fundamentals of electrical engineering

Houssem Rafik El-Hana BOUCHEKARA
2009/2010 1430/1431
KINGDOM OF SAUDI ARABIA
Ministry Of High Education
Umm Al-Qura University
College of Engineering & Islamic Architecture
Department Of Electrical Engineering
Fundamentals of Electrical
Engineering
2. Basic Electrical Circuits

2
2 BASIC ELECTRICAL CIRCUITS ..........................................................................................4
2.1 INTRODUCTION...............................................................................................................4
2.2 BASIC CONCEPTS AND DEFINITIONS.....................................................................................5
2.2.1 Charge ...................................................................................................................5
2.2.2 Current...................................................................................................................5
2.2.3 Voltage ..................................................................................................................7
2.2.4 Power and Energy .................................................................................................8
2.2.5 Circuit elements...................................................................................................11
2.2.5.1 Passive elements (loads)............................................................................................11
2.2.5.2 Active elements .........................................................................................................11
2.2.5.3 Sign convention..........................................................................................................12
2.2.6 Resistor................................................................................................................13
2.2.6.1 Ohm’s Law..................................................................................................................13
2.2.6.2 Conductance ..............................................................................................................14
2.2.7 Capacitor .............................................................................................................17
2.2.8 Inductor...............................................................................................................20
2.3 CIRCUIT THEOREMS.......................................................................................................23
2.3.1 Introduction.........................................................................................................23
2.3.2 Definitions and Terminologies.............................................................................23
2.3.3 Ohm’s Law...........................................................................................................26
2.3.4 Kirchhoff’s Laws...................................................................................................26
2.3.4.1 Series resistors and voltage division ..........................................................................30
2.3.4.2 Parallel resistors and current division........................................................................32
2.3.4.3 Series and parallel capacitors.....................................................................................32
2.3.4.4 Series and parallel inductors......................................................................................32
2.3.5 Electric Circuits Analysis ......................................................................................35
2.3.5.1 Mesh Analysis ............................................................................................................35
2.3.5.2 Nodal Analysis............................................................................................................35
2.3.6 Superposition Theorem .......................................................................................38
2.3.7 Thvenin’s Theorem ..............................................................................................38
2.3.8 Norton’s Theorem................................................................................................38
2.3.9 Source Transformation........................................................................................38
2.3.10 Maximum Power Transfer Theorem..................................................................38
2.3.11 Mesh and Nodal Analysis by Inspection ............................................................38
2.4 SINUSOIDS AND PHASORS...............................................................................................39
2.4.1 Introduction.........................................................................................................39
2.4.2 Sinusoids..............................................................................................................39
2.4.1 Phasors................................................................................................................42
2.4.2 Phasor Relationship for Circuit Element..............................................................46
2.4.2.1 Resistor ......................................................................................................................46
2.4.2.2 Inductor......................................................................................................................48
2.4.2.3 Capacitor....................................................................................................................49
2.4.3 Impedance and Admittance ................................................................................50
2.4.4 Kirchhoff laws in the frequency domain..............................................................52
2.4.5 Other Sinusoidal Parameters...............................................................................52
2.4.5.1 Mean or Average Value..............................................................................................52
2.4.5.2 Effective or RMS Value...............................................................................................53
2.4.6 Power in AC Circuits.............................................................................................53
2.4.7 Power Factor .......................................................................................................55
2.4.8 Power Factor Correction......................................................................................57
2.5 THREE PHASE CIRCUITS...................................................................................................58

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2.5.1 objectives.............................................................................................................58
2.5.2 Three-Phase Circuits Overview............................................................................58
2.5.3 WYE CONNECTION ..............................................................................................60
2.5.4 DELTA CONNECTIONS..........................................................................................62
2.5.5 THREE-PHASE POWER .........................................................................................64
2.5.6 THREE-PHASE CIRCUIT CALCULATIONS ...............................................................65
2.5.7 SUMMARY ...........................................................................................................69

