Fundamental of E E

8/16/2019 Fundamental of E E

1/128

CLB 10402 FUNDAMENTAL OF E & E

1

ELECTRICAL TECHNOLOGY

Topic 1: Basic Concept of Electrical

Overview

Electricity can be considered from two points of view. The scientist is concerned with

what happens in electric system and seeks to explain its mysteries. The engineer accepts

that electricity is there and seeks to make use of its properties without the need to fullyunderstand them.

A basic electrical system has four constituent parts as shown in Figure 1, i.e. source, load,

transmission system and control apparatus. A source may usually be thought of as a

battery or a generator, although for simplicity we might even think of a socket outlet as asource. The function of a load is to absorb the electrical energy supplied by the source.

The transmission system conducts energy from the source to the load, and the control

apparatus will control the transmission.

Figure 1: A Basic Electrical System

An Electrical system generally transmits energy due to the movement of electric charge.

Electricity appears in two forms which are called negative and positive electricity.

Electric charge is the excess of negative and positive electricity on a body or in space. If

the excess is negative, the body is said to have a negative charge and vice versa.

Theory

The most basic electrical quantity is a property of atomic particle called charge (positive

and negative). There is a force between charged particles, called Coulomb force, whichcauses like charge particles to be repelled from each other and unlike charge to be

attracted. This force is the basis for electrical behavior.

Transmission system

Source Control Load


2/128


2

Matter is composed of units called atoms, which may be pictured as being composed of a

nucleus contained positively charged particles called protons, surrounded by an equal

number of negatively charged particles, called electrons. Normally the number of positiveand negative charges in a given quantity of matter is equal and we say that it is unchargedor neutral. If electrons are moved from a neutral quantity of matter, the result is

negatively charged matter.

An electron is an elementary particle charged with a small and constant quantity of

electricity. A proton is similarly defined but charged with positive electricity, while the

neutron is uncharged and is therefore neutral. All electrons have certain potential energy.Given enough energy, they move freely between one energy level to another and this

movement is called an electric current flow. Current flows from a point of high

energy/potential to a point of low energy/potential. Conventional current flow opposite to

that of electron current.

System International (SI) standard defined unit of charge as the Coulomb. This is theamount of charge that results from an electrical current flow from one ampere for onesecond. It is also the charge possessed by 6.24 x 10

18 electrons. As a variable, the change

is described by the variable label Q for fixed charge or q(t) for time changing charge.

1.1 Current

An electrical current exists whenever charge, q(t), is transferred from one point in the

conductor to another. The SI system defines the basic unit of current as the ampere (A).

This represents a current for which charge is being transferred at the rate of one coulomb

in one second. In an equal form, we have the current defined as the rate of charge, i.e. the

derivative.

td dq

t i =)(

Where i(t) = current in ampere (A)Q(t) = charge in CoulombsT = time in seconds (s)

if the flow is uniform, i.e. a constant current, then the equation reduces to the form

t Q

I =

Here, I is the steady current and Q is the amount of charge transferred in a time t. Notethat the capital letter I is used for constant current and lowercase i(t) for varying current.


3/128


3

The polarity of the current really indicates direction of flow and depends on the polarity

of the charges being transferred. Two standards are used to describe the direction. The

most common called conventional current, defines positive current direction as the

motion of positive charges, thus opposite to the direction in which electron flow, asshown in Figure 1.1a. Some treatment use the standard called electron flow wherein

current direction is defined with a positive sign for the direction of electron flow.

For current flow in a circuit, there must be a complete circuit and enough driving

influence or force. The driving influence is called electromotive force (e.m.f). Current is

the rate of flow of charge through a section of the circuit. The factor, which causes acurrent to flow, is the e.m.f. It is similar to the value of potential difference (p.d) between

two points is a circuit. This relationship is governed by the Ohm’s law, which related the

value of current, resistance and voltage.

Figure 1.1a: Conventional current flow direction is opposite that of actual electron flow

Self assessment 1.1:

Thirty coulombs of charge pass a given in a wire in 5 s. How many ampere of current are

flowing?

Solution

From equation, for a constant current, we have

( )( ) A sC s

C

t

Q

I 6/65

30

====

1.2 Energy

The law of conservation of energy states that energy cannot be created or destroyed, but

only transformed. The electrical form of energy may be produced from many other typesof energy, such as chemical (as in battery), mechanical (as in hydroelectric generator),

and atomic (as in a nuclear reactor). The SI unit of energy is the joule (J). The lettersymbol W or w(t) is often used for energy.

Connecting Wire Connecting Wire

Actual Electron Flow

Conventional Current Flow


4/128


4

1.3 Voltage

If energy is expanded (as work) on a quantity of change, then the ratio of the energyexpended to the quantity of charge is called the voltage. Voltage is the force that makesthe charge move i.e. voltage produce the current, which does work. For example a battery

uses chemical processes to do work on charged particles to make them move through a

conductor. Thus, a voltage exists across the battery terminals to force the charges to

move. The unit of voltage in the SI system is the volt (V). The letter symbol V (for fixed),or v(t) (for varying), is often used for the voltage.


If 100 J of energy is used to move 20 C of charge through an element, what is the voltageacross the element?

Solution:

Because voltage is the ratio of energy to change, we have

( )( )

V C J

QW

V 520

100===

1.4 Ohm’s Law

A resistor is a two-terminal component with the property that the ratio of voltage across

its terminals to current through it is a constant. That constant is called its resistance.

Ohm’s law defines this fact in equation form, as

I

V R =

Where R = the resistance in ohms (ohm)V = the voltage drop in volts (V) I = the current passed in amperes (A)

Note that the voltage drop polarity is always such that the voltage is positive on theterminal of the resistor into which the conventional current enters.


5/128


5

Self assessment 1.4a:

Determine the resistance of the resistor given voltage supply is 25 V and current is 2 A/

Solution:

Using Ohm’s law,( )( )

Ω== 5.122

25

AV

R

Equation R=V/I serves as a definition of the unit of resistance, which is called the ohm,and is symbolized by a capital Greek omega, Ω. One ohm equals volt per ampere, i.e. a

current of 1 ampere through a resistance of 1 ohm will produce a voltage across the

resistor of 1 volt. Figure 1.4a shows the schematic symbol for the resistor and the voltage

across and current through the element. Note the relationship between voltage polarity

and current direction. The voltage across or dropped across the resistor is positive on theend into which the current enters. The schematic symbol used in Figure 1.4a is the most

common. In some cases, however, a small rectangle is also used for the resistor. The ideal

resistor has zero inductance and zero capacitance. The current versus voltage ( I-V ) curveshown in Figure 1.4a is a straight-line (linear) curve of I versus V with a slope of ( I/R).

Figure 1.4a: Schematic Symbol for a resistor and the IR characteristic curve

Self assessment 1.4b:

A resistor in the middle of a TV UHF tuner is found to have a voltage across it of 3.45 V

and a current through it of 6.9 mA at a certain instant of time. What is the resistance?

Solution:

The fact that the resistor I-V does not depend on time, we fine

Ω=== 5009.6

45.3

mA

V

I

V R

Slope = 1/RIR

VR

+

IR VR


6/128


6

1.5 Resistance and Conductance

The physical definition of resistance is in terms of the material size and its resistivity.Figure 1.5 shows a piece of material of constant cross section with area A and of lengthL. The resistance of these materials is given by

A

L R ρ =

Where L = the length of the wire A = its cross-surface area ρ = resistivity

Figure 1.5: Physical definition of electrical resistance

This is the same value that would be obtained if the ration of voltage across and current

through the device were measured and divide, as in Figure 4.1


For ammeter shunt, it is necessary to provide a resistance of exactly 0.356 Ω. How can

this be constructed from copper wire?

