# EEC-125-Electrical-Engineering-Science-2-Theory-1

```UNESCO-NIGERIA TECHNICAL
&amp; VOCATIONAL EDUCATION
REVITALISATION PROJECTPHASE II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERING TECHNOLOGY
ELECTRICAL SCIENCE II
COURSE CODE: EEC125
YEAR I- SEMESTER II
THEORY
Version 1: December 2008
1
Department
Electrical Engineering Technology
Subject
Electrical Engineering Science II
Year
1
Semester
2
Course Code
EEC 125
Credit Hours
2
Theoretical
1
Practical
2
CHAPTER 1 :
•
To Study the Characteristics of an Inductor
Week 1
CHAPTER 2 :
•
Determination of Series and parallel Resistors
Weeks 2
connection Using Kirchoff’s Law
CHAPTER 3 :
•
Determination of Series and parallel Resistors
Weeks 3
connection Using Kirchoff’s Law
CHAPTER 4 :
Weeks 4
To demonstrate Faraday’s and Lenz’s Laws
CHAPTER 5 :
•
Demonstration of Self and Mutual Inductance
Weeks 5
2
CHAPTER 6 :
•
Determination of inductance of a coil by Voltage
Weeks 6
and Current Measurements
CHAPTER 7 :
•
Magnetic Effect of Current Carrying Conductor
Weeks 7
CHAPTER 8 :
•
Determination of Mutual Inductance and Polarity
of Magnetically coupled coils.
Weeks 8
CHAPTER 9 :
•
Determination of Mutual Inductance and Polarity
of Magnetically coupled coils
Weeks 9
CHAPTER 10 :
•
The Effects of Saturation in a Magnetic Circuit
CHAPTER
•
Weeks 10
11 :
Determination of Hysteresis loop of an iron
Weeks 11
sample
CHAPTER 12 :
•
To Study the Effect of LR and Undriven LRC
Circuits
Weeks 12
CHAPTER 13 :
•
To Study Free Oscillations of the RLC Circuit
Weeks 13
3
CHAPTER 14 :
•
To study the charging and discharging of a
capacitor.
Weeks 14
CHAPTER 15 :
•
Demonstrate the application of EM in Transformer
Weeks 15
4
Week 1
On completion of this Topic the student should be able to:
•
Define magnetic flux, magnetic flux density, magnetomotive force,
magnetic field strength, reluctance, permeability of free space
(magnetic constants), relative permeability.
•
State the symbol, units and relationships of the terms
•
State analogies between electrical and magnetic circuits.
•
Draw the electrical equivalent of a magnetic circuit with or without
air-gap.
•
Solve simple magnetic circuit problems.
•
Distinguish between soft and hard magnetic materials
•
•
State examples of soft and hard magnetic materials.
Explain the B - H Curves for soft and hard magnetic materials.
1.1
Basic definitions in magnetism and magnetic
Circuits
1.2 Introduction
For the benefit of readers who will need to study and understand well
materials in this chapter and the subsequent two chapters, it is advisable to
revise some of the basic facts that have been previously learnt in physics
under magnetism. in this connection, we do know that simple experiments on
magnets and magnetism have revealed the following facts:
(a) the magnetic effects of a magnet appear to emanate from its poles which,
in the case of a bar magnet, are located near each end. For this reason,
iron filings cling mainly round the ends of a bar magnet as shown in
Fig. 6.1 (a)
5
(a) bar magnet with cluster of
(b) bar magnet suspended by a
Fig. 1.1
(b) a bar magnet suspended freely by a thread in an horizontal place, always
comes to rest with its axis in a north - south direction- The pole which points
towards the north is called the north - seeking (N) pole; while the pole which
points to the south is called south - seeking (S) pole, as shown in Fig. 1.1(b).
(c) If a N pole of a magnet (say, bar magnet) is brought near to the N pole of
another magnet repulsion of the magnets takes place. However, if the N pole
of a magnet is brought near to the S pole of another magnet, attraction of the
magnets takes place.
In other words, like poles (that is, two N poles or two S poles) repel, and
unlike poles (that is, an N pole and S pole) attract.
(d) the influence of a magnet can pass both through space and through nonmagnetic media. If a magnet, for example, is placed on one side of a sheet of
paper, it can attract an iron pin or a steel pin positioned on the other side of
the paper. This shows that the influence of the magnet passes through the
paper. In conclusion, the region of space within which the influence of the
magnet can be detected is called the magnetic field of the magnet.
1.3 Magnetic fields
A region in space around a magnet where magnetic influence is felt is called
a magnetic field. A compass needle is a device which can be used to find the
direction of the magnetic field at different points. Conventionally, a
magnetic field consists of lines of m force.
Studies of magnetic fields produced by magnets under various conditions
reveal the following:
(a) lines of magnetic field never cross one another, (in other words,
every line of magnetic field is a closed line).
(b) the magnetic lines of force (which may be called magnetic flux) can
continue through the body of the magnet which produces the field, and
can consequently form a complete loop.
6
(c) part of the magnetic field which happens to be outside the magnet
normally traces its path from a north - to a south - pole.
Week 2
Fig- 2.2 shows different patterns of magnetic fields produced by some
arrangements of magnets (bar magnets and a Horse - shoe magnet)
Fig. 2.2 Different magnetic fields
At this juncture, it may interest readers to observe that in all the arrangements
in Fig. 2.2 above the magnetic lines are oval in shape and come out of the N
pole into the air and enter the S pole. Since the lines arc continuous, one
Should imagine them inside the magnet, passing from the south (S) pole to the
north (N) pole through the magnetic material.
2.3 The definitions
Magnetic flux,Φ (sounds phi-in Greek) is the number of lines of magnetic
force which emanate from or entering into a magnet. It is therefore not
surprising that a magnet with a strong magnetic field produces more magnetic
lines of flux than does a magnet with a weak field. The SI unit of magnetic
flux is the weber (symbol iVh), named after a German physicist, Weber (W).
7
2.4 Magnetic flux density
Magnetic flux density, B is a measure of the magnetic flux passing through a
unit area in a plane at right angles to the flux. The SI unit of magnetic flux
density is Tesla (T) or. Wb/m2,
That is. Magnetic flux density, B = •'-.[^/m2] or[Tesla,T]
(2.1)
where A is the area (in m2) through which flux &lt;)&gt; (Wb) passes•i. Example 6.1 Calculate the value of the magnetic flux density when a flux
of 50p,^&amp; passes through an area of 2cm2.
2.5 Magnetomotive Force (m.m.f)
This is the force which drives magnetic flux through a magnetic circuit (i.e.
the route or path which is followed by magnetic flux) and corresponds to
electromotive force (e.m.f.) in an electric circuit. It is usually measured in
ampere - turns. -—^&gt;^
From our basic knowledge of magnetism in the study of physics, we do
know that when a current flows through a wire conductor it causes a magnetic
flux to be established around the wire. Besides, when the current through the
conductor wire is increased, the magnetic field around it also increases.
Furthermore, if current. /. passes in a coil of N turns wound on a solenoid
(or with an air core), it causes not only magnetic field (magnetic flux) to be
established but it causes an increase in the magnetic flux.
For this reason, we can express mathematically that magnetomotive force
(m.m.f) is given by
m.m.f.. F = !\. Ampere - turns or Ampere [A] \
(6.2)
8
where /&quot;is the current flowing in the conductors of the coil and iVis the
number of turns on the coil. Here, we note that since the m.m.f. is equal to the
product (current x number of turns) its unit is, therefore Ampere - turn (AT}.
However, since the number of turns is merely a figure (which is a
dimensionless quantity), then the m.m.f can just have the dimension of current
(i.e. Ampere).
Example 2,2
Determine the m.m.f produced by a coil of 600 turns if a current of 5A flows
in it.
Solution
7=5.4, N =600 turns.
From Eq. (6.2),
m.m.f, F^ 5x600=3000/&lt;
2.6 Magnetic Field Intensity
The magnetic field intensity is the m.m.f. per unit length, / (in metre) of the
magnetic circuit.
Mathematically, it can be expressed as
Ampere/metre [A/m].
IN F Magnetic field intensity H = — = —,
(6.3)
Other names by which magnetic field intensity are known are magnetic field
strength or magnetising force.
Example 2.3
A coil of length 0 25m is wound with 1000 turns of wire and carries a
current of 5A. Determine the magnetic field intensity at the centre of coil.
2.6 Permeability
The permeability of a medium (such as air, magnetic materials and nonmagnetic materials) is a measure of how easy it is to set up a magnetic field in
9
that medium. At any point in a magnetic field, the ratio between the magnetic
flux density B and
the magnetic field strength H is called the permeability (\\) of the medium in
which the field exits.
we note that the above unit can also be deduced to mean Henry/metre, (i.e.
