Intermediate 2 NATIONAL Electronic and QUALIFICATIONS ...sensitive device termed a light dependent...

©

100 marks are allocated to this paper.

Attempt all questions in Section A (50 marks).

Attempt any two questions from Section B (50 marks).

Intermediate 2 T i m e : 2 h o u r s 3 0 m i n u t e s

Electronic andElectrical FundamentalsSpecimen Question Paper

NATIONALQUALIFICATIONS

[C025/SQP068]

[C025/SQP068] 1

Page two[C025/SQP068] 2

Section A

Attempt ALL questions in this Section (50 marks).

1. The system illustrated below in Figure Q1 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 V1.

(b) Determine the expected readings on voltmeters V2 and V3 and ammeters A1 and A2.

Marks

1

4

(5)

Figure Q1

12V

A1V2

V1

V3

A2

0V

1kΩ

300 Ω

hfe = 50

Page three[C025/SQP068] 3

2. The circuit shown below in Figure Q2 represents a zener diode controlling the operation of a

small cassette recorder from the cigar lighter socket on a car. The recorder requires 7.5V dc.

The car battery voltage is nominally 12V, but when travelling at speed the alternator is charging

and the terminal voltage could be as high as 14.6V. When stopped at traffic lights, the terminal

voltage could be as low as 11V.

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 RS.

Marks

2

2

2

(6)

CR

Figure Q2

VS

7.5

V

33 Ω

RS

CR = Cassette Recorder

Page four[C025/SQP068] 4

3. The electric motor shown below in Figure Q3 has an output power of 100 kW and operates at an

efficiency of 80%.

Determine:

(a) the electrical energy in joules delivered in 15 minutes;

(b) the electrical energy in kWh delivered in 7 hours.

Marks

3

2

(5)

Pin

LOAD

Pout

Motor

Figure Q3

Page five[C025/SQP068] 5

4. An electronic security lock is shown in Figure Q4 below, with 5 volts providing a logic 1 and

0 volts a logic 0. The lock opens on receiving a logic 1 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.

(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.

Marks

2

3

(5)

A

B

C

D

Logic 1to open

lock

Z

Figure Q4

5V 0V

Page six[C025/SQP068] 6

5. In the network shown below in Figure Q5, determine:

(a) the total network resistance;

(b) the supply current;

(c) the potential difference across the 10 Ω resistor.

Marks

3

2

2

(7)

Figure Q5

40 Ω

60 Ω

12V

20 Ω

30 Ω

50 Ω10 Ω

Page seven[C025/SQP068] 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 1).

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.

Marks

3

1

3

(7)

Page eight[C025/SQP068] 8

7. Convert the following numbers.

(a) Binary to decimal 1011

(b) Hexademical to decimal F316

(c) Binary to hexadecimal 10100110

Marks

1

2

2

(5)

Page nine[C025/SQP068] 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 10MΩ in total darkness and approximately 150 Ω in bright light.

(a) State the function of the following:

(i) resistor 2.2 kΩ;

(ii) variable resistor 100 kΩ;

(iii) transistor T1 and T2.

(b) Describe the operation of the circuit shown in Figure Q8.

Marks

1

1

2

3

(7)

Figure Q8

100kΩ 1 kΩ

680 ΩLDR

2.2 kΩ

9V

0V

T1 T2

6V

Filament

Lamp

Page ten[C025/SQP068] 10

9. A sinusoidal voltage measured in an amplifier circuit is represented by the expression

v = 25sin α mV.

Determine:

(a) maximum value of the voltage;

(b) instantaneous value of the voltage, when α = 0.785 radians.

Marks

1

2

(3)

Page eleven[C025/SQP068] 11

Section B

Attempt any TWO questions in this Section (50 marks). Each question is worth 25 marks.

10. A temperature sensing circuit uses the switching transistor circuit, shown in Figure Q10, 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, VR1, whose resistance falls when the

temperature falls.

