ELECTRICAL TECHNOLOGY WORK BOOK print

Electrical Technology Work Book 51

EXCERCISE

Question 1

Define a capacitor and state its symbol, unit and basic formula.

Answer

A capacitor is an electrical device that is used to store electrical energy.

The symbol of capacitance is C.

The unit of capacitance is Farad (F).

The basic formula

= ( ) ( )
( )
Question 2

What is a capacitance?

Answer

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

Question 3
Draw and label the basic construction of capacitor.
Answer

52 Electrical Technology Work Book
Question 4
Sketch a symbol of unpolarised, polaries and variable capacitors.
Answer

Question 5
Give TWO (2) types of capacitor are polarized capacitor.
Answer
Aluminium
Tantalum

Question 6
Give four types of capacitor are unpolarized capacitor.
Answer
Mica
Ceramic
Film
Paper

Electrical Technology Work Book 53

Question 7
Sketch and explain a capacitor construction.
Answer

A capacitor consists of two metal plates separated by a certain distance d, in between the plates
lies a dielectric material with dielectric constant =εoε, where εo is the dielectric of air.
The dielectric material allows for charge to accumulate between the capacitor plates. Air (actually
vacuum) has the lowest dielectric value of εo = 8.854 x 10-12 Farads/meter.
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.

Question 8

Give THREE (3) factor that effecting capacitance.

Answer

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

54 Electrical Technology Work Book
Question 9
Find the total capacitor the circuits in Figure 38.

Figure 36

Answer

= 120 + 60 + 20 = 200 ( )

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1 = 1 + 1 + 1 = 1 + 1 + 1 = 42.5k (series)
40 80 40 80

Therefore

= 1 = 23.53μ
42.5
Then, the total Capacitance, CAB:

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

56 Electrical Technology Work Book
Question 10
Determine the total capacitor the circuits in Figure 39.

Figure 37

Answer

Figure 38 = 350 + 400 = 750 ( )
Then
Therefore 1 = 1 + 1 + 1 = 1 + 1 + 1 = 8.33
200 500 200 500

= 120 × 10−6

Electrical Technology Work Book 57
Question 11
Calculate the total capacitor the circuits for Figure 41a and 42b

Figure 39a

Figure 40b

Answer

58 Electrical Technology Work Book

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

Find Cb

Electrical Technology Work Book 59

1 = 1 + 1 = 1 + 1 = 0.61 ( )
9 2
1
= 0.61 = 1.63

b) Simplify the circuit first = 4 + 1.63 = 5.63

Find Ca and Cb

60 Electrical Technology Work Book

Find Ctotal = 30 + 20 = 50
= 40 + 50 = 90

1 = 1 + 1 + 1 = 1 + 1 + 1 = 131.11
10 10 50 90

= 7.627 × 10−6 = 7.627

Electrical Technology Work Book 61

Question 12

Three capacitors, 100pF, 200 pF and 500 pF are connected in series.Find the total capacitance,
CT and electric charge when 15 V is applied across its plates.

Answer

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

Question 13

Calculate the equivalent capacitance of two capacitors of 10μF and 620μF connected:
a) in parallel
b) in series

Answer = 10 + 620 = 630
in parallel
in series

1 = 1 + 1 = 101.61
10 620

= 1 = 9.84 × 10−6 = 9.84
101.6

62 Electrical Technology Work Book

Question 14

A 0.4μF capacitor is charged to 250 V before being connected across a 6 kΩ resistor. Determine:

a) The initial discharge current
b) The time constant of the circuit

Answer
Given

= 0.4 , = 250 , = 6 Ω

(a) The initial discharge current

= = 250
6

(b) The time constant of the circuit

 = = 4 × 6 = 0.024

Question 15

Refer to Figure 43, if the switch is closed, determine

a) Initial current charge
b) Initial potential difference through capacitor.
c) Time constant
d) Time taken for capacitor fully charges

Figure 41

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Answer

a) Initial current charge

= = 200 = 2
100

b) Initial potential difference through capacitor.

