Eec 125 Note 2021 - 240226 - 141917

ELECTRICAL
ENGINEERING
SCIENCE II (EEC 125)
Engr. A.
Mohammed
This lecture note is divided into three chapters. Chapter 1 introduces the concepts and laws associated with
magnetic fields and their application to magnetic circuits and materials. Chapter 2 concerns the principles
and laws governing electromagnetic induction and the concepts of self and mutual inductance. chapter 3
deals with the concepts, terms and definitions associated with alternating quantities. Although an A.C. can
have any waveshape, the most common waveform is a sinewave. For this reason, unless specified otherwise,
you may assume that sinusoidal waveforms are implied.
Electrical Engineering Science II (EEC 125)
CHAPTER 1
MAGNETIC FIELD AND CIRCUIT
On completion of this chapter you should be able to:
i. Describe the forces of attraction and repulsion between magnetised bodies.

ii. Understand the various magnetic properties and quantities, and use them to solve
simple series and parallel magnetic circuit problems.
iii. Appreciate the effect of magnetic hysteresis, and the properties of different types of
magnetic material.
1.1 INTRODUCTION
A magnet is a material that has two poles. These are the north seeking pole and the south
seeking pole (S-pole). A simple experiment with a pair of magnet reveals two important facts:
i. Like magnetic poles repel one another

ii. Unlike magnetic poles attract one another
The phenomena of magnetism and electromagnetism are dependent upon a certain property of
the medium called its permeability.
Every medium is supposed to possess two permeabilities:
(i) Absolute permeability (μ) and

(ii) Relative permeability (μr).
For measuring relative permeability, vacuum or free space is chosen as the reference medium.
It is allotted an absolute permeability of μ0 = 4π × 10− 7 henry/metre. Obviously, relative
permeability of vacuum with reference to itself is unity. Hence, for free space, absolute
permeability μ0 = 4π × 10− 7 H/m relative permeability μr = 1. Now, take any medium other
than vacuum. If its relative permeability, as compared to vacuum is μr, then its absolute
permeability is μ = μ0 μr H/m.
1.2 MAGNETIC FIELD

A permanent magnet is a piece of ferromagnetic material (such as iron, nickel or cobalt) which
has properties of attracting other pieces of these materials. A permanent magnet will position
itself in a north and south direction when freely suspended. The north-seeking end of the
magnet is called the north pole, N, and the south-seeking end the south pole, S.
Engr. A. Mohammed Page 1

The area around a magnet is called the magnetic field and it is in this area that the effects of
the magnetic force produced by the magnet can be detected. A magnetic field cannot be seen,
felt, smelt or heard and therefore is difficult to represent. Michael Faraday suggested that the
magnetic field could be represented pictorially, by imagining the field to consist of lines of
magnetic flux, which enables investigation of the distribution and density of the field to be
carried out.
Figure 1
1.3 WEBER AND EWING’S MOLECULAR THEORY

The basic assumption of this theory is that molecules of all substances are inherently magnets
in themselves, each having a N and S pole. In an unmagnetised state, it is supposed that these
small molecular magnets lie in all sorts of haphazard manner forming more or less closed loops
According to the laws of attraction and repulsion, these closed magnetic circuits are satisfied
internally, hence there is no resultant external magnetism exhibited by the iron bar. But when
such an iron bar is placed in a magnetic field or under the influence of a magnetising force,
then these molecular magnets start turning round their axes and orientate themselves more or
less along straight lines parallel to the direction of the magnetising force. This linear
arrangement of the molecular magnets results in N polarity at one end of the bar and S polarity
at the other.
1.4 CURIE POINT

As a magnetic material is heated, its molecules vibrate more violently. As a consequence,
individual molecular magnets get out of alignment as the temperature is increased, thereby
reducing the magnetic strength of the magnetised substance.

The temperature at which the vibrations of the molecular magnets become so random and out
of alignment as to reduce the magnetic strength to zero is called Curie point. More accurately,
it is that critical temperature above which ferromagnetic material becomes paramagnetic.
1.5 CLASSIFICATION OF MAGNETIC MATERIALS
Although all material have magnetic properties of some kind being either diamagnetic ,
paramagnetic or ferro-magnetic, the term magnetic material is customarily applied only to
substances which exhibit ferro-magnetism.
Diamagnetism: is a weak form of magnetism which arises only when an external field is
applied. It arises due to change in the orbital motion of electrons on application of a magnetic
field. There are no magnetic dipoles in the absence of a magnetic field and when a magnetic
field is applied the dipole moments are aligned opposite to field direction. The magnetic
susceptibility is negative i.e. B in a diamagnetic material is less than that of vacuum. Examples
are zinc, mercury, sulphur, copper, silver etc.
Paramagnetism: In a paramagnetic material the cancellation of magnetic moments between

electron pairs is incomplete and hence magnetic moments exist without any external magnetic
field. However, the magnetic moments are randomly aligned and hence no net magnetization
without any external field. When a magnetic field is applied all the dipole moments are aligned
in the direction of the field. The magnetic susceptibility is small but positive. i.e. B in a
paramagnetic material is slightly greater than that of vacuum. Examples are tin, aluminium,
magnesium, platinium, manganese etc.
Ferromagnetism: Certain materials possess permanent magnetic moments in the absence of

an external magnetic field. This is known as ferromagnetism. Permanent magnetic moments in
ferromagnetic materials arise due to uncancelled electron spins by virtue of their electron
structure. The coupling interactions of electron spins of adjacent atoms cause alignment of
moments with one another. Examples are iron, steel, nickel, cobalt and some of their alloys.
Ferromagnetic materials are of two types: soft magnetic material and hard magnetic material
a. Soft magnetic materials do not retain their magnetism for any appreciable time after
the magnetisation force has been removed. Soft magnets have a narrow hysteresis
curve and high initial permeability and hence easy to magnetize and demagnetize.
Soft magnets like Fe, Fe-Si are useful when rapid magnetization and
demagnetization is required as in transformer cores.

b. Hard magnetic materials retain their magnetism for a long time after the
magnetisation force has been removed hence they are used for making permanent
magnet. Hard magnets are used in all permanent magnets in applications such as
power drills, motors, speakers.
1.6 LAW OF MAGNETIC FORCE

Coulomb was the first to determine experimentally the quantitative expression for the magnetic
force between two isolated point poles. The law states that the force of attraction or repulsion
between two magnetic poles is directly proportional to their pole strengths and inversely
proportional to the square of the distance between them.
For example, if m1 and m2 represent the magnetic strength of the two poles, r the distance
between.
𝑚1 𝑚2
𝐹∝
𝑟2
𝑚1 𝑚2
𝐹=𝐾
𝑟2
In the S.I. system of units, the value of the constant k is = 1/4πμ0μr.
Where μ0 = absolute permeability and μr = relative permeability
If, in the above equation, m1 = m2 = m (say); r = 1 metre; and the medium is air
F = 1/4πμ0 N
Hence, a unit magnetic pole may be defined as that pole which when placed in vacuum at a
distance of one metre from a similar and equal pole repels it with a force of 1/4πμ0 newton.
1.7 MAGNETIC FIELD STRENGTH (H)

Magnetic field strength at any point within a magnetic field is numerically equally to the force
experienced by a N-pole of one weber placed at that point. Hence, unit of H is N/Wb.
Magnetic field strength is also defined as the mmf per metre length of the magnetic circuit. The
quantity symbol for magnetic field strength is H, the unit of measurement being ampere
turn/metre.

𝐹 𝑁𝐼
𝐻= = ampere turn/metre
𝑙 𝑙
Where l is the mean or average length of the magnetic circuit.
It would be helpful to remember that following terms are sometimes interchangeably used with
field intensity: Magnetising force, strength of field, magnetic intensity and intensity of
magnetic field.
1.8 MAGNETIC POTENTIAL

The magnetic potential at any point within a magnetic field is measured by the work done in
shifting a N-pole of one weber from infinity to that point against the force of the magnetic field.
It is given by
It is a scalar quantity.
1.9 FLUX PER UNIT POLE

Magnetic flux is the amount of magnetic field (or the number of lines of force) produced by a
magnetic source. A unit N-pole is supposed to radiate out a flux of one weber. Its symbol is
Φ. Therefore, the flux coming out of a N-pole of m weber is given by
𝛷 = 𝑚 𝑊𝑏
1.10 FLUX DENSITY (B)

It is given by the flux passing per unit area through a plane at right angles to the flux. It is
usually designated by the capital letter B and is measured in weber/meter2. It is a Vector
Quantity. If Φ Wb is the total magnetic flux passing normally through an area of A m2, then
𝛷
𝐵 = 𝑊𝑏/𝑚2 𝑜𝑟 𝑡𝑒𝑠𝑙𝑎 (𝑇)
𝐴
1.11 ABSOLUTE PERMEABILITY (μ) AND RELATIVE PERMEABILITY (μr)

In Fig. 2 is shown a bar of a magnetic material, say, iron placed in a uniform field of strength
H N/Wb. Suppose, a flux density of B Wb/m2 is developed in the rod.

Figure 2
Then, the absolute permeability of the material of the rod is defined as
μ = B/H henry/metre or B = μH = µ0 µr H Wb/m2…………………………………….. (i)
When H is established in air (or vacuum), then corresponding flux density developed in air is
B0 = µ0 H
Now, when iron rod is placed in the field, it gets magnetised by induction. If induced pole
strength in the rod is m Wb, then a flux of m Wb emanates from its N-pole, re-enters its S-pole
and continues from S to N-pole within the magnet. If A is the face or pole area of the
magentised iron bar, the induction flux density in the rod is
Bi = m/A Wb/m2
Hence, total flux density in the iron rod consists of two parts [Fig. 2 (b)].
(i) B0 –flux density in air even when rod is not present

(ii) Bi –induction flux density in the rod
B = B0 + Bi = µ0 H + m/A
Eq. (i) above may be written as B = µr µ0 H = µr B0
𝐵 𝐵(𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙)
𝜇𝑟 = =
𝐵0 𝐵(𝑣𝑎𝑐𝑢𝑢𝑚)
Hence, relative permeability of a material is equal to the ratio of the flux density produced in
that material to the flux density produced in vacuum by the same magnetising force.
1.12 INTENSITY OF MAGNETISATION (I)

It may be defined as the induced pole strength developed per unit area of the bar. Also, it is
the magnetic moment developed per unit volume of the bar.