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2 BASIC ELECTRICAL CIRCUITS
2.1 INTRODUCTION
Electric circuit theory and electromagnetic theory are the two fundamental theories
upon which all branches of electrical engineering are built. Many branches of electrical
engineering, such as power, electric machines, control, electronics, communications, and
instrumentations, are based on electric circuit theory. Therefore, the basic electric circuit
theory is the most important course for an electrical engineering student and always an
excellent starting point for a beginning student in electrical engineering education. Circuit
theory is also a valuable to students specializing in other branches of the physical sciences
because circuits are good model for the study of energy systems in general, and because the
applied mathematics, physics, and topology involved.
In electrical engineering, we are often interested in communicating or transferring
energy from one point to another. To do this requires an interconnection of electrical
devices. Such interconnection is referred to as an electric circuit and each component of the
circuit is known as an element.
A simple electric circuit is shown in Fig. 2.1. It consists of three basic components: a
battery, a lamp and connecting wires. Such a simple circuit can exist by itself; it has several
applications, such as a torch light, a search lights and so forth.
Figure 2. 1 : A simple electric circuit
An electric circuit is an interconnection of electrical elements.

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2.2 BASIC CONCEPTS AND DEFINITIONS
2.2.1 CHARGE
The concept of electrical charge is the underlying principle for explaining all
electrical phenomena. Also the most basic quantity in an electric circuit is the electric
charge. We all experience the effect of electric charge when we try to remove our wool
sweater and have it stick to our body or walk across a carpet and receive a shock.
We know from elementary physics that all matter is made of fundamental buildings
blocks known as atoms and that each atom consists of electrons, protons, and neutrons. We
also know that the charge e on an electron is negative and equal in magnitude to 1.602×10-19
C, while a proton carries a positive charge of the same magnitude as the electron. The
presence of equal numbers of protons and electrons leaves an atom neutrally charged.
2.2.2 CURRENT
The unit of current is the ampere abbreviated as (A) and corresponds to the quantity
of total charge that passes through an arbitrary cross section of a conducting material per
unit second. (The name of the unit is a tribute to the French scientist André Marie Ampère.)
Mathematically,
𝐼 =
𝑄
𝑡
(2. 1)
Or
Charge is an electrical property of the atomic particles of which matter
consists, measured in coulombs (C). Charge, positive or negative, is denoted by the
letter q or Q.
Current can be defined as the motion of charge through a conducting
material, measured in Ampere (A). Electric current, is denoted by the letter i or I.

6
𝑄 = 𝐼𝑡 (2. 2)
Where 𝑄 is the symbol of charge measured in Coulombs (C), I is the current in
amperes (A) and t is the time in second (s).
The current can also be defined as the rate of charge passing through a point in an
electric circuit i.e.
𝑖 =
𝑑𝑞
𝑑𝑡
(2. 3)
A constant current (also known as a direct current or DC) is denoted by symbol I
whereas a time-varying current (also known as alternating current or AC) is represented by
the symbol 𝑖 or 𝑖(𝑡). Figure 2.2 shows direct current and alternating current.
Figure 2. 1 : Two common type of current: (a) direct current (DC), (b) alternative current (AC).
Example 2. 1:
Determine the current in a circuit if a charge of 80 coulombs © passes a given point
in 20 seconds (s).
Solution:
𝐼 =
𝑄
𝑡
=
80
20
= 4𝐴
Example 2. 2:
How much charge is represented by 4.600 electrons?
Solution:
Each electron has -1.602×10-19
C. Hence 4.600 electrons will have:
1.602 × 10−19
× 4.600 = −7.369 × 10−16
𝐶