= 0

c) Time constant
 = = 20 × 100 = 2

d) Time taken for capacitor fully charges

5 = 5 2 = 0.01

Question 16

A circuit connect in series, consists of a resistor with a 0.8μF capacitor and has a time constant of
10 ms. Determine:

a) The value of the resistor
b) The capacitor voltage, 10 ms after connecting the circuit to a 15 V supply

Answer
Given

= 0.8 ,  = 10 , = 15

(a) The value of the resistor

 =
10 = (0.8) ( )

= 12.5 Ω

(b) The capacitor voltage, 10 ms after connecting the circuit to a 15 V supply

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= (1 − − )

= 15(1 − −1100)


= 15( 1 – 0.37 )


= 15(0.63)


= 9.45

Question 17

A 40μF capacitor is connected in series with a 60 kΩ resistor and the circuit is connected to a 25
V, D.C. supply. Calculate:

a) The initial value of the current,
b) The time constant of the circuit,
c) The value of the current one second after connection,
d) The value of the capacitor voltage three seconds after connection,
e) The time after connection when the capacitor voltage is 18 v

Answer

Given

= 40 , = 60 Ω, = 25

a) The initial value of the current,

= = 25 = 0.416
60

b) The time constant of the circuit,

 =
 = 40 × 60 = 2.4


c) The value of the current one second after connection,

= ( −  )

= ( −21.4 )

= 0.41 (0.659)
= 0.27


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

Electrical Technology Work Book 65

= (1 − − )

= 25(1 − −23.4 )
= 25( 1 – 0.287 )

= 25(0.713)
= 17.825

e) The time after connection when the capacitor voltage is 18 v

= (1 − − )

18 = 25(1 − −2 . 4 )



18 = (1 − −2 . 4 )
25


0.72 = 1 − −2 . 4



−2 . 4 = 1 – 0.72



−2 . 4 = 0.28



−2 . 4 = 0.28


2.4
− = − 1.273



= 3.055

66 Electrical Technology Work Book

Question 18

A capacitor is charged to 200 V and then discharged through a 70 kΩ resistor. If the time constant
of the circuit is 0.5 s, determine:

a) The value of the capacitor,
b) The time for the capacitor voltage to fall to 30 v,
c) The current flowing when the capacitor has been discharging for 0.8s

Answer
Given

= 200 , = 70 KΩ,  = 0.5S

(a) The value of the capacitor,

 =


0.5 = ( )(70 )


= 7.14
The time for the capacitor voltage to fall to 30 v,

= (1 − − )

30 = 200(1 − −0 . 5 )



30 = (1 − −0 . 5)
200


0.15 = 1 − −0 . 5



−0 . 5 = 1 – 0.15



−0 . 5 = 0.85



−0 . 5 = 0.85


0.5
− = − 0.163



= 0.082

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(c) The current flowing when the capacitor has been discharging for 0.8s

= = 200 = 2.86
70
= ( −  )

= ( −00.0.882 )

= 2.86 (0.06 )
= 0.17

Question 19

An 10F capacitor is connected in series to a 0.5MΩ resistor across the dc voltage supply of
240 V. Determine:

a) Time constant
b) Initial charging current
c) Time for capacitor voltage increases to 150 V
d) The current flowing through the capacitor after 4 seconds.
e) Energy stored in the capacitor when it is fully charge
f) Sketch the current and voltage curve to show the process of charging the capacitor.

Answer = 10 , = 0.5 , = 240
Given
a) Time constant =
= 0.5 (10µ) = 5
b) Initial charging current

= = 240 = 0.48
0.5

c) Time for capacitor voltage increases to 150 V
= (1 − − )

150 = 240 (1 − −5 )

150 = (1 − −5 )
240
0.625 = 1 − −5

−5 = 1 − 0.625

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−5 = 0.375

ln( −5 ) = ln(0.375)

− 5 = −0.981

= 4.9

d) The current flowing through the capacitor after 4 seconds.

= ( − / )
= 0.48 ( −4/5)

= 0.48 (0.449)

= 0.216
e) Energy stored in the capacitor when it is fully charge

= 1 2
2
1
= 2 (10 )(240)2 = 0.288

f) Sketch the current and voltage curve to show the process of charging the capacitor.

V

VMax

Voltage across

capacitor:

vc = Vmax (1 – e –t/)

t

Electrical Technology Work Book 69

Question 20

A capacitor with a capacitance of 30F which is connected in series to a 200KΩ resistor is being
placed a 200 DC voltage supply. Calculate initial current, initial potential different across capacitor,
the time constant during charging the time taken to be fully charge and the energy stored in the
capacitor.