Let m = pole strength induced in the bar in Wb
A = face or pole area of the bar in m2
Then I = m/A Wb/m2
Hence, it is seen that intensity of magnetisation of a substance may be defined as the flux
density produced in it due to its own induced magnetism.
If l is the magnetic length of the bar, then the product (m × l) is known as its magnetic moment
M.
𝑚 𝑚×𝑙 𝑀
𝐼= = = = 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑚𝑜𝑚𝑒𝑛𝑡/𝑣𝑜𝑙𝑢𝑚𝑒
𝐴 𝐴×𝑙 𝑉
1.13 SUSCEPTIBILITY (K)

Susceptibility is defined as the ratio of intensity of magnetisation I to the magnetising force H.
Therefore K = I/H henry/metre.
1.14 Relation between B, H, I And K

It is obvious from the above discussion that flux density B in a material is given by
B = B0 + m/A = B0 + I ∴ B = μ0 H + I
Now absolute permeability is
𝐵 𝜇0 𝐻 + 𝐼 𝐼
𝜇= = = 𝜇0 +
𝐻 𝐻 𝐻
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝜇 = 𝜇0 + 𝐾
Also
µ = µ0 𝜇𝑟
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
µ0 𝜇𝑟 = 𝜇0 + 𝐾 𝑜𝑟 𝜇𝑟 = 1 + 𝐾/ µ0
For ferro-magnetic and para-magnetic substances, K is positive and for diamagnetic

substances, it is negative. For ferro-magnetic substance (like iron, nickel, cobalt and alloys
like nickel-iron and cobalt-iron) μr is much greater than unity whereas for para-magnetic

substances (like aluminium), µr is slightly greater than unity. For diamagnetic materials
(bismuth) µr < 1.
Example 1: The magnetic susceptibility of oxygen gas at 20ºC is 167 × 10 −11 H/m. Calculate
its absolute and relative permeability.
Solution
𝜇 = 𝜇0 + 𝐾
𝜇 = 4𝜋 × 10−7 + 167 × 10−11 = 1.258 × 10−7 𝐻/𝑚
µ 1.258 × 10−7
𝜇𝑟 = = ≈1
µ0 4𝜋 × 10−7
Example 2: The pole of a bar magnet measures 1.2cm x 2.1cm and the associated flux is
10mWb. Calculate the flux density.
Solution
𝛷 10 × 10−3
𝐵 = = = 39.68 𝑤𝑏/𝑚2
𝐴 1.2 × 10−2 × 2.1 × 10−2
Example 3: Two magnetic north poles of strength 0.005 and 0.01 Wb are placed at a distance
of 5cm in air. Determine the force between them with its nature.
Solution
𝑚1 𝑚2
𝐹=𝐾
𝑟2
1 1
𝐾= = = 63325.74
4𝜋𝜇0 𝜇𝑟 4𝜋 × 4𝜋 × 10−7 × 1
0.005 × 0.01
𝐹 = 63325.74 × = 1266.51𝑁
(5 × 10−2 )2
1.15 MAGNETIC CIRCUIT
It may be defined as the route or path which is followed by magnetic flux. The law of magnetic
circuit are quite similar to (but not the same as) those of the electric circuit. Consider a solenoid
or a toroidal iron ring having a magnetic path of l metre, area of cross section A m 2 and a coil

of N turns carrying I amperes wound anywhere on it as in Fig. 1 Field strength inside the
solenoid is
Figure 3
𝑁𝐼
𝐻= 𝐴𝑇/𝑚
𝑙
𝜇0 𝜇𝑟 𝑁𝐼
𝐵 = 𝜇0 𝜇𝑟 𝐻 = 𝑊𝑏/𝑚2
𝑙
𝜇0 𝜇𝑟 𝐴𝑁𝐼
The total flux produce Φ = 𝐵 × 𝐴 = 𝑊𝑏
𝑙
𝑁𝐼
Φ= 𝑊𝑏
𝑙/𝜇0 𝜇𝑟 𝐴
The numerator ‘NI’ which produces magnetization in the magnetic circuit is known as
magnetomotive force (m.m.f.). Obviously, its unit is ampere-turn (AT). It is analogous to
e.m.f. in an electric circuit.
The denominator 𝑙/𝜇0 𝜇𝑟 𝐴 is called the reluctance of the circuit and is analogous to resistance
in electric circuits.
Therefore flux = m.m.f/reluctance or Φ = 𝐹/𝑆
Sometimes, the above equation is called the “Ohm’s Law of Magnetic Circuit” because it
resembles a similar expression in electric circuits i.e.
current = e.m.f. / resistance or I = V/ R
1.16 DEFINITIONS CONCERNING MAGNETIC CIRCUIT

1. Magnetomotive force (m.m.f.). It drives or tends to drive flux through a magnetic
circuit and corresponds to electromotive force (e.m.f.) in an electric circuit. M.M.F. is

equal to the work done in joules in carrying a unit magnetic pole once through the entire
magnetic circuit. It is measured in ampere-turns.
F = ΦS = NI
In fact, as p.d. between any two points is measured by the work done in carrying a unit
charge from one points to another, similarly, m.m.f. between two points is measured by
the work done in joules in carrying a unit magnetic pole from one point to another.
2. Ampere-turns (AT). It is the unit of magnetometre force (m.m.f.) and is given by the
product of number of turns of a magnetic circuit and the current in amperes in those
turns.
3. Reluctance. It is the name given to that property of a material which opposes the
creation of magnetic flux in it. It, in fact, measures the opposition offered to the passage
of magnetic flux through a material and is analogous to resistance in an electric circuit
even in form. Its units is AT/Wb.
𝑙 𝐹
𝑆= =
𝜇0 𝜇 𝑟 𝐴 𝜙
In other words, the reluctance of a magnetic circuit is the number of amp-turns required
per weber of magnetic flux in the circuit. Since 1 AT/Wb = 1/henry, the unit of
reluctance is “reciprocal henry.”
4. Permeance. It is reciprocal of reluctance and implies the case or readiness with which
magnetic flux is developed. It is analogous to conductance in electric circuits. It is
measured in terms of Wb/AT or henry.
5. Reluctivity. It is specific reluctance and corresponds to resistivity which is ‘specific
resistance’.
1.17 COMPARISON BETWEEN MAGNETIC AND ELECTRIC CIRCUITS

Similarities

Differences
1. Strictly speaking, flux does not actually ‘flow’ in the sense in which an electric
current flows.
2. If temperature is kept constant, then resistance of an electric circuit is constant and
is independent of the current strength (or current density). On the other hand, the
reluctance of a magnetic circuit does depend on flux (and hence flux density)
established in it. It is so because μ (which equals the slope of B/H curve) is not constant
even for a given material as it depends on the flux density B. Value of μ is large for low
value of B and vice-versa. Hence, reluctance is small (S = l/μA) for small values of B
and large for large values of B.
3. Flow of current in an electric circuit involves continuous expenditure of energy but
in a magnetic circuit, energy is needed only creating the flux initially but not for
maintaining it.
Example : A toroid of mean radius 40 mm, effective csa 3cm2 , and relative permeability 150,
is wound with a 900 turn coil that carries a current of 1.5 A. Calculate (a) the mmf, (b) the
magnetic field strength and (c) the flux and flux density.

Solution
𝐹 = 𝑁𝐼 = 900 × 1.5 = 1350𝐴𝑇

𝐹 1350
𝐻= = = 5371.48 𝐴𝑇/𝑚
𝑙 2𝜋 × 40 × 10−3
𝐵 = 𝜇𝐻 = 𝜇0 𝜇𝑟 𝐻 = 4𝜋 × 10−7 × 150 × 5371.48 = 1.0125 𝑇

Φ = BA = 1.0125 × 3 × 10−4 = 303.75µWb
Example: A steel toroid of the dimensions shown in Fig. example 5 is wound with a 500 turn
coil of wire. What value of current needs to be passed through this coil in order to produce a
flux of 250 μWb in the toroid, if under these conditions the relative permeability of the toroid
is 300. Ans: I = 0.55 amp
Figure example 5
Example: A solenoid with a core of csa of 15 cm2 and relative permeability 65, produces
a flux of 200 μWb. If the core material is changed to one of relative permeability 800 what
will be the new flux and flux density? Ans: B = 1.641T and Φ = 2.462 mWb
Example : A mild steel ring having a csa of 5cm2 and mean circumference of 40 cm has a
coil of 200 turns wound uniformly around it. Calculate (i) reluctance of the ring (ii) current
required to produce a flux of 800 µWb in the ring. Take relative permeability of mild steel
as 380 and µo = 4π * 10 -7 Ans: S = 1.675 * 106 AT/Wb I = 6.7 A
1.18 B-H CURVES (MAGNETIZATION CURVES)

For ferromagnetic materials, µ is not constant but varies with flux density and there is no easy way to
compute it. In reality, however, it isn’t µ that you are interested in: What you really want to know is,
given B, what is H, and vice versa. A set of curves, called B-H or magnetization curves, provides this
information. (These curves are obtained experimentally and are available in handbooks. A separate
curve is required for each material.) Figure shows typical curves for cast iron, cast steel, and sheet steel

Figure 4
1.19 AMPERE’S CIRCUITAL LAW:
If we apply the analogy to Kirchhoff’s voltage law (for magnetic circuits).
States that the algebraic sum of the rises and drops of the mmf around a closed loop of a
magnetic circuit is equal to zero; that is, the sum of the rises in mmf equals the sum of the drops
in mmf around a closed loop. Equation is referred to as Ampère’s circuital law. When it is
applied to magnetic circuits, sources of mmf are expressed by the equation. The summation is
algebraic and terms are additive or subtractive, depending on the direction of flux and how the
coils are wound.
Consider the magnetic circuit appearing in Figure constructed of three different ferromagnetic
materials. Applying Ampère’s circuital law, we have

Which states that the applied mmf NI is equal to the sum of the drops HL around the loop. The
path to use for the HL terms is the mean (average) path.
You now have two magnetic circuit models (Figure below). While the reluctance model (a) is
not very useful for solving problems, it helps relate magnetic circuit problems to familiar
electrical circuit concepts. The Ampere’s law model, on the other hand, permits us to solve
practical problems.

1.20 COMPOSITE SERIES MAGNETIC CIRCUIT
Figure 5
In Fig. 5 is shown a composite series magnetic circuit consisting of three different magnetic
materials of different permeability and lengths and one air gap (μr = 1). Each path will

have its own reluctance. The total reluctance is the sum of individual reluctances as they
are joined in series.
𝑙
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑒 = ∑
𝜇0 𝜇𝑟 𝐴
𝑙1 𝑙2 𝑙3 𝑙𝑔
= + + +
𝜇0 𝜇𝑟 𝐴1 𝜇0 𝜇𝑟 𝐴2 𝜇0 𝜇𝑟 𝐴3 𝜇0 𝐴𝑔
𝑚. 𝑚. 𝑓
𝑓𝑙𝑢𝑥 Φ =
𝑙
𝜇0 𝜇𝑟 𝐴
How to Find Ampere-turns
𝑎𝑚𝑝𝑒𝑟𝑒_𝑡𝑢𝑟𝑛𝑠 𝐴𝑇 = 𝐻 × 𝑙
Hence, following procedure should be adopted for calculating the total ampere turns of a
composite magnetic path.
(i) Find H for each portion of the composite circuit. For air, H = B/μ 0, otherwise H =
B/μ0μr.
(ii) Find ampere-turns for each path separately by using the relation AT = H × l.
(iii) Add up these ampere-turns to get the total ampere-turns for the entire circuit.
Example: A series magnetic circuit comprises of three sections (i) length of 80 mm with cross-
sectional area 60 mm2, (ii) length of 70 mm with cross-sectional area 80 mm2 and (iii) and
airgap of length 0.5 mm with cross-sectional area of 60 mm2. Sections (i) and (ii) are of a
material having magnetic characteristics given by the following table.