(a) Calculate the resistance of the temperature dependent resistor at the instant the buzzer starts

to sound.

(b) If the buzzer is rated at 6V and has a resistance of 600Ω:

(i) calculate the value of the resistor R3 required to ensure the correct operation of the

circuit;

(ii) determine the power dissipated in the resistor R3, assuming Vce = 0.5V.

(c) Explain the effect on the circuit operation if the 100 Ω resistor was replaced by a 47 Ωresistor. Justify your answer with calculations.

Marks

4

3

3

6

Figure Q10

R3

Buzzer

TR1

VR1

12V

0V

R2

100 Ω

12V

0V

Page twelve

10. (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 Q10(d).

(i) Determine the gain of the amplifier and calculate the output voltage when the input

voltage is 690mV.

(ii) Explain the purpose of the variable resistor connected to pins 1 and 5.

(iii) Explain the purpose of Rf.

Marks

5

3

1

(25)

12kΩ

0VVR1

2.2kΩ

100Ω

Rf10kΩ

V out

VR21Ω

ideal

12V

0V

1

5

6

Figure Q10(d)

Rf

Vout

VR2

0V

0V

12VVR1 100 Ω

[C025/SQP068] 12

Page thirteen[C025/SQP068] 13

11. An experiment was set up as illustrated in Figure Q11. A conductor was placed at 90° to the

magnetic field and the length of the conductor in the magnetic field was 0.1m. When the

current was allowed to flow, the conductor was deflected to the left. The magnetic field

strength is 0.5T.

(a) If the force required to pull the conductor back to its original position is 2N, determine

the magnitude and direction of the current in the conductor.

(b) If the conductor resistance is 0.1 Ω and the power supply is set at 200 volts, calculate:

(i) the value of the variable resistance VR1 when the conductor is deflected;

(assume the resistance of the connecting wires is negligible)

(ii) the energy used, in joules and kWh, by the variable resistor, VR1, if the experiment

lasts for 15 minutes.

(c) Describe, with the aid of a suitable (magnetic field) diagram, why the conductor was

forced to the left.

(d) In order to reduce the energy wasted in the variable resistor, it was decided to reduce the

current in the conductor to 20A. Describe two modifications that could be carried out

on the experimental set to achieve this and still leave the force at 2N.

(e) Calculate the power drawn from the supply in parts (b) and (d).

Marks

4

3

2, 2

6

2, 2

2, 2

(25)

200VDC

V

NS

A

B

R1

Figure Q11

Conductor in

Magnetic Field

l = ⋅0 1 m

200V

DC

0.1 m

Page fourteen[C025/SQP068] 14

12. (a) A dual seven segment display, such as the one shown below in Figure Q12(a), keeps

changing between the following values, 7A16 and C916. Determine the decimal and

binary equivalents of these numbers.

(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 Q12(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.

Marks

2, 2

2, 2

2, 2

3, 4

Figure Q12(a)

A

Z

B C D

Figure Q12(c)

Page fifteen[C025/SQP068] 15

12. (continued)

(d) A circuit is shown in Figure Q12(d). Mark on the diagram, provided on worksheet

Q12(d), the TTL logic chip number and the pin numbers that could be used to

construct the circuit. Use datasheet Q12(d) to assist your response.

[END OF QUESTION PAPER]

Marks

6

(25)A

Z

B C D

Figure Q12(d)

[C025/SQP068] 16

©


Electronic andElectrical FundamentalsDatasheet Q12(d)


[C025/SQP068]

[C025/SQP068] 19

©


Electronic andElectrical FundamentalsWorksheet Q12(d)


[C025/SQP068]

[C025/SQP068] 17

Day Month Year Number of seat Candidate number

Fill in these boxes and read what is printed below.

Full name of centre Town

First name and initials Surname

Date of birth

To be inserted in the front cover of the candidate’s answer book and returned with it only bycandidates who attempt Question 12.

