ELECTRICAL TECHNOLOGY WORK BOOK print

Electrical Technology Work Book 101

Find the reluctance for the air gap:

Reluctance,

2 = 2 = 2
2 2
(Since =1 for air)

= (4 × 1 × 10−3 10−4) = 3979000H
10−7)(2 ×

Total reluctance,

= 1 + 2 = 1172000 + 3979000 = 5151000H

Find the flux,

∅ = = 0.80 × 2 × 10−4 = 1.6 × 10−4
Finally, use the reluctance’s formula to find the current

=
∅
Then

= ∅

Note that

=
So,

= ∅

The current is

= ∅

(5151000)(1.6 × 10−4)
= 5000 = 0.165 A

102 Electrical Technology Work Book

BIBLIOGRAPHY

Alexander, C., & Sadiku, M. (2016). Fundamentals of Electric Circuits (6th ed.). Mc Graw Hill
Education.
Allan H Robbins, & Miller, W. C. (2013). Circuit Analysis - Theory and Practice (3rd ed.). Cengage
Learning.
Bird, C. M. eld J., Bolton, M. A. L. W., Sueker, A. L. R. S. K., Moura, T. W. M. T. L., & Warne, I. D. W.
K. A. B. D. (2008). Electrical Engineering Know It All. Newnes.
Bird, J. (2007). Electrical Circuit Theory and Technology. Routledge.
doi:10.4324/9780080549798
Chapman, S. J. (2002). Electric Machinery and Power System Fundamentals. Mc Graw Hill.
Schultz, M. E. (2010). Grob ’ s Basic Electronics (11th ed.). Maney Publishing Education.

Electrical Technology Work Book 103

ANSWERS

104 Electrical Technology Work Book

CHAPTER 1____________________________ Q6
Q1
Given

1. Carbon-zinc cell = 20 2, = 800 , = 0.02 Ω
2. Alkaline cell
3. Nickel Cadmium cell How to convert mm2 to m2
4. Edison cell
5. Mercury cell 20 2 = _______ 2

Q2 Since

( ) = 1 = 0.001

103 (1 )2 = (0.001 )2
= 10 ×
10 × 103
= = 250 = 40 1 2 = 0.0012 2

Q3 Therefore

a. Total emf in series 20 2 = 20 × 10−6 2

( ) = 12 × 1.5 = 18 Find the resistance

Total internal resistance in = = 0.02 × 10−6Ω × 800 = 0.8Ω
series 20 × 10−6 2

( ) Q7
=
× Given
= 12 × 0.24 = 2.88Ω
= 100 , = 2Ω, = 0.03 × 10−6Ω

b. Total emf in parallel Since, the resistance

( ) = 1.5 =

Total internal resistance in
series Then

( ) =01.224 = 0 . 0 2 Ω = = 0.03 × 10−6Ωm × 100m
= = 1.5 2  2

Q4 Q8

Given

Total e.m.f., = 2.6 2, = 500 , = 5Ω

= 1 + 2 + 3 + 4 Since, the resistance
=2+2+2+2
= 8 =

Q5

Total e.m.f. Then, the resistivity,

= 1 + 2 = 8 − 6 = 2

Electrical Technology Work Book 105

= = 5Ω × 2.6 × 10−6 2 = 6Ω = 15 = 2.5
500m 6Ω 6
= 0.026 × 10−6Ωm
= 0.026μΩm b. Apply Kirchhoff’s Law to find IR

Q 9 Voltage across 5  resistor 11.5 − − 3 − 2.5 = 0
= 11.5 − 3 − 2.5 = 6

5Ω = 1 − 2 = 30 − 10 = 20 Therefore

Hence, current in 5  resistor, i.e. = = 15 = 2.5Ω
6
reading on the ammeter
5Ω 20
5 = 5 = 4 Q 12

To find the total resistance,

Note that, Total resistance between terminals A
= 1 = = 3 4 0 = 7.5Ω
and B 18) +
= =11++66(+×6 ∕∕ + 1.5 1.5
18
Hence 18
= 7.5 − 5 = 2.5Ω = 1 + 4.5 + 1.5
= 7Ω
Q 10
Q 13

Given Rtotal = 36 and the supply voltage,
Vsupply = 18V

18 = 1 + 5 + 3

1 = 18 – 5 – 3 = 10

Apply Ohm’s Law to find total current

Note that,

=

= = 18 = 0.5
36

Apply Ohm’s Law to find R1, R2 and R3

1 = 1 = 10 = 20Ω Firstly, find RTotal
0.5
= 4 + +
2 = 2 = 5 = 10Ω The value of Ra
0.5
1 1 1 1
3 = 3 = 3 = 6Ω = 2 + 3 + 6 = 1
0.5
1
Q 11 = 1 = 1Ω