H (AT/m) 100 210 340 500 800 1500

B (Tesla) 0.2 0.4 0.6 0.8 1.0 1.2
Determine the current necessary in a coil of 4000 turns wound on section (ii) to produce
a flux density of 0.8 Tesla in the air-gap. Neglect magnetic leakage.
Solution.
Section (i) It has the same cross-sectional area as the air-gap. Hence, it has the same flux
density i.e. 0.8 Tesla as in the air-gap. The value of the magnetising force H corresponding to
this flux density of 0.8T as read from the B-H plot is 500 AT/m.
m.m.f. reqd = H × l = 500 × (80 × 10−3) = 40 AT

Section (ii) Since its cross-sectional area is different from that of the air-gap, its flux density
would also be different even though, being a series circuit, its flux would be the same.
Air-gap flux = B × A = 0.8 × (60 × 10−6) = 48 × 10−6 Wb
Flux density in this section = 48 × 10−6/80 × 10−6 = 0.6 T
The corresponding value of the H from the given graph is 340 AT/m
m.m.f. reqd, for this section = 340 × (70 × 10−3) = 23.8 AT.
Air-gap
H = B/μ0 = 0.8/4π × 10−7 = 0.6366 × 106 AT/m
∴ m.m.f. reqd. = 0.6366 × 106 × (0.5 × 10−3) = 313.31 AT
Total m.m.f. reqd. = 40 + 23.8 + 313.31 = 377.11
NI = 377.11 or I = 377.11/4000 = 0.094 A
1.21 PARALLEL MAGNETIC CIRCUITS
Figure 6

Fig. 6 (a) shown a parallel magnetic circuit consisting of two parallel magnetic paths ACB and
ADB acted upon by the same m.m.f. Each magnetic path has an average length of 2 (l1 + l2).
The flux produced by the coil wound on the central core is divided equally at point A between
the two outer parallel paths. The reluctance offered by the two parallel paths is = half the
reluctance of each path.
Fig. 6 (b) shows the equivalent electrical circuit where resistance offered to the voltage source
is = R//R = R/2
1.22 SERIES-PARALLEL MAGNETIC CIRCUITS
Figure 7
Such a circuit is shown in Fig. 7 (a). It shows two parallel magnetic circuits ACB and ACD
connected across the common magnetic path AB which contains an air-gap of length lg. As
usual, the flux Φ in the common core is divided equally at point A between the two parallel
paths which have equal reluctance. The reluctance of the path AB consists of
(i) air gap reluctance and

(ii) the reluctance of the central core which comparatively negligible.
Hence, the reluctance of the central core AB equals only the air-gap reluctance across which
are connected two equal parallel reluctances. Hence, the m.m.f. required for this circuit
would be the sum of
(i) that required for the air-gap and

(ii) that required for either of two paths (not both).
The equivalent electrical circuit is shown in Fig. 7 (b) where the total resistance offered to
the voltage source is = R1 + R//R = R1 + R/2.

Example: A magnetic circuit made of mild steel is arranged as shown in Fig. 8 The central
limb is wound with 500 turns and has a cross-sectional area of 800 mm2. Each of the outer
limbs has a cross-sectional area of 500 mm2. The air-gap has a length of 1 mm. Calculate the
current required to set up a flux of 1.3 mWb in the central limb assuming no magnetic leakage
and fringing. Mild steel required 3800 AT/m to produce flux density of 1.625 T and 850 AT/m
to produce flux density of 1.3 T.
Figure 8
Solution.
Flux in the central limb is = 1.3 mWb = 1.3 × 10−3 Wb
Cross section A = 800 mm2 =800 × 10−6 m2
∴ B = Φ/A = 1.3 × 10−6 /800 × 10−6 = 1.625 T
Corresponding value of H for this flux density is given as 3800 AT/m.
Since the length of the central limb is 120 mm.
m.m.f. required is AT = H × l = 3800 × (120 × 10−3) = 456 AT/m.
Air-gap
Flux density in the air-gap is the same as that in the central limb.
H = B/μ0 = 1.625/4π × 10−7 = 0.1293 × 107 AT/m
Length of the air-gap = 1 mm = 10−3 m
m.m.f. reqd. for the air-gap = H × l = 0.1293 × 107 × 10−3 = 1293 AT.
The flux of the central limb divides equally at point A in figure along the two parallel path
ABCD and AFED. We may consider either path, say ABCD and calculate the m.m.f. required
for it. The same m.m.f. will also send the flux through the other parallel path AFED.
Flux through ABCD = 1.3 × 10−3/2 = 0.65 × 10−3 Wb
Flux density B = 0.65 × 10−3/500 × 10−6 = 1.3 T
The corresponding value of H for this value of B is given at 850 AT/m.

∴ m.m.f. reqd. for path ABCD = H × l = 850 × (300 × 10−3) = 255 AT

As said above, this, m.m.f. will also send the flux in the parallel path AFED.
Total m.m.f. reqd. = 456 + 1293 + 255 = 2004 AT
Since the number of turns is 500,
I = 2004/500 = 4A.
Example: A cast steel D.C. electromagnet shown in Fig. 9 has a coil of 1000 turns on its central
limb. Determine the current that the coil should carry to produce a flux of 2.5 mWb in the air-
gap. Neglect leakage. Dimensions are given in cm. The magnetisation curve for cast steel is
as under:
Flux density (Wb/m2): 0.2 0.5 0.7 1.0 1.2
Amp-turns/metre: 300 540 650 900 1150
Figure 9
Solution.
Two points should be noted (i) there are two (equal) parallel paths ACDE and AGE across the
central path AE. (ii) flux density in either parallel path is half of that in the central path because
flux divides into two equal parts at point A. Total m.m.f. required for the whole
electromagnet is equal to the sum of the following three m.m.fs.
i. that required for path EF
ii. that required for air-gap
iii. that required for either of the two parallel paths; say, path ACDE
Flux density in the central limb and air gap is
= 2.5 × 10−3/ (5 × 5) × 10−4 = 1 Wb/m2
Corresponding value of H as found from the given data is 900 AT/m.
∴ AT for central limb = 900 × 0.3 = 270
H in air-gap = B/μ0 = 1/4π × 10−7 = 79.56 × 104 AT/m
AT required = 79.56 × 104 × 10−3 = 795.6
Flux density in path ACDE is 0.5 Wb/m2 for which corresponding value of H is 540 AT/m.

∴ AT required for path ACDE = 540 × 0.6 = 324

Total AT required = 270 + 795.6 + 324 = 1390
Current required = 1390/1000 = 1.39 A
1.23 MAGNETIC HYSTERESIS

Hysteresis comes from a Greek word meaning ‘to lag behind’. It is found that when the
magnetic field strength in a magnetic material is varied, the resulting flux density displays a
lagging effect.
Consider such a specimen of magnetic material that initially is completely unmagnetised. If no
current flows through the magnetising coil then both H and B will initially be zero. The value
of H is now increased by increasing the coil current in discrete steps. The corresponding flux
density is then noted at each step. If these values are plotted on a graph until magnetic saturation
is achieved, the dotted curve (the initial magnetisation curve) shown in Fig. 10 results.

Figure 10
Let the current now be reduced (in steps) to zero, and the corresponding values for B again
noted and plotted. This would result in the section of graph from A to C. This shows that when
the current is zero once more (so H = 0), the flux density has not reduced to zero. The flux
density remaining is called the remanent flux density (OC). This property of a magnetic
material, to retain some flux after the magnetising current is removed, is known as the
remanence or retentivity of the material.
Let the current now be reversed, and increased in the opposite direction. This will have the
effect of opposing the residual flux. Hence, the latter will be reduced until at some value of ˗H
it reaches zero (point D on the graph). The amount of reverse magnetic field strength required
to reduce the residual flux to zero is known as the coercive force. This property of a material
is called its coercivity.
If we now continue to increase the current in this reverse direction, the material will once more
reach saturation (at point E). In this case it will be of the opposite polarity to that achieved at
point A on the graph.
Once again, the current may be reduced to zero, reversed, and then increased in the original
direction. This will take the graph from point E back to A, passing through points F and G on
the way.
Note that residual flux density shown as OC has the same value, but opposite polarity, to that
shown as OF. Similarly, coercive force OD = OG.
In taking the specimen through the loop ACDEFGA we have taken it through one complete
magnetisation cycle. The loop is referred to as the hysteresis loop. The degree to which a

material is magnetised depends upon the extent to which the ‘molecular magnets’ have been
aligned.
Thus, in taking the specimen through a magnetisation cycle, energy must be expended. This
energy is proportional to the area enclosed by the loop, and the rate (frequency) at which the
cycle is repeated.
Magnetic materials may be subdivided into what are known as ‘hard’ and ‘soft’ magnetic
materials. A hard-magnetic material is one which possesses a large remanance and coercivity.
It is therefore one which retains most of its magnetism, when the magnetising current is
removed. It is also difficult to demagnetise. These are the materials used to form permanent
magnets, and they will have a very ‘fat’ loop as illustrated in Fig. 11(a). A soft magnetic
material, such as soft iron and mild steel, retains very little of the induced magnetism. It will
therefore have a relatively ‘thin’ hysteresis loop, as shown in Fig. 11 (b). The soft magnetic
materials are the ones used most often for engineering applications. Examples are the magnetic
circuits for rotating electric machines (motors and generators), relays, and the cores for
inductors and transformers.
Figure 11
When a magnetic circuit is subjected to continuous cycling through the loop a considerable
amount of energy is dissipated. This energy appears as heat in the material. Since this is
normally an undesirable effect, the energy thus dissipated is called the hysteresis loss. Thus,
the thinner the loop, the less wasted energy. This is why ‘soft’ magnetic materials are used for
the applications listed above.

CHAPTER 2
ELECTROMAGNETISM
This chapter concerns the principles and laws governing electromagnetic induction and the
concepts of self and mutual inductance.
On completion of this chapter you should be able to use these principles to:
i. Understand the basic operating principles of motors and generators.
ii. Carry out simple calculations involving the generation of voltage, and the production
of force and torque.
iii. Determine the value of inductors, and apply the concepts of self and mutual
inductance to the operating principles of transformers.
2.1 INTRODUCTION
It is well known that whenever an electric current flows through a conductor, a magnetic field
is immediately brought into existence in the space surrounding the conductor. It can be said
that when electrons are in motion, they produce a magnetic field. The converse of this is also
true i.e. when a magnetic field embracing a conductor moves relative to the conductor, it
produces a flow of electrons in the conductor. This phenomenon whereby an e.m.f. and hence
current (i.e. flow of electrons) is induced in any conductor which is cut across or is cut by a
magnetic flux is known as electromagnetic induction.
Most applications of magnetism involve magnetic effects due to electric currents. The
current, I, creates a magnetic field that is concentric about the conductor, uniform along its
length, and whose strength is directly proportional to I.
2.2 FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION
It is mainly due to the pioneering work of Michael Faraday, in the nineteenth century, that the
modern technological world exists as we know it. Without the development of the generation
of electrical power, such advances would have been impossible. Thus, although the concepts
involved with electromagnetic induction are very simple, they have far-reaching influence.
Faraday’s law is best considered in two interrelated parts:
1 The value of emf induced in a circuit or conductor is directly proportional to the rate of
change of magnetic flux linking with it.
2 The polarity of such an emf, induced by an increasing flux, is opposite to that induced
by a decreasing flux.

Figure 1
−𝑁𝑑Φ
𝑒= 𝑣𝑜𝑙𝑡 − − − − − −(1)
𝑑𝑡
Equation 1 forms the basis for the definition of the unit of magnetic flux, the weber, thus:
The weber is that magnetic flux which, linking a circuit of one turn, induces in it an emf of one
volt when the flux is reduced to zero at a uniform rate in one second. In other words, 1 volt =
1 weber/second or 1 weber = 1 volt second.
2.3 Lenz’s Law

This law states that the polarity of an induced emf is always such that it opposes the change
which produced it. This is similar to the statement in mechanics that for every force there is an
opposite reaction.
2.4 Fleming’s Righthand Rule

This is a convenient means of determining the polarity of an induced emf in a conductor. Also,
provided that the conductor forms part of a complete circuit, it will indicate the direction of the
resulting current flow.
The first finger, the second finger and the thumb of the righthand are held out mutually at right
angles to each other (like the three edges of a cube as shown in Fig 2a). The First finger
indicates the direction of the Flux, the thuMb the direction of Motion of the conductor relative
to the flux, and the sECond finger indicates the polarity of the induced Emf, and direction of
Current flow. This process is illustrated in Fig. 2b, which shows the cross-section of a
conductor being moved vertically upwards at a constant velocity through the magnetic field.