H/m), since 1 ATfWb = \j Henry . The permeability of free space or a vacuum
(with symbol, ^o )
has the value 4n x 10~7 ///m; that is, Ug =4n x 10~7 ///M. All non - magnetic
10
Week 3
3.1 Symbols, units and analogy between magnetic and electric
circuits
The magnetic terms so far encountered are presented in Table 6.1 with their
symbols, units and their electrical equivalents.
Table 3.1 Analogy between magnetic and electric circuits
S/N
Magnetic Circuit
Electric Circuit
Qty. Symbo Unit 1
Qty. Symbol Unit
1. 2.
3.
m.m.f. F Ampere-turn
e.m.f. &pound; Volt(^) current
Amp.(^) resistance R Ohm
flux 4&gt; Weber, (Wb)
{n}
Reluctance S Amp./weber 1 / (A^b) ( Q
4.
H.Ho A
Magneti- H Ampere-sing force
turn/metre
/
Is ^J
Elec. field E VolV strength
metre (F7m)
9
5.
6.
Flux density B Webers/sq. metre
(1 tesia3S
•
1 Wb/m2 )
current f) .4/m2 density
Permeability H webers/Amp .-turn- Conduc. CT Siemens/ metre
metre
RELATIONAL STATEMENTS/EQUATIONS
1.
Flux = m.m.f./reluctance
Current = e.m.f./resistance
2.
3.
Once the magnetic flux is set up it
requires no further supply of
energy
Total m.m.f.= ΦS1+
Energy supply is required
continuously to maintain the
flow of current.
Total e.m.f. = . IR.1
+IR2+IR3+...
3.4
Electrical Equivalent of Magnetic Circuits
3.4.1 Introduction
In section 3.3 above, we discussed the analogy between magnetic and electric
circuits. Now, we are in a better position to draw electrical equivalent diagram
of any given magnetic circuit. In the succeeding sections we shall discuss the
electrical equivalent diagram for the following magnetic circuits^ (i) seriesconnected magnetic circuits (ii) parallel-connected magnetic circuits (iii)
composite magnetic circurt6.4.2 Series Magnetic Circuits
Consider the magnetic circuit shown in Fig. 6.3(a) below.
Fig. 3.3(a) is a series-connected magnetic circuit because the same flux is
established in all parts of the circuit, assuming there is no magnetic leakage.
The circuit consists of an iron part of length ^ and air-gap of length /^. We
note the iron part of length /[ = AB + BC + CD + DE + EF and the air-gap is
10
made up of FA = l-i. A coil is uniformly wound on the iron part, and produces
an m.m.f. ofF=NI. The equivalent electric circuit diagram is shown in Fig.
6.3(b). The procedure for calculating the equivalent reluctance of a seriesconnected magnetic circuit is the same as for calculating the equivalent
resistance of a series-connected electric circuit, which is as follows:
Equivalent reluctance, S, =Si +S,[A/Wb],
(3.8)
where Si and S^ are the respective reluctances of the iron path and the air gap.
Hence,
following the Kirchhoffs law for a series-connected magnetic circuit, we
have;
F=F,+F,,
(3.9)
where, F^ == &lt;|)S, == m.m.f required to produce flux &lt;j&gt; in the iron
path, F-i = ^2 == m-m-i required to produce flux ^ in the airgap. FJ- = ^S^ = m.m.f required to produce flux ^ in the whole
circuit.
i.e.
or,
FT =
F^ + F^ == &lt;t&gt;5i + &lt;fr5z
&lt;t&gt;5^=({&gt;5i +^
or, Sy. = S^ + S^.
This final result explains the proof of Eq.(6.8) above.
N.B. Example 3.5 provided ahead illustrates this concept.
The assumption that there is no flux leakage in the air-gap is not accurate. In
practice there is some leakage flux, and the method of coping with this
problem is discussed immediately below.
Magnetic Leakage and Fringing
Two of the most disturbing phenomena in magnetic circuits are magnetic
leakage and fringing. Magnetic leakage has to do with magnetic flux which
leak away from its normal magnetic path in a magnetic material. This is
illustrated clearly in Fig. 6.4. Fringing concerns the phenomena whereby
some of the flux produced by the wi\ fringes or bypasses the air-gap at the
edges of the gap. This is also illustrated in Fig. 6.4. The ratio of the total flux
to useful flux is given by the magnetic leakage coefficient, which is
11
Leakage coefficient, =
total flux
AT
useful flux 4&gt;u
From Eq.(3.10) it is clear that leakage coefficient X , has a value greater than
unity, since ^^ &gt; &lt;t&gt;u. In pratice, the typical value of the leakage coefficient
ranges between 1.05 to 1.4.
3.4 ParaIlel-connccted Magnetic Circuits
Fig. 6.5 is a parallel magnetic circuit consisting of two parallel magnetic
paths, acted upon by the same m.m-f. The parallel magnetic circuit can be
dealt with in much the same way as a parallel electric circuit. The magnetic
circuit in Fig. 6.5(a) can be represented by the equivalent electric circuit in
Fig. 6.5(b).
Fig. 6.5 Parallel-connected Magnetic Circuit
12
3.6 Example of Soft and Hard Magnetic Materials
1.
Materials with relative permeabilities very close to 1 in value are
sometimes classified as non-magnetic. For this reason, both paramagnetic
and diamagnetic materials can be classified as non-magnetic. Glass, air,
aluminium, wood, and many other materials are non-magnetic. For practical
purposes, they are neither attracted to nor repelled by a magnetic field.
2.
Ferromagnetic materials are commonly used in the construction of
motors, transformers, relays and in other devices where a strong magnetic flux
is needed.
3.7 B - H Curves for soft and hard magnetic
materials
6.8.1 Introduction
If a graph of the magnetic flux density (B) is plotted against the magnetising
field strength. (//) for a magnetic material, the resulting curve is known as the
B - H curve. Fig. 6-9 shows a typical graph of the BH curve or magnetisation
curve. Although for any magnetic material B = uH . however this will not lead
us to a straight line graph since p is not a constant number. In practical terms,
it is evident that p.r (relative permeability of. say, iron) is not constant. From
the graph it can be observed that at the initial stage (between the origin 0 and
the &quot;Knee&quot; of the curve), as the magnetic field strength (//) increases gradually
the flux density (B) increases rapidly- The knee of the curve marks the onset
of saturation (see position of the knee marked on the curve).
23
Week 4
On completion of this chapter the student should be able to:
• Explain the magnetic effect of electric current.
• Draw magnetic fields around straight conductors, adjacent par all el
conductors and solenoids.
• Demonstrate by experiment the magnetic effect of a current carrying
conductor in a magnetic field.
• Explain the force on a current carrying conductor in a magnetic field.
• State the direction of the force
• Derive the expression for the magnitude of the force (i.e. F = BIL.
Newtons}.
• Explain the concept of electromagnetic induction
• State Faraday's laws of electromagnetic induction.
• State Lenz 's laws of electromagnetic induction.
• Verify by experiments Faraday's law and Lenz's law.
• Derive the expressions for magnitude of e.m.f. induced in a conductor
or a coil.
• State the applications of electromagnetic induction.
4.1
ELECTROMANETISM
4.1.1 Magnetic effect of electric current
A conductor carrying an electric current produces a magnetic field around
itself (conductor). This relationship between electricity and magnetism is
known as electromagnetism. This phenomenon was discovered in 1820 by
Oersted in Copenhagen, (now the capital city of Denmark). Oersted
discovered that if a wire (conductor) carrying an electric current is placed
above a magnetic compass needle (as shown in Fig. 4.1) and in line with the
normal direction of the latter, the needle deflected is clockwise or
anticlockwise, depending upon the direction of the current.
24
4,2 Magnetic fields around straight conductors, adjacent parallel
It is discovered that if we look along the conductor, and if the current is
flowing away from the reader into the paper (as marked by the symbol &reg; )
inside the conductor (as shown in Fig. 4.2) the magnetic field has a clockwise
direction and the lines of magnetic flux can be represented by concentric
circles around the wire (as shown in Fig. 4.2)
If the current is reversed, the magnetic field will remain as concentric circles
but in anticlockwise direction. In that case, the conductor with current
flowing away from the paper towards the direction of the reader is represented
usually by concentric circles with a dot in the centre. In other words, the
direction of the magnetic field previously shown in Fig. 4.3 becomes anticlockwise.
Suppose we have a small rectangular cardboard pierced about its centre point,
with a straight conductor carrying current passing through the centre point, If
25
we sprinkle some iron filings fairly uniformly on the cardboard around the
conductor, we see that the magnetic field pattern (formed by the iron filings)
round the straight current -carrying conductor consists of concentric circles
with the conductor as centre. We notice the concentric arrangement of the iron
fillings tend to be most pronounced in the vicinity of the conductor and the
intensity of the field decreases as the distance from the conductor increases.