Since the resistors are connected in parallel, The value of Rb
the voltage for every junction is same value.
1 = 1 + 1 = 0.83
6Ω = 5Ω = 3 × 5Ω = 3 × 5 = 15 2 3

a. Apply Ohm’s Law to find the
reading on the ammeter

106 Electrical Technology Work Book

= 1 = 1.2Ω (b)
0.83

Therefore

= 4 + 1 + 1.2 = 6.2Ω

Then, find each current I1, I2, I3, I4, I5 and I6 by
using Ohm’s Law

Find I1 and V1 5 × 5
5 + 5
Note that, = 5 ∕∕ 5 = = 2.5Ω

= (c)

1 = = 30 = 4.84
6.2Ω

1 = 1 × 4Ω = 4.84 × 4 = 19.36V

Find V2 and I2 I3 and I4

2 = 1 × = 4.84 × 1 = 4.84

2 = 2 = 4.84 = 2.42 = 6 ∕∕ 4 = 6 × 4 = 2.4Ω
2Ω 2Ω 6 + 4

2 4.84 (d)
3Ω 3Ω
3 = = = 1.61

4 = 2 = 4.84 = 0.81
6Ω 6Ω

Find V3 and I5 and I6

3 = 1 × = 4.84 × 1.2 = 5.81 = 1.6 + 2.4 = 4Ω

5 = 3 = 5.81 = 1.94 From total current 24
3Ω 3Ω 24 4
= = = 6
3 5.81
6 = 2Ω = 2Ω = 2.91 Q 15

Q 14 1T o t a l=re s1 1is+tan 1 c 2e+of 1t h3e=ci41rc+uit16, R+to81tal
= 0.542
The circuit is reduced step by step as 1
shown in diagrams (a) to (d) below. 0.542Ω

(a)

= = 1.846Ω

Current, I 20
1.846Ω
= = = 10.833

= 3 + 2 = 5Ω Voltage drop across resistor 8Ω,
VR3

3 = = 20

Current through resistor 4Ω, IR1

Electrical Technology Work Book 107

= = 20 = 5 = 9
1 4Ω
Therefore, the total of resistance is

Q 16

Find the total resistance of the circuit,
Rtotal

= 1 + 2 + 3
=4+6
+8
= 18Ω

Find the current, I using Ohm’s Law

= = 15 = 2 + + 20
18Ω = 2 + 9 + 20 = 31
= 0.833
Find the current from supply, Is

Voltage drop across resistor 6Ω, VR2
using Ohm’s Law

2 = × 2 = 0.833 × 6
= 5

Q 17

Find the equivalent resistance of the
circuit, Rtotal

= 20 = 645.16
31

Q 18

The delta-connected network is shown labelled,
together with the equivalent star-connected
network is shown below.

= 4 + 8 + 6 = 18

1 = 1 + 1 = 1 + 1 = 1
18 18 18 9

108 Electrical Technology Work Book

1 = = 100 × 100
+ + 100 + 100 + 100

= 33.33

2 = = 100 × 100
+ + 100 + 100 + 100

= 33.33

3 = = 100 × 100
+ + 100 + 100 + 100

= 33.33

Q 20

1 = = 4 4×1 5 = 0.4Ω Converting from Delta to Star
+ + +1+ connection

2 = = 4 4×5 5 = 2Ω = 4 x 8 = 1.78Ω
+ + +1+ +8+
4 6
1×5
3 = + + = 4 +1+ 5 = 0.5Ω = 8 x 6 = 2.67Ω
+8+
Q 19 4 6

= 4 4 x 6 6 = 1.33Ω
+8+

= 1.33 + 12 = 13.33
= 2.67 + 10 = 12.67

= 13.33 ∕∕ 12.67 = 13.33 × 12.67 = 6.5Ω
13.33 + 12.67

Electrical Technology Work Book 109

Therefore − 1 − 2 = 0
− 8 − 4 = 0
= 1.78 + 6.5 = 8.28 = 8 + 4 = 12
CHAPTER 2____________________________
Q1 Q6

Apply KVL Theorem Apply KVL Theorem
Σ = 0
Σ = 0
20 + 10 + 3 − 45 = 0 3 + 4 − 1.5 − 3 = 0
3 = 45 − 30 = 15 3 = 7 − 1.5 = 5.5

Q2 Q7

Apply KVL Theorem To calculate I1, we need to find Rtotal.
Σ = 0
1 = 1 + ( 2 1 3)
20 + 2 + 15 − 45 = 0 1 +
3 = 45 − 35 = 10
1 = 1 + (1 1 2) = 1 + 1
Q3 2 + 2 3