(a) (b)
Figure 2
Example 1: The flux linking a 100 turn coil changes from 5mWb to 10mWb in a time of 5ms.
Calculate the average emf induced in the coil.
Example 2: A 250 turn coil is linked by a magnetic flux that varies as follows: an increase
from zero to 20mWb in a time of 0.05s; constant at this value for 0.02s; followed by a decrease
to 4mWb in a time of 0.01s. Assuming that these changes are uniform, draw a sketch graph
(i.e. not to an accurate scale) of the variation of the flux and the corresponding emf induced in
the coil, showing all principal values.
2.5 EMF INDUCED IN A SINGLE STRAIGHT CONDUCTOR

Consider a conductor moving at a constant velocity v metre per second at right angles to a
magnetic field having the dimensions shown in Fig. 3. The direction of the induced emf may
be obtained using Fleming’s righthand rule, and is shown in the diagram. Equation 1 is
applicable, and in this case, the value for N is 1.
Figure 3

Thus
dΦ
𝑒=
𝑑𝑡
But flux is constant
Φ
𝑒=
𝑡
Φ = 𝐵𝐴
BA
𝑒=
𝑡
The cross sectional area of the field 𝐴 = 𝑙 × 𝑑
Bld
𝑒=
𝑡
𝑑
𝑉 = , 𝑒 = 𝐵𝑙𝑣 − − − − − −(2)
𝑡
The above equation is only true for the case when the conductor is moving at right angles to
the magnetic field. If the conductor moves through the field at some angle less than 90°, then
the ‘cutting’ action between the conductor and the flux is reduced. This results in a consequent
reduction in the induced emf. Thus, if the conductor is moved horizontally through the field,
the ‘cutting’ action is zero, and so no emf is induced. To be more precise, we can say that only
the component of the velocity at 90° to the flux is responsible for the induced emf. In general
therefore, the induced emf is given by
𝑒 = 𝐵𝑙𝑣 𝑠𝑖𝑛 𝜃 𝑣𝑜𝑙𝑡

where v sinθ is the component of velocity at 90° to the field, as illustrated in Fig. 4 .
Figure 4

Example 3: A conductor is moved at a velocity of 5m/s at an angle of 60° to a uniform

magnetic field of 1.6mWb. The field is produced by a pair of pole pieces, the faces of which
measure 10cm by 4cm. If the conductor length is parallel to the longer side of the field,
calculate the emf induced.
This section, covering the induction or generation of an emf in a conductor moving through a
magnetic field, forms the basis of the generator principle. However, most electrical generators
are rotating machines, and we have so far considered only linear motion of the conductor.
Consider the conductor now formed into the shape of a rectangular loop, mounted on to an
axle. This arrangement is then rotated between the poles of a permanent magnet. We now have
the basis of a simple generator as illustrated in Fig. 5
Figure 5
The two sides of the loop that are parallel to the pole faces will each have an effective length l
metre. At any instant of time, these sides are passing through the field in opposite directions.
Applying the right hand rule at the instant shown in Fig. 5, the directions of the induced emfs
will be as marked, i.e. of opposite polarities. However, if we trace the path around the loop, it
will be seen that both emfs are causing current to flow in the same direction around the loop.
This is equivalent to two cells connected in series as shown in Fig. 6
The situation shown in Fig. 5 applies only to one instant in one revolution of the loop (it is
equivalent to a ‘snapshot’ at that instant).

Figure 6
If we were to plot a graph of the total emf generated in the loop, for one complete revolution,
it would be found to be one cycle of a sinewave, i.e. an alternating voltage.
2.6 FORCE ON A CURRENT-CARRYING CONDUCTOR

Figure 7(a) shows the field patterns produced by two pole pieces, and the current flowing
through the conductor. Since the lines of flux obey the rule that they will not intersect, then the
flux pattern from the poles will be distorted as illustrated in Fig. 7(b). Also, since the lines of
flux tend to act as if elastic, then they will try to straighten themselves. This results in a force
being exerted on the conductor, in the direction shown.
Figure 7
The direction of this force may be more simply obtained by applying Fleming’s left hand rule.
This rule is similar to the right hand rule. The major difference is of course that the fingers and
thumb of the left hand are now used. In this case, the First finger indicates the direction of the
main Flux (from the poles). The seCond finger indicates the direction of Current flow. The
thuMb shows the direction of the resulting force and hence consequent Motion. This is shown
in Fig. 8

Figure 8
We can derive an expression for the magnetic force on a current by taking a sum of the magnetic
forces on individual charges.
The force on an individual charge moving at the drift velocity Vd is given by

𝐹 = 𝑞𝑉𝑑 𝐵 sin 𝜃
𝐹 = 𝑁𝑞𝑉𝑑 𝐵 sin 𝜃 where N is number of charge carriers in the section of the wire of length l
𝐹 = 𝑛𝐴𝑙𝑞𝑉𝑑 𝐵 sin 𝜃 where n = number of charge carriers per unit volume
𝐹 = (𝑛𝑞𝐴𝑉𝑑 )𝑙𝐵 sin 𝜃
𝑛𝑞𝐴𝑉𝑑 = 𝐼
𝐹 = 𝐵𝐼𝑙 sin 𝜃 − − − − − − − (3)
Any conductor extending beyond the main field does not contribute to the force exerted.
Example 4: A pair of pole pieces 5cm by 3cm produce a flux of 2.5mWb. A conductor is placed
in this field with its length parallel to the longer dimension of the poles. When a current is
passed through the conductor, a force of 1.25N is exerted on it. Determine the value of the
current.
If the conductor was placed at 45° to the field, what then would be the force exerted?
2.6 FORCE BETWEEN PARALLEL CONDUCTORS
When two parallel conductors are both carrying current their magnetic fields will interact to
produce a force of attraction or repulsion between them. This is illustrated in Fig. 9

Figure 9
In order to determine the value of such a force, consider first a single conductor carrying a
current of I ampere. The magnetic field produced at some distance d from its centre is shown
in Fig. 10.
Figure 10
𝑁𝐼
In general, 𝐻 = 𝑎𝑚𝑝𝑒𝑟𝑒 𝑡𝑢𝑟𝑛/𝑚𝑒𝑡𝑟𝑒
𝑙
But in this case, N = 1 and l = 2πd, so
𝑁𝐼
𝐻=
2𝜋𝑑
Now, flux density 𝐵 = 𝜇0 𝜇𝑟 𝐻 tesla, and as the field exists in air, then µr =1. Thus, the flux
density at distance d from the centre is given by
𝜇𝑜 𝐼
𝐵=
2𝜋𝑑

Consider now two conductors Y and Z carrying currents I1 and I2 respectively, at a distance of
d metres between their centres as in Fig 11
Figure 11
we can say that the flux density acting on Z due to current I1 flowing in Y is:
𝜇𝑜 𝐼1
𝐵=
2𝜋𝑑
And the force exerted on Z = B1I2l newton, or B1I2 newton per meter length of Z
Hence force/meter length acting on Z
𝜇𝑜 𝐼1 𝐼2
=
2𝜋𝑑
4𝜋 × 10−7 𝐼1 𝐼2
=
2𝜋𝑑
2 × 10−7 𝐼1 𝐼2
𝑑
Now, the current I2 flowing in Z also produces a magnetic field which will exert a force on Y.
Using the same reasoning as above, it can be shown that:
force/metre length acting on Y
2 × 10−7 𝐼1 𝐼2
𝑑
If I1 = I2 = 1A and d = 1m
Force exerted on each conductor =2 × 10−7 𝑛𝑒𝑤𝑡𝑜𝑛

This value of force forms the basis for the definition of the ampere, namely: that current, which
when maintained in each of two infinitely long parallel conductors situated in vacuum, and
separated one metre between centres, produces a force of 2 × 10−7 𝑛𝑒𝑤𝑡𝑜𝑛 per metre length
on each conductor.
Example 5: Two long parallel conductors are spaced 35mm between centres. Calculate the
force exerted between them when the currents carried are 50 A and 40 A respectively.

2.7 SELF AND MUTUAL INDUCTANCE

The effects of self and mutual inductance can be demonstrated by another simple experiment.
Consider two coils, as shown in Figure 1. Coil 1 can be connected to a battery via a switch.
Coil 2 is placed close to coil 1, but is not electrically connected to it. Coil 2 has a galvo
connected to its terminals.
Figure 1
When the switch is closed, the current in coil 1 will rapidly increase from zero to some steady
value. Hence, the flux produced by coil 1 will also increase from zero to a steady value. This
changing flux links with the turns of coil 2, and therefore induces an emf into it. This will be
indicated by a momentary deflection of the galvo pointer.
Similarly, when the switch is subsequently opened, the flux produced by coil 1 will collapse to
zero. The galvo will again indicate that a momentary emf is induced in coil 2, but of the
opposite polarity to the first case. Thus, an emf has been induced into coil 2, by a changing
current (and flux) in coil 1. This is known as a mutually induced emf.
If the changing flux can link with coil 2, then it must also link with the turns of coil 1. Thus,
there must also be a momentary emf induced in this coil. This is known as a self-induced emf.
Any induced emf obeys Lenz’s law. This self-induced emf must therefore be of the opposite
polarity to the battery emf. For this reason, it is also referred to as a back emf. Unfortunately,
it is extremely difficult to demonstrate the existence of this back emf. If a voltmeter was
connected across coil 1, it would merely indicate the terminal voltage of the battery.
2.8 SELF-INDUCTANCE
Self-inductance is that property of a circuit or component which causes a self-induced emf to
be produced, when the current through it changes. The unit of self-inductance is the henry,
which is defined as follows:

A circuit has a self-inductance of one henry (1 H) if an emf of one volt is induced in it, when
the circuit current changes at the rate of one ampere per second (1 A/s).
The quantity symbol for self-inductance is L. From the above definition, we can state the
following equation
−𝑒
𝐿= ℎ𝑒𝑛𝑟𝑦
𝑑𝑖⁄
𝑑𝑡
Example: A coil has a self-inductance of 0.25 H. Calculate the value of emf induced, if the
current through it changes from 100 mA to 350 mA, in a time of 25ms.
Example Calculate the inductance of a circuit in which an emf of 30 V is induced, when the
circuit current changes at the rate of 200 A/s.
2.9 SELF-INDUCTANCE AND FLUX LINKAGES
Consider a coil of N turns, carrying a current of I amp. Let us assume that this current produces
a flux of фweber. If the current now changes at a uniform rate of di /dt ampere per second, it
will cause a corresponding change of flux of dф /dt weber per second. Let us also assume that
the coil has a self-inductance of L henry.Theself-induced emf may be determined from
𝑑𝑖
𝑒 = −𝐿
𝑑𝑡
However, the induced emf is basically due to the rate of change of flux linkages
𝑑Φ
𝑒 = −𝑁
𝑑𝑡
Since both equations represent the same induced emf, then
𝑑Φ 𝑑𝑖
𝑁 =𝐿
𝑑𝑡 𝑑𝑡
𝑑Φ
𝐿=𝑁
𝑑𝑖
A coil which is designed to have a specific value of self-inductance is known as an inductor
If the current I through an N turn coil produces a flux of ф weber, then its self-inductance is
given by the equation

