Q8 (a) State the function of the following: (i) (ii) resistor. kω; variable resistor 00 kω; (iii) transistor T and T. (b) Describe the operation of the circuit shown in Figure Q8. (7) [C05/SQP068] 9 Page nine

10 9. A sinusoidal voltage measured in an amplifier circuit is represented by the expression v = 5sin α m V. Determine: (a) (b) maximum value of the voltage; instantaneous value of the voltage, when α = radians. () [C05/SQP068] 0 Page ten

11 Section B Attempt any TWO questions in this Section (50 marks). Each question is worth 5 marks. 0. A temperature sensing circuit uses the switching transistor circuit, shown in Figure Q0, to give an audible warning when the temperature falls below the set level. The temperature sensor used in this circuit is a temperature dependent resistor, VR, whose resistance falls when the temperature falls. V V R VR Buzzer TR R 00 Ω 0V Figure Q0 (a) Calculate the resistance of the temperature dependent resistor at the instant the buzzer starts to sound. 4 (b) If the buzzer is rated at 6 V and has a resistance of 600 Ω: (i) calculate the value of the resistor R required to ensure the correct operation of the circuit; (ii) determine the power dissipated in the resistor R, assuming V ce = 0. 5V. (c) Explain the effect on the circuit operation if the 00 Ω resistor was replaced by a 47 Ω resistor. Justify your answer with calculations. 6 [C05/SQP068] Page eleven

12 0. (continued) (d) It was decided to dispense with the buzzer circuit and replace it with a remote alarm system. Due to the distance between the temperature sensing circuit and the remote alarm, it was necessary to amplify the signal using the circuit shown below in Figure Q0(d). RfR f 0kΩ 0V. kω ideal 6 5 V out V out kω V R VR Ω VV V R VR 00Ω Ω 0V Figure Q0(d) (i) Determine the gain of the amplifier and calculate the output voltage when the input voltage is 690 m V. (ii) Explain the purpose of the variable resistor connected to pins and 5. (iii) Explain the purpose of R f. 5 (5) [C05/SQP068] Page twelve

13 . An experiment was set up as illustrated in Figure Q. A conductor was placed at 90 to the magnetic field and the length of the conductor in the magnetic field was 0. m. When the current was allowed to flow, the conductor was deflected to the left. The magnetic field strength is 0. 5T. V R 00VV DC N A S Conductor in Magnetic Field l = 0. m B Figure Q (a) (b) (c) (d) (e) If the force required to pull the conductor back to its original position is N, determine the magnitude and direction of the current in the conductor. If the conductor resistance is 0. Ω and the power supply is set at 00 volts, calculate: (i) the value of the variable resistance V R when the conductor is deflected; (assume the resistance of the connecting wires is negligible) (ii) the energy used, in joules and kw h, by the variable resistor, V R, if the experiment lasts for 5 minutes. Describe, with the aid of a suitable (magnetic field) diagram, why the conductor was forced to the left. In order to reduce the energy wasted in the variable resistor, it was decided to reduce the current in the conductor to 0 A. Describe two modifications that could be carried out on the experimental set to achieve this and still leave the force at N. Calculate the power drawn from the supply in parts (b) and (d). 4, 6,, (5) [C05/SQP068] Page thirteen

14 . (a) A dual seven segment display, such as the one shown below in Figure Q(a), keeps changing between the following values, 7A 6 and C9 6. Determine the decimal and binary equivalents of these numbers., Figure Q(a) (b) Draw the BS Symbol and construct the truth tables for the following Boolean expressions. (i) Z = A.B.C (ii) Z = R + S + T,, (c) The circuit shown in Figure Q(c), is used to switch off an automated process if the correct materials are not available. Determine the Boolean expression for the circuit and complete the truth table., 4 A B C D Z Figure Q(c) [C05/SQP068] 4 Page fourteen

15 . (continued) (d) A circuit is shown in Figure Q(d). Mark on the diagram, provided on worksheet Q(d), the TTL logic chip number and the pin numbers that could be used to construct the circuit. Use datasheet Q(d) to assist your response. A B C D 6 (5) Z Figure Q(d) [END OF QUESTION PAPER] [C05/SQP068] 5 Page fifteen