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Example 2. 3 :
The total charge entering a terminal is given by 𝑞 = 5𝑡 sin4𝜋𝑡 𝑚𝐶. Calculate the
current at 𝑡 = 0.5 𝑠.
Solution:
𝒊 =
𝒅𝒒
𝒅𝒕
=
𝒅
𝒅𝒕
𝟓𝒕 𝐬𝐢𝐧 𝟒𝝅𝒕 = 𝟓 𝐬𝐢𝐧 𝟒𝝅𝒕 + 𝟐𝟎𝝅𝒕 𝐜𝐨𝐬 𝟒𝝅𝒕 𝒎𝑨
At 𝑡 = 0.5 𝑠.
𝑖 = 31.42 𝑚𝐴
Example 2. 4 :
Determine the total charge entering a terminal between 𝑡 = 1 𝑠 and 𝑡 = 2 𝑠 if the
current passing the terminal is 𝑖 = 3𝑡2
− 𝑡 𝐴.
Solution:
𝒒 = 𝒊𝒅𝒕
𝟐
𝒕=𝟏
= 𝟑𝒕 𝟐
− 𝒕 𝒅𝒕 = 𝒕 𝟑
−
𝒕 𝟐
𝟐 𝟏
𝟐
= 𝟖 − 𝟐
𝟐
𝟏
− 𝟏 −
𝟏
𝟐
= 𝟓. 𝟓 𝑪
2.2.3 VOLTAGE
Charge moving in an electric circuit gives rise to a current, as stated in the preceding
section. Naturally, it must take some work, or energy, for the charge to move between two
points in a circuit, say, from point a to point b. The total work per unit charge associated
with the motion of charge between two points is called voltage. Thus, the units of voltage
are those of energy per unit charge; they have been called volts in honor of Alessandro
Volta.
We write:
𝑣 𝑎𝑏 =
𝑑𝑤
𝑑𝑞
(2. 4)
where 𝑤 is energy in joule (J) and 𝑞 is charge in coulombs (C).
1 volt = 1 joule/coulomb = 1 newton meter/coulomb
Like electric current, a constant voltage is called a DC voltage and it is represented
by V, whereas a sinusoidal time-varying voltage is called an AC voltage and it is represented
by 𝑣. The electromotive force (e.m.f) provided by a source of energy such as battery (DC
voltage) or an electric generator (AC voltage) is measured in volts.
Voltage (or potential difference) is the energy required to move charge from one
point to the other, measured in volts (V). Voltage is denoted by the letter v or V.

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2.2.4 POWER AND ENERGY
Although current and voltage are the two basic variables in an electric circuit, they
are not sufficient by themselves. For practical purposes, we need to know how much power
an electric device can handle. We also know that when we pay our bills to the electric utility
companies, we are paying for the electric energy consumed over a certain period of time.
Thus power and energy calculations are important in circuit analysis.
We write this relationship as:
𝑝 =
𝑑𝑤
𝑑𝑡
(2. 5)
Where p is power in watts (W), w is energy in joules (J), and t is time in seconds (s).
From voltage and current equations, it follows that:
𝑝 =
𝑑𝑤
𝑑𝑡
=
𝑑𝑤
𝑑𝑞
.
𝑑𝑞
𝑑𝑡
= 𝑣𝑖 (2. 6)
Or
𝑝 = 𝑣𝑖 (2. 7)
The power 𝑝 in this equation is a time-varying quantity and is called the
instantaneous power. Thus, the power absorbed or supplied by an element is the product of
the voltage across the element and the current through it.
It is important to realize that, just like voltage, power is a signed quantity, and that it
is necessary to make a distinction between positive and negative power. The electrical
engineering community uniformly adopts the passive sign convention, which simply states
that the power dissipated by a load is a positive quantity (or, conversely, that the power
generated by a source is a positive quantity).
By the passive sign convention, current enters through the positive polarity of the
voltage. In this case, 𝑝 = +𝑣𝑖 𝑜𝑟 𝑣𝑖 > 0 implies that the element is absorbing power.
However, if 𝑝 = −𝑣𝑖 𝑜𝑟 𝑣𝑖 < 0, as in Fig. 2.3, the element is releasing or supplying power.
.
Power is the time rate of expending or absorbing energy, measured in
watts (W). Power, is denoted by the letter p or P.
Passive sign convention is satisfied when the current enters through the positive
terminal of an element and 𝒑 = +𝒗𝒊. If the current enters through the negative
terminal, 𝒑 = −𝒗𝒊