𝑁Φ
𝐿= 𝐻𝑒𝑛𝑟𝑦
𝐼
Example: A coil of 150 turns carries a current of 10 A. This current produces a magnetic flux
of 0.01 Wb. Calculate (a) the inductance of the coil, and (b) the emf induced when the current
is uniformly reversed in a time of 0.1 s.
Example: A current of 8 A, when flowing through a 3000 turn coil, produces a flux of 4 mWb.
If the current is reduced to 2 A in a time of 100 ms, calculate the emf thus induced in the coil.
Assume that the flux is directly proportional to the current.
2.10 FACTORS AFFECTING INDUCTANCE
Consider a coil of N turns wound on to a non-magnetic core, of uniform cross sectional area A
metre2 and mean length L metre. The coil carries a current of I amp, which produces a flux of
ф weber. We know that the inductance will be
𝑁Φ
𝐿= ℎ𝑒𝑛𝑟𝑦, But ф = BA
𝐼
𝑁BA
Therefore 𝐿 =
𝐼
𝑁I Hl
𝑎𝑙𝑠𝑜 𝐻 = , 𝑠𝑜 𝐼 =
𝑙 𝑁
𝑁 2 BA
𝐿=
𝐻𝑙
𝐵
𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛 = 𝜇0 𝜇𝑟
𝐻
𝜇0 𝜇𝑟 𝑁 2 A
𝐿=
𝑙
𝑙
𝑏𝑢𝑡 𝑆 =
𝜇0 𝜇 𝑟 𝐴
𝑁2
𝐿= 𝐻𝑒𝑛𝑟𝑦
𝑆
Example: A 600 turn coil is wound on to a non-magnetic core of effective length 45 mm and
csa 4 cm2. (a) Calculate the inductance (b) The number of turns is increased to 900. Calculate

the inductance value now produced. (c) The core of the 900 turn coil is now replaced by an
iron core having a relative permeability of 75, and of the same dimensions as the original.
Calculate the inductance in this case
2.11 MUTUAL INDUCTANCE
When a changing current in one circuit induces an emf in another separate circuit, then the two
circuits are said to possess mutual inductance. The unit of mutual inductance is the henry, and
is defined as follows. Two circuits have a mutual inductance of one henry, if the emf induced
in one circuit is one volt, when the current in the other is changing at the rate of one ampere
per second. The quantity symbol for mutual inductance is M, and expressing the above
definition as an equation we have
𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑐𝑜𝑖𝑙 2

𝑀 =
𝑟𝑎𝑡𝑒𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑐𝑜𝑖𝑙 1
−𝑒2
𝑀=
𝑑𝑖1⁄
𝑑𝑡
−𝑀𝑑𝑖1
𝑒2 = 𝑣𝑜𝑙𝑡
𝑑𝑡
This emf may also be expressed in terms of the flux linking coil 2. If all of the flux from coil 1
links with coil 2, then we have what is called 100% flux linkage. In practice, it is more usual
for only a proportion of the flux from coil 1 to link with coil 2. Thus the flux linkage is usually
less than 100%. This is indicated by a factor, known as the coupling factor, k. 100% coupling
is indicated by k = 1. If there is no flux linkage with coil 2, then k will have a value of zero. So
if zero emf is induced in coil 2, the mutual inductance will also be zero. Thus, the possible
values for the coupling factor k, lie between zero and 1. Expressed mathematically, this is
written as
0≤𝑘≤1
Consider two coils possessing mutual inductance, and with a coupling factor less than 1. Let a
change of current di1 /dt amp/s in coil 1 produce a change of flux d φ1 /dt weber/s. The
proportion of this flux change linking coil 2 will be dφ2 /dt weber/s. If the number of turns on
coil 2 is N2, then

−𝑁2 𝑑Φ2
𝑒2 = 𝑣𝑜𝑙𝑡
𝑑𝑡
𝑀𝑑𝑖1 𝑁2 𝑑Φ2
=
𝑑𝑡 𝑑𝑡
𝑁2 𝑑Φ2
𝑀= ℎ𝑒𝑛𝑟𝑦
𝑑𝑖1
As with self-inductance for a single coil, mutual inductance is a property of a pair of coils.
They therefore retain this property, regardless of whether or not an emf is induced. Hence,
equation may be modified to
𝑁2 Φ2
𝑀=
𝐼1
Example: Two coils, A and B, have 2000 turns and 1500 turns respectively. A current of 0.5
A, flowing in A, produces a flux of 60Wb. The flux linking with B is 83% of this value.
Determine (a) the self-inductance of coil A, and (b) the mutual inductance of the two coils.
2.12 RELATIONSHIP BETWEEN SELF- AND MUTUAL-INDUCTANCE
Consider two coils of N1 and N2 turns respectively, wound on to a common non-magnetic core.
If the reluctance of the core is S ampere turns/weber, and the coupling coefficient is unity, then
𝑁12 𝑁22
𝐿1 = 𝑎𝑛𝑑 𝐿2 =
𝑆 𝑆
Therefore
𝑁12 𝑁22
𝐿1 𝐿2 = − − − − − − − − − (1)
𝑆2
𝑁2 Φ 𝑁1
𝑀= ℎ𝑒𝑛𝑟𝑦, 𝑎𝑛𝑑 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑦
𝐼1 𝑁1
𝑁1 𝑁2 Φ
𝑀=
𝑁1 𝐼1
1 Φ
=
𝑆 𝑁1 𝐼1
𝑁1 𝑁2
𝑀=
𝑆

2
𝑁12 𝑁22
𝑀 = − − − − − − − − − − − − − (2)
𝑆2
Comparing equation 1 and 2
𝑀2 = 𝐿1 𝐿2
Therefore
𝑀 = √𝐿1 𝐿2
The above equation is correct only provided that there is 100% coupling between the coils; i.e.
k = 1. If k is less than 1, then the general form of the equation, shown below
𝑀 = 𝑘√𝐿1 𝐿2
Example: A 400 turn coil is wound onto a cast steel toroid having an effective length of 25 cm
and csa 4.5 cm2. If the steel has a relative permeability of 180 under the operating conditions,
calculate the self-inductance of the coil.
2.13 INDUCTANCES IN SERIES
Let the two coils be so joined in series that their fluxes (or m.m.fs) are additive i.e., in the same
direction.
Figure 2
Let M = coefficient of mutual inductance

L1 = coefficient of self-inductance of 1st coil
L2 = coefficient of self-inductance of 2nd coil.
𝑑𝑖
Then, self induced e.m.f. in A is 𝑒1 = − 𝐿1 𝑑𝑡
𝑑𝑖
Mutually-induced e.m.f. in A due to change of current in B is = e′1 = − 𝑀 𝑑𝑡

𝑑𝑖
Self-induced e.m.f. in B is = 𝑒2 = − 𝐿2 𝑑𝑡
𝑑𝑖
Mutually-induced e.m.f. in B due to change of current in A is = 𝑒 ′ 2 = − 𝑀. 𝑑𝑡
𝑑𝑖
Total induced e.m.f. in the combination = − 𝑑𝑡 (𝐿1 + 𝐿2 + 2𝑀)
If L is the equivalent inductance then total induced e.m.f. in that single coil would have been
𝑑𝑖
= −𝐿 𝑑𝑡
𝐿 = 𝐿1 + 𝐿2 + 2𝑀
When the coils are so joined that their fluxes are in opposite directions
Figure 3
𝑑𝑖 𝑑𝑖 𝑑𝑖 𝑑𝑖
𝑒1 = − 𝐿1. 𝑑𝑡 , e1′ = 𝑀. 𝑑𝑡, 𝑒2 = − 𝐿1. 𝑑𝑡, 𝑒2′ = 𝑀. 𝑑𝑡
𝑑𝑖
Total induced e.m.f. = − 𝑑𝑡 (𝐿1 + 𝐿2 − 2𝑀)
𝐿 = 𝐿1 + 𝐿2 − 2𝑀
Example: Two coils with a coefficient of coupling of 0.5 between them, are connected in series
so as to magnetise (a) in the same direction (b) in the opposite direction. The corresponding
values of total inductances are for (a) 1.9 H and for (b) 0.7 H. Find the self-inductances of the
two coils and the mutual inductance between them.

2.14 INDUCTANCE IN PARALLEL
Figure 4
In Fig. 4, two inductances of values L1 and L2 henry are connected in parallel. Let the
coefficient of mutual inductance between the two be M. Let i be the main supply current and
i1 and i2 be the branch currents Obviously, i = i1 + i2
𝑑𝑖 𝑑𝑖1 𝑑𝑖2
= +
𝑑𝑡 𝑑𝑡 𝑑𝑡
In each coil, both self and mutually induced e.m.fs. are produced. Since the coils are in parallel,
these e.m.fs. are equal. For a case when self-induced e.m.f., we get
𝑑𝑖1 𝑑𝑖2 𝑑𝑖2 𝑑𝑖1 𝑑𝑖

𝑒 = 𝐿1 +𝑀 = 𝐿2 +𝑀 =𝐿
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
Where L is the equvalent inductance of the two inductor
Hence
𝑑𝑖1 𝑑𝑖2 𝑑𝑖2 𝑑𝑖1

𝐿1 +𝑀 = 𝐿2 +𝑀
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑖1 𝑑𝑖2
(𝐿1 − 𝑀) = (𝐿2 − 𝑀)
𝑑𝑡 𝑑𝑡
𝑑𝑖1 (𝐿2 − 𝑀) 𝑑𝑖2

=
𝑑𝑡 (𝐿1 − 𝑀) 𝑑𝑡
Therefore
𝑑𝑖 (𝐿2 − 𝑀) 𝑑𝑖2 𝑑𝑖2

= +
𝑑𝑡 (𝐿1 − 𝑀) 𝑑𝑡 𝑑𝑡
𝑑𝑖 (𝐿2 − 𝑀) 𝑑𝑖2
=( + 1) − − − − − −(1)
𝑑𝑡 (𝐿1 − 𝑀) 𝑑𝑡

Also
𝑑𝑖 𝑑𝑖1 𝑑𝑖2
𝐿 = 𝐿1 +𝑀
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑖 1 𝑑𝑖1 𝑑𝑖2
= (𝐿1 +𝑀 )
𝑑𝑡 𝐿 𝑑𝑡 𝑑𝑡
𝑑𝑖 1 (𝐿2 − 𝑀) 𝑑𝑖2 𝑑𝑖2

= (𝐿1 +𝑀 )
𝑑𝑡 𝐿 (𝐿1 − 𝑀) 𝑑𝑡 𝑑𝑡
𝑑𝑖 1 (𝐿2 − 𝑀) 𝑑𝑖2
= (𝐿1 + 𝑀) − − − − − − − (2)
𝑑𝑡 𝐿 (𝐿1 − 𝑀) 𝑑𝑡
equating (1) and (2)
(𝐿2 − 𝑀) 𝑑𝑖2 1 (𝐿2 − 𝑀) 𝑑𝑖2

( + 1) = (𝐿1 + 𝑀)
(𝐿1 − 𝑀) 𝑑𝑡 𝐿 (𝐿1 − 𝑀) 𝑑𝑡
𝐿1 𝐿2 − 𝑀2
𝐿= − − − − − (3)
𝐿1 + 𝐿21 − 2𝑀
Equation 3 is when mutual field assists the separate fields.
For when the two fields oppose each other
𝐿1 𝐿2 − 𝑀2
𝐿=
𝐿1 + 𝐿21 + 2𝑀
Example: Two coils of inductances 4 and 6 henry are connected in parallel. If their mutual
inductance is 3 henry, calculate the equivalent inductance of the combination if (i) mutual
inductance assists the self-inductance (ii) mutual inductance opposes the self-inductance
2.15 ENERGY STORED
As with an electric field, a magnetic field also stores energy. When the current through an
inductive circuit is interrupted, by opening a switch, this energy is released. This is the reason
why a spark or arc occurs between the contacts of the switch, when it is opened.
Consider an inductor connected in a circuit, in which the current increases uniformly, to some
steady value I amp. This current change is illustrated in Fig. 2. The magnitude of the emf
induced by this change of current is given by