4.2
Direction of the Magnetic Field due to an Electric Current
in a straight conductor
Several rules are known for the determination of the direction of the magnetic
field around a straight current-carrying conductor.
A good rule of representing the relationship is to grip the conductor with
the right hand, with the thumb pointing in the direction of the current, the
fingers then point in the direction of the magnetic field around the conductor.
This rule may be referred to as the right-handgrip rule. This is illustrated in
Fig. 4.4 below.
Fig. 4.4 Illustrating the right-hand grip rule
There is an alternative way of representing the relationship between the
direction of current in a conductor and that of the magnetic field produced by
it (the current) alongside the conductor carrying the current.
Imagine that the screw is to travel inside the cork in the same direction as the
current (i.e. towards the right as shown in Fig. 7.5), it must be turned
clockwise when viewed from the left-hand side. By similar reasoning, the
direction of the magnetic field is clockwise around the conductor, as shown by
the curved arrow.
26
(a) Screw moving in current direction
Fig. 4.5 Right-hand cork-screw rule
27
Week 5
5.1
Magnetic Field of a solenoid
For a start, let us define a solenoid as a number of rums of a wire wound in the
same direction. All the turns in a solenoid are in the same direction so that
they can assist one another in producing a magnetic field. We note that if a rod
of iron is inserted inside the solenoid, only the magnetic field becomes
intensified. When an electric current is passed through a solenoid or long
cylindrical coil, the magnetic flux produced is very similar to that of a bar
magnet. In this case, one end of the solenoid acts like a N pole and the other a
S pole. Consequently, magnetic lines (shown in dotted lines in Fig. 7.6) move
from the N pole to the S pole.
The rule for the polarity of a solenoid carrying a current can be stated as
follows:
WJien viewing one end of the solenoid, it will be of N polarity if the current is
flowing in an anticlockwise direction, and of S polarity if the current is/lowing
in a clockwise direction.
Fig. 5.1 Magnetic effect of a solenoid
Fig 5.1 illustrates an easy method of remembering the rule. by indicating
arrows on the letters N and S. At end A of the solenoid, the current direction is
anticlockwise and so, that end serves as the N pole. At end B, the reverse is
the case.
5.2
Magnetic Field around Two long Parallel Conductors
In order to understand the resulting magnetic field around two long parallel
conductors carrying current we need first to draw the magnetic fields
produced by each conductor and then combining these fields.
28
Fig. 5.1 shows two parallel conductors, A and B, each carrying current
towards the paper- The magnetic field due to current in A .ilone is represented
by the dotted circles in Fig. 7.7(a), and that due to B alone is represented by
the fairly uniformly spaced curved dotted lines.
29
Fig. 5.3 Magnetic fields due to two long parallel current – carrying
conductors
As a result of the influence of magnetic flux of A on B, and vice - versa we
produce a resultant magnetic field configured as shown in 1-iy- 7.7(b). In this
case, the two fluxes tend to neutralise each other in the space between
conductors A and B, but they tend to assist each other in the space outside A
and B. In addition, the resultant magnetic flux behaves like a stretched elastic
material. This gives the impression that conductors A and B are attracted
towards each other. In other words, there is a force of attraction between A
and B.
If the current in A is reversed, i.e. if the two conductors carry currents in
opposite direction, as shown in Fig 7.7(c). The two fields assist each other in
the space between the conductors, A and B, resulting in a lateral pressure
between the lines of magnetic force. The circular line offerees are no longer
symmetrical about each conductor, but appear displaced as shown.
Consequently, there is a strengthened field between the conductors A and B,
and a weakened field to the left of A and the right ofB. We notice that the
lateral pressure referred to above is repulsive in nature, therefore there exist in
the space between the two conductors lines offeree of repulsion.
30
Week 6
6.1 Force on a Current Carrying Conductor in a Magnetic
Field
Under this topic we shall consider the interaction between magnetic field due
to the current in a conductor and the magnetic field in which the conductor is
placed.
For this purpose, we have the cross-section of conductor carrying current
towards an observer together with its magnetic field shown in 1. A 1 so
shown m Fig. 7.9(b) is another magnetic field considered to be uniform.
Fig. 6.1
If the conductor is placed in the field, it will be seen that the resultant
magnetic flux has been distorted so much so that it partially surrounds the
conductor (wire), as shown in Fig. 6.1(c). This distorted field acts like a
stretched elastic string bent out of the straight (like a catapult) and the flux
exerts a force F urging the conductor out of the way. In this case the conductor
A will move from left to right, which is the direction from the strong part of
the field (where the lines are very dense) to the weaker part. The brief
explanation of the phenomenon is that if we compare the field in Fig. 6.1(a)
with that on the left hand side of A in Fig. 6.1(a), we see that on the left side
31
the two fields (i.e. the arrows) are in the same direction, whereas on the lower
side they are in opposite directions. Consequently, the combined effect is to
strengthen the magnetic field on the left side and weaken it on the right side,
thus producing the distribution shown in Fig. 6.1(c).
However, reversing the direction of the current reverses the direction of the
resultant force, as shown in Fig. 6.1(d).
6.2 State the Direction of the Force on a Current-carrying
Conductor in a Magnetic Field.
In the previous week a brief mention is made about the force F which acts on
a current-carrying conductor in a magnetic field without much emphasis on
how to determine the direction of the force, F.
In this section we shall discuss the direction of the force on a currentcarrying conductor placed in a perpendicular magnetic field. Generally, the
direction of the force can be determined using Fleming's left-hand rule. It
states that: If the left hand is held with the thumb, first finger, second finger
held mutually perpendicular to each other with the first finger in the direction
of the field B and the second (middle) finger in the direction of the current /,
then the motion or force on the conductor is in the direction of the thumb M.
Field (B)
(b) Fleming's left hand rule
Fig. 6.2 Fleming’s left hand rule to determine force direction on a
conductor
The rule is applicable only if the magnetic field (B) and current (I) are
perpendicular or inclined, to each other. We should observe that if the field
and current are parallel to each other, no force acts on the conductor. As an
exercise, we can use the above rule to verify the direction of the force acting
on the current-carrying conductor in Fig- 7.10(b). Furthermore, we can also
verify both by Fleming's left hand rule and by drawing magnetic flux patterns,
the direction of the force on the current-carrying conductors shown in Fig.
6.1(c) &amp; (d).
27
7.6 Derive the Expression for the Magnitude of the Force in a
Current-carrying Conductor in a Magnetic Field
Experiment has shown that the force F (Newton s)ac ting on a conductor
carrying a current / (Amperes) at right angles to a magnetic field of flux
density B (Tesla) is directly proportional to (a) the magnitude of current in the
conductor (b) magnetic flux density and (c) length of conductor [metre].
Hence, F = 5/?[Newtons]
Furthermore, we recall (already treated in chapter 6) that if for a magnetic
field
having a cross-sectional area A [metre 2 ] and uniform flux density B (Tesia),
then the total flux can be represented
Example 6.1
Draw the resultant magnetic flux structure and determine the direction of the
force acting on a current-carrying conductor placed perpendicular to the
magnetic field in the following diagrams shown in Fig. 7.11 (a) and (b) below.
Fig. 6.3
Solution
(a) The direction of the current/and that of the magnetic flux are known and
shown. Therefore using Fleming left hand rule the direction of the force F^
is perpendicularly upward as shown in Fig 6.3(a).
28
Week 7
60 turns. It is situated in a radial magnetic field of 0.47'. Determine the
force (in milli newtons) on the coil when the current is \OmA.
Hint: Using usual notations', F = Nx BIl, -where I = n x D.
1. 75^iWb 2. 200 N/m 3. \AT A. 0.575. 7.54mN
7.1
ELECTROMAGNETIC INDUCTION
7.2 Explain the Concept of Electromagnetic Induction
Electromagnetic induction is a method of obtaining an electric current with
the aid of magnetic flux. It is on record that Michael Faraday, one of the
greatest British scientists made the discovery of electromagnetic induction in
1831. hi otherwords, he discovered that electric current could be produced
without batteries but solely by using magnets or magnetic fields.
In explaining the concept of electromagnetic induction, we will find that
when a permanent magnet NS is moved towards a coil C as shown in Fig.
7.13(a) galvanometer G deflects in one direction and when the magnet is
moved away from the coil the galvanometer deflects in the opposite direction
(sec Fig. 7.13(b)).
Magnet moved
towards the coil —&gt;
. ^&quot;&quot;^s
Magnet moved away
from the coil &lt;——
(~'^,/rww
Fig. 7.1 Electromagnetic Induction illustrated
When the magnet is placed stationarily (at rest) near the coil the galvanometer
remains stationarily at zero position. We should also note that when the
magnet is moved nearer more magnetic lines link the coil.