Solution:

Resistivity of copper is 1.7 x 108 ohm-m. Let say the copper wire has a diameter of 0.010

inch. We need to determine the proper length to obtain 0.356 Ω. First we find the cross-

sectional area in square meters. Diameter D, is

( )( )

m xinm

in D 41054.2

/0254.0

010.0==

Then,

272

10067.54

m x D

A =Π

=

L

A


7/128


8/128


8

Solution:

The equivalent resistance can be found from

Ω=Ω+Ω+Ω= k k k k Req56.812

1

47

1

82

11

The current is found from Ohm’s Law, the 15 V source, and the equivalent resistance:

mAk V

RV

I eq

net 75.156.8

15=

Ω==

The current through is found from Ohm’s Law directly since each has the 15 V sourceacross it:

mA R

V I 18.0

15

1

1 ==

mA R

V I 32.0

15

2

2 ==

mA R

V I 25.1

15

3

3 ==

1.6

Power

When work is done over a period of time, a definition of the rate at which work is done,

is more useful than the amount of work. This is called the power and is describe by the SIunit of watt (W), which is work done are a rate of one joule In one second. The lettersymbol P (for fixed) or p(t) (for varying) is often used for the power label. Since it is arate, power is defined by the derivative

dt

dwt p =)(

or, for fixed power,

t W

P =

if the power is known, the energy can be found, in general, by the relation

τ τ d pt w )()( =

or, for constant power, W=pt. The variable of integration is τ .


9/128


9

The specification of condition at t = - infinity is not always convenient, so an alternative

to previous equation makes use of known conditions at some initial time, to.

)()()( ot wd pt w += τ τ

Since current is the time rate of change of charge and voltage is the rate of change of

work with charge, we see that p(t) may be generalized for electrical variables as

)()()( t vt it p =

or, when the voltage and current do not vary with time, as P=IV.


i. An amount of energy equal to 200 J is used in 10 s by electrical motor. What is the power?

Solution:

( )( )

W s J

t W

p 2010

200===

ii. A 60 W light bulb operates on 120 V. How much current does it requires?

( )

( ) AV

W

V

P

I 5.012060

===

1. Power Dissipation

The fact that the current and dropped voltage of a resistor are in phase indicates that the

resistor is taking energy from the circuit at a constant rate. This rate is called the

electrical power dissipated and is measured is joules/second (J/s) or watts (W). This

power shows up as heating of the resistor as it dissipates the energy to its surroundings.

The basic relation gives the amount of power

IV P =

Where P = power dissipated in watts (W) I = current through the resistor (A)V = voltage across the resistor (V)

The above equation can be expressed in alternative forms by using Ohm’s Law,

R

V P or R I P

22 ==


10/128


10

Summary

Current is directly related to voltage, and inversely related to the circuit resistance.

Ohm’s Law states this as I=V/R. The direction of current flow through a circuit, externalto the voltage source, is from the source’s negative side, through the circuit, and back tothe source’s positive side. Arrows are frequently drawn on schematic diagrams to

illustrate this. The direction electrons would move through a circuit is the standard being

used in this textbook. Conventional current is considered to flow in the direction that a

positive charge would move through the circuit. That is, from positive to negative.

Electrical power is the rate of using electrical energy to do electrical work. Power is

generally dissipated in a circuit or in a component in the form of heat. One watt of power

is the performance of electrical work at the rate of 1 Joule per second. A joule is the

energy used in moving one coulomb of charge between two points that have a difference

of potential of one volt between them. Electrical energy usage is computed by

multiplying the power used times the amount of time it was used (W=PxT). Units of

measure for usage of electrical energy are; wattseconds (Ws); watthour (Wh); and forlarge amounts of energy usage, kilowatthour (kWh).

Tutorials

1. A current in a circuit is due to a p.d of 10 V applied to a resistor of resistance 100 Ω.

What resistance would permit the same current to flow if the supply voltage were

100V?

2. Calculate the current in a circuit due to p.d of 10 V applied to a 10 kΩ resistor. If thesupply voltage is doubled while the circuit resistance is trebled, what is the newcurrent in the circuit?

3. A p.d of 12 V is applied to a 7.5 ohm resistor for a period of 5 s. Calculate the electric

charge transferred in this time?

Further Reading

The principle of electricity is important is the understanding of circuit operation. Solid

knowledge in the relation for various parameters in electric circuit will enhance further

analysis and assist in design problems. Ensure proper understanding of relevant parameters in SI units is important.

References

1. Edward Hughes, “Electrical Technology”, 7th

. Edition, Prentice Hall

2. Charles K. Alexander & Matthew N.O. Sadiku, “Fundamental of Electric Circuit”,McGraw-Hill.


11/128


11

Summary of important formulae

F [newtowns] = m [kg] x a [ms-2] F = ma

Torque, T =Fr (newtown-meters)

Work, W = Fl (joules)

Kinetic energy, W = ½ mu2

Power, P = Fu (watts)= Tω = Mω = 2πnT

Efficiency, η = Po / Pin

Electric charge, Q = It (Coulombs)

Voltage, V = P/I (volts)= W/Q = IR

Power, P = IV = I 2 R = V 2 /R (watts)

In a series circuit

voltage,1 2 3 ( )V V V V volts= + +

resistance,1 2 3 ( ) R R R R ohms= + +

In a parallel network

current,1 2 3

1 2 3

( )

1 1 1 1

I I I I amperes

R R R R

= + +

= + +

Effective resistance of two parallel resistors:

1 2

1 2

R R

R R R= +

Current division rule for two resistors:

21

1 2

R I I

R R= ×

+

Energy, W = I 2 Rt

Resistance, R = ρl/A


12/128


12

Solution Tutorials

1. Ω===

=

Ω

==

10001.0

100

1.0

100

10

I

V R

AV

R

V I

2.

A xk

V

R

V I

A xk

V

R

V I

4

3

1067.630

20

100.110

10

−

−

=Ω

==

=Ω

==

3. ( ) C s A IT Q

AV

R

V I

856.1

6.1

5.7

12

===

=

Ω

==


13/128


13

Topic 2: Circuit Theorems

Overview

An electrical network is an interconnection of elements called sources and components.

Each element has a voltage across its terminals and a current passing through its

terminals. The objective of network analysis and the use of network theorems are to

determine the magnitude and polarity of the voltages and currents as a function of time.

The elements of a linear network are of two types; passive and active. Passive elements

which inject no energy into the networks are components such as resistors, capacitors,and inductors. Active elements, which may inject energy into the network are real and

controlled sources of voltage and current. A passive element receives power from the

network, while an active element delivers power to it.

Theory

Electrical networks can be defined by the following five classifications:

• Linear or non-linear

• Time Invariant and Time Varying

• Passive and Active

• Lumped and Distributed

• Pattern (refer to Figure 2a)

Linear networks use ideal sources that either maintain a specified voltage between theirterminals, regardless of current, or maintain a specified current regardless of voltage. Fig.

2b shows a network that helps define the following terms:

• A node is a point in a network at which two or more elements are joined. If there arethree or more elements connected at a node, that node is called a. junction.

• A branch of a network extends from one junction to another and may consist of asingle element or a series of elements. Thus, there is a node at the ends of each

element and a junction at the end of each branch.

• A loop is a closed path for current in a network, while a mesh is a combination ofloops.

Figure 2a: Networks Patterns; ladder, lattice and bridged-T


14/128


14

Passive Elements

Node

Source

Figure 2b: General schematics to define networks terms

Following are five types of network or circuit techniques widely used in electrical circuits

analyses.

2.1) KIRCHOFF'S LAW

Kirchoff's current law states that

At any instant the algebraic sum of the currents at a junction in a network is zero.Different signs are allocated to currents held to flow towards the junction and to those

away from it.

Kirchoff's voltage law states that

At any instant in a closed loop, the algebraic sum of the e.m.f.s acting around the loopis equal to those the algebraic sum of the p.d.s round the loop.

Self-Assessment 2.1:

i) In Fig.2.1a, all currents are given in both magnitude and direction except for that

through R 3. Find the current through R 3.

Solution:

To apply KCL, we first note the interconnection of resistors at the top forms one junction

of four elements as illustrated in Figure 2.1b. Solving for the unknown, we solve l3.

mA I

I

I I I I T

80

5070200

3

3

213

=

−−=

−−=

Loop

Node and Junction Branch

Branch


15/128


15

((a)(a) (b)

IT = 200 mA

Figure 2.1: Example circuit showing how KCL is applied

ii) For the network shown in Figure in Figure 2.1c, determine the unknown voltage drop,

V2.

Figure 2.1c: Example circuit showing how KVL is applied

Solution:

1 2 3

2 1 3

2 45

V V V E

V V V E

V V

− + =

= + −

= −

+

-

+ +- -


16/128


16

2.2) THEVENIN’S LAW

The current through a resistor R connected across any two points A and B of and activenetwork (i.e. network containing one or more sources of e.m.f) is obtained by dividing

the p.d between, with R disconnected by (R + r), where r is the resistance of the network

measured between points A and B with R disconnected and the sources of e.m.f replaced

by their internal resistances.

