Electrical and Electronic Technology (9th Edition) by Edward Hughes, John Hiley & Keith Brown | Complete Solution Manual | A+ Study Guide

Solution Manual For
Hughes Electrical and Electronic Technology 9th Edition by Edward Hughes John Hiley
Keith Brown
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Solutions to Exercises
Section One: Electrical Principles
Exercises 1
1. F  ma
F
 a   80  0.4 m/s2
m 200
2. 2 kN
3. F  ma
2
Here the acceleration is that of gravity, i.e. 9.81 m/s
 F  10  9.81  98.1 N
4. 9.81 m/s, 19.62 m/s, 29.43 m/s
5. F  ma

 a    1  10
F 3
 0.1 m/s2
m 10  10 3

v  at
v
 t   5  50 s
a 0.1
6. 625 kN
7. The velocity has no effect on the weight, hence
F  ma  10  9.81  98.1 N
The acceleration requires extra force, hence
F  ma  10  (9.81  3)  128.1 N
2
8. 0.98 m/s downwards
9. (a) F  5  103  300  10  9.81
3


 14 715 N  14.7 kN
90  10
3

(b) W  Fd  14 715   10
60
 221  10 J  221 MJ
6


221  10
6
  61.3 kWh
3.6  10
6


90  10
3
(c) P  Fu  14 715   14 715  25
60  60
 368 000 W  368 kW
(d) W  mu   300  10  25
1 2 1 3 2
2 2

 93.75  10 J  26 kWh
6


10. 16 370 Nm



Exercises 2

1. 0.000 005 A  5  106 A  5 A
2. 3 MV

, V
3. R   6  10 
I 0.6
4. 40 V

legal principle or case, exploring various perspectives or discussing its implications in practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a
student’s ability to identify relevant legal issues, apply appropriate laws and precedents, and reach conclusions




V 240
5. I    0.25 A
R 960
6. 2.9 k
V 240
7. I    28.2  10 A  28.2 kA
3

R 8.5  103
8. 5.1 mA
9. V  IR  22  5  110 V
10. 1 mA, 0.67 mA
V1
11. I  10  0.1 A
R1 100
V2 100
R    1000   1 k
2
I 0.1
12. 1 k
13. V/V
5

4

3

2

1

I/A
0.2 0.4 0.6 0.8 1.0 1.2

14. R/
1000


800


600


I/mA
10 20 30 40 50 60

15. Q  It  2.5  8  20 C
16. 8 C
17. V  IR  2  40  80 V
18. 212 Nm
Po
 20  10  22 727 W  VI
3
19. (a) P 
i
 0.88

, 22 727
 I  94.7 A
240
(b) W  Pt  22.727  6  136.4 kWh
Cost  136.4  8  1091 p
20. 83.8%, 146 kJ; 67.2 
3
21. 1000 cm has a mass of 1 kg, hence, for the pump
10  9.81  40  9810 W
6
(a) Po  1.5  3
10 60 legal principle or case, exploring various perspectives or discussing its implications in practice.2.3. Skills
Tested in Law ExamsLegal Analysis and Application: Law exams test a student’s ability to identify relevant
legal issues, apply appropriate laws and precedents, and reach conclusions

, For motor, P Pump output 9810
 Pump input  
p
o
0.9
 10 900 W
10 900 10 900
and P    12 824 W  12.82 kW

i
0.85
12 824
(b) P  VI  I   26.7 A
480
(c) W  Pt  12.82  8  102.6 kWh
22. 59.5 , 0.65 p
23. Energy to raise 1 kg of aluminium from 12 °C to 660 °C
 950  (660  12)  615 600 J
Latent heat for 1 kg
 450 000 J
 Heat energy for 1 kg is 615 600  450 000  1 065 600 J
40  1 065 600
(a) Po   11 840 W = 11.84 kW
3600
P 11.84
P  o   13.93 kW

i
0.85
(b) Cost  13.93  8  20  2229 p
legal principle or case, exploring various perspectives or discussing its implications in practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a
student’s ability to identify relevant legal issues, apply appropriate laws and precedents, and reach conclusions




Exercises 3

1. Across 2  resistor, V2  IR2  2  2  4 V
Across 3  resistor, V3  IR3  2  3  6 V
Across 5  resistor, V5  IR5  2  5  10 V
Total voltage V  V2  V3  V5  4  6  10  20 V
2. 500 , 0.48 A
V 120
3. R   80 
I 1.5
 R1  R2  R3  30  30  R3
 R3  20 
4. 60 
30
5. VAB   100  30 V
30  70
VBC  VLN  VAB  100  30  70 V
VAC  VLN  100 V
VBN  VBC  70 V
6. 35 V, 5 V, 35 V, 50 V, 10 V
9
7. (a) I6    1.5 A
R6 6
9
I   1.0 A
9
9

I15   0.6 A
15

10