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Figure 2. 2: The passive sign convention.
The law of conservation of energy must be obeyed in any electric circuit. For this
reason the algebraic sum of power in a circuit, at any instant of time, must be zero:
𝑝 = 0 (2. 8)
This again confirms the fact that the total power supplied to the circuit must balance
the total power absorbed.
From Eq.(2. 6), the energy absorbed or supplied by an element from 𝑡0to time 𝑡 is :
𝑤 = 𝑝 𝑑𝑡
𝑡
𝑡0
= 𝑣𝑖 𝑑𝑡
𝑡
𝑡0
(2. 9)
Although the unit of energy is the joule, when dealing large amounts of energy, the
unit used is the kilowatt hour (kWh) where 1 Wh=3600 J.
Example 2. 5:
A source e.m.f. of 5 V supplies a current of 3A for 10 minutes. How much energy is
provided in this time?
Solution:
𝑊 = 𝑉𝐼𝑡 = 5 × 3 × 10 × 60 = 9 𝑘𝐽
Example 2. 6:
An electric heater consumes 1.8Mj when connected to a 250 V supply for 30
minutes. Find the power rating of the heater and the current taken from the supply.
Energy is the capacity to do work, measured in joules (J).

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Solution:
𝑃 =
𝑊
𝑡
=
1.8 × 106
30 × 60
= 1000 𝑊
I.e. power rating of heater = 1kW.
𝑃 = 𝑉𝐼
Thus
𝐼 =
𝑃
𝑉
=
1000
250
= 4𝐴
Hence the current taken from the supply is 4A.
Example 2. 7:
An energy sources forces a constant current of 2A for 10 s to flow through a
lightbulb. If 2.3 kJ is given off in the form of light and heat energy, calculate the voltage drop
across the bulb.
Solution:
The total charge is:
∆𝑞 = 𝑖∆𝑡 = 2 × 10 = 20 𝐶
The total voltage drop is:
𝑣 =
∆𝑤
∆𝑞
=
2.3 × 103
20
= 115 𝑉
Example 2. 8:
Find the power delivered to an element at 𝑡 = 3 𝑚𝑠 if the current entering its
positive terminals is:
𝑖 = 5 cos60𝜋𝑡 𝐴
And the voltage is:
(a) 𝑣 = 3𝑖,
(b) 𝑣 = 3
di
dt
.
Solution:
(a) The voltage is 𝑣 = 3𝑖 = 15 cos 60𝜋𝑡 𝐴; hence, the power is :
𝑝 = 𝑣𝑖 = 75 cos2
60𝜋𝑡 𝑊
At 𝑡 = 3 𝑚𝑠,
𝑝 = 75 cos2
60𝜋𝑡 × 3 × 10−3
= 53.48 𝑊

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(b) We find the voltage and the power as
𝑣 = 3
𝑑𝑖
𝑑𝑡
= 3 −60𝜋 5 sin60𝜋𝑡 = −900𝜋 sin 60𝜋𝑡 𝑉
𝑝 = 𝑣𝑖 = −4500𝜋 sin60𝜋𝑡 cos 60𝜋𝑡 𝑊
At 𝑡 = 3 𝑚𝑠,
𝑝 = −4500𝜋 sin0.18𝜋 cos 0.18𝜋 = −6.396𝑊
2.2.5 CIRCUIT ELEMENTS
As we discussed in the Introduction, an element is the basic buildings block of a
circuit. An electric circuit is simply an interconnection of elements there are two types of
elements found in electric circuits: passive elements and active elements. An active element
is capable of generating energy while a passive element is not. Our aim in this section is to
gain familiarity with some important passive and active elements.
2.2.5.1 Passive elements (loads)
A load generally refers to a component or a piece of equipment to the output of an
electric circuit. In its fundamental form, the load is represented by one or a combination of
the following circuit elements:
1. Resistor (R).
2. Inductor (L).
3. Capacitor (C).
A load can either be resistive, inductive or capacitive nature or a blend of them. For
example, a light bulb is a purely resistive load whereas a transformer is both inductive and
resistive.
2.2.5.2 Active elements
The most important active elements are voltage or current sources that generally
deliver power to the circuit connected to them. There are two kinds of sources: independent
and dependent sources.
An ideal independent source is an active element that provides a specified voltage or
current that is completely independent of other circuit variables.
An ideal dependent (or controlled) source is an active element in which the source
quantity is controlled by another voltage or current.
It should be noted that an ideal voltage source (dependent or independent) will
produce any current required to ensure that the terminal voltage is as stated; whereas an
ideal current source will produce the necessary voltage to ensure the stated current flow.
Table 2.1 shows the basic circuit elements along with their symbols and schematics
used in an electric circuit.