An alternative way of stating Thevenin’s theorem is as follows. An active network havingtwo terminals A and B can be replaced by a constant-voltage source having an e.m.f and

internal resistance r. The value of E is equal to the open-circuit p.d between A and B, and

r is the resistance of the network measured between A and B with the load disconnected

and sources of e.m.f replaced by their internal resistances.

Suppose A and B in Figure 2.2a to be the two terminals of a network consisting ofresistors having resistances R 2 and R 3 and a battery having an e.m.f. El and an internal

resistance Rl. It is required to determine the current through a load of resistance Rconnected across AB. With the load disconnected as in Figure 2.2b.

Figure 2.1: Networks to illustrate Thevenin’s Theorem

Current through31

13 R R

E R

+=

and

p.d. across31

31

3 R R

R E R

+=


17/128


17

R L = 100 Ω

R 2 = 50 Ω b

R 1 = 100 Ω R 4 = 25 Ω

R R eq

R 2 = 50 Ω

Since there is no current through R 2, p.d. across AB is

31

31

R R

R E V

+=

Figure 2.2c shows the network with the load disconnected and the battery replaced by its

internal resistance Rl. Resistance of network between A and B is

31

31

2 R R

R R Rr

++=

Thevenin's theorem merely states that the active network enclosed by the dotted line in

Figure 2.2a can be replaced by the very simple circuit enclosed by the dotted line in

Fig.2.2d and consisting of a source having an e.m.f. E equal to the open-circuit potential

difference V between A and B, and internal resistance r, where v and r have the valuesdetermined above. Hence,

Current through Rr

E I R

+==


Use Thevenin's theorem to find the equivalent circuit between points a and b in Figure

2.2.la. Use this circuit to find the load voltage across the external load R L.

R 1 = 100 Ω R 4 = 25 Ω a

(a)R eq = 75 Ω a

b

(b) (c)

Figure 2.2.1: Example circuit for Thevenin’s Theorem

R 3 = 75 Ω

R 3 = 75 Ω VT = 8.33 VR 3

=100 Ω


18/128


18

Solution:

Figure 2.2.Ib shows the network for finding the resistance. The Thevenin voltage is

implying the voltage dropped across the 75Ω resistor in Figure 2.2.1a. Since the 25Ωresistor carries no current (open circuit from a to b), the Thevenin's voltage is,

( )

( ) ( )

V R I V

mA RT

V I

V V R R R

RV

R R R

R R R R R

L L L

L EQ

T L

T

eq

76.4

6.47

33.8

7575150

7515025

321

3

321

213

4

==

=+

=

=++

=

Ω=Ω+Ω

Ω+Ω+Ω=

++

++=

Figure 2.2.Ic shows the equivalent circuit with the load attached. The load voltage isshown above.

2.3) NORTON’S LAW

When a branch in a circuit is open-circuit, the remainder of the circuit can be represented

by one source of e.m.f in series with a resistor. Norton's Theorem is therefore a

restatement of Thevenin's theorem using an equivalent current-generator source insteadof the voltage-generator source. It therefore can be stated that:

The current which flows in any branch of a network is the same as that which wouldflow in the branch if it were connected across a source of electrical energy, the short-

circuit current of which is equal to the current that would in short-circuit across the

branch, and the internal resistance of which is equal to the resistance which appears

across the open-circuited branch terminals.

Norton's theorem is illustrated in Figure 2.3.


The model of a physical battery is an ideal voltage source in series with a resistor, asshown in Fig 2.3 a. Find the Norton equivalent circuit to show how it can be modeled as a

current source.


19/128


19

R = 0.01 Ω a a

I N = 1200 A

(a) (b) b b

Figure 2.3: Example circuit for Norton’s Theorem

Solution:

The Norton current will be the current through a short between output terminals a and b.

AV

R

V I I SC N 1200

01.0

12=

Ω===

If the ideal voltage source is replaced by a short to find the Norton resistance, the, valuewill simply be the series resistance, R N = R = 0.01Ω. Thus the Norton equivalent circuit

is shown in Figure 2.3b.

2.5) SUPERPOSITION THEOREM

The Superposition Theorem states that in any network containing more than one source,

the current in, or the p.d across, any branch can found by considering each source

separately and adding their effects; omitted sources of e.m.f are replaced by resistances

equal to their internal resistances. Self-assessment in Figure 1.4 illustrates Superpositiontheorem, based on the network in Figure 2.4a. Since there are two sources of e.m.f. in the

network, then two separate network need to be considered, each having one source of

e.m.f. as in Figure 2.4b and Figure 2.4c.

R eq = 0.01 Ω


20/128


20


For Figure 2.4b,

A I I

A I I I

A I I I

A x I

A I

A x

A I I

A x I

A I

x

cb

cb

c

c

bb

b

b

72.057.385.2

57.378.621.3

85.242.657.3

36.042.678.6181

18

78.695.2

20

78.6181

1812

36.021.357.3

21.357.3182

18

57.38.2

10

8.2182

1821

21

222

111

1

2

21

2

1

=+−=+

=+−=+=

−=−=+=

=−+

−=

==

=+

+

=−=+

−=+

−=

==

Ω=+

+

Figure 2.4: Example circuit for the Superposition Theorem

thus

and

also

For Figure 2.4c

thus

and

also

thus

and

also


21/128


21

2.5) DELTA-STAR TRANSFORMATION

Figure 2.5a shows three resistor R 1, R 2 and R 3 connected in a closed mesh or delta tothree terminals A, B and C, their numerical subscripts 1,2 and 3 being opposite to the

terminals A,B and C respectively. It is possible to replace these delta-connected resistors

by three resistors R a, R b and R e connected respectively between the same terminals A, B

and C and a common point S, as in Figure 2.5b. Such arrangement is said to be star-

connected. It will be noted that the letter subscripts are now those of the terminals to

which the respective resistor are connected. If the star-connected network is to beequivalent to the delta-connected network, the resistance between any two terminals in

fig b must be the same as that between the same two terminals in figure a. Thus, if we

consider terminals A and B in figure a, we have a circuit having a resistance R 3 in parallel

with a circuit having resistances R 1 and R 2 in series, hence

For figure 4.3 l(b), we have

( )

ba AB

AB

R R R

R R R

R R R R

+=

++

+=

321

213

Figure 2.5: Example circuit for the Star-Delta Transformation

In order that both networks may be equivalent to each other, R AB must have equal value.

∆( )

321

213

R R R

R R R R R ba ++

+=+

Similarly,

321

3121

R R R

R R R R R R cb

++

+=+

and321

3221

R R R

R R R R R R ca

++

+=+


22/128


22

Similarly,

321

13

R R R

R R Rb

++

=

and321

21

R R R

R R Rc

++=

Subtracting,

321

2132

R R R

R R R R R R ca ++

−=−

Adding and dividing by 2,

321

32

R R R R R Ra ++

=

The equivalent star resistance connected to a given terminal is equal to the product of the

two delta resistance connected to the same terminal divided by the sum of the deltaresistance.

Summary

• Most circuit problems can be solved by applying Kirchoff’s laws to producesimultaneous equations; the solution of these equations is often unnecessarily difficult.

• The superposition theorem states that we can solve a circuit problem one source at atime, finally imposing the analyses one on another.

• Thevenin's theorem states that any network supplying a load can be replaced by aconstant-voltage source in series with an internal resistance.

• Norton's theorem states that any network supplying a load can be replaced by aconstant-current source in parallel with an internal resistance.

• The delta-star transformation permits us to replace any three loads connected in delta by an equivalent three load connected in star. The star-delta transformation permits the

converse transfer.


23/128


23


For delta-star transformation

2 3

1 2 3

a R R R

R R R=

+ +

For star-delta transformation

1b c

b ca

R R R R R

R= +


24/128


24

Tutorials

1. A network is arranged as shown in Figure A. Calculate the value of the current in the

8Ω resistor by (a) the Superposition theorem, (b) Kirchoff’s Laws, and (c) Thevenin’s

theorem.