As illustrated and proved by Faraday himself, an induced e.m.f always
produces an induced current when there is a relative movement between the
magnet and the coil and it is always obtained when a change occurs in the
number of magnetic lines linking the coil. Besides, Faraday also showed that
the magnitude of the induced e.m.f. is proportional to the rate at which the
29
magnetic lines passing through the coil is varied. Alternatively, we can deduce
that when a conductor cuts or is cut hy magnetic lines, an e.m.f. is generated
in the conductor and the magnitude of the generated e.m.f. is proportional to
the rate at which the conductor cuts or is cut by the magnetic flux.
7.3 Faraday's Laws of Electromagnetic Induction
All the foregoing facts gathered from the explanation above sum up to what is
known as Faraday's Laws of Electromagnetic induction. The laws can be
stated as follow:
First law states that whenever there is a change m the magnetic lines
linking a circuit. there will always be an induced e.m.f. in the circuit, or,
whenever a conductor cuts a magnetic flux or is cut by a magnetic flux. an
e.m.f. is induced in that conductor.
Second Law states that the magnitude of the induced e.m.f. in a circuit is
directly proportational to the rate of change of the magnetic flux (or magnetic
lines) linking the circuit.
7.9 Lenz’s Laws of Electromagnetic Induction
Two methods are known for finding the direction of the induced or generated
e.m.f. They are:
(i) Fleming's Right Rule, and
(ii) Lenz's Law.
Fleming's Right-hand Rule
This rule states that if the first finger of the right hand is pointed in the
direction of the magnetic flux, (as in Fig. 7.14). and if the thumb is pointed m
the direction of motion of the conductor at right angle to the magnetic field,
then the second finger, held at right angles to both the thumb and the first, will
represent the direction of the induced e.m.f-
Fig. 7.2 Fleming's Righthand Rule Illustrated
30
NB The right-hand rule is used/or induced current or e.m.f. but the left-hand
rule.
7.92 Lenz's Law
In 1835 Heinrich Lenz, a German physicist stated a rule, now known as
Lenz's law, which can be used to determine the direction of induced e.m.f.
Lenz 's law states that the direction of an induced e. m.f. is always such
that it tends to set a current opposing the motion or the change of flux
responsible for inducing that e.m.f.
At this juncture we may wish to note that Fleming's righthand rule is
normally used where induced e.m.f. is due to magnetic flux-cutting, say a
conductor (i.e. when there is a dynamically induced e.m.f.) and Lenz's law
is used when there is a change by magnetic flux-linkages (i.e. statically
induced e.m.f.)
In other words, induced e.m.f. can be either dynamically induced or
statically induced. If it is dynamically induced, usually the field is
stationary and the conductors cut across it (the field). For example, this is
akin to the structure of d.c. generator. However, if the e.mf. Is statically
induced, usually the conductors or the coil remains stationary and flux
linked with it is changed by simply increasing or decreasing the current
producing the flux, for example as in transformers.
We know that the direction of the e.m.f. induced in a moving conductor
depends on its direction of motion and also on the direction of the magnetic
field lines. Applying the right-hand rule, the direction of e.m.f. is as shown in
Fig. 7.18, having known the direction of flux of density B, Tesla and direction
of motion of the conductor. Suppose is the length of the conductor, lying
within the field and let it move a distance x in a time;, then area swept by the
conductor within the flux is, (length x breadth) = l.x.
Hence, flux cut, F=Blx
(7.3)
Furthermore, we know from Faraday's law that whenever a conductor passes
through magnetic flux, an e.m.f. is induced in the conductor and its magnitude
depends on how fast the conductor passes through the flux. In other words, the
magnitude of e.m.f. induced in this way is given.
31
Week 8
Symbolically, this becomes
where, V •= velocity of the conductor.
11 the conductor AA&quot; moves at an angle with the direction of flux, then the
induced c.m.f. becomes.
E = B IV sin 9, [volt]
(7.6)
I
Example 7.4
One pole of an electric generator produces a magnetic flux of 0.2 Wb. An
armature conductor passes through this flux at a uniform rate, taking 0.05 seconds
to pass completely through the flux. Calculate the e.m.f. induced in the
conductor.
Solution
In this case, &lt;(&gt;=0.2Wb. t= 0.05s.
Using Eq.(7.4), we obtain the e.m.f. induced
as E^-^
T 0.05 E=4V.
31
Example 7.5
A conductor of length 0.5 metre moves in a uniform magnetic field of flux
density 2
Wfr/ffi2 at a uniform velocity of 40 metres/second. Calculate the induced e.m.f,
under
the following conditions:
(a) The conductor moves at right angle to the magnetic field,
(b) The conductor moves at an angle of 30&deg; to the direction of the field,
When the coil is positioned initially perpendicular to 5, the flux linkage = NAB
(as stated above). However, when the coil is turned through 60&deg; , the
flux density normal to the coil is now B cos60&deg;. And so, flux change
through the
coil == NAB -NAB cos60&deg;
^^xlO^-UxlO-^xO^ =6xl0-3
. , , „ flux change 6 x 10&quot;3 .'. average induced
e.m.t. == ——————— = ————
time
0.2
= 30xl0-3^.
7.12 Solve Problems on Electromagnetism and Electro-magnetic Induction
^ Example 7.8
Exa A current-carrying conductor is situated at right angles to a
uniform magnetic field having a flux density of 0.57. Calculate the
r
current in the conductor if the force per metre length of the
conductor is 2QN.
Solution
Data given are:
5=0.57', F/\=20N, /=? Using
Eq.(7.1), F= BH
i.e.
15
r- i
i
Current required /= — • — = 20 • — = 40/1. ^N
0.5
JT Example 7.9
Calculate e.m.f. generated in the axle of a car traveling at 90 k/h, assuming
the length of the axle to be 1.8/n and the vertical component of the earth's
magnetic field to be 50^i T
Solution
Data provided are:
32
V==90km/h, l=1.8m. B^SOx^T
Expressing V in m/s, we have
90xl000fm1 V^————
J-J=25OT/s. 60x60[^]
Using the data given in Eq.(7.5), we obtain the required e.m.f.
generated as E = B I V == 50 x 10~6 x 1.8 x 15V,
E =2250 x 10'6 = 2250 ^V.
Example 7.10
A current-carrying conductor of length 400/nm is moved at a uniform speed
at right-angles to its length and to a uniform magnetic field having a density
of 0.5 T. If the e.m.f. generated in the conductor is 3 V and the conductor
forms part of a closed circuit having a resistance of 0.5H , calculate; (a)
amount of current in the conductor; (b) the velocity of the conductor; (c) the
force acting on the conductor; (d) the work done (in Joules) when the
conductor has moved 500mm.
Solution
Using the usual notations, the data given are:
/=400fl&raquo;&raquo;=0.4m; B=0.57; E=3^; R=0.5^?;
d = 500w/n = 0.5m.
E 3
(a) Current in the conductor, / = — = — = 6A.
R 05
(b) Velocity of the conductor, can be obtained from the expression, E =
BIV. where
i.e. V = -&pound;- = ——3—— = 15
m/s B\ 0.5 x 0.4
(c) Force acting on the conductor,
F= 5/1=0.5x6x0.4 =1.2 N
(d) Work done by the conductor,
W= Force x distance = F x d
^=1.2x 0.5 =6 Joules.
Example 7.11
A six-pole motor has a magnetic flux of 0.05Wh per pole and the armature
is rotating at 600 rev/min. Calculate the average e-m-f. generated per
armature conductor.
Solution
Using the usual notations, the data given are:
37
B = 0.05 Wh per pole, v = speed of armature (conductor). It should be noted
that each time the armature conductor passes under a pole it cuts a flux of
0.05 Wb. Hence, the total flux cut in one revolution is 0.05 x 6 = Q3Wb .
V = speed of the armature (conductor) = 600 rev./min.
600 ,
= —— rev.s.
60 '
=70
rev./s.
Using the expression, E = BIV we can obtain the average e-m.f- generated
per (length of a conductor) in one revolution, i.e. E/\=BV
&pound;/l=0.3xl0=3y (with/=! assumed).
Example 7.12
A coil of 500 turns is wound on an iron core and a certain current produces
a flux of 4000 \^Wb- When the current is opened, the remaining flux in the
iron is 2900 uW&amp;-Tfthis reduction process of flux takes 0.2?, find the
average value of the induced e.m.f.
Solution
m this case, using the usual notations the data given are:
7v==500, •,=4000xlO-A^yb, 4*2 = 2900 xW-^Wb
,1
t=Q.2s.