Apply KVL Theorem = 1.2Ω
Σ = 0
Then, we use Ohm’s Law to find I1
20 + + 15 − 45 = 0
= 45 − 35 = 10 =

Q4 1 = = 8 = 6.67
1.2
Apply KVL Theorem
Σ = 0 To find, I2 we use Ohm’s Law

20 + 10 + 3 − 45 = 0 2 = = 8 = 4
3 = 45 − 30 = 15 1 2

Q5 Apply Kirchhoff’s Current Law (KCL) to find I3

Apply KVL Theorem Σ = 0
Σ = 0 1 − 2 − 3 = 0
6.67 − 4 − 3 = 0

3 = 2.67

Q8

To find I1, we can apply KCL at node a

Σ = 0
6 + 2 − 1 = 0

110 Electrical Technology Work Book Step 2:

1 = 8 Apply KVL around each closed loop of the circuit
To find I2, we can apply KCL at node b
Loop 11
Σ = 0
1 − 2 − 1 = 0 20 − 3 2 − 4( 2 + 1) = 0
8 − 2 − 1 = 0 20 − 3 2 − 4 2 − 4 1 = 0

2 = 7 20 − 7 2 − 4 1 = 0
To find I5, we can apply KCL at node c 7 2 + 4 1 = 20 … (1)

Σ = 0 Loop 12
1 − 3 + 5 = 0
10 − 1 − 4( 2 + 1) = 0
5 = 2 10 − 1 − 4 2 − 4 1 = 0
To find I3, we can apply KCL at node f
10 − 4 2 − 5 1 = 0
Σ = 0 4 2 + 5 1 = 10 … (2)
3 + 3 − 12 = 0 Step 3:

3 = 9 Solving the simultaneous linear equations (1) and
To find I4, we can apply KCL at node e (2) using matrix

Σ = 0 7 2 + 4 1 = 20 … (1)
2 + 4 − 3 = 0 4 2 + 5 1 = 10 … (2)
7 + 4 − 9 = 0 Matrix form

4 = 2 [54 47] [ 12] = [1200]
To find I5, we can apply KCL at node d
∆= |54 47| = 16 − 35 = −19
Σ = 0
6 − 4 − 2 = 0 ∆ 1 = |2100 47| = 80 − 70 = 10
6 − 2 − 2 = 0
∆ 2 = |54 1200| = 40 − 100 = −60
6 = 4

Q9

Step 1:
Assign a direction of current to each closed loop of
the circuit

2 = ∆ 2 = −60 = 3.16
∆ −19

Electrical Technology Work Book 111

1 = ∆ 1 = 10 = −0.53 Nodal at Va
∆ −19
20 − 10 −
Apply KCL to find I3 3 − 4 + 1 = 0

3 = 1 + 2 = −0.53 + 3.16 = 2.63 20 − − + 10 − = 0
3 3 4 1 1
Alternative solution for simultaneous linear
equations 50 − (31 + 1 + 1) = 0
3 4

Equations (1) X 5 (7 2 + 4 1 = 20) × 5 (1129) 50
3
Equations (2) X 4 (4 2 + 5 1 = 10) × 4 =

35 2 + 20 1 = 100 … (3) = 50 × 12 = 600 = 10.53
3 × 19 57

16 2 + 20 1 = 40 … (4) 1 = 10 − = 10 − 10.53 = −0.53
1 1
Equations (3) - equations (4)

35 2 + 20 1 = 100 2 = 20 − = 20 − 10.53 = 3.16
16 2 + 20 1 = 40 3 3

19 2 = 60 3 = = 10.53 = 2.63
2 = 3.16 4 4
Replace I2 = 3.16A into equation (1)
Q 11

7(3.16) + 4 1 = 20

4 1 = 20 − 22.12

1 = −2.12 = −0.53
4
Apply KVL around each closed loop of the circuit
Apply KCL to find I3, Loop 11

3 = 1 + 2 = 3.16 + (−0.53) = 2.63 10 − 10 1 − 40( 1 − 2) = 0
10 − 10 1 − 40 1 + 40 2 = 0
Q 10
10 − 50 1 + 40 2 = 0
Apply KCL at Va 50 1 − 40 2 = 10 … (1)
2 − 3 + 1 = 0 Loop 12
−20 2 − 30 2 − 40( 2 − 1) = 0
−50 2 − 40 2 + 40 1 = 0
40 1 − 90 2 = 0 … (2)
Solving equation 1 and equation 2 using matrix
50 1 − 40 2 = 10 … (1)