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Circuit Element Symbol Schematic
Resistor R
Inductor L
Capacitor C
Independent voltage
source
V
Independent current
source
I
Dependent voltage
source
V
Dependent current
source
I
Table 2. 1 : Common circuit elements and their representation in an electric circuit.
2.2.5.3 Sign convention
It is common to think of current as the flow of electrons. However, the standard
convention is to take the flow of protons to determine the direction of the current.
Figure 2. 3: Sign convention.
In a given circuit, the current direction depends on the polarity of the source
voltage. Current always flow from positive (high potential) side to the negative (low
potential) side of the source as shown in the schematic diagram of Figure 2.4(a) where Vs is

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the source voltage, VL is the voltage across the load and I is the loop current flowing in the
clockwise direction.
Please observe that the voltage polarity and current direction in a sink is opposite to
that of the source.
In Source current leaves from the positive terminal
In Load (Sink) current enters from the positive terminal
2.2.6 RESISTOR
Materials in general have a characteristic behavior of resisting the flow of electric
charge. This physical property, or ability to resist current, is known as resistance and is
represented by the symbol R.
The resistance of any material with a uniform cross sectional area A depends on A
and its length 𝑙, as shown in Fig. 2.3. In mathematical form:
𝑅 = 𝜌
𝑙
𝐴
(2. 10)
Where 𝜌 is known as the resistivity of the material in ohm-meters.
To describe the resistance of a resistor and hence its characteristics, it is important
to define the Ohm’s law.
2.2.6.1 Ohm’s Law
Mathematically
𝑉 𝛼 𝐼
𝑉 = 𝑅𝐼 (2. 11)
The resistance R of an element denotes its ability to resist the flow of electric
current, it is measured in ohms (𝛀). The circuit element used to model the current-
resisting behavior of a material is the resistor.
Ohm’s law states that the voltage v across a resistor is directly proportional to the
current i flowing through the resistor.

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𝐼 =
𝑉
𝑅
(2. 12)
Where the constant of proportionality R is called the resistance or electrical
resistance, measured in ohms (Ω).
Example 2. 9:
Find R if the voltage V and current I are equal to 10 V and 5 A respectively.
Solution:
Using Ohm’s law
𝑅 =
𝑉
𝐼
=
10
5
= 2 𝐴
2.2.6.2 Conductance
A useful quantity in circuit analysis is the exact opposite of resistance R, Known as
conductance and denoted by G:
𝐺 =
1
𝑅
=
𝑖
𝑣
(2. 13)
Where G is measured in Siemens (𝑆) and sometimes also represented by the unit
mho (ohm spelled back-ward), with the symbol (Ω)(upside-down omega).
The power disspated by a resistor can also be expressed in terms of R using previous
equations:
A short circuit is a circuit element with resistance approaching zero.
An open circuit is a circuit element with resistance approaching infinity.
.
Conductance is the ability of an element to conduct electric current; it is measured
in mhos or Siemens (S).