Page two

A

Z

B C D

Figure Q12(d)

[C025/SQP068] 18

INTEGRATED-CIRCUIT DIAGRAMS—7400 SERIES

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

14 13 12 11 10 9 8

1 2 3 4 5 6 7

Gnd

Vcc

00 Quadruple 2 input NAND

gate

01 Quadruple 2 input NAND

gate with open collector output

02 Quadruple 2 input NOR gate 03 Quadruple 2 input NAND

gate—open collector inputs

04 Hex inverter 05 Hex inverter—open collector

outputs

06 Hex inverter with high

voltage open collector output

07 Hex driver with open

collector output

08 Quadruple 2 input AND gate 09 Quad 2 input AND gate—

open collector outputs

10 Triple 3 input NAND gate 11 Triple 3 input AND gate

13 Dual 4 input NAND gate

Schmitt trigger

14 Hex Schmitt Trigger 15 Triple 3 input AND gate—


16 Hex Inverter with open

collector output

20 Dual 4 input NAND gate 21 Dual 4 input AND gate 22 Dual 4 input NAND gate—


25 Dual 4 input NOR gate with

strobe

26 Quad 2 input NAND buffer


27 Triple 3 input NOR gate 28 Quad 2 input NOR buffer 32 Quad 2 input OR gate

Grateful acknowledgement is given to R.S. Components for permission to reproduce this sheet.



1. (a) V1 = 0.6V to 0.7V

(b) V2 = 500 × 10−6 × 1000 = 0.5V

V3 = Ie × Re = 25.5 × 10−3 × 300 = 7.65V

A1 = Ic = Ib × hfe = 500 × 10−6 × 50 = 25mA

A2 = Ib + Ic = Ie = 25 × 10−3 + 500 × 10−6 = 25.5mA

2. (a) Maximum Current = Vs max − Vz/Rs

= 14.6 − 7.5/33

= 215mA

Minimum Current = Vs min − Vz/Rs

= 11 − 7.5/33

= 106mA

(b) Power Rating Zener = 7.5 × 215 × 10−3

= 1.61W

(c) Power Rating Rs = (215 × 10−3)2 × 33

= 1.526W

3. (a) Pin = Pout/Efficiency

= 100/0.8 = 125kW

Electrical Energy delivered = 125 × 103 × 15 × 60 = 112.5MJ

(b) Electrical Energy = 125 × 7 = 875kWh

Marks

1

1

1

1

1

(5)

1

1

2

2

(6)

3

2

(5)

Section A

Page three[C025/SQP068] 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. (a) Total resistance is obtained from 100 Ω (20 Ω, 30 Ω and 50 Ω in series) in parallel with

100 Ω (40 Ω and 60 Ω in series), ie 50 Ω which is series with 10 Ω giving a total of 60 Ω.

(b) Is = Vs/Rt = 12/60 = 0.2A

(c) Potential difference = 0.2 × 10 = 2V

6. (a)

(b) Z = A. B. C

(c)

Marks

2

3

(5)

3

2

2

(7)

3

1

3

(7)

A B C Z

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 0

A

B

C

Z

Marks

Page four

7. (a) 1011 = 23 + 2 + 1 = 1110

(b) F316 = 15 × 16 + 3 = 24310

(c) 10100110 = A616

8. (a) (i) Limit the base current to transistor T1.

(ii) Transistor biasing allows the circuit to operate at different light levels.

(iii) Transistor T1: Switch

Transistor T2: Amplifier

(b) Suitable description required.