Answer

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

= = 200 = 1
200

The initial potential different across capacitor
= 0

The time constant during charging the time taken to be fully charge

Fully charge equal to five time  =
 = 30 × 200 = 6

5 = 5 × 6 = 30

The energy stored in the capacitor.

= 1 2
2
1
= 2 (30 )(200)2 = 0.6

70 Electrical Technology Work Book

CHAPTER FOUR: INDUCTOR AND INDUCTANCE

SHORT NOTE

 An inductor is made of a coil of conducting wire and a passive element designed to store
energy in the magnetic field while a capacitor stores energy in the electric field.

 The unit ([Henry] or [H]) is named for Joseph Henry, and is equal to a [Volt-second/Ampere].
 Inductance is the physical property of a circuit that opposes any change in current flow.
 Symbol for inductance is the letter L.
 Unit of measurement for Inductance is henry and symbol for henry is the letter h or H. There

are two types of Inductor i.e. fixed inductor and variable inductor.

v  d  L di
dt dt

 When inductors are connected in series, the total inductance is;

Leq = L1 + L2 + L3 + … + Ln
 When inductors are connected in parallel, the total inductance is;

1/Leq = 1/L1 + 1/L2 + 1/L3 + … + 1/Ln

 Electromagnetic Induction is when Faraday discovered that a changing magnetic flux leads to
a voltage in a wire loop. Induced voltage (emf) causes a current to flow.

Faraday’s Law of Induction:

 N  B
t

induced rate of change
emf of flux with time

number
of loops

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 The faster the change, the larger the induced emf where is the the induced emf is a voltage.

Self-Inductance for a solenoid (toroidal inductor) :

L  N 2A  N 2ro A
 

N is the number of turns of wire

A is the cross-sectional area of the toroid in m2.

mr is the relative permeability of the core material

mo is the vacuum permeability (4π × 10-7 H/m)

l is the length of the wire used to wrap the toroid in meters

 Factors Affecting Inductance are inductance (L), number of turns (N), cross sectional area of
the core (A), permeability of core material (μ) and length of the coil (l).

Energy stored in inductor: 2

E = ½ LI

Graph and formula of Rise and Decay of Current

The Voltage VL and Current iL
during the Storage Phase and
Discharge (Decay) Phase

 The larger the Resistance, the smaller the Time Constant, the faster the Inductor
stores the energy and decays the energy, and vice versa.

72 Electrical Technology Work Book

EXCERCISE

Question 1
State the symbol and unit of inductance
Answer
Symbol of inducdor is L and unit of inductance is Henry(H)
Question 2
By using suitable diagram,explain briefly about first Faraday’s Law.
Answer

Faraday’s law says that:
 an emf is induced in a loop when it moves through an electric field
 the induced emf produces a current whose magnetic field opposes the original change
 the induced emf is proportional to the rate of change of magnetic flux

Question 3
a. List FOUR (4) factors that influence inductances.
b. Give the definition of inductances and state its unit.

Answer
Number of turns
PermeabilityCross-sectional area of coil
Length of the coil
Inductance is the physical property of a circuit that opposes any change in current flow and unit of
measurement for inductance is Henry (H).

Electrical Technology Work Book 73

Question 4

State the symbol for inductor and unit of inductance.

Answer

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

Question 5

First Law of Faraday’s Electromagnetic Induction state that emf is induced whenever a conductor
is rotated in magnetic field which are induced emf. Explain briefly the Second Law of Faraday’s
Electromagnetic Induction.

Answer

Second Law of Faraday’s Electromagnetic Induction state the induced emf is equal to the rate
change of flux linkages which is the product of turns i.e n of the coil and the flux associated with
it.

Question 6

State the current and voltage equation of capacitor when it is charging.

Answer

Current equation : ( ) = (1 − − )
Voltage equation : ( ) = (1 − − )

( ) = − /

74 Electrical Technology Work Book
Question 7
State the formula of the total inductance LT in Figure 44 and 45

Figure 42

Figure 43

Answer

For Figure 1a (parallel circuit) :

1 = 1 + 1 + 1 + 1
1 2 3 4

= 1

1 + 1 + 1 + 1
1 2 3 4

For Figure 1b (series circuit) :

= 1 + 2 + 3

Electrical Technology Work Book 75

Question 8

Based on Figure 46, calculate :
a) the total inductance
b) energy stored in the circuit.