Figure A

2. Calculate the voltage across AB in the network shown in Figure B and indicate the

polarity of the voltage, using (a) Kirchoff’s Laws, and (b) Delta-star transformation.

Figure B

Further Reading:

There are other techniques that can be utilized for circuit or network analysis. The stated

techniques are the most common and widely used. A proper understanding of network

analysis will be acquired after applying the available techniques in actual circuit and after

further analysis.


25/128


25

References:

1. Edward Hughes, “Electrical Technology”, 7th. Edition, Prentice Hall

2. Charles K. Alexander & Matthew N.O. Sadiku, “Fundamental of Electric Circuit”,

McGraw-Hill.


26/128


26

Solution Tutorials:

1 a) Superposition Theoremi2 i4

( )( )

( )

12 815

205 12.450

12 815

20

T R

= + = Ω +

0.3213T T

V i A

R= =

( )2

150.19278

25

T ii A= = ( )2

3

120.1157

22

ii A= =

i6

5

17.06

0.3518

0.1954

T

T

R

i A

i A

= Ω

=

=

∆3 5

0.1157 0.1954 0.32i i i

A A= += + =

i1i3

i5


27/128


28/128

ELECTRICAL TECHNOLOGY ENG 2022

28

1 c) Thevenin’s theorem

( )[ ]

( ) [ ]

15 510 12

15 5

15 5 10 1215 5

6.41

TH

TH

R

R

×+

+ =

× + + +

= Ω

Loop 1

1 1 2

1 2

11

0 4 5 15 15

0 4 20 15

4 151

20

i i i

i i

ii

= − + + +

= − + +

−= −

Loop 2

2 2 1

2 1

0 6 22 15 15

0 6 37 15 2

i i i

i i

= − + + +

= − + + −

1 in 2

22

2 2

2

4 150 6 37 15

20

0 6 37 3 11.25

0.1165

ii

i i

i A

− = − + +

= − + + −

=

26 12

4.6

TH

TH

V i

V V

= −

=

R TH

VTH

Loop 1 Loop 2

R TH

VTH


29/128


29

∆

4.6

14.41

0.32

V i

i A

=Ω

=

2 a) Kirchoff’s laws

1 1 3 1 2

1 2 3

1 3 2

3 2 3 1 3

1 2 3

0 10 6 2 2 2 2

0 10 10 2 2 1

0 6 5 2 2

0 5 6 6 2 2

0 2 6 13 3

i i i i i

i i i

i i i

i i i i i

i i i

= − + + − + +

= − + + − −

= + − −

= + + − +

= − + + −

(2 x 2) + (3x6)

2 3

32 3

0 32 88

882.75 5

32

i i

ii i

= +

= − = − −

(2 x 6) + (3x2)

1 3

32 3

0 32 56

561.75 6

32

i ii

i i

= +

= − = − −

(5 & 6) in 1

3 3 3

3 3 3

3

3

0 10 10( 1.75 ) 2( 2.75 ) 20 10 ( 17.5 ) ( 5.5 ) 2

10 25

0.4

i i ii i i

i

i A

= − + − + − −

= − + − + − −

− =

= −

R across i3 = 5 ohm

∆ Vi3 = (0.4)(5) = 2 V above A

a

b

i1

6 Ohm


30/128


30

2 b) Delta-star transformations

=

6 20.92

13

6 52.3

135 2

0.7713

×= Ω

×= Ω

×= Ω

( )( )

6.77 4.3092 2

6.77 4.3

5.55

T

T

R

R

×= + +

+

= Ω

110

1.80

5.55

V i A= =

Ω

2

6.77 1.81.10

11.07

Ai A

Ω ×= =

Ω

3

4.3 1.80.70

11.07

Ai A

Ω ×= =

Ω

1.10 2.3 2.53

0.7 0.77 0.539

A

B

V A V

V A V

= × Ω =

= × Ω =

∆ 2.53 0.539

2

AB A B

AB

AB

V V V

V

V V above A

= −

= −

=

0.92 Ω0.77 Ω

2.3 Ω

B

A


31/128


31

Topic 3: Electrostatic Current and Capacitors

Overview

A capacitor is a component of electrical and electronic circuits that exhibits the property

of capacitance. In this module, capacitance is defined and the characteristics of

capacitors are given. The basic definition of a capacitor is found from the measured behavior of the component in an electric circuit. Consider two metal plates separated by

an insulator and connected to a source of some voltage, V, as shown in Figure 3. Current

will flow in the circuit to build up charges on the plates in response to the voltages. A t

every instant of time the ratio of charge on the plates, Q (in coulombs) to voltage across

the plates, V (in volts), is a constant. Thus, if the voltage changes, the charge on the plates will also change such that the ratio will remains constant. This ratio is called thecapacitance. The capacitance or capacity of the component is usually denoted by the variable, C.

Figure 3: A basic capacitor construction

Theory

3.1 Types of capacitors and their application

Capacitors differ in two ways, its construction and type of dielectric. There are different

types of capacitors available, namely:

• Stacked capacitors• Tubular capacitors

• Disc capacitors

• Electrolytic capacitors• Variable capacitors

a. Stacked capacitors

A series of stacked plates is employed. The dielectric insulators separate these plates.

This capacitor, often called multilayer, is illustrated in Figure 3.1a. Each layer acts like a

Plate

Charge, Q

Plate

Voltage

V


32/128


32

single capacitor and the entire stack is like a parallel arrangement. Since capacitors in

parallel add the net capacitance is the sum of the individual members. Assembly isencapsulated in glass, ceramic, or a high-quality plastic. Two types of stacked capacitorare based on mica and ceramic.

Figure 3.1a: A stacked capacitor

b. Tubular capacitors

If the capacity must be large, we need a large plate area and small separation between the plates with a high dielectric constant material. A very common technique for making the

area large, yet keeping the total size down, is to form the plates as two long strips

separated by an insulator. This assembly is then rolled in to a tubular form. Figure 3.1b

illustrates this basic concept. The two strips extend beyond the insulator, one on each

side. This allows secure connections to the plates with, typically, axial leads. Thus such acapacitor often looks like a resistor, although it is usually larger. The tubular form is then

encapsulated in plastic or ceramic. Examples of tubular capacitors are based on paper and plastic film

Figure 3.1b: A tubular capacitor


33/128


33

c. Disc capacitors

A disc capacitor is formed by depositing metal film on each side of a ceramic dielectric

insulator. Since the ceramic can be made very thin, it is possible to make the effective

plate separation small to increase the capacitance. In some case several such discs are

stacked to increase the capacitance. The complete package is then encapsulated in aceramic or plastic insulation. Figure 3.1c shows an example of disc capacitors.

Figure 3.1c: A disc capacitor

d. Electrolytic capacitors

In general, all capacitor types are limited to a maximum capacity of a few microfarads.It is possible to produce larger capacitance in such units but only if they are made very

large. In many applications, such as power supply filtering, it is desirable to have

capacitance values of hundreds or even thousands of microfarads. The electrolytic type

of capacitor can provide these values in a modest size. Examples of electrolytic

capacitors are made polarized, AC electrolytic and Tantalum-based capacitors. Refer to

Figure 3.1d.

Figure 3.1d: An electrolytic capacitor


34/128


34

e. Variable Capacitors

There are many instances when the value of capacitance must be varied to alter circuitry

performance. We classify two modes of such variation. A tuning capacitor is one which

regular and repeated variation of capacity is required in some applications. An example isthe tuning of different stations in an AM radio receiver. A trimmer variable capacitor is

one, which only a single time, or infrequent variations in capacitance, is necessary. Refer

to Figure 3.1e and 3.lf for examples of this type of capacitors.

Figure 3.1e & Figure 3.1f: Examples of variable capacitor

Basic Constructions of a Capacitor:

A simple capacitor can be made from two strips of metal foil sandwiched with two thin

layers of insulation. Waxed paper is a suitable insolent; the wax needed to keep damp out

of the paper which otherwise would quickly cease being an insulator. Thus we have a

device bringing two conductors of large area into a very close proximity with one anotheryet which are insulated, and this would provide a practical capacitor which can be used to

hold electric charge.


35/128


35

3.2 Capacitance, Charge

The property of a capacitor to store an electric charge when its plate is at different potentials is referred to as its capacitance.

The unit capacitance is termed the farad (F) which may be defined as the capacitance of a

capacitor between the plates of which there appears a potential difference of 1 volt when

it is charge by 1 coulomb of electricity.