9. (a) maximum = 25mV

(b) v = 25 × 10−3

sin 0.785

v = 17.68mV

1

2

2

(5)

1

1

2

3

(7)

1

2(3)

[C025/SQP068] 24

Marks

Page five

10. (a) For the transistor to be on the voltage at the base, R1–R2 junction must be 0.7V therefore

the voltage drop across R1 will be 12 − 0.7 = 11.3V

Then R1 = VR1/VR2 × R2 = 11.3/0.7 × 100 = 1614 Ω

or

I = V/R = 0.7/100 = 7mA so R1 = VR1/I = 11.3/0.007 = 1614 Ω

(b) (i) The voltage drop across the buzzer, for correct operation, should be 6V and

therefore the voltage to be dropped across R3 = Vs − Vbuzzer − Vce

= 12 − 6 − 0.5 = 5.5V

so R3 = VR3/Vbuzzer × Rbuzzer = 5.5/6 × 600 = 550 Ω

or

Ibuzzer = V/R = 6/600 = 10mA so R3 = VR3/I = 5.5/10mA = 550 Ω

(ii) P = I2R

= 10mA × 10mA × 550 Ω = 55mW

or

P = V2/R = 5.5 × 5.5/550 = 55mW

(c) If the 100 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.

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.

4

3

3

6

[C025/SQP068] 25

R3

Buzzer

TR1

12V

0V

Section B

VR

R RVs V

R2

2

1 2

47 1661 12 0 34=+

× = × = ⋅( / )47/1661

Marks

Page six

10. (continued)

(d) (i) The gain of the non-inventing amplifier can be found from

A = 1 + Rf/Ri = 1 + 10 × 103/2.2 × 103 = 5.55

Vout = A × Vin = 690mV × 5.55 = 3.83V

(ii) The variable resistor connected between pins 1 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.

(iii) Rf is the feedback resistor and it provides this feedback by producing a small part of

the output voltage at the inverting input.

11. (a) F = BLI so I = F/BL = 2/0.5 × 0.1 = 40A

Using Fleming’s left hand rule, the direction of the current is from A to B or downwards

on the diagram.

(b) (i) I = 40A, Vs = 200V Rconductor = 0.1 Ω

Rt = V/I = 200/40 = 5 Ω

R variable = Rt − Rconductor = 5 − 0.1 = 4.9 Ω

(ii) Power in the variable resistor can be found from

P = I2 × R = 402 × 4.9 = 7840W

W(joules) = P × t = 7840 × 15 × 60 = 7056000J or 7.056MJ

W(kWh) = P(kW) × t(hours) = 7.84 × 0.25 = 1.87kWh

(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.

5

3

1

(25)

4

3

2, 2

6

[C025/SQP068] 26

movement

strengthened

field

weakened

field

Marks

Page seven

11. (continued)

(d) Using F = BLI the current I has to be reduced by a factor of 2 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 2;

(ii) increase the magnetic field strength by a factor of 2.

(e) The power drawn from the power supply can be found from:

(i) P = V × I = 200 × 40 = 8000W or 8kW

(ii) P = V × I = 200 × 20 = 4000W or 4kW

12. (a) C916 = 110010012 = 20110

7A16 =011110102 =12210

(b)

2

2

2

2

(25)

2

2

2, 2

2, 2

[C025/SQP068] 27

A

B

C

Z&

R

S

T

Z≥ 1

A B C Z

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 0

R S T Z

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

Marks

Page eight

12. (continued)

(c) The Boolean expression and completed truth table are:

Z = A.B.C + B.C.D

(d) The TTL logic chip number and the pin numbers that could be used to construct the

circuit, are shown below.

[END OF MARKING INSTRUCTIONS]

3

4

6

(25)

[C025/SQP068] 28

A B C D Z

0 0 0 0 1

0 0 0 1 1

0 0 1 0 1

0 0 1 1 1

0 1 0 0 1

0 1 0 1 1

0 1 1 0 1

0 1 1 1 0

1 0 0 0 1

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 0

A

Z

B C D

1 2

04 7

5

4 116

9

11

10 118

1

13

2 1112

Intermediate 2 NATIONAL Electronic and QUALIFICATIONS ...sensitive device termed a light dependent...

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Transcript of Intermediate 2 NATIONAL Electronic and QUALIFICATIONS ...sensitive device termed a light dependent...