Figure 44

Answer

the total inductance
Simplify the circuit first

= 2 + 5 + 10 = 17

= 2// = 2 ×
2 +
3 × 17
= 3 + 17 = 2.55

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= 20 + 2.55 + 4 = 26.55

energy stored in the circuit.

= 1 2
2
1
= 2 (26.55 )(3)2 = 1.912

Question 9
State the formula of self inductance.
Calculate the total inductance for each parallel circuit in Figure 47 and 48.

Figure 45

Figure 46

Answer

The formula of self inductance

L  N 2A  N 2r o A


Electrical Technology Work Book 77

The total inductance for figure 6a

= 1 = 1 × 2 = 10 × 50 = 8.33
2 1 + 2 10 + 50

The total inductance for figure 6b

Simplify the circuit

= 2// 3

= 2 × 3 = 10 × 20 = 6.67
2 + 3 10 + 20

= 1// = 1 × = 40 × 6.67 = 5.72
1 + 40 + 6.67

Question 10
Calculate the value of Lx if the equivalent inductance for the circuit in Figure 49 is 3.5 H.

Figure 47

Answer
Simplify the circuit first

= (3.5// ) + 1.5
× 3.5
3.5 = + 3.5 + 1.5

3.5 − 1.5 = × 3.5
+ 3.5
× 3.5
2= + 3.5

2 ( + 3.5) = × 3.5

2 + 7 = 3.5

7 = 3.5 − 2

7 = 1.5

∴ = .

Question 11

State the current and voltage equation of capacitor when it is discharging.

Answer

Current equation : ( ) = ( − / )
Voltage equation : ( ) = − − /

Question 12

Calculate the equivalent inductance, LT as shown in Figure 50 below.

Figure 48

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

= 3 + 7 = 100 + 50 = 150

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Find the Lb (pallel circuit between La and L6)

1 = 1 + 1
6
× 6 150 × 25
= + 6 = 150 + 25 = 21.43

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


= + 2 = 21.43 + 200 = 221.43
Find the Ld (parallel circuit between L5 and Lc)

= 5 × = 10 × 221.43 = 9.57
5 + 10 + 221.43

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

Electrical Technology Work Book 81

= + 1 = 9.57 + 40 = 49.57
Therefore, the total inductance is

= × 4 = 49.57 × 30 = 24.96
+ 4 49.57 + 30

Question 13

Define electromagnetic induction.

Refer to Figure 51 below. Calculate the equivalent inductance.

Figure 49

82 Electrical Technology Work Book

Answer

Electromagnetic induction is where a voltage or current is produced in a conductor by a changing
magnetic flux.
Find the equavalent circuit

Simplify the circuit

= 1 + 2 + 3 = 80 + 10 + 60 = 150
= 4 + 5 = 50 + 20 = 70

= //

= × = 150 × 70 = 47.73
+ 150 + 70

Therefore, the total inductance is

= 6//
× 47.73 × 40
= + = 47.73 + 40 = 21.76

Electrical Technology Work Book 83

Question 14

By referring to Figure 52, with switch in position 1. Sketch the graph for rise of current in inductor
circuit. Calculate :

a) Time constant
b) Time taken for current achieve maximum value
c) Instantaneous current at t = 0.5s
d) Energy stored in inductor

Figure 50

Answer

Sketch the graph for rise of current in inductor circuit.

Calculate :

Time constant

= = 15 = 1.5
10 Ω

Time taken for current achieve maximum value
5 = 5 (1.5 ) = 7.5

84 Electrical Technology Work Book

Instantaneous current at t = 0.5s

( ) = (1 − − )

( ) = (1 − − )

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

( ) = 10 ( 1 − −0.33 )
( ) = 10 ( 1 − 0.719 ) = 10 − 7.19

∴ ( ) = 2.81

Energy stored in inductor = 1 2
Question 15 2
1
= 2 (15)(10)2 = 750

Sketch the waveform of rise and decay of current.
An inductor of 20 mH has a current of 2.5 A flowing in it. Find the energy stored in the magnetic
field of the inductor.

Answer

Electrical Technology Work Book 85

b) Given = 20 , = 2.5
Question 16
= 1 2
2
1
= 2 (20 )(2.5)2 = .