The average value of the induced e.m.f. can be obtained using the
expression in
- A^, - •,) - 500(2900 - 4000) x 10&quot;6
Eq.(7.7),as &pound;=
/
02
-6
. ^500x1100x10 ^^ 0-2
7.13 State applications of electro magnet ism and electromagnetic
induction
For the purpose-of slating applications of electromagnetism and
electromagnetic induction, the following examples are possible:
(a) Electric Bell. •
(b) Magnetic circuits of generators and motors.
(d) Moving - iron ammeter and voltmeter
(e) Moving - coil loudspeeker.
(f) Ignition coil.
N.A The principle of operations of none of the above-Hsted items will be
discussed because they are found in details in other courses such as
38
Electrical/Electronics Instrumentation. Electrical Measurements, and
Electrical Machines.
39
Week 9
On completion of this should be able to:
Define self inductance and mutual inductance
state the symbols and units of the terms stated above
• State the expression for the equivalent inductance in inductances
connected in series and in parallel.
•
•
9.1
9.2
State the expression/or the induced voltage across an inductor.
Slate the expression for inductance in inductive coupled coils
connected in series aiding or opposing.
Self - inductance and Mutual inductance
introduction
Inductance is that property of an electric circuit that opposes any changes
(increase or decrease) in current flow (and hence flux change). We note
that physical components known as inductors exhibit the property of
inductance. Inductors exist in different sizes and shapes. Most practical
inductors are made up of conductor wire formed into a coil of several or
many turns.
Any circuit in which a change of current is accompanied by a change of
magnetic flux and which consequently produces an induced counter e.m.f.
is said to possess self' - inductance. It is quantitatively measured in terms of
coefficient of self induction L. In other words, whenever there is an
increase of current (and hence flux) through an inductor (i.e. a coil), it is
always opposed by the instantaneous production of counter e.m.f. of self induction.
Whenever we have an electric circuit magnetically coupled to another
circuit (as evident in transformers), as the current changes (in a transformer)
so does the flux. The changing flux induces a voltage into the other electric
circuit. Circuits that are linked by magnetic flux exhibit mutual inductance.
Inductance
The effect of self - inductance in an electrical circuit is to produce e.m.fs
which always oppose any changes in current value. In an R - L d.c. circuit
such as shown in Fig. 8.1 (a) these effects are observable mainly when
switching ON or OFF the current. The d.c. circuit contains a resistor of
resistance R and an inductor (i.e. coil) of self - inductance L in series. In the
circuit when the switch S is closed, the current rises from zero
(exponentially) to the maximum value expected according to Ohm's law (i =
40
E/R). While the current is rising, the self- inductance of the coil causes an
e.m.f to be induced which opposes the current build - up. '
Fig. 9.1 Illustrating Current Rise and Induced E.M.F fall in a RL d.c.
Circuit, when S is closed.
Consequently, the current does not immediately achieve its maximum
(final) value, but rises gradually as shown in Fig. 8.1 (a). Specifically, we
note that at time t) =0 (say), the current / = 0 while the induced e.m.f. value
e = E. As time progresses (i.e. t] &gt; 0), i rises (exponentially) while e falls.
Fig. 8.2 shows what happens when the switch S is suddenly opened at time
/y.
In this case, the current falls gradually from its previous maximum value /
(= E/R) to zero, and this fall of current causes a sudden high induced e.m.f.
hi other words as the switch S is opened at time / = ^ , current begins to
decrease (exponentially) from its maximum value (/) while the induced
e.m.f. immediately attains its maximum value e == E and immediately later
starts to fall until it reaches zero value. We note especially in Fig. 8.2(a)
that when the current is decreasing, the induced e.m.f. tends to prevent the
decrease of the current and its direction is therefore the same as that of the
current.
41
Week 10
Method I
(a) Self inductance in terms of flux ' linkages per ampere
Self inductance of a coil can be defined as the weber - turns per ampere in
the coil. We note that weber - turns, Mt&gt; = flux - linkage.
Suppose we have a solenoid having N turns and carrying a current of /
ampere, then it produces a flux of^ webers. Thus, its weber - turns are /V^i
, and its weber -turns per ampere are N^/I.
By definition, L = ^M
(8-1)
Its unit of dimension is Henry [//] in commemoration of the famous
American Physicist, Joseph Henry (1797 - 1878) who discovered
electromagnetic induction independently about a year after Michael
Faraday's discovery. From the above relation, in Eq. (8.1), if A^i = 1 Whtum,I= 1 ampere, then L =^1
henry (H). m words, this can be defined
as follows:
A coil is said to have self- inductance (L) of one henry (H) if a current of
I ampere flowing through it produces flux - linkage off Wb - turn in it.
Example 8.1
A coil of 250 turns produces a flux of 0.01 Wb when carrying current of
5A. For this current, calculate the inductance of the coil.
Solution
•• •
In this case. A'=250, ^=0.02&raquo;^&raquo;, /^5A, Z,=TT • c /o ^ r M^ 250x0.01
Using Eq. (8.1), L == —- = ————— = Q5H .
Method II
(b)
Self- inductance (L) terms of average induced e.m.f. and rate of
change of current.
Generally speaking, it can be stated that if a coil has an inductance L,
henry's and if the current through it increases from;/ to iy amperes in /
seconds, then average rate of change of current = ———, amperes/second
h and, average induced e.m.f. e^ = -L x rate of change of current .'. average
induced e.m.f, e^ = -L x '-^——/, volts
(10.2)
44
From Eq. (8.3), we can state that if— = 1 ampere/sound dt
and e^ == 1 .volt, then L = \H
Hence, in words, we can state that if a coil has a self- inductance of
one henry if e.m.f. of one volt is induced in it when current through it
changes at the rate of one ampere/second.
Corollary
If the LHS and RHS ofEq. (8.1) are cross - multiplied and the resulting
expression
is divided by (on both sides, we obtain
Consequently, we can state that
average e.m.f. induced in a coil = - Ll/t
or, average e.m.f. induced in a coil = - N^/t. If Eq.(8-2) is expressed in
differential calculus form, then we can obtain.
-L^-N^
(8.5) dt
Example 8.2
A coil of 500 turns is wound on a non - magnetic core and a current of 2.5A
through the coil produces a magnetic flux of 150 ^Wb. Calculate (a) the
inductance of the
coil, and (b) the average value of the induced e.m.f. if the given current is
reversed in 0.1.?.
Solution
(a)
The data given are: 7v=500, /=2.5A. ifr=l50x lO&quot;6^.t^OAs. Hence,
from Eq.(8.1), we have
Method I
(b) We note that the current changes from 2.5A to - 25A in O.ls,
.-. average rate of current change, —=-(2.5x2)/0.1 dt
= -50A/S Hence, from Eqn (8.3).
.-. average e.m.f. induced in coil, e^ = -L— = -0.03 x (50) dt CL = 1-5^.
Method II (Alternatively, using expression in Eq. (8.5))
From the given data, it implies the flux changes from 150 u^Z&gt; to 150uW&amp; in O.ls..'. average rate of change of flux, -•= -(150x 10~6 x 2)/0.1
dt
Hence, from the expression, e^ = -N—' dt •
average induced e.m.f. in the coil, e^ = -500 x (-0.003)
=5V .
45
N.B. Thee^ comes with a positive sign because of the e.m.f acts in the
same direction as the original current, initially Hying to prevent the current
decreasing to zero value and then opposing its growth in the reverse
direction.
Method III
(c) Self inductance &lt;U in terms of the dimensions of the solenoid
that if / be the length of a magnetic circuit (i.e. coil) (in metres), and A its
cross - sectional area (rn ), then for a coil ofN turns with a current /
amperes:
\
H = IN//{Amperes/metre]
and, total flux, 4&gt; =
BA = pp//^ . (Remember, u == B/H and u = U (for non-magnetic core, u^
= 1).
\.e.
4=47i x 1Q-7 x (IN/l)A Wb (where, Ho =4rt x 10-7)
Substituting for ()) in Eq. (8.1), we have
inductance, L = (4n x 10~7 x /4A^2//, henrys
(8.6) However, if the coil is wound on a magnetic core then 4 ^ 1 - In
that case,
Inductance, L = (4n x 10~7 x u,. x AN2/I), henrys.
(8.7)
inductance is proportional to the square of the number of turns and the
cross-sectional area, and is inversely proportional to the length of the
magnetic material.
Example 8.3
A coil of 500 turns is wound uniformly on an iron ring which has a mean
diameter
of 10cm, cross - sectional area of 5cm2 and a relative permeability of 350,
calculate the inductance of the coil
Solution
The data given are as follows:
^o=4TtxlO'\ ^,.=350. D= 10cm = O.lm, A^xlO'^m2 .