𝐿𝐼
𝑒= 𝑣𝑜𝑙𝑡
𝑡
Figure 5
The average power input to the coil during this time is:
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 = 𝑒 × 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
From the graph, it may be seen that the average current over this time is I/2 amp.
Therefore,
1 𝐿𝐼𝐼 𝐿𝐼 2
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 = 𝑒𝐼 = =
2 2𝑡 2𝑡
But, energy stored = average power × time
𝐿𝐼 2
𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑜𝑟𝑒𝑑 𝑊 = ×𝑡
2𝑡
1
𝑊 = 𝐿𝐼 2 𝑗𝑜𝑢𝑙𝑒𝑠 − − − − − −(1)
2
Equation 1 applies to a single inductor. When two coils possess mutual inductance, and are
connected in series, both will store energy. In this situation, the total energy stored is given by
the equation
1 1
𝑊= 𝐿1 𝐼12 + 𝐿2 𝐼22 ± 𝑀𝐼1 𝐼2 𝑗𝑜𝑢𝑙𝑒𝑠
2 2

When the coils are connected in series such that the two fluxes produced act in the same
direction, the total flux is increased and the coils are said to be connected in series aiding. In
this case the total energy stored in the system will be increased, so the last term in the above
equation is added, i.e. the positive sign applies. If, however, the connections to one of the coils
are reversed, then the two fluxes will oppose each other, the total flux will be reduced, and the
coils are said to be in series opposition. In this case the negative sign is used. These two
connections are shown in Figure .
Example: Two inductors, of inductance 25 mH and 40mH respectively, are wound on a

common ferromagnetic core, and are connected in series with each other. The coupling
coefficient, k, between them is 0.8. When the current flowing through the two coils is 0.25 A,
calculate (a) the energy stored in each, (b) the total energy stored when the coils are connected
(i) in series aiding, and (ii) in series opposition.
2.16 AUTOMOBILE IGNITION SYSTEM

Practical application of mutual induction is found in the single-spark petrol-engine ignition
system extensively employed in automobiles and air engines. Fig.7 shows the circuit diagram
of such a system as applied to a 4-cylinder automobile engine. It has a spark coil (or induction
coil) which consists of a primary winding (of a few turns) and a secondary winding (of a large
number of turns) wound on a common iron core (for increasing mutual induction). The primary
circuit (containing battery B) includes a ‘make and break contact’ actuated by a timer cam.
The secondary circuit includes the rotating blade of the distributor and the spark gap. The timer
cam and the distributor are mounted on the same shaft and are geared to rotate at exactly half
the speed of the engine shaft. It means that in the case of automobile engines (which are
fourcycle engines) each cylinder is fired only once for every two revolutions of the engine shaft

Figure 7
When timer cam rotates, it alternately closes and opens the primary circuit. During the time
primary circuit is closed, current through it rises exponentially, and so does the magnetic field
of the primary winding. When the cam suddenly opens the primary circuit, the magnetic field
collapses rapidly thereby producing a very large e.m.f. in secondary by mutual induction.
During the time this large e.m.f. exists, the distributor blade rotates and connects the secondary
winding across the proper plug and so the secondary circuit is completed except for the spart
gap in the spark plug. However, the induced e.m.f. is large enough to make the current jump
across the gap thus producing a spark which ignites the explosive mixture in the engine
cylinder. The function of capacitor C connected across the ‘make and break’ contact is two-
fold :
(i) to make the break rapid so that large e.m.f. is induced in secondary and
(ii) to reduce sparking and burning at the ‘make-and-break’ contact thereby prolonging
their life.

Chapter 3
A.C. FUNDAMENTALS
This chapter deals with the concepts, terms and definitions associated with alternating
quantities. The term alternating quantities refers to any quantity (current, voltage, flux, etc.),
whose polarity is reversed alternately with time. For convenience, they are commonly
referred to as a.c. quantities.
On completion of this chapter you should be able to:
i. Explain the method of producing an a.c. waveform.
ii. Define all of the terms relevant to a.c. waveforms.
iii. Obtain values for an a.c., both from graphical information and when expressed in
mathematical form.
iv. Understand and use the concept of phase angle.
v. Use both graphical and phasor techniques to determine the sum of alternating
quantities.
3.1 INTRODUCTION
Alternating voltage may be generated by rotating a coil in a magnetic field, as shown in Fig. 1
(a) or by rotating a magnetic field within a stationary coil, as shown in Fig. 1 (b).
Figure 1
The value of the voltage generated depends, in each case, upon the number of turns in the coil,
strength of the field and the speed at which the coil or magnetic field rotates. Alternating
voltage may be generated in either of the two ways shown above, but rotating-field method is
the one which is mostly used in practice.
3.2 EQUATIONS OF THE ALTERNATING VOLTAGES AND CURRENTS

Consider a rectangular coil, having N turns and rotating in a uniform magnetic field, with an
angular velocity of ω radian/second, as shown in Fig. 2. Let time be measured from the X-

axis. Maximum flux Φm is linked with the coil, when its plane coincides with the X-axis. In
time t seconds, this coil rotates through an angle θ = ωt. In this deflected position, the
component of the flux which is perpendicular to the plane of the coil, is Φ = Φm cos ω t. Hence,
flux linkages of the coil at any time are N Φ = N Φm cos ω t. According to Faraday’s Laws of
Electromagnetic Induction, the e.m.f. induced in the coil is given by the rate of change of flux-
linkages of the coil. Hence, the value of the induced e.m.f. at this instant (i.e. when θ = ω t) or
the instantaneous value of the induced e.m.f. is
Figure 2
Similarly, the equation of induced alternating current is i = Im sin ω t ...(iv) provided the coil
circuit has been closed through a resistive load. Since ω = 2πf, where f is the frequency of
rotation of the coil, the above equations of the voltage and current can be written as

Figure 3
3.3 SIMPLE WAVEFORMS

The shape of the curve obtained by plotting the instantaneous values of voltage or current as
the ordinate against time as abscissa is called its waveform or wave-shape.
Figure 4
An alternating voltage or current may not always take the form of a systematical or smooth
wave such as that shown in Fig. 3. Thus, Fig. 4 also represents alternating waves. But while
it is scarcely possible for the manufacturers to produce sine-wave generators or alternators, yet
sine wave is the ideal form sought by the designers and is the accepted standard. The waves
deviating from the standard sine wave are termed as distorted waves. In general, however, an
alternating current or voltage is one the circuit direction of which reverses at regularly recurring
intervals.
Time Period: The time taken by an alternating quantity to complete one cycle is called its time
period T. For example, a 50-Hz alternating current has a time period of 1/50 second.
Frequency: The number of cycles/second is called the frequency of the alternating quantity.
Its unit is hertz (Hz).

f = PN/120 where N = revolutions in r.p.m. and P = number of poles
Amplitude: The maximum value, positive or negative, of an alternating quantity is known as

its amplitude.
Phase: By phase of an alternating current is meant the fraction of the time period of that
alternating current which has elapsed since the current last passed through the zero position of
reference.
Root-Mean-Square (R.M.S.) Value

The r.m.s. value of an alternating current is given by that steady (d.c.) current which when
flowing through a given circuit for a given time produces the same heat as produced by the
alternating current when flowing through the same circuit for the same time. It is also known
as the effective or virtual value of the alternating current, the former term being used more
extensively.
The standard form of a sinusoidal alternating current is i = Im sin ω t = Im sin θ. The mean of
the squares of the instantaneous values of current over one complete cycle is (even the value
over half a cycle will do).
Hence, we find that for a symmetrical sinusoidal current
𝑟. 𝑚. 𝑠. 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 0.707 × 𝑚𝑎𝑥. 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡

The r.m.s. value of an alternating current is of considerable importance in practice, because the
ammeters and voltmeters record the r.m.s. value of alternating current and voltage respectively.
In electrical engineering work, unless indicated otherwise, the values of the given current and
voltage are always the r.m.s. values
Average Value
The average value Iav of an alternating current is expressed by that steady current which
transfers across any circuit the same charge as is transferred by that alternating current during
the same time. In the case of a symmetrical alternating current (i.e. one whose two half-cycles
are exactly similar, whether sinusoidal or non-sinusoidal), the average value over a complete
cycle is zero. Hence, in their case, the average value is obtained by adding or integrating the
instantaneous values of current over one half-cycle only. But in the case of an unsymmetrical
alternating current (like half-wave rectified current) the average value must always be taken
over the whole cycle.
𝜋
𝑖 𝐼𝑚 𝜋
𝐼𝑎𝑣 = ∫ 𝑑𝜃 = ∫ 𝑠𝑖𝑛𝜃𝑑𝜃
0 𝜋−0 𝜋 0
𝐼𝑚 𝐼𝑚 2𝐼𝑚
|− cos 𝜃|𝜋0 = [(1 − (−1)] =
𝜋 𝜋 𝜋
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 0.637 × 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒
Note. R.M.S. value is always greater than average value except in the case of a rectangular
wave when both are equal.
Form Factor
It is defined as the ratio, Kf = r.m.s. value /average value
0.701𝐼𝑚
𝑘𝑓 = = 1.11
0.637𝐼𝑚
Crest or Peak or Amplitude Factor

It is defined as the ratio Ka = maximum value / rms value
𝐼𝑚
𝑘𝑎 = = √2 = 1.414
𝐼𝑚 /√2

Example 1: An alternating current of frequency 60 Hz has a maximum value of 120 A. Write down the
equation for its instantaneous value. Reckoning time from the instant the current is zero and is becoming
positive, find (a) the instantaneous value after 1/360 second and (b) the time taken to reach 96 A for the
first time.
Solution.
The instantaneous current equation is 𝑖 = 120 sin 2 𝜋 𝑓𝑡 = 120 sin 120 𝜋 𝑡
Now when t = 1/360 second, then
(a)
1
𝑖 = 120 sin (120 × 𝜋 × ) … 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
360
1
= 120 sin (120 × 180 × ) … 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒
360
= 120 sin 60 º = 103.9 𝐴
(b)
96 = 120 × 𝑠𝑖𝑛 2 × 180 × 60 × 𝑡 … 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒
96
sin(360 × 60 × 𝑡) = = 0.8
120
360 × 60 × 𝑡 = 𝑠𝑖𝑛−1 0.8 = 53º (𝑎𝑝𝑝𝑟𝑜𝑥)
∴ 𝑡 = 𝜃/2𝜋𝑓 = 53/360 × 60 = 0.00245 𝑠𝑒𝑐𝑜𝑛𝑑.
Example 2: An alternating current varying sinusoidally with a frequency of 50 Hz has an RMS
value of 20 A. Write down the equation for the instantaneous value and find this value (a)
0.0025 second (b) 0.0125 second after passing through a positive maximum value. At what
time, measured from a positive maximum value, will the instantaneous current be 14.14 A?
Example 3: A rectangular coil, measuring 25 cm by 20 cm, has 80 turns. The coil is rotated,
about an axis parallel with its longer sides, in a magnetic field of density 75 mT. If the speed
of rotation is 3000 rev/min, calculate, from first principles, (a) the amplitude, r.m.s. and average
values of the emf, (b) the frequency and period of the generated waveform, (c) the
instantaneous value, 2ms after it is zero.

3.4 PHASE DIFFERENCE
When two alternating quantities of the same frequency have different zero points, they are said
to have a phase difference. The angle between the zero points is the angle of phase difference.
In Phase
Two waveforms are said to be in phase, when the phase difference between them is zero. That
is the zero points of both the waveforms are same. The waveform, phasor and equation
representation of two sinusoidal quantities which are in phase is as shown. The figure shows
that the voltage and current are in phase.
𝑣 = 𝑉𝑚 sin 𝑤𝑡
𝑖 = 𝐼𝑚 sin 𝑤𝑡
Lagging
In the figure shown, the zero point of the current waveform is after the zero point of the voltage
waveform. Hence the current is lagging behind the voltage. The waveform, phasor and equation
representation is as shown.