16 [C05/SQP068] 6

17 [C05/SQP068] Intermediate Electronic and Electrical Fundamentals Datasheet Q(d) Time: hours 0 minutes NATIONAL QUALIFICATIONS [C05/SQP068] 9

18 [C05/SQP068] Intermediate Electronic and Electrical Fundamentals Worksheet Q(d) Time: hours 0 minutes NATIONAL QUALIFICATIONS Fill in these boxes and read what is printed below. Full name of centre Town First name and initials Surname Date of birth Day Month Year Candidate number Number of seat To be inserted in the front cover of the candidate s answer book and returned with it only by candidates who attempt Question. [C05/SQP068] 7

19 A B C D Z Figure Q(d) [C05/SQP068] 8 Page two

20 INTEGRATED-CIRCUIT DIAGRAMS 7400 SERIES 00 Quadruple input NAND gate 0 Quadruple input NAND gate with open collector output 0 Quadruple input NOR gate 0 Quadruple input NAND gate open collector inputs Hex inverter 05 Hex inverter open collector outputs 06 Hex inverter with high voltage open collector output 07 Hex driver with open collector output Quadruple input AND gate 09 Quad input AND gate open collector outputs 0 Triple input NAND gate Triple input AND gate Dual 4 input NAND gate Schmitt trigger 4 Hex Schmitt Trigger 5 Triple input AND gate open collector outputs 6 Hex Inverter with open collector output Dual 4 input NAND gate Dual 4 input AND gate Dual 4 input NAND gate open collector outputs 5 Dual 4 input NOR gate with strobe Quad input NAND buffer open collector outputs 7 Triple input NOR gate 8 Quad input NOR buffer Quad input OR gate Grateful acknowledgement is given to R.S. Components for permission to reproduce this sheet. [C05/SQP068] 0 Page two

21 [C05/SQP068] Intermediate Electronic and Electrical Fundamentals Specimen Marking Instructions NATIONAL QUALIFICATIONS [C05/SQP068]

22 Section A. (a) V = 0. 6 V to 0. 7V (b) V = = 0. 5V V = I e R e = = V A = I c = I b h fe = = 5 ma A = I b + I c = I e = = 5. 5mA (5). (a) Maximum Current = Vs max Vz/Rs = / = 5 ma Minimum Current = Vs min Vz/Rs = 7. 5/ = 06 ma (b) Power Rating Zener = =. 6 W (c) Power Rating Rs = (5 0 ) =. 56 W (6). (a) P in = P out /Efficiency = 00/0. 8 = 5 kw Electrical Energy delivered = =. 5MJ (b) Electrical Energy = 5 7 = 875 kw h (5) [C05/SQP068] Page two

23 4. (a) The lock will only open when switches A and C are pressed but will not open if B or D are pressed. (b) Z = (A.C).(B + D) (5) 5. (a) Total resistance is obtained from 00 Ω (0 Ω, 0 Ω and 50 Ω in series) in parallel with 00 Ω (40 Ω and 60 Ω in series), ie 50 Ω which is series with 0 Ω giving a total of 60 Ω. (b) Is = Vs/Rt = /60 = 0. A (c) Potential difference = 0. 0 = V (7) 6. (a) A B C Z (b) (c) Z = A. B. C A B Z C (7) [C05/SQP068] Page three

24 7. (a) 0 = + + = 0 (b) F 6 = = 4 0 (c) 0000 = A6 6 (5) 8. (a) (i) Limit the base current to transistor T. (ii) (iii) Transistor biasing allows the circuit to operate at different light levels. Transistor T: Switch Transistor T: Amplifier (b) Suitable description required. (7) 9. (a) maximum = 5 mv (b) v = 5 0 sin v = mv () [C05/SQP068] 4 Page four