112 Electrical Technology Work Book

40 1 − 90 2 = 0 … (2)
Matrix Form

[5400 −−9400] [ 12] = [100] = 1 ∕∕ ( 2 + 4)
Determinant = 1 ∕∕ (1 + 2)

∆= |5400 −−9400| = −4500 − (−1600) = −2900 1 = 1 + 1
Determinant for I1 1 3
1 × 3 3
∆ 1 = |100 −−4900| = −900 − (0) = −900 = 1 + 3 = 4 = 0.75Ω
Determinant for I2
iii. Draw the Thevenin equivalent circuit
∆ 2 = |5400 100| = 0 − 400 = −400
Therefore,

2 = ∆ 2 = −400 = 0.138
∆ −2900

1 = ∆ 1 = −900 = 0.310
∆ −2900

Apply KCL to find I3

3 = 1 − 2 = 0.310 − 0.138 = 0.172 iv. Find the the ITH (Use Ohm’s Law)

Q 12 = 3 = 6 2 = 2.18
+ 0.75 +

i. Terminate the load resistor, R3 and Q 13
find the Thevenin’s source, VTH
i. Terminate the load resistor, RL and
find the Thevenin source, VTH

= 1 (1 + 2) × 8
+ (1 + 2)

= 6 4
+
= 4 1 × 10

ii. Find the Thevenin resistance, RTH by = 8
removing all power source (voltage
source is shorted)

Electrical Technology Work Book 113

ii. Find the Thevenin resistance, RTH by ii. Find the Thevenin resistance, RTH by
removing all power sources (voltage removing all power source (current
source is shorted) source is opened)

= 2 + ( 3|| 1) = 3 + 2
= 3 + (4||1) = 2 + 3 = 5Ω
iii. Draw the Thevenin circuit

= 3 + (4 × 1) = 3.8Ω
(4 + 1)

iii. Draw the Thevenin circuit

iv. Find the Thevenin’s current ITH (Use
Ohm’s Law)

iv. Find the the ITH (Use Ohm’s Law) = = 5 6 4 = 0.67
+ +

= = 8 2 = 1.38 Q 15
+ 3.8 +

Q 14 i. Terminate RL and find IN

i. Terminate the load resistor, RL and
find the Thevenin source, VTH

Firstly, find the Rtotal

= 3 = 1 + ( 3 ∕∕ 2)
Apply Ohm’s Law to find I3
= 1 + (3 ∕∕ 2)
3 = 1 = 3
3 = 3 3 = 3 × 2 = 6 = 1 + 3× 2 = 2.2Ω
3+ 2

Next, find I1 by using Ohm’s Law

114 Electrical Technology Work Book

1 = = 8 = 3.64
2.2

To find IN, apply Current Divider Rule (CDR)

= 2 = 3 3 2 × 3 = 3 3 2 × 3.64 = 2.184
+ +

ii. Find RN by removing all power Apply Current Divider Rule (CDR) to find IN
sources (shorted voltage and opened
current)

= 2 = 3 × 3 = 2 2 3 × 3 = 1.2
3 + 2 +

ii. Find RN by removing all supply
(shorted voltage and opened current)

= 2 + ( 3 ∕∕ 1)

= 2 + (3 ∕∕ 1)

= 2 + 3 × 1 = 2.75Ω
3 + 1
= 2 + 3
iii. Draw the Norton’s equivalent circuit = 2 + 3 = 5Ω
iii. Draw the Norton’s equivalent circuit

iv. Find IL iv. Find IL
Apply CDR to find IL

= × = 2.75 4 × 2.184 = 0.89 Apply CDR to find IL
+ 2.75 +

Q 16 = × = 5 5 4 × 1.2 = 0.67
+ +
i. Terminate RL and find IN

Electrical Technology Work Book 115

CHAPTER 3____________________________ Q5
Q1  Aluminium
 Tantalum
 A capacitor is an electrical device that
is used to store electrical energy. Q6
 Mica
 The symbol of capacitance is C.  Ceramic
 Film
 The unit of capacitance is Farad (F).  Paper

 The basic formula Q7
( )
= ( ) ( )

Q2
 Capacitance is defined to be the
amount of charge, Q stored in
between the two plates for a potential
difference or voltage, V exists across
the plates

Q3

Q4

116 Electrical Technology Work Book Q9

 A capacitor consists of two metal = 120 + 60 + 20
plates separated by a certain = 200 ( )
distance d, in between the plates lies
a dielectric material with dielectric 1 = 1 + 1 + 1 = 1 + 1 + 1
constant =εoε, where εo is the 40 80 40 80
dielectric of air.
= 42.5k (series)
 The dielectric material allows for
charge to accumulate between the Therefore
capacitor plates. Air (actually vacuum)
has the lowest dielectric value of εo = = 1 = 23.53μ
8.854 x 10-12 Farads/meter. 42.5

 All other materials have higher
dielectric values, since they are higher
in density and can therefore
accumulate more charge.