The circuit are consist of one inductor of 10 mH which is connected in series with the resistor of
1kΩ and the input of DC voltage is 150 V. Determine :

a) The time constant
b) The initial current

Answer = = 10 = μ
The time constant 1 Ω

The initial current

= = 150 = .
1 Ω

86 Electrical Technology Work Book

Question 17

The current 5A is flowing in a solenoid with 250 turns and 20mm length. Given that the solenoid
diameter is 10mm, µr is 1000 and µo is 1.256 x 10-6. Find the value of inductance that exists in
the solenoid.

Answer

Given

= 250

= 5

= 1000
0 = 1.256 10−6
= 20 = 20 × 10−3 = 0.02
= 10 = 10 × 10−3 = 0.01
= 2 = (0.201)2 = 78.54
Find the value of inductance

L  N 2A  N 2ro A
 

= ( )( . × − )( )( . )
× −
∴ = .

QUESTION 18

An 80 mH inductor is connected in series with 5Ω resistor. The circuit is applied with 15 dc power
supply. Determine the energy stored in the inductor.

Answer

= = 15 =
5Ω
1
= 2 2

= 1 (80 )(3)2 = 0.36
2

Electrical Technology Work Book 87

Question 19

A circuit has a 30Ω resistor that is connected in series with a 1.5 H inductor. The circuit is
connected to 120V power supply. Determine :

a) The time constant
b) The current at 0.05 s
c) Time taken to increase to 2.5 A

Answer = = 1.5 = 0.05
The time constant 30Ω

The current at 0.05s

( ) = (1 − − )

= = 120 = 4
30
( ) = 4 (1 − −00..0055 )

( ) = 4 ( 1 − −1 )
( ) = 4 ( 1 − 0.368 ) = 4 (0.632)

∴ ( ) = .

Time taken to increase to 2.5 A

( ) = (1 − − )

2.5 = 4 (1 − −0. 0 5)

2.5 = 1 − −0. 0 5
4
0.625 − 1 = − −20

−0.375 = − −20

ln(0.375) = ln( −20 )

ln 0.375 = − 20

− 0.981 = −20

∴ = .

88 Electrical Technology Work Book
Question 20
From the Figure 53 below, find :

Figure 51

a) τ, time constant
b) IL when open switch
c) IL when switch closed (starting current)
d) IL when t = 5 second
e) Voltage drop at resistor, VR

f) Voltage drop at inductor, VL
g) Energy stored when archive maximum.

Answer

τ, time constant

= = 10 =
5Ω

IL when open switch

=

IL when switch closed (starting current)

= = 20 =
5

IL when t = 5 second

( ) = (1 − − )

( ) = 4 (1 − −25 )

( ) = 4 ( 1 − −2.5 )
( ) = 4 ( 1 − 0.082 ) = 4 (0.918)

∴ ( ) = .

Voltage drop at resistor, VR
= × = 3.672 × 5 = .

Electrical Technology Work Book 89

Voltage drop at inductor, VL

= − = 20 – 18.36 = .
Energy stored when archive maximum.

= 1 2
2
1
= 2 (10 )(4)2 = 80

90 Electrical Technology Work Book

CHAPTER FIVE: MAGNETIC CIRCUIT ELECTROMAGNETISM
& ELECTROMAGNETIC INDUCTION

SHORT NOTE

The magnetism terms and definition

i. Magnetomotive Force @ mmf, Fm

 The current in a conductor produces a magnetic field. The cause of magnetic field

is called the magnetomotive force (mmf)

 The unit of mmf, the ampere turns (AT) is established on the basic of the current

in a single loop (turn) of wire.

=
ii. Magnetic Field Intensity @ Magnetizing force, H

 Identifies the magnetic flux density per unit length of a coil

 The unit is ampere turns/meter (AT/m)

= = =

= ℎ, = 2 = ℎ

iii. Magnetic Flux Density, B

 The amount of flux per unit area, perpendicular to the magnetic field.

 The unit is weber/meter2 (Wb/m2) or Tesla(T)

= ∅

iv. Reluctance, S

 The opposition to the production of flux in a material, which corresponds to

resistance

 The unit is Ampere turns / Weber (AT/Wb)

= = =
∅ ∅ ∅

Electrical Technology Work Book 91

v. Permeability, µ

 The permeability is the ability of a material to concentrate magnetic lines of

flux. The higher the permeability, more easily a magnetic field can be

established.