Using Eq. (8.7), the value of the inductance of the coil is
given as, HpU.^v2 _47ixl0-7 x350x5xl0&quot;4 x 5002
- 47ixl0-7 x350x5xl0-4x5002 „,,,., LK= U.I
where nD=l
10.2 Determination of Mutual Inductance (M)
If two coils X and Y are placed relatively close to each other as shown in
Fig. 8.3 below then we can notice that when switch S is closed, current
flows in X and produces a flux which becomes linked with Y and the e.m.f.
46
induced in Y causes a momentary current to flow through galvanometer
G. (for the fig see week 11)
47
Week 11
Fig. 11 Illustrating mutual inductance
We may also notice that when S is opened, the collapse of the flux induces
an e.m.f. in the reverse direction in Y. In conclusion, a change of current in
coil X is accompanied by a change of flux linked with coil Y and therefore
by an e.m.f. induced in Y. Consequently the two coils are said to have
mutual inductance.
Mutual inductance can also be determined in three ways as given for
self-inductance. The three ways are:
(a) Mutual inductance (M) in terms of average induced e.m.f. and rate of
change of current.
(b) Mutual inductance in terms of flux - linkages per ampere.
(c) Mutual inductance (M) in terms of the dimensions of the two coils.
Next, we shall discuss each of these three methods in turns.
Method I
Mutual inductance in terms of average induced e. m.f. and rate of change
of current If, say, two coils, X and Y, have a mutual inductance ofM henrys
and if the current in coil X increases fi-onu'/ to i^ amperes in t seconds:
.'. average e.m.f. induced in coil ~M(i,-i,} volts
(8.8)
Y=
From Eq. (8.8) we can deduce the following:
• the expression is similar to that of Eq. (8.2).
• the minus sign shows that the e.m.f. induced in coil Y tends to make
current flow in such a direction as to oppose the increase of flux
due to the growth of current in coil X.
• two coils have a mutual inductance of 1 Henry if an e.m.f. of 1 Volt is
48
induced in one coil when the current in the other coil varies uniformly at
the rate of 1 ampere per second.
Example 11
Two coils A and B have a mutual inductance of 0.2H. If the current in coil
A is varied from 4 to 2A in 0-2s, calculate the average e.m.f. induced in
coil B,
Solution
The data given are:
M=0.2H, t/=4/t, i^=2A, t=0.2s. Using Eq. (8.8) we can
obtain the average e.m.f. induced as
(2-4) average e.m.f. induced in coil B = -0.2 -0.2 =2V.
Method II
Mutual inductance in terms of flux - linkages per ampere Suppose^, and
((ie represent the flux in webers linked with coil Y due to currents and
amperes respectively in coil X (sometimes called the primary) and if N^
represents the number of turns on coil Y (called the secondary)
(8.9)
average e.m.f. induced in r=(T2 '1) 2 , volts
t
Equating expression in Eq. (8,8) and Eq(8.9) we have:
change of flux - linkages with secondary
change of current in primary
N..B. By comparison, the expression in Eq. (8.10) is similar inform to that
obtained in Eq. (8.5) for self inductance except that the change influx takes
place in the secondary coil Y and the change in current takes place in the
primary coil X.
Furthermore, if we know only flux - linkages with secondary and current in
primary (but not change in flux - linkages and change in primary) as
expressed in Eq. (8.10), then the mutual inductance
Example 8.5
Refer to the question in Example 8.4 and calculate the change of flux
linked with coil B, assuming that coil B has a winding of 250 turns.
Solution
In addition to the data provided in Example 8.4 we have N^ = 250.
Using Eq.(8.9) we get the required change of flat (&lt;(^ -((^)=({i, having
known that
the average induced e.m.f. =1V. (see solution to Example 8.4)
2=i^0(^,) Q2
~ '
^=OA-=0.00\6Wb
250
Method m Mutual inductance in terms of the dimensions of the two coils
49
For illustration let us refer to Fig.8.3 and assume that coil X and coil Y
have windings TV&quot;) and N^ turns respectively.
Without bothering ourselves for proofs, we can state that mutual
inductance in
respect of the two coils can be given as,
M= NIN2 = Jv1^2 , W
(8.12) //UoU^^ reluctance L J
(N.B. From our studies of magnetic circuits in chapter 6 we know that
reluctance =
//HoHrA).
Example 8.6
Two coils X and Y having 40 and 400 turns respectively are wound side by – side on an iron core of cross - sectional area of 100c/n2 and mean
length 160cin. Calculate the mutual inductance between the coils if the
relative permeability of the iron is 1600.
•Solution
The data given are; A^=40 =400 A = 100 x 10~4m2 ^O'W. / = 1.6/n .
Using the expression of Eq. (8.10), mutual inductance can be obtained as
JViJv; 40x400
1.6/4n x 10'7 x 1600 x 10.'. M= 0-20
11.1 Coefficient of Coupling or Coupling Coefficient
If two coils, A and B, are magnetically coupled and each has self inductances Li and L,2 respectively and if the mutual inductance between
the coils is M, then we note that if the coils are placed close together,
almost all the flux produced by current in one coil passes through the other
coil and ihc coils are said to the tightly coupled. In that case K re 1.
However, if the coils arc well spaced apart, only a small fraction of the flux
in the primary is linked with the secondary, then the coils are said to be
loosely coupled. Consequently, either K &laquo; 1 or K &raquo; 0.
Example 8.7
Two coils, X and Y, have self inductances ofl50u// and 300u// respectively.
A
current of 2A through coil X produces flux linkages of 80^ Wb - turns in
coil Y.
Calculate (i) the mutual inductance between the coils (ii) coefficient of
coupling of the coils.
50
Solution
The data given aie:7L, =150uff. N3 = 300&raquo;Aff. i, = 2A A^&lt;t&gt;,
^SOxlO'6^. (assuming that coils X and Y have windings/V, and N^
respectively, and current;/
leads to the production of flux (^i initially in coil Y).
flux - linkage of coil Y yv,A,
(i)
Using Eq. (8.10), M=-—————, „ ——^ current in
coil X
(ii)
t,
.. 80xl0-6 M = 40u//,
2M
40xl0-6
Coefficient of coupling, K=0.189.
11.2 Symbols of Inductors and units of Inductance
11.2.1 Symbols
The common circuit symbols for inductors are shown in Fig. 8.4. The
names given to the symbols are known as air - core lype, iron - core type
and variable iron - core type as illustrated in Fig. 8.4.
(a) air - core
(b) iron - core
(c) variable iron - core Fig.
11.3 Circuit Symbols for Inductors
11.3.1 Units of Inductance
We recall that there are two types of inductance: namely, self inductance
and mutual inductance. Whether we refer to either self inductance or
mutual inductance, the unit of measurement is the Henry [H]. However, we
should note that sub - units of Henry are possible: such as microhenry) and
milli henry {mH).
11.4 Expression for equivalent Inductance of inductances
connected in Series and in Parallel
11.4.1 Equivalent Inductance of Series - connected Inductors
Consider a circuit consisting of N inductors, connected in series, as shown
in Fig. 8.5(a), with the equivalent circuit shown in Fig. 8.5(b). Like in the
case of resistors connected in series, the inductors have the same current
through them.
51
Fig. 8.5 Electric Circuit diagram of Inductors connected in series and their
equivalent Circuit
52
FUNDAMENTALS OF AC. THEORY
Week 12
Week 12
12.1
Describe the production of an alternating e.m.f.
Alternating e.m.f. may be produced by rotating a coil in a magnetic field or by
rotating a magnetic field within a stationary coil. A typical circuit arrangement
whereby an alternating e.m.f- can be produced is shown in Fig. 9.1. Fig. 9.1 shows
a loop DABC rotated at a constant speed in a clockwise direction in a uniform
magnetic field due to poles NS. The ends of the loop are connected to two slip
rings C| and C^. Bearing on these rings are carbon brushes &pound;, and E-^ which are
connected to either an electric lamp L or an external resistor R.
In the present position of the rotating coil (loop) of Fig. 9.1, me plane of the coil is
parallel to the field. In this case, sides DA and CB are moving at right angles to
the field.
According to Fleming's right hand rule, an e.m.f. is produced along DA and CB in
the direction shown on the loop. In this case when the coil is horizontal, e.m.f.
attains its maximum value. However, no induced e.m.f. is produced along the sides
AB or DC because these sides do not &quot;cut&quot; the field lines as they rotate.
A coil of loop
DABC C1C2
two slip rings C1
and C2. E1]E2 carbon brushes
Fig. 12.1 Product of an alternating e.m.f.
When the coil has turned through 90&deg;, that is: when the coil is in vertical
position, e.m.f. produced is zero. At this instant, the sides DA and CB are both
moving parallel to the field. Consequently no induced e.m.f. is obtained.