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𝑝 = 𝑣𝑖 = 𝑖2
𝑅 =
𝑣2
𝑅
(2. 14)
The power dissipated by a resistor may also be expressed in terms of G as:
𝑝 = 𝑣𝑖 = 𝑣2
𝐺 =
𝑖2
𝐺
(2. 15)
We should note two things from the two previous equations:
1. The power dissipated in a resistor is a nonlinear function of either current or
voltage.
2. Since R and G are positive quantities, the power dissipated in a resistor is always
positive. Thus, a resistor always absorbs power from the circuit.
Example 2. 10:
An electric iron draws 2 A at 120 V. find its resistance.
Solution:
From Ohm’s law
𝑅 =
𝑉
𝐼
=
120
2
= 60 Ω
Example 2. 11:
A light bulb draws 0.5 A current at an input voltage of 230 V. determine the
resistance of the filament and also the power dissipated.
Solution:
From Ohm’s law
𝑅 =
𝑉
𝐼
=
230
0.5
= 460 Ω
Since a bulb is purely resistive load, therefore all the power is dissipated in the form
of heat. This can be calculated using any of three power relationships shown above
𝑃 = 𝑉𝐼 = 230 × 0.5 = 115 𝑊
𝑃 = 𝐼2
𝑅 = 0.5 2
× 460 = 115 𝑊
𝑃 =
𝑉2
𝑅
=
(230)2
460
= 115 𝑊
Example 2. 12:
In the circuit shown in Fig. 2.5, calculate the current I, the conductance G, and the
power p.

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Figure 2. 4: Electric circuit of example 2.12.
Solution:
The voltage across the resistor is the same as the source voltage (30V) because the
resistor and the voltage source are connected to the same pair of terminals. Hence, the
current is:
𝑖 =
𝑉
𝑅
=
30
5 × 103
= 6 𝑚𝐴
The conductance is
𝐺 =
1
𝑅
=
1
5 × 103
= 0.2 𝑚𝑆
We can calculate the power in various ways:
𝑃 = 𝑉𝐼 = 30 6 × 10−3
= 180 𝑚𝑊
𝑃 = 𝐼2
𝑅 = 6 × 10−3 2
5 × 103
= 180 𝑚𝑊
𝑃 =
𝑉2
𝑅
=
(30)2
5 × 103
= 180 𝑚𝑊
Example 2. 13:
For the circuit shown in Fig 2.6, calculate the voltage v, the conductance G, and the
power p.
Figure 2. 5 : Electric circuit of example 2.13.

17
Solution:
Answer: 20 V, 100µS, 40mW.
Example 2. 14
A voltage source of 20 sin 𝜋𝑡 𝑉 is connected across a 5 𝑘Ω resistor. Find the current
through the resistor and the power dissipated.
Solution:
𝑖 =
𝑣
𝑅
=
20 sin 𝜋𝑡
5 × 103
= 4 sin 𝜋𝑡 𝑚𝐴
Hence,
𝑝 = 𝑣𝑖 = 80 sin2
𝜋𝑡 𝑚𝑊
2.2.7 CAPACITOR
A capacitor is a passive element designed to store energy in its electric field. Besides
resistors, capacitors are the most common electrical components. Capacitors are used
extensively in electronics, communications, computers, and power systems. For example,
they are used in the tuning circuits of radio receivers and as dynamic memory elements in
computer systems.
In many practical applications, the plates may be aluminum foil while the dielectric
may be air, ceramic, paper, or mica.
When a voltage source v is connected to the capacitor, the source deposits a
positive charge q on one plate and a negative charge −q on the other. The capacitor is said to
store the electric charge.
The amount of charge stored, represented by q, is directly proportional to the
applied voltage v so that:
𝑞 = 𝐶𝑣 (2. 16)
where C, the constant of proportionality, is known as the capacitance of the
capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael
Faraday.
The relationship between voltage and current for a capacitor is governed by the
following equation:
A capacitor consists of two conducting plates separated by an insulator (or
dielectric).