25 Section B 0. (a) For the transistor to be on the voltage at the base, R R junction must be 0. 7 V therefore the voltage drop across R will be 0. 7 =. V Then R = V R /V R R =. / = 64 Ω or I = V/R = 0. 7/00 = 7mA so R = V R /I =. / = 64 Ω V R Buzzer TR 0V 4 (b) (i) The voltage drop across the buzzer, for correct operation, should be 6 V and therefore the voltage to be dropped across R = V s V buzzer V ce = = 5. 5V so R = V R /V buzzer R buzzer = 5. 5/6 600 = 550 Ω or I buzzer = V/R = 6/600 = 0 m A so R = V R /I = 5. 5/0 m A = 550 Ω (ii) P = I R = 0mA 0mA 550 Ω=55 mw or P = V /R = /550 = 55 mw (c) If the 00 Ω resistor is replaced by a 47 Ω resistor this will change the voltage at the base of the transistor to a value determined as follows, assuming that the temperature level remains constant. V R = R R + R Vs 0 4V = ( 47/66/ ) = This means that the transistor will be in the off condition for this temperature level and the buzzer will not sound. Therefore this has had the effect of requiring the temperature level to be at a lower value before the buzzer will sound. 6 [C05/SQP068] 5 Page five

26 0. (continued) (d) (i) The gain of the non-inventing amplifier can be found from A = + R f /R i = /. 0 = V out = A V in = 690 mv =. 8 V 5 (ii) (iii) The variable resistor connected between pins and 5 is to allow adjustment of the amplifier output voltage to ensure that for 0V in the output is 0V. It is for the adjustment of the offset null. R f is the feedback resistor and it provides this feedback by producing a small part of the output voltage at the inverting input. (5). (a) F = BLI so I = F/BL = / = 40 A Using Fleming s left hand rule, the direction of the current is from A to B or downwards on the diagram. 4 (b) (i) I = 40 A, Vs = 00 V R conductor = 0. Ω R t = V/I = 00/40 = 5 Ω R variable = R t R conductor = 5 0. = 4. 9 Ω (ii) Power in the variable resistor can be found from P = I R = = 7840 W W(joules) = P t = = J or MJ W(kWh) = P(kW) t(hours) = =. 87 kwh, (c) The lines of magnetic force on the left hand side of the conductor tend to cancel out and weaken the field, while the lines of force on the right hand side are all in the same direction strengthening the field and hence the force on the conductor is to the left. movement strengthened field weakened field 6 [C05/SQP068] 6 Page six

27 . (continued) (d) Using F = BLI the current I has to be reduced by a factor of then two separate modifications that could be carried out are: (i) increase the length of the conductor that is in the magnetic field by a factor of ; (ii) increase the magnetic field strength by a factor of. (e) The power drawn from the power supply can be found from: (i) P = V I = = 8000 W or 8kW (ii) P = V I = 00 0 = 4000 W or 4kW (5). (a) C9 6 = 0000 = 0 0 7A 6 = 000 = 0 (b) A R B C & Z S T Z, A B C Z R S T Z , [C05/SQP068] 7 Page seven

28 . (continued) (c) The Boolean expression and completed truth table are: Z = A.B.C + B.C.D A B C D Z (d) The TTL logic chip number and the pin numbers that could be used to construct the circuit, are shown below. A B C D Z 6 (5) [END OF MARKING INSTRUCTIONS] [C05/SQP068] 8 Page eight

X025/11/01 X025/11/01 ELECTRONIC AND ELECTRICAL FUNDAMENTALS INTERMEDIATE 2 NATIONAL QUALIFICATIONS 2015 WEDNESDAY, 3 JUNE 9.00 AM 11.

*X025/11/01* X025/11/01 ELECTRONIC AND ELECTRICAL FUNDAMENTALS INTERMEDIATE 2 NATIONAL QUALIFICATIONS 2015 WEDNESDAY, 3 JUNE 9.00 AM 11. X05//0 NATIONAL QUALIFICATIONS 05 WEDNESDAY, JUNE 9.00 AM.0 AM ELECTRONIC AND ELECTRICAL FUNDAMENTALS INTERMEDIATE 00 marks are allocated to this paper. Answer all questions in Section A (50 marks). Answer