When the coil has turned another 90&deg; and its plane is now horizontal or parallel
to the field but in an opposite direction to the first horizontal position we started
with. So the e.m.f. along DA and CB are in opposite directions, according to
50
FUNDAMENTALS OF AC. THEORY
Week 12
Fleming's right angle rule. In summary, when the coil is turned 180&deg;, the e.m.f. is
reversed. 12,2 Equation of the Alternating Voltages
12.2 Stating relationship between instantaneous and peak values
of a sinusoidal wave
in Fig. 12.1 let us imagine a rectangular coil (loop) of length / metres of each of
the parallel sides DB and CB, having N turns and rotating in a magnetic field of
flux density B W/m1. Suppose the peripheral speed of each side of the loop to be
V metres per second. This velocity can be resolved into two components,
perpendicular and parallel respectively to the direction of the magnetic flux, as
shown in Fig. 9.2 below. When the coil has turned through angle 0, its velocity V
can be resolved into two mutually perpendicular components, viz (i) VCosO
component - which is parallel to the direction of the magnetic flux and (ii) VsinO
component which is perpendicular to the direction of the magnetic flux
Fig. 9.2
e.m.f. generated in one side of the loop which contains N turns, e = Nv-BlV
sin 6.And so, total e.m.f. generated in, both sides of the coil is
e -= 2BNI V sin 9
(9.1)
we note that when 0 = 90&deg;, e has maximum value of
En, (say) =1BNIV (Volts)
(9.2)
Equation (9.1) can be rewritten as e = E^ sin 6 .
If b = AB = width of the coil (in metres),
and,
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FUNDAMENTALS OF AC. THEORY
Week 12
f~= frequency of rotation of coil (in
Hz), then, V=nbf.
If A = / x b = area of loop in square metres, then
Em=2BNl x nbf=l7iBANf (volts)
(9.3)
However, the instantaneous value of e.m.f. (generally at any value of 9)
generated in the coil can be expressed as
e=E,,, sin 8 = InBANf sin Q volts)
(9.4)
The e.m.f. can be represented by a sine wave as in Fig. 9.3, where &pound;„ represents
the maximum value of the e.m.f. and e is the value after the loop has rotated
through an angle 6 from the position of zero e.m.f. at ^=0&deg;.
50=1x n
.'. Speed = 50 revolution/second or. Speed = 50 x 60 rev./min == 3000
rev./min.
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Fundamentals of A.C. Theory
Week 13
Week 13
13.1 Another form of e.m.f. equation
Apart from the form the e.m.f. equation was expressed in Eq. (9.4), yet another
form could be as
e=En, srnQ^E^ sincotsE^ sin2nft=En, sin—t,
(9.6)
where, 6= cot, o)=2Jcf and period, T==l/f.
From Eq. (9.6), we can deduce the following:
(i) the peak value or amplitude of an alternating e-m.f. is given by the coefficient
of die
sine of the time angle. (ii) the frequency/is given by the
coefficient of time divided by Ix.
For example, if the equation of an alternating e.m.f. is given by e= 10
^n314?.then
314 its peak value (i.e. maximum value) E,n == 10F, its frequency / = —— =
50Hz . and e is
In
the instantaneous value, (e is the e.m.f. value at every time instant.)
13.2 Defining instantaneous, average, r.m.s, form factor and
peak value alternating current or voltage
13.2.1
Instantaneous value of an alternating
voltage or cuiirent.
As explained earlier on, the instantaneous value is the value of the e.m.f. or
current at any time instant.
For an alternating e.m.f., its instantaneous value e = Em sin wt, whereas for an
alternating current its instantaneous value / = A,, sin wt.
Example 9.1 A square coil of side 10cm, having 200 turns, is rotated at 1200
rev/min about an axis through the centre and parallel with two sides in a uniform
magnetic field of density 0.57'. Calculate: (a) The frequency (b) The induced
e.m.f.
Solution
60 sec
sec (ii)
be obtained as,
Using Eq (9.3), the instantaneous e.m.f can
e - 27rBANn. In our case, B = 0.5T,
A = 0.1 x Q.lm2 = 0.01m-'
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Fundamentals of A.C. Theory
Week 13
N= 200, n = 20 (obtained form (i) above) Consequently, the
r.m.s. value of the e.m.f. can be obtained as
Crms = 0-707e = 0.707 x 2w x 0.5 x 0.01 x 200 x20
e^= 88.84^
Example 9.2 Determine the number of poles of an alternator driven at a speed of
375 ^/mm an&lt;* itg&reg;&quot;61^1^ an e.m.f having a frequency of50//z.
Solution
Using Eqn. (9.5).
/^=——- =8pairs
[&quot;^J
P=8x2 = 16poles
Example 9.3
Ife= 125 sin 2nft is the instantaneous value of an alternating e.m.f. with
periodic time 0.01s. (a) what will be its value 0.002s after passing through zero?
(b) if the voltage is applied across a 20 - ohm resistor, what is its instantaneous
current value at; = 0.002s
Solution
(a) Frequency, /=-=——=io0 c/s or Hz
.'. instantaneous value, e=125 sin27txl00xt
i.e.
e=125 sin2007Tt •At t= 0.002s.
e=125 sin 20071 x 0.002=125 sin0.4
bearing in mind that, 2n radians = 360&deg; (i.e. K rad
= 180&deg;), . (. . 180 e = 125 x sin 0.47C x =125
sin 72&deg; =125x0.951^119V (b)
when the
instantaneous voltage is applied across a 20 - ohm
resistor, its instantaneous current,
. e 125 sin 2007rt
After t = 0.002s and using the same notation as before,
Example 9.4
(a)
Find the amplitude, phase, period, and frequency of the sinusoidal
waveform e=10cos(50t+20&deg;)
(b) An e.m.f. waveform is given by V = 200sin628t. How long does it take this
waveform to complete one - half cycle?
Solution
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Fundamentals of A.C. Theory
Week 13
(a)
Amplitude, emf = 10V; phase, (&amp; == 20&deg;; angular frequency, ro = W^y
Period,
T 0.1257
22 f=l =7.95g Hz 50
(b) Period, T = 10ms, which is the time for 1 cycle. w = 28
Consequently, the time for one-half cycle = .5 x 10= 5ms
13.3 Average value of an alternating voltage or current
Suppose we consider the case of current which is a non - sinusoidal waveform as
shown in Fig. 9.4.
Fig. 9.4 Illustrating how to find the average value
if we consider generally n equally - spaced mid - ordinates /,, i^ ..., in taken over
either the positive or the negative half- cycle
then, average value of current over half a cycle
This approach is known as the mid - ordinate method. This method may be used
both for sinusoidal and non ~ sinusoidal waveforms. If we have a sinusoidal
waveform i = !„, sin 9, it can be proved that (for half of a cycle):
average value of current, ;,„ = 0.637 x maximum
value
(9.8)
We note that average value over a complete cycle for a sinusoidal waveform is
zero. There is another approach known as the analytical method by which the
average
value can be determined. This will be discussed immediately below.
With the aid of integral calculus, the average value can be determined by the
expression
As an illustration, we consider an alternating current i = /„ sin 6, shown in Fig.
9.5 and show how to determine its average current value.
55
Fundamentals of A.C. Theory
Week 13
Fig. 9.5 Full - wave rectified sine wave
The average current value of the full - wave rectified sine wave can be
determined as follows (bearing in mind that the period T =- n):
iav-0.6371,
In other words, the average current, i^ = 0.637 x maximum current value. If we
consider voltage quantity we shall state that the average voltage V^= 0.637 x
maximum voltage value.
N.B. ffwe consider a whole cycle of a sinusoidal signal i = 1^ sin cot d (o)t)
(such as in Fig. 9.3} and the period T = In, we shall be able to prove according to
Eq. 9.9(b) that its average current i^ = 0.
13.4 Root - Mean - Square (R.M.S) Value
The r.m.s value of an alternating current is the d-c. current value flowing through
a given circuit for a given time produces the same heat quantity (heating effect)
as produced by the alternating current when flowing the same circuit for the
same time duration.
For the purpose of computing the r.m.s value either the mid - ordinate method
or analytical method may be used.
Let us consider a current having the waveform shown in Fig. 9.6 (a). If this
current
flows through a circuit having resistance R ohms, the heating effect of i, is i^R,
That of ^ is ij R, etc. as shown in Fig. 9.6(b). we observe that the heating effect is
positive during both the positive and negative half cycles. In general, if there are
n equally - spaced mid - ordinates in half a cycle, then:
average heating effect during half a cycle
Suppose we have a direct current of I amperes flowing through the same
resistance R it will produce a d.c. heating effect equal to the average heating
56
Fundamentals of A.C. Theory
Week 13
effect of the alternating current, and thus to produce the same quantity of heat in
half a cycle:
Then,
or,
In other words, it can be stated that the root- mean-square ( or r.m.s.) value of a
sinusoidal current is measured in terms of the direct current that produces the
same heating effect in the same resistance. Hence, if Im be the maximum or peak
value of the sinusoidal current, the average heating effect over a cycle (or half a
cycle ) is half the maximum heating effect,
Thus far, we have been discussing the mid - ordinate method. The next method
is the analytical method of calculating the r.m.s. value.