CV Q

C

V

Q

F cecapaciV d p Applied

C eCh

=

=

= )(tan

)(.

)(arg

In practice, the farad is found to be inconveniently large and the capacitance is usuallyexpressed in microfarads (µF) or in picofarads (ρF).

1 µF = 10-16

F

1 ρF = 10-12

F

3.3 Capacitors in series, parallel

3.3.1 Capacitors in series

Suppose C1 and C2 in Figure 3.3.1 to be two capacitors connected in series with suitablecentre-zero ammeters Ai and A2, a resistor R and a two-way switch S. When S is put

over to position a, A1 and A2 are found to indicate exactly the same charging current,

each reading decreasing simultaneously from a maximum to zero. Similarly when S is put

over to position b, A1 and A2 indicate similar discharges.

If V1 and V2 are the corresponding p.d.s across C1 and C2 respectively, then from

equation

1 1 2 2Q C V C V = =

So that

2

2

1

1 C Q

V and C Q

V ==

or

∆


36/128


36

If we were to replace C1 and C2 by a single capacitor of capacitance C farads such that itwould have the same charge Q coulombs with the same p.d. of V volt, then

C Q

V or CV Q ==

But it is evident from Figure 3.3.1 that V = V1 + V2. Substituting for V, V1, and V2 wehave.

Hence the reciprocal of the resultant capacitance of capacitors connected in series is thesum of the reciprocal of their respective capacitances.

21 C

Q

C

Q

C

Q+=

∆21

111

C C C +=

Figure 3.3.1: Capacitors in series

3.3.2 Capacitors in parallel

Suppose two capacitors, having capacitances C1 and C2 farads respectively, to be

connected in parallel, Figure 3.3.2 across a p.d of V volts. The charge on C1 is Q1

coulombs and that on C2 is Q2 coulombs, where

Q1 = C1V and Q2 = C2V

If we were to replace C1 and C2 by a single capacitor of such capacitance C farads that

the same total charge of (Q4 + Q2) coulombs would be produced by the same p.d.,

then Q1 + Q2 = CV

Substituting for Q1 and Q2, we have

C1V + C2V = CV

C=C1 + C22


37/128


37

Hence the resultant capacitance of capacitors in parallel is the arithmetic sum of theirrespective capacitance.

Figure 3.3.2: Capacitors in parallel

3.4 Energy stored in capacitors and di-electric strength

3.4.1 Energy stored

Suppose the p.d. across a capacitor of capacitance C farads to be increased from v to (v+

dv) volts in dt seconds.

The charging current, i amperes is given by

dt

dvC i .=

Instantaneous value of power to capacitor is

wattsdt

dvvC wattsiv .=

And energy supplied to capacitor during interval dt is

joulesdvCvdt dt

dvvC ... =

Hence total energy supplied to capacitor when p.d. is increased from 0 to V volts is

[ ]∫ ==v v

joulesCV vC dvCv0

2

0

2

2

1

2

1.


38/128


38

∆ 2

2

1CV W =

alsoC

Q

C

QC W

22

.2

1

2

1=

=

For a capacitor with dielectric of thickness d metres and area A square metres, energy percubic metre is

joules D

DE

E d

V

Ad V

d A

Ad CV

ε

ε ε

ε

2

2

2

22

2

1

2

1

2

1

2

1

..2

1.

2

1

==

=

=

=

3.4.2 Dielectric strength

If the p.d. between the opposite sides of a sheet of solid insulating material is increased

beyond a certain, the material breaks down resulting a tiny hole or puncture through the

dielectric so that it can not be used as an insulator anymore.

The potential gradient necessary to cause the breakdown is called dielectric strength ofthe material. The value of the dielectric strength of a given material decreases with

increase of thickness.

Material Thickness

(mm)

Dielectric strength

(MV/m)

Air (at normal pressure andtemperature)

0.20.6

1.0

6.010

5.754.92

4.46

3.272.98

Mica 0.01

0.1

1.0

200

176

61

Glass (density 2.5) 1.0

5.0

28.5

18.3

Ebonite 1.0 50

Paraffin-waxed paper 0.1 40-60

Transformer oil 1.0 200

Ceramics 1.0 50


39/128


39

Summary

• Capacitance is a measure of the ability to store electric charge• Capacitance is also a measure of the ability to store energy in an electric field.• Charging is the process of increasing the charge held in a capacitor.• Discharging is the process of reducing the charge held in a capacitor• Farad is the capacitance of a capacitor which has a potential difference of 1V

when maintaining a charge of 1 C

• Leakage current is the rate of movement of charge through a dielectric• Permittivity is the ratio of electric flux density to electric field strength measured

in farads per meter.


Q [coulombs] = C [farads] x V [volts] Charging current of capacitor i is

1 µF = 10-6 Fdq dv

C dt dt

= ×

1 pF = 10-12

F

Energy stored in capacitor is

For capacitors in parallel W = ½ CV 2 joulesC = C1 + C2 +…

Energy per cubic meter of dielectric is

For capacitors in series2

21 1 1

2 2 2

D E DE ε

ε = = joules

1 2

1 1 1...

C C C = + +

For C1 and C2 in series

21

1 2

C V V

C C = ×

+

Electric field strength in dielectric,V

E d

=

Electric flux density,Q

D

A

=

Capacitance, o r A

C d

ε ε =

Absolute permittivity, o r D

E ε ε ε = =

Permittivity of free space is128.85 10o

F mε

−= ×

Capacitance of parallel-plate capacitor with n plates is( 1)n A

d

ε −


40/128


40

Self-Assessment

1. A 50 µF capacitor charged to 250 V. What is the stored energy?

Solution:

The stored energy can be found using J CV W 5625.12

1 2 ==

2. Three capacitors with values 270, 47 and 1200 pF are connected in series, what is the

equivalent series capacity?

Solution:

101055.21200

1

47

1

270

11×=++=

pF pF pF Req

Thus, pF Req 39109.31055.2

1 1110

=×=×

= −

3. What would be the net capacitance if the three capacitors of example 2 previously are

arranged in parallel?

Solution:

pF pF pF pF Rnet 1517120047270 =++=

4. A capacitor having a capacitance of 80 µF is connected across a 500 V d.c supply,

calculate the charge.

Solution:

From Q = CV

Charge = (80x10-6

)F x 500 V = 0.04 C = 40 mC

5. A capacitor is made with seven metal plates parallel (i.e. multi-plate capacitor) andseparated by sheets of mica having a thickness of 0.3 mm and a relative permittivity

of 6. The area of one side of each plate is 500 cm2. Calculate the capacitance in

microfarads.

Solution:

Using the equation; ( )[ ] Faradsd AnC r o /1−= ε ε

C = 0.0531 x 10-6

F = 0.0531 µF


41/128


41

Tutorials

1. Three capacitors have capacitances of 10 µF, 15 µF, and 20 µF respectively. Calculate

the total capacitance when they are connected (a) in parallel (b) in series

2. A certain capacitor has a capacitance of 3 µF. A capacitance of 2.5 µF is required by

combining this capacitance with another. Calculate the capacitance of the second

capacitor and state how it must be connected to the first.

3. Two capacitors, A and B are connected in series across a 200 V dc supply. The p.d

across A is 120 V. This p.d is increased to 140 V when a 3 µF capacitor is connected

in parallel with B. Calculate the capacitance of A and B.

4. A parallel-plate capacitor has a capacitance of 300 pF. It has 9 plates, each 40 mm x

30 mm separated by mica having a relative permittivity of 5. Calculate the thickness

of the mica.

Further Readings

Capacitor is a passive-type component which is being widely used in microelectronic

circuits and systems. Time constant/delay device, Filter and Oscillator circuits are someexample of the use of capacitors. Further reading of the function of capacitors in

electronic circuits will enhance our understanding of the importance of capacitance effect

in circuits.

References




McGraw-Hill.


42/128


42

Solution Tutorials:

1 a) in parallel

1 2 3

10 15 20 45

T

T

C C C C

C F F F F µ µ µ µ

= + +

= + + =

1 b) in series

1 2 3

1 1 1 1

1 1 1 14.615

10 15 20

T

T

C C C C

F C F F F

µ µ µ µ

= + +

= + + =

21

2

2.5

3

?