 The Physical meaning of capacitance
can be seen by relating it to the
physical characteristics of the two
plates, so that, the capacitance is
related to the dielectric of the material
in between the plates, the square area
of a plate and the distance between
the plates.

Q8
 Capacitance between two plates
proportional to the surface area
 Capacitance between two plates
inversely proportional to the thickness
of dielectric
 Increasing the dielectric constant of
the material between the plates

Then, the total Capacitance, CAB:

= + 85 × 10−6
= 23.53 × 10−6 + 85 × 10−6
= 108.53 × 10−6
= 108.53

Electrical Technology Work Book 117

Q 10

Find Cb

Figure 1

= 350 + 400 = 750 ( )
Then

1 = 1 + 1 + 1
200 500
1 1 1
= 200 + + 500

= 8.33

Therefore 1 = 1 + 1 = 1 + 1 = 0.61 ( )
9 2
= 120 × 10−6
1
= 0.61 = 1.63

Q 11 = 4 + 1.63 = 5.63

a)Find Ca and simplify the circuit
= 4 + 5 = 9 ( )

118 Electrical Technology Work Book = 7.627 × 10−6 = 7.627
b) Simplify the circuit first Q 12

Find Ca and Cb Find the total capacitance

1 = 1 + 1 + 1 = 0.017
100 200 500

= 58.82 × 10−12 = 58.82

Find the electric charge when 15 V is applied
across its plates using this formula

=


= = 58.82 × 10−12 × 15 = 882.35

Q 13

(a) in parallel

= 10 + 620 = 630

= 30 + 20 = 50 (b) in series
= 40 + 50 = 90
Find Ctotal 1 = 1 + 1 = 101.61
10 620

= 1 = 9.84 × 10−6
101.6
= 9.84

Q 14

Given

= 0.4 , = 250 , = 6 Ω

(a) The initial discharge current

1 = 1 + 1 + 1 = 1 + 1 + 1 = = 250
10 10 50 90 6

= 131.11

Electrical Technology Work Book 119

(b) The time constant of the circuit a) The initial value of the current,

 = = 4 × 6 = 0.024 = = 25 = 0.416
Q 15 60

a) Initial current charge b) The time constant of the circuit,
 =
= = 200 = 2
100  = 40 × 60 = 2.4

b) Initial potential difference through
capacitor. c) The value of the current one second after
connection,
= 0
c) Time constant = ( −  )

 = = 20 × 100 = 2 = ( −21.4 )
d) Time taken for capacitor fully charges
= 0.41 (0.659)
5 = 5 2 = 0.01 = 0.27

Q 16
d) The value of the capacitor voltage three
Given seconds after connection,

= 0.8 ,  = 10 , = 15 = (1 − − )

(a) The value of the resistor = 25(1 − −23.4 )
 = = 25( 1 – 0.287 )

10 = (0.8) ( ) = 25(0.713)
= 12.5 Ω = 17.825

(b) The capacitor voltage, 10 ms after e) The time after connection when the
connecting the circuit to a 15 V supply capacitor voltage is 18 v

= (1 − − ) = (1 − − )

= 15(1 − −1100) 18 = 25(1 − −2 . 4 )

18
= 15( 1 – 0.37 ) 25
= (1 − −2 . 4 )

= 15(0.63)
0.72 = 1 − −2 . 4

= 9.45
−2 . 4 = 1 – 0.72
Q 17

Given −2 . 4 = 0.28

= 40 , = 60 Ω, = 25
−2 . 4 = 0.28

120 Electrical Technology Work Book

− = − 1.273 (c) The current flowing when the capacitor has
2.4
been discharging for 0.8s

= 3.055 = = 200 = 2.86
70

Q 18 = ( −  )
= ( −00.0.882 )
Given
= 2.86 (0.06 )
= 200 , = 70 KΩ,  = 0.5S = 0.17

(a) The value of the capacitor, Q 19

 = Given

= 10 , = 0.5 , = 240
0.5 = ( )(70 )
a) Time constant

= 7.14
(a) The time for the capacitor voltage to

fall to 30 v,

= (1 − − ) =
= 0.5 (10µ) = 5

30 = 200(1 − −0 . 5 ) b) Initial charging current

30 = = 240 = 0.48
200 0.5
= (1 − −0 . 5)

c) Time for capacitor voltage increases to 150
0.15 = 1 − −0 . 5 V


−0 . 5 = 1 – 0.15
= (1 − − )
150 = 240 (1 − −5 )
−0 . 5 = 0.85
150 −5 )
240 = (1 −
−0 . 5 = 0.85
0.625 = 1 − −5
−5 = 1 − 0.625