 The absolute permeability, = , for air, vacuum and other non-magnetic

material, = 1, so =

 Permeability of free space, = 4 × 10−7, (H/m), for air =


 Relative permeability, = ( ) =
( )

Composite series magnetic circuit

For a series magnetic circuit having n parts, the total reluctance S is given by:
= 1 + 2 + 3 + ⋯ +
(This similar to resistor connected in series in a electrical circuit)

Comparison between electrical and magnetic quantities

Electrical circuit Magnetic circuit

e.m.f, E (V) mmf, Fm (A)

Current, I (A) Flux, ϕ (Wb)

Resistance, R (Ω) Reluctance, S ( −1)

I = E ϕ =
R
ρl l
R = A S = μo μr A

92 Electrical Technology Work Book

EXERCISE

Question 1

Define the Magnetism in Electrical Technology.

Answer

A Magnetism is defined as the force produced by charge particles (electrons) of magnet.

Question 2

Define the permanent magnet in magnetism.

Answer

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

Question 3

Give the definition of Magnetomotive Force (mmf) in Electrical Technology

Answer

The current in a conductor produces a magnetic field. The cause of magnetic field is called the

magnetomotive force (mmf)

The unit of mmf, the ampere turns (AT) is established on the basic of the current in a single loop

(turn) of wire.

=

Question 4

State FIVE (5) characteristics of magnetic field in Electrical Technology

Answer

Forming a closing loop
Did not crossed against each other
Has a certain direction
Repel between one another
Has a tension along its distance where it will tends to make them as short as possible.

Electrical Technology Work Book 93

Question 5

List THREE (3) methods which are used to determine the magnetic field direction.

Answer

Compass
Screw rule
The right hand rules

Question 6

“The direction of an induced e.m.f is always such that it tends to set up a current opposing the
motion or the change of flux responsible for inducing that e.m.f”

The above statement is related to the definition of _________________________.

Answer

“The direction of an induced e.m.f is always such that it tends to set up a current opposing the
motion or the change of flux responsible for inducing that e.m.f”

The above statement is related to the definition of Lenz’s Law.

Question 7

Draw the laws of magnetic attraction and repulsion by using two bar magnets.

Answer

Magnetic attraction – unlike poles

94 Electrical Technology Work Book

Magnetic repulsion – similar poles/two likes poles

Question 8
Define permanent magnet and give TWO (2) examples of magnet
Answer
A permanent magnet is a piece of ferromagnetic material which has properties of attracting other
pieces of these materials.
Two (2) examples of magnet:
Iron
Nickel
Question 9
List and draw the characteristics of magnetic field line.
Answer
Forming a closing loop
Did not crossed against each other
Has a certain direction
Repel between one another
Has a tension along its distance where it will tends to make them as short as possible.

Electrical Technology Work Book 95

Magnetic field characteristics

Question 10

A coil of 200 turns is wound uniformly over a wooden ring which has mean circumference of
600mm. If current through the coil is 4A, calculate the magnetic field strength, H.

Answer

Given

Number of turns = 200, Dimension, D = 600mm, Current, I = 4A.

=

= 4 × 200

= 800

=

= (600 × 10−3)

= 1.88

= = 800 = 424.36 /
1.88

96 Electrical Technology Work Book

Question 11

Ilustrate the magnetic lines of two fields for parallel conductors in the following condition
i) ii)

Answer ii. Current flows in the opposite direction

i. Current flows in the opposite direction

Question 12

Explain first Faraday’s Law by using suitable diagram.

Answer

Faraday’s First Law stated that electromotive force will be induced whenever a conductor is placed
in a varying magnetic field or whenever a conductor is rotated in magnetic field.

Question 13
Calculate the reluctance of an iron ring which has a length of 300mm, cross sectional area of
150mm2 and relative permeability of the iron ring is 2000. (Assume:μo = 4 × 10−7H/m)

Answer

Reluctance = = 300 = 795.77 × 106 /
(4 ×10−7)(200)(150×10−6)

Electrical Technology Work Book 97

Question 14

A closed magnetic circuit of cast steel contains a 6cm long path of cross-sectional area 1cm2 and
a 2cm path of cross-sectional area 0.5cm2. A coil of 200 turns is wound around the 6cm length of
the circuit and a current of 0.4A flows. Determine the flux density in the 2cm path, if the relative
permeability of the cast steel is 750.