18
𝑖 = 𝐶
𝑑𝑣
𝑑𝑡
(2. 17)
𝑣 =
1
𝐶
𝑖𝑑𝑡 + 𝑣(0)
𝑡
0
(2. 18)
where C is the capacitance measured in Farads (F) and v(0) is the initial voltage or
initial charge stored in the capacitor.
When v = V (constant DC voltage),
𝑑𝑣
𝑑𝑡
= 0 and𝑖 = 0. Hence a capacitor acts as an
open circuit to DC.
Previous equation shows that capacitor voltage depends on the past history of the
capacitor current. Hence, the capacitor has memory—a property that is often exploited.
The instantaneous power delivered to the capacitor is:
𝑝 = 𝑣𝑖 = 𝑐𝑣
𝑑𝑣
𝑑𝑡
(2. 19)
𝑡
−∞
= 𝐶 𝑣
𝑑𝑣
𝑑𝑡
𝑑𝑡
𝑡
−∞
= 𝐶 𝑣 𝑑𝑣
𝑡
−∞
=
1
2
𝐶𝑣2
𝑡=−∞
𝑡
(2. 20)
We note that 𝑣 −∞ = 0, because the capacitor was unchanged at 𝑡 = −∞. Thus,
𝑤 =
1
2
𝐶𝑣2 (2. 21)
𝑤 =
𝑞2
2𝐶
(2. 22)
This energy is stored in the electric field of the capacitor which is supplied back to
the circuit when the actual source is removed.
Example 2. 15:
(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the
energy stored in the capacitor.
Solution:
(a) since 𝑞 = 𝐶𝑣,
𝑞 = 3 × 10−12
× 20 = 60 𝑝𝐶
(b) The energy stored is
𝑤 =
1
2
𝐶𝑣2
=
1
2
× 3 × 10−12
× 400 = 600 𝑝𝐽
Example 2. 16:
The voltage across a 5*F capacitor is:

19
𝑣 𝑡 = 10 cos6000𝑡 𝑉
Calculate the current through it.
Solution:
By definition, the current is:
𝑖 𝑡 = 𝐶
𝑑𝑣
𝑑𝑡
= 5 × 10−6
𝑑
𝑑𝑡
10 cos 6000𝑡
𝑖 𝑡 = −5 × 10−6
× 6000 × 10 sin6000𝑡 = −0.3 sin6000𝑡 𝐴
Example 2. 17:
Determine the voltage across a 2*F capacitor if the current through it is:
𝑖 𝑡 = 6𝑒−3000𝑡
𝑚𝐴
Assume that the initial capacitor voltage is zero.
Solution:
Since
𝑣 =
1
𝐶
𝑖𝑑𝑡 + 𝑣(0)
𝑡
0
And
𝑣 0 = 0
𝑣 =
1
2 × 10−6
6𝑒−3000𝑡
𝑑𝑡
𝑡
0
. 10−3
𝑣 = 1 − 𝑒−3000𝑡
𝑉
Example 2. 18:
Determine the current through a 200-μF capacitor whose voltage 𝑣(𝑡)is shown in
Fig. 2.7.
Figure 2. 6: For example 2.18.

20
Solution:
The voltage waveform can be described mathematically as
𝑣 𝑡 =
50𝑡 𝑉 0 < 𝑡 < 1
100 − 50𝑡 𝑉 1 < 𝑡 < 3
−200 + 50𝑡 𝑉 3 < 𝑡 < 4
0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒
The current waveform is a shown in Fig .2.8 and can be described mathematically as
𝑖 𝑡 =
10 𝑚𝐴 0 < 𝑡 < 1
−10 𝑚𝐴 1 < 𝑡 < 3
10𝑚𝐴 3 < 𝑡 < 4
0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒
2.2.8 INDUCTOR
An inductor is a passive element designed to store energy in its magnetic field.
Inductors find numerous applications in electronic and power systems. They are used in
power supplies, transformers, radios, TVs, radars, and electric motors.
Any conductor of electric current has inductive properties and may be regarded as
an inductor. But in order to enhance the inductive effect, a practical inductor is usually
formed into a cylindrical coil of a ferromagnetic material with many turns of conducting
wire.
In an inductor, the relationship between voltage and current is given by the
following differential equation:
An inductor consists of a coil of conducting wire.