T C F

C F

C

µ

µ

=

=

=

2

2

2

1 1 1

2.5 3

166666.67

15

F F C

F C

C F

µ µ

µ

µ

= +

=

=

∆ in series.

3.

21

1 2

2

1 2

1 2 2

1 2

2 1

120 200

0.6( )

0.6 0.4

1.5 1

C V V

C C

C C C

C C C

C C

C C

= ×+

= ×+

+ =

=

= −

2

2 1

2

2 1

2 1 2

21

3140 2003

30.7

3

0.7 0.7 0.9

0.9 0.32

0.7

C C C

C

C C

C C C

C C

µ µ

µ

µ

µ

µ

+= × + +

+=

+ +

+ = +

+= −

120 V

C1 C2

C1 C2

140 V


43/128


43

2 in 1

22

2 2

2

1

0.9 0.31.5

0.71.9286 0.64286

5.4

3.6

C C

C C

C F

C F

µ

µ

µ

µ

+ =

= +

=

=

4.( )

2

12

0

12 2

300

9

1200

58.85 10

1

8.85 10 5(9 1)1200

300

1.416

r

o r

C pF

n

A mm

n AC Farads

d

mmd

pF

d mm

ε ε

ε ε

−

−

=

=

=

== ×

− =

× × − =

=


44/128


44

Topic 4: Magnetism and Magnetic Circuits

Overview

Magnet is a certain form of iron ore that attracts iron. Iron fillings are used to indicate the pattern of the magnetic field due to effect of magnet. Magnetic field is the region around

the magnet within which a force will be exerted upon magnetic materials. The places on

the magnet where the magnetic field appears to be most concentrated are referred to asthe poles of the magnet (north and south-seeking poles). Similar poles will repel, unlike

poles will attract. This is similar to the concept of electric charges, repelling and

attracting depending on the types of charge. The pattern that the iron fillings take in

Figure 4.1 is the magnetic field. Magnetic effect has been utilised in various electrical

and electronic devices/systems. Electrical system such as transformer and electrical

motor operations depend on the principles of magnetic circuits.

Figure 4.1: Magnetic field lines

Theory

4.1 Magnetism

When a magnet is created, there is an area of magnetic influence near the magnet called a

magnetic field. A magnetic field is establish either by alignment of internal magnetic

forces within a magnetic material, or by organised movement of charges (current flow)through conductor materials. This magnetic field is composed of magnetic lines of force,

called flux. The symbol for flux is Greek letter phi (Φ). A magnetic field is a vectorquantity, i.e., it has direction and strength. The strength of a magnetic field is measured inthe number of lines of magnetic force, or magnetic flux per unit area, B. The SI unit isTesla (T) or weber per square meter (Wb m2). 1 T = 1 Wb m2.

A

Φ=Β

Where B = flux density (T)Φ = flux (Wb)

A = cross-sectional area (m2 )


45/128


45

The lines of flux have certain properties (refer to Figure 4.1);

• Lines of flux always form closed loops, beginning from N pole and finishing fromS pole.

• Lines of flux cannot cross one another• Lines of flux take the possible path• Parallel lines of flux in the same direction repel one another

4.2 Magnetic Circuit

The presence of a magnetic filed is often illustrated through the concept of lines of force.

The total number of lines is a measure of flux, Φ, or the strength of the filed. We use the

concept of the number of flux lines passing perpendicularly through a given cross-

sectional area. i.e. B.

4.2.1 Magnetic field due to the flow of an electric current

Two properties of the magnetic field are important, i.e. pattern of the field and itsstrength. Figure 4.2.1 shows the magnetic field due to a long straight wire. The directionof the field follows the ‘right-hand grip’ rule. According to this rule, the thumb of theright hand points along the direction of the current and the direction of the fingers give

the direction of the magnetic field. The magnetic field strength ( B) depends linearly withthe current ( I ) through the wire, i.e.

I B α

Figure 4.2.1 shows the magnetic field due to a long straight wire

As we move away from the wire, the magnetic field gets weaker, if r represents distance,

then

r I B α

So we have; ( )r B o π µ 21=


46/128


46

The constant µo is referred lo as permeability of free space and measures the ease with

which magnetic forces are transmitted through space. (µo = 4п x 10-7

). The suitable unitfor µo is Tesla meter per ampere (Tm A

-1).

4.2.2 Magnetic field due to a short coil of wire

Consider the magnetic field due to a single circular loop. Figure 4.2.2 shows the pattern

of the magnetic field produced from above. We can think of the magnetic field as beingdue to the result of the magnetic fields of two wires carrying currents in opposite

directions. At points near the center, the magnetic fields of these two wires add to make

the total magnetic field stronger. The strength of the field is given by;

( )r B o 21µ =

Figure 4.2.2: Pattern of magnetic field due to a single circular loop (from above)

4.2.3 Magnetic field due to a long coil of wire

The magnetic field for a long coil of wire is shown Figure 4.2.3 where the magnetic field

inside the coil is approximately uniform and is given by:

( )l n B o 1µ =

where l is the length of the coil and n is the number of turns.

4.2.4 Force acting on a current-carrying conductor in a magnetic field

The force acting on a conductor has two parts; direction and magnitude. The direction

tells us which way the conductor tends to move and the magnitude tells us the strength

of the force exerted on the conductor. Strictly, the force only acts on the moving


47/128


47

charges, but because they are bound to the conductor, the whole conductor will move

(refer to Figure 4.2 4a). A convenient way of knowing the direction in which force actsis given by ' Fleming’s left-hand rule’ (see Figure 4.1.41b). The first finger point in thedirection of the magnetic field, while the second finger points in the direction of the

current flow. The direction of the thumb indicates the direction of the force acting on the

conductor. The magnitude of the force depends on;

• The strength of the magnetic fields

• The amount of current flowing in the conductor

• The length of the conductor in the magnetic field

So, F = Bil (F is measured in Newtons). Force acting at an angle is governed by:

( )θ sin Bil F =

Figure 4.2.3: The magnetic field for a long coil of wire

Figure 4.2.4a: Force acting on the moving charge in a conductor


48/128


48

Figure 4.2.4b: Fleming’s left hand rule

4.2.5 The magnetic force between two current-carrying conductors

It is important to consider the force between two current-carrying parallel conductors.

Suppose we have two thin, very long conductors which we place a distance r apart.Suppose current I 1 flows in the first conductor and current I 2 flow in the secondconductor. There will magnetic fields interaction due to these two current-carrying

conductors. Both conductors will experience a uniform magnetic field with strength, B1

and B2, with current I 1 and I 2 flowing respectively. The magnitude of force acting on alength l of the second conductor is determined by the formula

( ) r l l l l l B F o π µ 2/21211,2 ==

Now, we need to determine the direction in which it acts. This is done by using the

“ Fleming’s left-hand rule’ .

Special case:

Suppose the current flowing in each conductor, l 1 and l 2, is one ampere, theconductors are separated by 1 meter and the length of each conductor is 1 meter

then the force between the conductors is:

N r F o7

1,2 1022/ −×== π µ

The operation of the magnetic force between two current-carrying conductors is shown in

Figure 4.2.5.


49/128


49

Figure 4.2.5: Magnetic force between two current-carrying conductors

4.2.6 Magnetic Toroid

To produce a magnetic field in most magnetic devices, a current is established in a coil ofwire (refer to figure 4.2.6). The flow of current produces a magnetic field around the

wire. Consider a coil of wire where the coil is formed into a toroid as shown. When a

current flows in the coil, a magnetic field is formed within the toroid. This current

produces a magneto motive force (mmf) directly proportional to the amount of current,

that flows and the number of turns in the coil. The basic formula for magneto motiveforce is given by:

Figure 4.2.6: Magnetic field in a coil of wire

NI F = Where F = Magneto motive force (mmf) in amp turns

N = number of turns in the coil

I = coil current (A)


50/128


51/128


51

R F Φ=

A L R

r o µ µ =

Because of the definition of µ, the expression for R can also be written as

π F R = or R F π =

This latter expression is frequently referred to as Ohm’s law for magnetic circuits. The

reciprocal or reluctance is called permeance (P).