Answer
Given
Cast steel 1

ℎ, 1 = 6 , − , 1 = 1 2, , = 0.4
Cast steel 2

ℎ, 2 = 2 , − , 2 = 0.5 2

, = 750, 0 = 4 × 10−7

Find the reluctance of cast steel first.

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

1 = 1 = (4 × 6 × 10−2 × 10−4 ) = 6.366 × 105 /
1 10−7) (750)(0.5

For the 2 cm long path, the reluctance is

2 = 2 = (4 × 2 × 10−2 × 10−4 ) = 4.244 × 105 /
2 10−7) (750)(0.5

Total circuit reluctance

= 1 + 2 = (6.366 + 4.244) × 105 = 10.61 × 105 /
Then, calculate the flux using this formula;

, ∅ = = 0.4 × 200 = 7.5 × 10−5
10.61 × 105
Therefore, the flux density, B in the 2cm path

= ∅ = 7.5 × 10−5 = 1.51
0.5 × 10−4
Question 15

If the total flux in a magnetic circuit is 2mWb and the cross-sectional area of the circuit is 10cm2,
calculate the flux density.

Answer

Given,

, ∅ = 2 − , = 10 2

Calculate the flux density, B using this formula

= ∅ = 2 × 10−3 = 2
10 × 10−4

98 Electrical Technology Work Book

Question 16

A coil of turns is wound uniformly over of 200 turns is wound uniformly over a wooden ring which
has a mean circumference of 600mm. If current through the coil is 4A, calculate the magnetic field
strength, H.

Answer

Given

= 200, , = 600 , , = 4
Firstly, determine the Magneto motive force, Fm

= = (4)(200) = 800

Then, find the length of coil using this formula

=

= = (600 × 10−3) = 1.88

Therefore, the magnetic field strength, H is

= = 800 = 424.36 /
1.88

Question 17

Given the value of the current = 0.3A, coil = 500 turns, flux, ∅ = 536 , crossectional area (A)
= 1800mm2 and length, = 150 . Calculate magnetomotive force (mmf), reluctance of the
circuit, magnetic flux density (B) and permeability, µ by referring to figure 1.

Figure 52

Answer
Given

current = 0.3A, coil = 500 turns, flux, ∅ = 536 , crossectional area (A)
= 1800mm2 and length, = 150 , 0 = 4 × 10−7

Find the magnetomotive force (mmf) for the magnetic circuit using this formula
= = 0.3 × 500 = 150

Find the reluctance of the circuit, if the flux (∅)given is536μWb.

Electrical Technology Work Book 99

= = 150 = 2.8 × 105 /
∅ 536

Find the magnetic flux density (B), if the cross sectional area (A) of the conductor is 1800mm2.

= ∅ = 536 = 298
1800 × 10−6 2

Find the permeability, if the length, of conductor is 150mm using this formula

=


=

= (4 × 150 × 10−3 × 10−6) = 237 −1
10−7)(2.8 × 105)(1800

= = (4 × 10−7)(237) = 0.298 × 10−3 /

Question 18

The flux density in an iron-core is 5T. If the area of the area core is 40cm2. Calculate the magnetic
flux in weber units.

Answer

Given

, = 5
Find the magnetic flux using this formula

= ∅

∅
5 = 40 × 10−4

∅ = 5(40 × 10−4) = 0.02

Question 19

With a flux of 400µWb through an area of 0.0005m2, what is the flux density B in Tesla units?

Answer

Given

, ∅ = 400 × 10−6 , − , = 0.0005 2

Find the density, B using this formula

= ∅ = 400 × 10−6 = 0.8
0.0005

100 Electrical Technology Work Book
Question 20.

Figure 53

A section through a magnetic circuit of uniform cross – sectional area 2cm2. The cast steel core
has a mean length of 25cm. The air gap is 1mm wide and the coil has 5000 turns. The B-H curve
for cast steel is shown in Figure 55. Determine the current in the coil to produce a flux density of
0.80T in the air gap, assuming that all the flux passes through both parts of the magnetic circuit.

Answer

Given
= 2 2, = 25 , , = 5000 , , = 0.8

Find the reluctance for cast steel core:

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

Reluctance of core,

1 = 1 =
1

Then =

Thus 1 = 1 = 1 = (25 × 10−2)(750) = 1172000
( ) (0.8)(2 × 10−4)