𝑖 = 𝐼𝑚 sin 𝑤𝑡 − 𝜑
Lagging
In the figure shown, the zero point of the current waveform is after the zero point of the voltage
waveform. Hence the current is lagging behind the voltage. The waveform, phasor and equation
representation is as shown.
𝑖 = 𝐼𝑚 sin 𝑤𝑡 + 𝜑
3.5 PHASOR REPRESENTATION OF ALTERNATING QUANTITIES

A phasor is a rotating vector. Apart from the fact that a phasor rotates at a constant velocity, it
has exactly the same properties as any other vector. Thus its length corresponds to the
magnitude of a quantity. It has one end arrowed, to show the direction of action of the quantity.
Consider two such rotating vectors, v1 and v2, rotating at the same angular velocity, rad/s. Let
them rotate in a counter clockwise direction, with v2 lagging behind v1 by π/6 radian (30°).
This situation is illustrated in Fig.7 .

Figure 7
Notes
• Any a.c. quantity can be represented by a phasor, provided that it is a sine wave.
• Any number of a.c. voltages and/or currents may be shown on the same phasor diagram,
provided that they are all of the same frequency.
• Figure 7 shows a counter-clockwise arrow, with ω rad/s. This has been shown here to
emphasise the point that phasors must rotate in this direction only. It is normal practice
to omit this from the diagram.
• When dealing with a.c. circuits, r.m.s. values are used almost exclusively. In this case,
it is normal to draw the phasors to lengths that correspond to r.m.s. values.
Example 4: Four currents are as shown below. Draw to scale the corresponding phasor
diagram.
𝑖1 = 2.5 sin(𝜔𝑡 + 𝜋/4) 𝑎𝑚𝑝, 𝑖2 = 4 sin(𝜔𝑡 − 𝜋/3) 𝑎𝑚𝑝
𝑖3 = 6 sin 𝜔𝑡 𝑎𝑚𝑝, 𝑖4 = 3 cos 𝜔𝑡 𝑎𝑚𝑝
Addition of Alternating Quantities

Consider two alternating currents, i1= Im1 sinωt amp and i2=Im2 sin ( ωt + π/4) amp, that are to
be added together. There are three methods of doing this, as listed below.
➢ Plotting them on graph paper. Their ordinates are then added together, and the resultant
waveform plotted. This is illustrated in Fig. 8. The amplitude, Im , and the phase angle,
θ, of the resultant current are then measured from the two axes.

➢ Drawing a scaled phasor diagram, as illustrated in Fig.9. The resultant is found by

completing the parallelogram of vectors. The amplitude and phase angle are then
measured on the diagram.
➢ Resolving the two currents, into horizontal and vertical components, and applying
Pythagoras theorem. This method involves using a sketch of the phasor diagram,
followed by a purely mathematical process. This phasor diagram, including the
identification of the horizontal and vertical components, is shown in Fig. 9.
The triangle of H.C., V.C., and the resultant current, is shown in Fig.10 . From this, we can
apply Pythagoras theorem to determine the amplitude and phase angle, thus

Figure 10
The graphical technique is very time-consuming (even for the addition of only two quantities).
The accuracy also leaves much to be desired, a phasor diagram, drawn to scale, can be the
quickest method of solution. However, it does require considerable care, in order to ensure a
reasonable degree of accuracy, the use of the resolution of phasors is, with practice, a rapid
technique, and yields a high degree of accuracy. Unless specified otherwise it is the technique
you should use.
Example 5: Determine the phasor sum of the two voltages specified below.
𝑉_1 = 25 sin(𝜔𝑡 + 𝜋/2)
𝑉_2 = 15sin(𝜔𝑡 − 𝜋/6)
Example 6: Calculate the phasor sum of the three currents listed below.
𝐼1 = 6 sin 𝜔𝑡 𝑎𝑚𝑝
𝐼2 = 8 sin( 𝜔𝑡 − 𝜋/2) 𝑎𝑚𝑝
𝐼3 = 4 sin( 𝜔𝑡 + 𝜋/6) 𝑎𝑚𝑝

3.6 A.C. THROUGH RESISTANCE, INDUCTANCE AND CAPACITANCE
A.C. Through Pure Ohmic Resistance Alone
The circuit is shown in Fig. 13. Let the applied voltage be given by the equation.
𝑉 = 𝑉𝑚 sin 𝜔𝑡
Let R = ohmic resistance; i = instantaneous current Obviously, the applied voltage has to supply
ohmic voltage drop only. Hence for equilibrium v = iR;
𝑉 = 𝑉𝑚 sin 𝜔𝑡 = 𝑖𝑅
𝑉𝑚
𝑖= sin 𝜔𝑡
𝑅
𝑉𝑚
𝐼𝑚 =
𝑅
Hence,
𝐼 = 𝐼𝑚 sin 𝜔𝑡
Instantaneous power, 𝑝 = 𝑣𝑖 = 𝑉𝑚 𝐼𝑚 𝑠𝑖𝑛2 𝜔𝑡
𝑉𝑚 𝐼𝑚
𝑉𝑚 𝐼𝑚 𝑠𝑖𝑛2 𝜔𝑡 = (1 − 𝑐𝑜𝑠2𝜔𝑡)
2
Average value of cos 2wt is zero
𝑉𝑚 𝐼𝑚 𝑉𝑚 𝐼𝑚
𝑃= = × = 𝑉 × 𝐼 𝑤𝑎𝑡𝑡
2 √2 √2
Where V = r.m.s. value of applied voltage.
I = r.m.s. value of the current.

Example 6
An ac circuit consists of a pure resistance of 10Ω and is connected to an ac supply of 230 V,
50 Hz. Calculate the (i) current (ii) power consumed and (iii) equations for voltage and current
Solution
i.
𝑉 230
𝐼= = = 23𝐴
𝑅 10
ii.
𝑃 = 𝑉𝐼 = 230 × 23 = 5260 𝑊
iii.
𝑉𝑚 = 𝑉√2 = 230√2 = 325.27
I𝑚 = 𝐼√2 = 23√2 = 32.52
𝜔 = 2𝜋𝑓 = 2 × 50 × 3.14 = 314𝑟𝑎𝑑/𝑠𝑒𝑐
v = 325.27 sin 314𝑡 V
𝑖 = 32.52 sin 314𝑡
Example 7: A 60-Hz voltage of 115 V (r.m.s.) is impressed on a 100 ohm resistance : (i) Write
the time equations for the voltage and the resulting current. Let the zero point of the voltage
wave be at t = 0 (ii) Show the voltage and current on a time diagram. (iii) Show the voltage and
current on a phasor diagram. Answer v (t) = 1.63 sin 377 t and i (t) = 1.63 sin 377 t
A.C. Through Pure Inductance Alone
Whenever an alternating voltage is applied to a purely inductive coil, a back e.m.f. is produced
due to the self-inductance of the coil. The back e.m.f., at every step, opposes the rise or fall of
current through the coil. As there is no ohmic voltage drop, the applied voltage has to overcome
this self-induced e.m.f. only. So at every step

𝑑𝑖
𝑉=𝐿
𝑑𝑡
𝑉 = 𝑉𝑚 𝑠𝑖𝑛𝜔𝑡
𝑑𝑖
𝑉𝑚 𝑠𝑖𝑛𝜔𝑡 = 𝐿
𝑑𝑡
𝑑𝑖 𝑉𝑚
= 𝑠𝑖𝑛𝜔𝑡
𝑑𝑡 𝐿
𝑉𝑚
𝑖= ∫ 𝑠𝑖𝑛𝜔𝑡 𝑑𝑡
𝐿
𝑉𝑚 𝑉𝑚 𝜋
𝑖= (−𝑐𝑜𝑠𝜔𝑡) = sin(𝜔𝑡 − 2 )
𝜔𝐿 𝜔𝐿
𝜋
𝑖 = 𝐼𝑚 sin(𝜔𝑡 − 2 )
Where
𝑉𝑚
𝐼𝑚 =
𝜔𝐿
Instantaneous power
The instantaneous power in the above circuit can be derived as follows

As seen from the above equation, the instantaneous power is fluctuating in nature.
Average power
From the instantaneous power we can find the average power over one cycle as follows
The average power in a pure inductive circuit is zero. Or in other words, the power consumed
by a pure inductance is zero.
The voltage, current and power waveforms of a purely inductive circuit is as shown in the
figure.
Example: A pure inductive coil allows a current of 10A to flow from a 230V, 50 Hz supply.
Find (i) inductance of the coil (ii) power absorbed and (iii) equations for voltage and current.

Solution
i.
𝑉 230
𝑋𝐿 = = = 23Ω
𝐼 10
𝑋𝐿
𝐿= = 0.073𝐻
2𝜋𝑓
ii. P=0
iii.
𝑉𝑚 = 𝑉√2 = 325.27 𝑉
𝐼𝑚 = 𝐼√2 = 14.14 𝐴
𝜔 = 2𝜋𝑓 = 314 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑣 = 325.27 sin 314𝑡 𝑉
𝑖 = 14.14 sin(314𝑡 − 𝜋/2) 𝐴
A.C. Through Pure Capacitance Alone

When an alternating voltage is applied to the plates of a capacitor, the capacitor is charged first
in one direction and then in the opposite direction.
𝑄 = 𝑐𝑣
𝑞 = 𝐶𝑉𝑚 𝑠𝑖𝑛𝜔
𝑑𝑞 𝑑𝐶𝑉𝑚 𝑠𝑖𝑛𝜔
𝑖= =
𝑑𝑡 𝑑𝑡
𝑉𝑚
𝑖= 𝑐𝑜𝑠𝜔𝑡
1
𝜔𝑐
𝑉𝑚 𝜋
𝑖= sin(𝜔𝑡 + 2 )
𝑋𝐶

The denominator XC = 1/ωC is known as capacitive reactance and is in ohms.
The maximum value of the instantaneous power is VmIm/2.
A 318μF capacitor is connected across a 230V, 50 Hz system. Find (i) the capacitive reactance
(ii) rms value of current and (iii) equations for voltage and current.
Solution
i.
1 1
𝑋𝐶 = = = 10Ω
2𝜋𝑓 2 × 𝜋 × 50
ii.
𝑉 230
𝐼= = = 23 𝐴
𝑋𝐶 10
iii.
𝑉𝑚 = 𝑉√2 = 325.27 𝑉
𝐼𝑚 = 𝐼√2 = 32.53 𝐴
𝜔 = 2𝜋𝑓 = 314𝑟𝑎𝑑/𝑠𝑒𝑐
𝑣 = 325.27 sin 314𝑡 𝑉
𝑖 = 32.53 sin(314𝑡 + 𝜋/2) 𝐴
Example: A 50-Hz voltage of 230 volts effective value is impressed on a capacitance of 26.5
μF. (a) Write the time equations for the voltage and the resulting current. Let the zero axis of
the voltage wave be at t = 0. (b) Show the voltage and current on a time diagram. (c) Show the
voltage and current on a phasor diagram.
Answer a) v (t) = 325 sin 314 t; i (t) = 2.71 sin (314t + π/2) = 2.71 cos 314 t.