− 0.5 = − 0.163 −5 = 0.375

ln( −5 ) = ln(0.375)

= 0.082

− = −0.981
5
= 4.9

Electrical Technology Work Book 121

d) The current flowing through the capacitor The time constant during charging the time
after 4 seconds. taken to be fully charge

= ( − / )  =
= 0.48 ( −4/5)  = 30 × 200 = 6
Fully charge equal to five time
= 0.48 (0.449) 5 = 5 × 6 = 30

The energy stored in the capacitor.

= 0.216 = 1 2
2
e) Energy stored in the capacitor when it is
fully charge = 1 (30 )(200)2 = 0.6
2

= 1 2 CHAPTER 4____________________________
2

= 1 (10 )(240)2 = 0.288 Q1
2
Symbol of inducdor is L and unit of inductance
f) Sketch the current and voltage curve to is Henry(H)
show the process of charging the capacitor. Q2

V
VMax

Voltage across
capacitor:

t

Q 20

Given

= 30 , = 200 Ω, V = 200V
Find the initial current

= = 200 = 1
200
The initial potential different across capacitor
Faraday’s law says that:
= 0  an emf is induced in a loop when

it moves through an electric field

122 Electrical Technology Work Book

 the induced emf produces a Q7
current whose magnetic field
opposes the original change For Figure 1a (parallel circuit) :

 the induced emf is proportional to 1 = 1 + 1 + 1 + 1
the rate of change of magnetic flux 1 2 3 4
1
Q3 =
1 + 1 + 1 + 1
a) FOUR (4) factors that influence 1 2 3 4
inductances For Figure 1b (series circuit) :
1. Number of turns
2. Permeability = 1 + 2 + 3
3. Cross-sectional area of coil
4. Length of the coil Q8 the total inductance
i.
b) Inductance is the physical property of a
circuit that opposes any change in Simplify the circuit first
current flow and unit of measurement
for inductance is Henry (H).

Q4 = 2 + 5 + 10 = 17

Symbol for inductor is the letter L (L) and
symbol for unit of inductance is the letter h or
H (h or H).

Q5 = 2// = 2 ×
2 +
Second Law of Faraday’s Electromagnetic 3 × 17
Induction state the induced emf is equal to the = 3 + 17 = 2.55
rate change of flux linkages which is the
product of turns i.e n of the coil and the flux
associated with it.

Q6

Current equation :

( ) = (1 − − ) = 20 + 2.55 + 4

( ) = (1 − − ) = 26.55

Voltage equation : ii. energy stored in the circuit.

( ) = − / = 1 2
2

Electrical Technology Work Book 123
Q 10
= 1 (26.55 )(3)2 = 1.912 Simplify the circuit first
2

Q9
a) The formula of self inductance

L  N 2A  N 2ro A
 

b) The total inductance for figure 6a = (3.5// ) + 1.5

3.5 = × 3.5 + 1.5
+ 3.5

3.5 − 1.5 = × 3.5
+ 3.5

2 = × 3.5
+ 3.5

= 1 = 1 × 2 = 10 × 50 2 ( + 3.5) = × 3.5
2 1 + 2 10 + 50

= 8.33 2 + 7 = 3.5

The total inductance for figure 6b 7 = 3.5 − 2

7 = 1.5

∴ = .

Simplify the circuit Q 11 ( ) = ( − / )
Current equation : ( ) = − − /
Voltage equation :

Q 12
Find the La by adding L3 and L7 (series circuit).

= 2// 3

= 2 × 3 = 10 × 20
2 + 3 10 + 20

= 6.67
1 × = 3 + 7 = 100 + 50
= 1// = 1 + = 150

= 40 × 6.67
40 + 6.67
= 5.72

124 Electrical Technology Work Book

Find the Lb (pallel circuit between La and L6)

1 = 1 + 1 = + 1 = 9.57 + 40
6 = 49.57

= × 6 = 150 × 25 Therefore, the total inductance is
+ 6 150 + 25

= 21.43

Find the Lc by adding L2 and Lb (series circuit)

= + 2 = 21.43 + 200 = × 4 = 49.57 × 30
= 221.43 + 4 49.57 + 30

Find the Ld (parallel circuit between L5 and Lc) = 24.96

Q 13
a) Electromagnetic induction is where a

voltage or current is produced in a
conductor by a changing magnetic flux.
b) Find the equavalent circuit
Simplify the circuit