4.3 Parallel Magnetic Circuits

4.3.1 Self-Inductance

Changing currents in an isolated electric circuit can induce voltages within the same

circuit, i.e. within itself. This is called self-inductance. A circuit component designedspecifically for the purpose of providing self-inductance is called an inductor. Theconcept of self-inductance arises when it is found that if the current is caused to vary in

time, a voltage will generated or induced across the coil terminals. This induced voltage

is related to the rate of change of current by a constant. This constant is called

inductance, and the unit is Henry ( H ), which is equal to one volt per amp per second, i.e.1 H = lV/ (A/s). In equation form,

( ) ( )dt di Lt v Ll =

Where v L(t) = voltage induced in volts (V) L = inductance in henrys (H)di L /dt = rate of change of current through coil in amperes per

second (A/s)

Figure 4.3.1 shows the physical structure of an inductor. The basic equation for inductor

is given by

( ) dt didid N dt d N t v L Φ=Φ=

c L A N

did

N L2µ

=Φ

=


52/128


52

Figure 4.3.1: The physical structure of an inductor

4.3.2 Mutual Inductance

Consider two coils arranged near each other so that the flux from either coil passes

through the other. They are magnetically linked. Magnetic field will be induced in each

coil due to the other coil. This is mutual inductance as a result of parallel magnetic

circuits. Figure 4.3.23 shows the relationship. In general the mutual inductance is given by;

1. 2k L L =

Where k = coupling coefficient value (between 0 and 1)

The mutual inductance has a polarity with respect to the self-inductance of the coils. Thatis, the mutual inductance can induce a voltage that is in opposite polarity to that induced

from self-inductance. This is generally specified by showing a dot on the schematic to

indicate the end of the coil which an entering current would induce a voltage to enhancethe self-induced voltage of the second coil. Figure 4.3.2b shows a coil for which current

introduced into the top of each coil produces an induced voltage of the same sense as the

self-inductance voltage. In Figure 4.3.2b, the opposite is true.


53/128


54/128


54

Summary

• A magnetic field can be described using lines of flux. Such lines form closed loops,do not cross and, when parallel, repel one another.

• Magnetic fields have North, and South poles. Like poles repel, unlike poles attract.• A current-carrying conductor lying in a magnetic field experiences a force.• The relative directions of the field, force and current are given by the left-hand rule.• The relative directions of the filed, motion and induced emf. are given by the right-

hand rule. (Fleming's right-hand rule)

• A magnetic flux is a created by a magneto motive force.• The magnetic field strength is the mmf gradient at any point in a field.• The magnetic field strength and flux density at any point in a field are related by the

permeability. For ferromagnetic material, the relative permeability varies according to

the magnetic field strength.

• The variation of flux density with magnetic field strength is illustrated by themagnetisation characteristics (B/H curve).

• The reluctance of a magnetic circuit is the ratio of the mmf to the flux.


Magnetomotive force,

F = NI (amperes or ampere-turns)

Magnetic field strength

H = F/l = NI/L (amperes per meter)

Flux density,

B = µ x H (teslas)

Permeability of free space,

µo = 4π x 10-7

(henrys per meter)

Reluctance of a magnetic circuit,

o r

F S

l S Aµ µ

= Φ

=

Induced e.m.f.

( )di

e L voltsdt d

e N dt

φ

= ×

= ×

Inductance,


55/128


55

N L I Φ= (webers per ampere or henrys)

Energy stored in an inductor,

W f = ½ LI 2 (joules)

Energy density in a magnetic field,w f = ½ BH (joules per cubic meter)

EMF induced by mutual inductance,

die M

dt = × (volts)

Mutual inductance,

2 2

1

1 2

N M I

N M

S

Φ

=

=

Coupling coefficient of a mutual inductor,

1 2( )k L L=

Effective inductance of a mutual inductor

1 1 2 L L L M = + ±

Self-Assessment

1. A conductor carries a current of 800 A at right angles to a magnetic field having a

density of 0.5 T. Calculate the force on the conductor in newtons per meter length.

Solution:

From formula: F = BIL ; Force per meter length = 0.5 T x 800 A = 400 N

2. A rectangular coil measuring 200 mm by 100 mm is mounted such that it can be

rotated about the midpoints of the 100 mm sides. The axis of rotation is at the right

angles to a magnetic field of uniform flux density 0.05 T. Calculate the flux in thecoil for the following conditions:

a. the maximum flux through the coil and the position at which it occurs.

b. the flux through the coil when the 100 mm sides are inclined at 45 degrees to the

direction of flux.


56/128


56

Solution:

a) The maximum flux will pass through the coil when the plane of the coil is at right

angles to the direction of the flux.

mWb BA 110200005.0 3 =××==Φ −

b) mWb BA o 71.045sin101sin 3 =××==Φ −θ

3. A coil of 200 turns is wound uniformly over a wooden ring having a meancircumference of 600 mm and a uniform cross-sectional area of 500 mm

2. If the

current through the coil is 4 A, calculate

a. the magnetic field strength

b. the flux density

c. the total flux

Solution:

a) Mean circumference = 600 mm = 0.6 m

H = 4x200 / 0.6 = 1.333 A/m

b) From formula: Flux density = µo H = 1675 µT

c) Cross-sectional area = 500 mm2 = 500x10

-6 m

2

Total flux = 1675 µT x (500x10-6)m2 = 0.8375 µWb

Tutorials

1. A mild steel ring has a mean diameter of 160 mm and a cross-sectional area of 300

mm2. Calculate

a. the mmf. to produce a flux of 400 µWb b. the corresponding values of the reluctance of the ring and the relative

permeability

2. A 1500-turn coil surrounds a magnetic circuit which has a reluctance of 6x106 A/Wb.

What is the inductance of the coil?

3. If an emf. of 5 V is induced in a coil when the current in an adjacent coil varies at a

rate of 80 A/s, what is the value of the mutual inductance of the two coils?

4. When a current of 2 A through a coil P is reversed, a deflection of 36 divisions is

obtained on a fluxmeter connected to a coil Q. If the fluxmeter constant is 150 µWb-turns/div, what is the value of the mutual inductance of coils P and Q?


57/128


57

Further Readings

Magnetic circuit’s concept is important in the understanding of many electrical/electronic

circuits and systems. The basis of motor operation, either DC or AC depends on the

characteristics of magnetic circuit relationship. More emphasis and understanding are

needed in the topic of inductance and how it is applied in electrical machines.

References




McGraw-Hill.


58/128


58

Solution Tutorials:

1 a)

6

6

6

400 10

400 101.33

300 10

Wb

B T A

−

−

−

Φ = ×

Φ ×= = =

×

b) From figure, 3

6

6

1750

1750 160 10 880

8802.2 10 /

400 10

A H m

F Hl A

F S H

π −

−

−

=

= = × × × =

= = = ×Φ ×

2. 2 2

6

15000.375

6 10

N L H

S −= = =

×

3. 5

65.280

die M

dt

mH

= =

= =

4.

1.35 mH


59/128


59

Topic 5: Single-Phase A.C Circuits

Overview

Many applications of electrical and electronics involve voltages and currents that areconstant in time. These networks are called direct current (dc), although the abbreviation

dc is used with voltage as well as current. Time-varying voltage and current sources are

also of great importance in electrical and electronics. Alternating current can be

abbreviated to ac, hence a system with an alternating current is known as an ac system.

The curves relating current to time are known as waveforms. Basic waveforms are such

as sinusoidal, square and triangular waves. Other waveforms can be more complicated.Phase is also important in an alternating voltages and currents circuits, e.g., single and

multiphase circuits or electrical motors.

Most homes and rural areas are supplied with single-phase ac electrical power. Con-sequent, most electrical home appliances and electrically driven farm machines employsingle-phase motors. They drive washing machines refrigeration and air-conditioning

compressors, grain dryers, fans, pumps, sewing machines vacuum cleaners, clocks,

phonographs, hand tools, and so on. Literally millions of single-phase motors aremanufactured each year. The difficulty with single-phase as a power source for motors is

that it does not lend itself to producing a rotating magnetic field. Several schemes have

been developed to circumvent this difficulty. Each result in a motor with specific

characteristics suitable for certain range of uses only. In this module, the analysis of accomponents and circuits will be conducted and power parameters are determined using

complex variable and phasor diagrams.