Analytical method involves of the application of the expression,
where
Y^ = r.m.s of current or voltage
y(t} = general (current or voltage) function
\y(t)\ = square of the (current or voltage)
function
r= period of a cycle
(Current)
•(a) A given current waveform
57
Fundamentals of A.C. Theory
Week 14
(b) a resulting (heating effect) wavef Fig. 9.6 Illustrating the r.m.s value
59
Fundamentals of A.C. Theory
Week 14
Week14
Example 9.5
(a) Determine the peak voltage value of a sinusoidal alternating voltage ofr.m.s
4.5V.
(b) Determine the r.m.s value of a rectangular current wave with an amplitude of
8.M.
(c) Determine the average value of a sinusoidal alternating current of ISA
maximum value.
Solution
(a) From Eq.(9.10), ^=0.707^
V_ = —1— x V. = -45- V = 6.36V
m
0.070 &quot;&quot; 0.707
(b) From Eq.(9.11). /^ =0.707 /,
=0.707 x 8.8/4 =6.22A
(c) From Eq.(9.IO), /„== 0.637 ^
=0.637x25/4=15.9.4
Example 9.6
A triangular current waveform has the following values over one - half cycle
Current (A) 0 2 4 68 10 86 42 0
Time (ms) 0 10 20 30 40 50 60 70 80 90 100
Determine (a) the average current value, and (b) the r.m.s current
value.
Solution
(a) By applying Eq. (9.7), the average current value.
where n = number of time intervals
(b) By applying Eq. (9.11), the r.m.s current value
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Fundamentals of A.C. Theory
Week 14
Example 9.7
Find the average and effective (rms) values of the rectangular voltage wave
shown in
Fig.9.7.
Fig. 9.7 Diagram for Example 9.7
Solution
Method I
(a
)
fT
Looking at Eq. 9.9(a) this method is based on the fact that eff)dt stands for
J t)
the area under the graph from limits 0 to T.
area under the graph from limits 0 to T
Time period (T)
This means that rinding (calculating) the average value of the waveform
implies calculating the net area of the waveform over one period. T= 0.03s.
If A^ == area of section abed and A^= area of section defg, then the net
area = A^{-A,).
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Fundamentals of A.C. Theory
Week 14
Bearing in mind that Fig. 9.7 enables us to determine we need to draw the
graph of V^(t)to find mmf- Since both negative and positive voltages
become positive when squared, therefore the negative portion in Fig. 9.7
becomes positive when (- 2V} is squared i.e. W1. Consequently the graph
of V2(t) lS as shown in Fig. 9.8 below.
Method II
With reference to Fig. 9.7, we have the following: (a) For the time interval ad, Vs1 \QV
For the time interval d~g, V=~2V The period T = 7^ +
7^ = 0.01 + 0.02 = 0.03s.[1 f r 0.
62
(b) In order to determine the mis value, we start by finding V2^) for each lime
interval (a - d) and (d - ^ I For the a - d interval.
For the d - g interval, Next, we apply Eq. (9.13) to get the
mis value of voltage,
63
Week 15
N.B.
as result a is the same as obtained by method I.
15.1 Form factor, Kf Form factor
is the ratio of mis value to average value. Generally it can be expressed as,
15.2 Crest or Peak or Amplitude Factor, K,
It can be defined as the ratio of the maximum value to the r.m.s. value.
64
15.3 Explain phase lag or phase lead as applied to a.c. circuits
15.3.1 Phase difference
Let us consider a general expression for the sinusoidal waveform,
e(t)=E,sin(cot+(()). where (cot + &lt;t&gt;) is the argument and (() is the
phase angle. Both argument and phase can
be measured in radians or degrees.
Next let us imagine we have two sinusoidal waveforms expressions as
C|(t)= E^sincdt and e^t) = E^ sin(cot +&lt;^&gt;) respectively and shown in Fig. 9.9
(t)=e,sin(cot + 9) Fig. 9.9 Two sinusoidal waveforms with the same amplitude
but different phases.
The starting point A, of e^t) occurs first in time. Consequently, we say that €3(1)
leads e|(t)orthat e,(t) lags 63(1) by &lt;|). Ifij)=0, then ei (t) and e;(t) are said to be
in phase; in other words, they reach their zero point, minima and maxima at
exactly the same time.
Generally speaking, when comparing two sinusoidal waveforms the leading
sinusoidalwaveform is one which reaches its maximum (or zero) value earlier
than the other one.
Similarly, the lagging sinusoidal waveform is one which reaches its maximum
(or zero) value later than the other waveform.
There are some general points to note when comparing two or more waveforms.
They include the following:
(a) a sinusoidal waveform can be expressed in either sine or cosine form.
However, when comparing two sinusoidal waveforms, it is advantageous to
express both as either sine or cosine with positive amplitudes. To achieve this,
the following trigonometric identities are useful;
65
We can transform a sinusoidal waveform from sine form to cosine form or vice
versa using these relationships stated above.
(b)
The phase difference between any pair of waveforms can be determined
either from their respective instantaneous equations or from the drawing of their
respective waveforms, (see example 9.8 as an illustration).
15.4 Differentiate between series and parallel resonance
An ac circuit is said to be in resonance when the applied voltage V (with constant
magnitude, but of varying frequency) and the resulting current / are in phase.
Series Resonance
Consider we are given an RLC series circuit of Fig. 9.29
Fig. 9.29 An RLC Series circuit
66
The RLC series circuit has a complex impedance, Z=R+j(X,-Xc)=R+J(coL-——}.
For resonance to occur,
(X,-X,-)=Q This implies that Z = R (from Eq. (9.21) i.e. inductive reactance, X=
capacitive reactance, X•
Consequently, when X\ = X^ This also implies that
or. Resonance frequency. /, =(9.22(a)) , (Hz)
2;r WLC
In regard to series resonance, there are few things of interest to note. These
may include the following:
(i) Graphical representation of impedance (Z) X^, X^ and R as functions of
frequency, co, is as shown in Fig. 9.30-
Fie. 9.30
N.B. The graphs of XL, Xc and
R are shown in broken lines
From Fig. 9-30 we see that when series
resonance occurs at w=cDr, the impedance
curve (Z) reaches its minimum value at X, and Z = R. In other words, the current;
in
V y
the circuit of Fig. 9.29 /=—=—, becomes maximum.
z R
(ii) The phase angle, (0) relationship with frequency (o) is as shown in Fig. 9.31.
It can be observed that before resonance occurs when, X^ &gt; X\. and the phase
angle (6) varies from above 0&deg; to 90&deg;. Conversely, when X^ &gt; X^ the phase angle
0
varies from below 0&deg; and -90&deg;- However, at 9=0&deg;, resonance occurs at &amp;&raquo;,..
Phase angle.
* f^&gt;\
67
Fig. 9.31 Phase angle curve as a function ofo
(iii)The phasor diagram showing the current and all the voltages in a series
resonant circuit is as shown in Fig. 9.32.
Fig. 9.32 Phaser diagram of current and the voltages in an RLC series circuit. At
resonance, (see also Fig. 9.29) ^ + V^ = 0 and V = V^ = Rl.
(iv)The graphical relation in a series RLC circuit between current I and
frequency co of the applied voltage is shown in Fig. 9.33. The graph is known as
the resonance curve.
68
At resonance, (X^ - X^ ) = 0 and when this happens (9.22) R
Before we reach resonance frequency, w^ the impedance Z is greater than A, •therefore
the current / flowing in the circuit is less than /^.
Example 9.25
A 120 - V ac source supplies a series circuit resistance and inductance of \0 and 25mff
respectively. The generator frequency is the resonance frequency of the circuit. Determine (a)
the resonance frequency and (b) the current.
69
Solution
(a) From Eq.(9.21), the resonance frequency,
/,=————=——,
'
=225/fe. WLC 2^25 x 10-3 x
20 x 10-6
(b) At resonance, Z = R. Therefore the required current,
Example 9.26
An RLC series circuit consists of a coil having resistance of 10/3 and inductance of Q.5H and
a variable capacitance. The circuit is connected across a 150^, 50C/S supply. Calculate the
capacitance at which the circuit resonates and the voltage across the capacitor at this
resonant frequency.
Solution
N.B. A coil is an element consisting of a resistor, R and an inductor, each of a fixed
resistance value and a fixed inductance value respectively.
Fig. 9.34 is the required circuit diagram
Fig. 9.34
70
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