= 5 × = 10 × 221.43
5 + 10 + 221.43

= 9.57

Find the Le by adding Ld and L1 (series circuit)

Electrical Technology Work Book 125

= 1 + 2 + 3 b. Time taken for current
= 80 + 10
+ 60 = 150 achieve maximum value

= 4 + 5 = 50 + 20 5 = 5 (1.5 ) = 7.5
= 70 c. Instantaneous current at t =

0.5s
( ) = (1 − − )

( ) = (1 − − )

= = 100 = 10
10
( ) = 10 (1 − −01..55 )

( ) = 10 ( 1 − −0.33 )

( ) = 10 ( 1 − 0.719 )

= 10 − 7.19

∴ ( ) = 2.81

= // a. Energy stored in inductor

× 150 × 70 = 1 2
+ 150 + 70 2
= = 1
= 2 (15)(10)2 = 750
= 47.73

Therefore, the total inductance is Q 15
a)
= 6//
×
= +

= 47.73 × 40 = 21.76
47.73 + 40

Q 14
i. Sketch the graph for rise of current in

inductor circuit.

b) Given

ii. Calculate : = 20 , = 2.5
1
a. Time constant = 2 2

= = 15 = 1.5 1
10 Ω 2
= (20 )(2.5)2 = .

126 Electrical Technology Work Book

Q 16 The time constant ( ) = 4 (1 − −00..0055 )
i)
( ) = 4 ( 1 − −1 )
= = 10 = μ ( ) = 4 ( 1 − 0.368 )
1 Ω
= 4 (0.632)
ii) The initial current ∴ ( ) = .
Q 17 iii. Time taken to increase to 2.5 A
Given = = 150 = .
1 Ω ( ) = (1 − − )

2.5 = 4 (1 − −0. 0 5)
2.5
= 250 4 = 1 − −0. 0 5

= 5 0.625 − 1 = − −20
= 1000 −0.375 = − −20
0 = 1.256 10−6 ln(0.375) = ln( −20 )
= 20 = 20 × 10−3 = 0.02
= 10 = 10 × 10−3 = 0.01 ln 0.375 = − 20
= 2 = (0.201)2 = 78.54
Find the value of inductance − 0.981 = −20

∴ = .

N 2A N 2ro A Q 20
 
L   i) τ, time constant

= = 10 =
5Ω
( )( . × − )( )( . ) ii) IL when open switch
= × −
=
∴ = . iii) IL when switch closed (starting

current) 20
5
Q 18 15 = = =
Q 19 5Ω
= = = iv) IL when t = 5 second
i.
= 1 2 ( ) = (1 − − )
2
1 ( ) = 4 (1 − −25 )
= 2 (80 )(3)2 = 0.36

The time constant ( ) = 4 ( 1 − −2.5 )
( ) = 4 ( 1 − 0.082 )
= = 1.5 = 0.05
30Ω = 4 (0.918)
∴ ( ) = .
ii. The current at 0.05s v) Voltage drop at resistor, VR
( ) = (1 − − )
= × = 3.672 × 5
= .

vi) Voltage drop at inductor, VL

120 = − = 20 – 18.36 = .
30
= = = 4

Electrical Technology Work Book 127

vii) Energy stored when archive Q6

maximum. “The direction of an induced e.m.f is
always such that it tends to set up a
= 1 2 current opposing the motion or the
2 change of flux responsible for inducing
1 that e.m.f”
= 2 (10 )(4)2 = 80
The above statement is related to the
CHAPTER 5____________________________ definition of Lenz’s Law.
Q1
Q7
A Magnetism is defined as the force
produced by charge particles
(electrons) of magnet.

Q2

A permanent magnet is a piece of Magnetic attraction – unlike poles
ferromagnetic material (such iron,
nickel or cobalt) which has properties
of attracting other pieces of these
materials.

Q3

 The current in a conductor Magnetic repulsion – similar poles/two likes
produces a magnetic field. poles
The cause of magnetic field is
called the magnetomotive Q8
force (mmf) A permanent magnet is a piece of
ferromagnetic material which has
 The unit of mmf, the ampere properties of attracting other pieces
turns (AT) is established on of these materials.
the basic of the current in a Two (2) examples of magnet:
single loop (turn) of wire. i. Iron
= ii. Nickel

Q4 Q9

i. Forming a closing loop i. Forming a closing loop
ii. Did not crossed against each ii. Did not crossed against each

other other
iii. Has a certain direction iii. Has a certain direction
iv. Repel between one another iv. Repel between one another
v. Has a tension along its v. Has a tension along its

distance where it will tends to distance where it will tends to
make them as short as
possible.
Q5

i. Compass
ii. Screw rule
iii. The right hand rules

128 Electrical Technology Work Book Q 12
make them as short as Faraday’s First Law stated that
possible. electromotive force will be induced
whenever a conductor is placed in a
varying magnetic field or whenever a
conductor is rotated in magnetic field.