A.C. Through Resistance and Inductance

A pure resistance R and a pure inductive coil of inductance L are shown connected in series in
Fig. 14
Figure 14
𝑉 = √𝑉𝑅2 + 𝑉𝐿2 = √(𝐼𝑅)2 + (𝐼. 𝑋𝐿 )2 = 𝐼 √𝑅 2 + 𝑋𝐿 2
𝑉
𝐼=
√𝑅 2 + 𝑋𝐿 2
The quantity √𝑅 2 + 𝑋𝐿2 is known as the impedance (Z) of the circuit.
𝑉𝐿 𝐼𝑋𝐿 𝜔𝐿
tan 𝜙 = = =
𝑉𝑅 𝐼𝑅 𝑅
The mean power consumed by the circuit is given by the product of V and that component of
the current I which is in phase with V.
So
𝑃 = 𝑉 × 𝐼 𝑐𝑜𝑠 𝜙
The term ‘cos φ’ is called the power factor of the circuit.
Active, Reactive and Apparent Power

Let a series R-L circuit draw a current of I when an alternating voltage of r.m.s. value V is
applied to it. Suppose that current lags behind the applied voltage by φ. The three powers
drawn by the circuit are as under:

i. Apparent power (S) It is given by the product of r.m.s. values of applied voltage and
circuit current. ∴ S = VI = (IZ)I = I2Z volt-amperes (VA)
ii. Active power (P or W) It is the power which is actually dissipated in the circuit
resistance. P = I2R = VI cos φ watts
iii. Reactive power (Q) It is the power developed in the inductive reactance of the circuit.
Q = I2XL = I2.Z sin φ = I.(IZ) sin φ = VI sin φ volt-amperes-reactive (VAR)
These three powers are shown in the power triangle of Fig 1 from where it can be seen that
S2 = P2 + Q2.
Q-factor of a Coil
Reciprocal of power factor is called the Q-factor of a coil or its figure of merit. It is also known
as quality factor of the coil.
Q factor = 1/power factor OR 1/cos φ
Also Q = 2π (maximum energy stored)/(energy dissipated per cycle)
Example : In a series circuit containing pure resistance and a pure inductance, the current and
the voltage are expressed as : i (t) = 5 sin (314 t + 2 π/3) and v (t) = 15 sin (314 t + 5 π/6) (a)
What is the impedance of the circuit ? (b) What is the value of the resistance? (c) What is the
inductance in henrys? (d) What is the average power drawn by the circuit? (e) What is the
power factor?
Solution.
𝑃ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 2 𝜋/3 = 2 𝑋 180°/3 = 120° 𝑎𝑛𝑑
𝑝ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 5 𝜋/6 = 5 × 180°/6 = 150°.
Hence, current lags behind voltage by 30°. It means that it is an R-L circuit.
𝑍 = 𝑉𝑚/𝐼𝑚 = 3 𝛺.
𝐴𝑙𝑠𝑜 314 = 2𝜋𝑓 𝑜𝑟 𝑓 = 50 𝐻𝑧.

𝑅 = 𝑍𝑐𝑜𝑠 30° = 2.6 𝛺 ; 𝑋𝐿 = 𝑍 𝑠𝑖𝑛 30° = 1.5
𝑋𝐿 = 2𝜋𝑓𝐿 = 314 𝐿 = 1.5, 𝐿 = 4.78 𝑚𝐻

2
2
5
𝑃 = 𝐼 𝑅 = ( ) × 2.6 = 32.5 𝑊
√2
𝑝. 𝑓. = 𝑐𝑜𝑠 30° = 0.866 (𝑙𝑎𝑔).
R-C Series circuit
Consider an AC circuit with a resistance R and a capacitance C connected in series as shown

in the figure (a).
The alternating voltage v is given by
The current flowing in the circuit is i. The voltage across the resistor is VR and that across the
capacitor is VC.
VR = IR is in phase with I
VC = IXC lags behind the current by 90 degrees
The current I is taken as the reference phasor. The voltage VR is in phase with I and the voltage
VC lags behind the current by 90⁰. The resultant voltage V can be drawn as shown in the figure.
From the phasor diagram we observe that the voltage lags behind the current by an angle Φ or
in other words the current leads the voltage by an angle Φ.
The waveform and equations for an RC series circuit can be drawn as below.

𝑖 = 𝐼𝑚 sin 𝑤𝑡 + 𝜑
From the phasor diagram, the expressions for the resultant voltage V and the angle can be
derived as follows.
𝑉 = √𝑉𝑅2 + 𝑉𝐶2
𝑉𝑅 = 𝐼𝑅, 𝑉𝐶 = 𝐼𝑋𝐶
𝑉 = √𝐼𝑅 2 + 𝐼𝑋𝐶 2
𝑉 = 𝐼√𝑅 2 + 𝑋𝐶2
𝑉 = 𝐼𝑍
Where the impedance Z is
𝑍 = 𝑅 2 + 𝑋𝐶2
The phase angle
𝑉𝑐
Φ = 𝑡𝑎𝑛−1 ( )
𝑉𝑅
𝐼𝑋𝑐
Φ = 𝑡𝑎𝑛−1 ( )
𝐼𝑅
1
Φ = 𝑡𝑎𝑛−1 ( )
𝜔𝐶𝑅

Average power
Hence the power in an RC series circuit is consumed only in the resistance. The capacitance
does not consume any power.
Impedance Triangle
We can derive a triangle called the impedance triangle from the phasor diagram of an RC series
circuit as shown
Phasor algebra for RC series circuit
Example: A Capacitor of capacitance 79.5μF is connected in series with a non-inductive

resistance of 30 Ω across a 100V, 50Hz supply. Find (i) impedance (ii) current (iii) phase angle
(iv) Equation for the instantaneous value of current.
Solution

R-L-C Series circuit
Consider an AC circuit with a resistance R, an inductance L and a capacitance C connected in

series as shown in the figure. The alternating voltage v is given by
The current flowing in the circuit is i. The voltage across the resistor is VR, the voltage across
the inductor is VL and that across the capacitor is VC.
VR = IR is in phase with I
VL = IXL leads the current by 90 degrees
VC = IXC lags behind the current by 90 degrees
With the above information, the phasor diagram can be drawn as shown. The current I is taken
as the reference phasor. The voltage VR is in phase with I, the voltage VL leads the current by
90⁰ and the voltage VC lags behind the current by 90⁰. There are two cases that can occur

VL>VC and VL<VC depending on the values of XL and XC. And hence there are two possible
phasor diagrams. The phasor VL-VC or VC-VL is drawn and then the resultant voltage V is
drawn
From the phasor diagram we observe that when VL>VC, the voltage leads the current by an
angle Φ or in other words the current lags behind the voltage by an angle Φ. When VL<VC,
the voltage lags behind the current by an angle Φ or in other words the current leads the voltage
by an angle Φ.
From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be
derived as follows.
Where impedance
Phase angle

From the expression for phase angle, we can derive the following three cases
Case (i): When XL>XC
The phase angle Φ is positive and the circuit is inductive. The circuit behaves like a series RL
circuit.
Case (ii): When XL<XC
The phase angle Φ is negative and the circuit is capacitive. The circuit behaves like a series RC
circuit.
Case (iii): When XL=XC
The phase angle Φ = 0 and the circuit is purely resistive. The circuit behaves like a pure resistive
circuit.
The voltage and the current can be represented by the following equations. The angle Φ is
positive or negative depending on the circuit elements.
Average power

Hence the power in an RLC series circuit is consumed only in the resistance. The inductance
and the capacitance do not consume any power.
Phasor algebra for RLC series circuit
Example 10: A 230 V, 50 Hz ac supply is applied to a coil of 0.06 H inductance and 2.5
resistance connected in series with a 6.8 μF capacitor. Calculate (i) Impedance (ii) Current (iii)
Phase angle between current and voltage (iv) power factor (v) power consumed
Example 11: A resistance R, an inductance L= 0.01 H and a capacitance C are connected in

series. When an alternating voltage v = 400sin(3000t-20º) is applied to the series combination,
the current flowing is 10√2sin(3000t-65º). Find the values of R and C.

Example 12: A coil of pf 0.6 is in series with a 100μF capacitor. When connected to a 50Hz
supply, the potential difference across the coil is equal to the potential difference across the
capacitor. Find the resistance and inductance of the coil.

Resonance in R-L-C Circuits
The frequency at which the net reactance of the series circuit is zero is called the resonant
frequency f0. Its value can be found as under :
XL − XC = 0 or XL = XC or ω0L = 1/ω0C
1
𝑓0 =
2𝜋√𝐿𝐶
When a series R-L-C circuit is in resonance, it possesses minimum impedance Z = R. Hence,
circuit current is maximum, it being limited by value of R alone. The current I0 =V/R and is in
phase with V.

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ELECTRICAL

i. Describe the forces of attraction and repulsion between magnetised bodies.

i. Like magnetic poles repel one another

Every medium is supposed to possess two permeabilities:

(i) Absolute permeability (μ) and

1.2 MAGNETIC FIELD

Engr. A. Mohammed Page 1

1.3 WEBER AND EWING’S MOLECULAR THEORY

1.4 CURIE POINT

Engr. A. Mohammed Page 2

Paramagnetism: In a paramagnetic material the cancellation of magnetic moments between

Ferromagnetism: Certain materials possess permanent magnetic moments in the absence of

Engr. A. Mohammed Page 3

1.6 LAW OF MAGNETIC FORCE

In the S.I. system of units, the value of the constant k is = 1/4πμ0μr.

Where μ0 = absolute permeability and μr = relative permeability

1.7 MAGNETIC FIELD STRENGTH (H)

Engr. A. Mohammed Page 4

Where l is the mean or average length of the magnetic circuit.

1.8 MAGNETIC POTENTIAL

1.9 FLUX PER UNIT POLE

1.10 FLUX DENSITY (B)

1.11 ABSOLUTE PERMEABILITY (μ) AND RELATIVE PERMEABILITY (μr)

Engr. A. Mohammed Page 5

Then, the absolute permeability of the material of the rod is defined as

μ = B/H henry/metre or B = μH = µ0 µr H Wb/m2…………………………………….. (i)

(i) B0 –flux density in air even when rod is not present

Eq. (i) above may be written as B = µr µ0 H = µr B0

1.12 INTENSITY OF MAGNETISATION (I)

Engr. A. Mohammed Page 6

Let m = pole strength induced in the bar in Wb

A = face or pole area of the bar in m2

Then I = m/A Wb/m2

1.13 SUSCEPTIBILITY (K)

Therefore K = I/H henry/metre.

1.14 Relation between B, H, I And K

Now absolute permeability is

For ferro-magnetic and para-magnetic substances, K is positive and for diamagnetic

Engr. A. Mohammed Page 7

𝜇 = 4𝜋 × 10−7 + 167 × 10−11 = 1.258 × 10−7 𝐻/𝑚

1.15 MAGNETIC CIRCUIT

Engr. A. Mohammed Page 8

Therefore flux = m.m.f/reluctance or Φ = 𝐹/𝑆

current = e.m.f. / resistance or I = V/ R

1.16 DEFINITIONS CONCERNING MAGNETIC CIRCUIT

Engr. A. Mohammed Page 9

1.17 COMPARISON BETWEEN MAGNETIC AND ELECTRIC CIRCUITS

Engr. A. Mohammed Page 10

Engr. A. Mohammed Page 11

𝐹 = 𝑁𝐼 = 900 × 1.5 = 1350𝐴𝑇

𝐵 = 𝜇𝐻 = 𝜇0 𝜇𝑟 𝐻 = 4𝜋 × 10−7 × 150 × 5371.48 = 1.0125 𝑇

1.18 B-H CURVES (MAGNETIZATION CURVES)

Engr. A. Mohammed Page 12

1.19 AMPERE’S CIRCUITAL LAW:

If we apply the analogy to Kirchhoff’s voltage law (for magnetic circuits).

Engr. A. Mohammed Page 13

Engr. A. Mohammed Page 14

1.20 COMPOSITE SERIES MAGNETIC CIRCUIT