60/128


60

Theory

5.1 Determination of r.m.s, mean and peak, shape factor and peak factor

If I m is the maximum values of a current which varies sinusoidal as shown in Fig. 5.1, theinstantaneous value i is represented by

θ sinm I i =

Where θ is the angle in radians from instant of zero current

Figure 5.1: Average and r.m.s values of a sinusoidal current

For very small interval dθ radians, the area of the shaded strip is i. dθ ampere-radians.The use of the unit 'ampere-radians' avoids converting the scale on the horizontal axisfrom radians to seconds. Therefore, total area enclosed by the current wave over half-

cycle is;

[ ]∫ ∫ −==π π

π θ θ θ θ

0 0

0cos.sin. mm I d I d i

[ ]11−−−= m I radiansampere I m −= 2

From, Average = (Area enclose ever half-cycle)/(Length of base over half-cycle)

Average value of current over a half-cycle is:

( ) ( )rad radsamp I I mav π /2 −=

i.e. amperes I I mav 637.0=


61/128


61

If the current is passed through a resistor having resistance R ohms, instantaneous heatingeffect = i2 R watts. The variation of i2 R during a complete cycle is shown in Figure 5.1during interval dθ , heat generated is i2 R . dθ watt-radians and is represented by the area ofthe shaded strip. Hence, heat generated during the first half-cycle is area enclosed by the

i2 R curve and is;

∫ ∫=π π

θ θ θ 0 0

222 .sin. d R I d Ri m

( ) θ θ π

d R I m

∫ −=0

2

.2cos12

π

θ θ 0

2

2sin2

1

2

−=

R I m

radianswatt R I m −=2

2

π

Average heating effect can be expressed as:

Average heating effect over half-cycle = (Area enclosed by i2R curve over half cycle)

Length of base

Hence, the heating effect is:

( ) ( )( )

2

212

2

m

m

I R watt radians I R watts

radians

π

π

−=

This result is really obvious from the fact that sin2 θ = 1/2 - 1/2 cos 2θ . This means thatthe square of a sine wave may be regarded as being made up of two components: a) a

constant quantity equal to half the maximum value of sin2 θ curve, and b) a cosine curvehaving twice the frequency of the sin θ curve. From Figure 5.1 it is seen that the curve of

the heating effect undergoes two cycles of change during on recycle of current. The

average value of component (b) over a complete cycle is zero; hence the average heatingeffect is 21 2 m I R . If I is the value of direct current through the same resistance to producethe same heating effect;

therefore,

0.70712

mm

I I = =

Where I = instantaneous current I m = peak current


62/128


62

Since the voltage across the resistor is directly proportional to the current, it follows that

the relationship derived for current also applies to voltage. Hence, in general, averagevalue of a sinusoidal current or voltage is 0.637 x maximum value.

therefore; I av = 0.637 I m

r.m.s value of a sinusoidal current or voltage is 0.707 x maximum value;

therefore; I = 0.707 I m

Given that form (shape) factor:. .r m s value

verage value, hence the form factor of a sine wave is;

0.707 max 1.110.637 max

f imum value k imum value

× = =×

and peak (crest factor) =max

. .

peak or imum valuer m s value

, hence for a sine wave:

max1.414

0.707 max p

imum valuek

imum value = =

×

5.2 R-L-C Circuit is series and parallel

a) Series R-L-C Circuit

Circuit of the contain elements of resistance, inductance, and capacitance. In aninductive AC circuit, current lags voltage by 90 degrees. In a AC capacitive circuit,current leads voltage by 90 degrees. It can be seen that inductance and capacitance

are 180 degrees apart. Since they are 180 degrees apart, one element will cancel out

all or part of the other element (Figure 5.2a).

An AC circuit is:

Resistive if XL and XC are equal.

Inductive if XL is greater than XC.

Capacitive if XC is grater that XL.


63/128


63

Figure 5.2a: Elements

• Calculating Total Impedance in a Series R-L-C Circuit

The following formula is used to calculate total impedance of a circuit containing

resistance, capacitance, and inductance:

( )22

L C Z R X X = + −

In the case where inductive reactance is greater than capacitive reactance,subtracting XC from XL results in a positive number. The positive phase angle is an

indicator that the net circuit reactance is inductive, and current lags voltage. In the

case where capacitive reactance is greater than inductive reactance, subtracting XC from XL results in a negative number. The negative phase angle is an indicator that

the net circuit reactance is capacitive and current leads voltage. In either case, the

value squared will result in positive number.

• Calculating Reactance and Impedance in a Series R-L-C Circuit

In the following Figure 5.2b, given a circuit with 120 volt, 60 hertz, resistance is

1000, inductance is 5 mH, and capacitance is 2 F.

Figure 5.2b: Example circuit to calculate total impedance


64/128


64

To calculate total impedance, first calculate the value of XL from XC, then impedancecan be calculated.

1

2

1

6.28 60 0.000002

1,327

C

C

C

X fC

X

X

π =

=× ×

= Ω

( )

( )

( )

22

22

2

1000 1.884 1,327

1,000,000 1,325.116

1,000,000 1,755,932.41

2,755,932.41

1,660.

L C Z R X X

Z

Z

Z

Z

Z

= + −

= + −

= + −

= +

=

=

• Calculating Circuit Current in a Series R-L-C CircuitOhm’s Law can be applied to calculate total circuit current.

120

1,660.1

0.072

E I

Z

I

I amps

=

=

=

b) Parallel R-L-C Circuit

• Calculating impedance in a parallel circuit

Total impedance (Zt) can be calculated in a parallel R-L-C circuit if values of

resistance and reactance are known. One method of calculating impedance involvesfirst calculating total current, then using the following formula:

t t

t

E Z

I =

Total current is the vector sum of current flowing through the resistance plus, the

difference between inductive current and capacitive current. This is expressed in the

following formula:


65/128


65

( )22

t R C L I I I I = + −

In the following Figure 5.2c, given a circuit with 120 volt, 60 hertz, capacitivereactance has been calculated to be 25 ohm and inductive reactance 50 ohm.

Resistance is 1000 ohm.

Figure 5.2c: Example circuit to calculate impedance in parallel

A simple application of Ohm’s Law will find the branch currents. Remember, voltage

is constant throughout a parallel circuit.

120

50

2.4

L L

L

L

E I

X

I

I amps

=

=

=

120

1000

0.12

R

R

R

E I

R

I

I amps

=

=

=

120

25

4.8

C C

C

C

E I

X

I

I amps

=

=

=

Once the branch currents are known, total current can be calculated.

( )( )

22

220.12 4.8 2.4

0.0144 5.76

5.7744

2.4

t R C L

t

t

t

t

I I I I I

I

I

I amps

= + −

= + −

= +

=

=

Impedance is now found with an application of Ohm’s Law


66/128


66

120

2.4

50

t t

t

t

t

E Z

I

Z Z

=

=

= Ω

5.3 Determination of power and power factor using complex variables and phasor

diagrams

5.3.1 Power and Power Factor in an AC Circuit

Power consumed by a resistor is dissipated in heat and not returned to the source. This istrue power. True power is the rate at which energy is used. Current in an AC circuit rises

to peak values and diminishes to zero many times a second. The energy stored in themagnetic field of an inductor, or plates of a capacitor, is returned to the source when

current changes direction. Power in an AC circuit is the vector sum of true power and

reactive power. This is called apparent power. True power is equal to apparent power in a

purely resistive circuit because voltage and current are in phase. Voltage and current arealso in phase in a circuit containing equal values of inductive reactance and capacitive

reactance. If voltage and current are 90 degrees out of phase, as would be in a purely

capacitive or purely inductive circuit, the average value of true power is equal to zero.There are high positive and negative peak values of power, but when added together the

result is zero.

5.3.2 True power and apparent power formulas

The formula for true power is:

cos P EI θ =

Apparent power is measured in volt-amps (VA). True power is calculated from another

trigonometric function, the cosine of the phase angle (cos). The formula for apparent power is:

P EI =

True power is measured in watts. In a purely resistive circuit, current and voltage are in

phase. There is a zero degree angle displacement between current and voltage. The cosineof zero is one. Multiplying a value by one does not change the value. In a purely resistive

circuit the cosine of the angle is ignored. In a purely reactive circuit, either inductive or

capacitive, current or voltage is 90 degrees out of phase. The cosine of 90 is zero.

Multiplying a value times zero results in a zero product. No power is consumed in a

purely reactive circuit.


67

Documents

Fundamental of E E