Magnetic field characteristics Q 13
Q 10

Given R(4e l ×u1c0ta−7n)c(e2300 0 0 )=(15 0 × 1 0 − 6=) = 795.77 ×
Number of turns = 200, Dimension, D = 106 /
600mm, Current, I = 4A.

= Q 14

= 4 × 200 Given

= 800 Cast steel 1

= ℎ, 1 = 6 ,

= (600 × 10−3) − , 1
= 1 2, , =
= 1.88 0.4

800 Cast steel 2
1.88
= = = 424.36 / ℎ, 2 = 2 ,

Q 11 − , 2
= 0.5 2

i. Current flows in the opposite direction , = 750, 0
ii. Current flows in the opposite direction = 4 × 10−7

Find the reluctance of cast steel first.

For the 6cm long path, the reluctance,
S is

1 = 1 × 10−2 × 10−4 )
= (4 × 1 (750)(0.5

6

10−7)
= 6.366 × 105 /

Electrical Technology Work Book 129

For the 2 cm long path, the reluctance =
is

2 = 2 = = (600 × 10−3) = 1.88
= 2
2 × 10−2 Therefore, the magnetic field
(750)(0.5 strength, H is
(4 × 10−7) × 10−4 )
= 4.244 × 105 / 800
= = 1.88 = 424.36 /

Total circuit reluctance

= 1 + 2 = (6.366 + 4.244) Q 17
× 105
= 10.61
× 105 / Given

Then, calculate the flux using this formula; current = 0.3A, coil = 500 turns, flux, ∅
= 536 , crossectional area (A)
, ∅ = = 0.4 × 200 = 1800mm2 and length,
10.61 × 105 = 150 , 0 = 4 × 10−7
= 7.5 × 10−5 Find the magnetomotive force (mmf) for the
magnetic circuit using this formula
Therefore, the flux density, B in the 2cm path

= ∅ = 7.5 × 10−5 = 1.51 = = 0.3 × 500 = 150
0.5 × 10−4
Find the reluctance of the circuit, if the
Q 15 flux (∅)given is536μWb.

Given, = = 150 = 2.8 × 105 /
∅ 536
, ∅ = 2
− , Find the magnetic flux density (B), if the cross
= 10 2 sectional area (A) of the conductor is
1800mm2.
Calculate the flux density, B using this formula
∅ 536
= ∅ = 2 × 10−3 = 2 = = 1800 × 10−6 2 = 298
10 × 10−4
Find the
Q 16 permeability,
if the length, of conductor is 150mm
Given using this formula

=
= 200, ,
= 600 , ,
= 4 =

Firstly, determine the Magneto
motive force, Fm 150 × 10−3
(4 × 10−7)(2.8 × 105)(1800 × 10−6)
= = (4)(200) = 800 =

Then, find the length of coil using this = 237 −1
formula
= = (4 × 10−7)(237)
= 0.298 × 10−3 /

130 Electrical Technology Work Book

Q 18 Thus 1 = ( = 1 ( 2()05 . 8×=)(120 ×− 1 2 1)(07−540))
= 1172000
Given

, = 5

Find the magnetic flux using this Find the reluctance for the air gap:
formula

= ∅ Reluctance,

2 2
5 = 40 ∅ 2 = 2 = 2
× 10−4

∅ = 5(40 × 10−4) = 0.02 (Since =1 for air)

Q 19 = (4 1 × 10−3
× 10−7)(2 × 10−4)
Given = 3979000H

, ∅ = 400 × 10−6 , Total reluctance,
− ,
= 0.0005 2 = 1 + 2 = 1172000 + 3979000
= 5151000H
Find the density, B using this formula
Find the flux,
∅ 400 × 10−6
= = 0.0005 = 0.8 ∅ = = 0.80 × 2 × 10−4
= 1.6 × 10−4
Q 20.
Finally, use the reluctance’s formula
Given to find the current

= 2 2, = =
25 , , = ∅
5000 , , = 0.8
Then

Find the reluctance for cast steel core: = ∅

When B= 0.80T, H= 750A/m: Note that

Reluctance of core, =

1 So,
1
1 = = ∅

= The current is

Then = = ∅


= (5151000)(1.6 × 10−4)
5000
= 0.165 A

Electrical Technology Work Book 131