21
𝑣 = 𝐿
𝑑𝑖
𝑑𝑡
(2. 23)
Or
𝑖 =
1
𝐿
𝑣𝑑𝑡
𝑡
0
+ 𝑖(0) (2. 24)
where L is the constant of proportionality called the inductance of the inductor. The
unit of inductance is the henry (H), named in honor of the American inventor Joseph Henry,
and 𝑖(0) is the initial current stored in the magnetic field of the inductor.
When i = I (constant DC current),
𝑑𝑖
𝑑𝑡
= 0, v = 0. Hence an inductor acts as a short
circuit to DC.
The inductor is designed to store energy in its magnetic field. The energy stored can
be obtained from previous equations. The power delivered to the inductor is:
𝑝 = 𝑣𝑖 = 𝐿
𝑑𝑖
𝑑𝑡
𝑖 (2. 25)
The energy stored is:
𝑡
−∞
= 𝐿
𝑑𝑖
𝑑𝑡
𝑖 𝑑𝑡
𝑡
−∞
(2. 26)
𝑤 = 𝐿 𝑖 𝑑𝑡
𝑡
−∞
=
1
2
𝐿𝑖2
𝑡 −
1
2
𝐿𝑖2
−∞ (2. 27)
Since, 𝑖 −∞ = 0
𝑤 =
1
2
𝐿𝑖2 (2. 28)
This energy is stored in the magnetic field of the inductor which can be supplied
back to the circuit when the actual source is removed.
Inductance is the property whereby an inductor exhibits opposition to the
change of current flowing through it, measured in henrys (H).

22
Example 2. 19:
The current through a 0.1 H inductors is 𝑖 𝑡 = 10𝑡𝑒−5𝑡
𝐴. Find the voltage across
the inductor and the energy stored in it.
Solution:
Since,𝑣 = 𝐿
𝑑𝑖
𝑠𝑡
and 𝐿 = 0.1𝐻
𝑣 = 0.1
𝑑
𝑑𝑡
10𝑡𝑒−5𝑡
= 𝑒−5𝑡
+ 𝑡 −5 𝑒−5𝑡
= 𝑒−5𝑡
1 − 5𝑡 𝑉
The energy stored is:
𝑤 =
1
2
𝐿𝑖2
=
1
2
0.1 100𝑡2
𝑒−10𝑡
= 5𝑡2
𝑒−10𝑡
𝐽
Example 2. 20:
Find the current through a 5 H inductor if the voltage across it is:
𝑣 𝑡 = 30𝑡2
𝑡 > 0
0 𝑡 < 0
Also find the energy stored within 0 < 𝑡 < 5 𝑠.
Solution:
Since 𝑖 =
1
𝐿
𝑣 𝑡 𝑑𝑡 + 𝑖(𝑡0)
𝑡
𝑡0
and 𝐿 = 5 𝐻,
𝑖 =
1
𝐿
30𝑡2
𝑑𝑡 + 0 = 2𝑡3
𝐴
𝑡
0
The power is:
𝑝 = 𝑣𝑖 = 60𝑡5
The energy stored is then:
𝑤 = 𝑝 𝑑𝑡 = 60𝑡5
𝑑𝑡
5
0
= 156.25 𝑘𝐽
Alternatively, we can obtain the energy stored by writing:
𝑤0
5
=
1
2
𝐿𝑖2
5 −
1
2
𝐿𝑖 0 =
1
2
5 2 × 53
− 0 = 156.25 𝐾𝐽

23
2.3 CIRCUIT THEOREMS
2.3.1 INTRODUCTION
This section outlines the most commonly used laws and theorems that are required
to analyze and solve electric circuits. Relationships between various laws and equation
writing techniques by inspection are also provided. Several examples are shown
demonstrating various aspects of the laws. In addition, situations are presented where it is
not possible to directly apply the concepts and potential remedies are provided.
2.3.2 DEFINITIONS AND TERMINOLOGIES
In the following, various definitions and terminologies frequently used in circuit
analysis are outlined.
Electric Network: a connection of various circuit elements can be termed as an
electric network. The circuit diagram shown in Figure 2.9 is an electric network.
In other words, a branch represents any two-terminal element. The circuit in Fig. 2.9
has five branches, namely, the 10-V voltage source, the 2-A current source, and the three
resistors.
A node is usually indicated by a dot in a circuit. If a short circuit (a connecting wire)
connects two nodes, the two nodes constitute a single node. The circuit in Fig. 2.9 has three
nodes a, b, and c. Notice that the three points that form node b are connected by perfectly
conducting wires and therefore constitute a single point. The same is true of the four points
forming node c. We demonstrate that the circuit in Fig. 2.9-(a) has only three nodes by
redrawing the circuit in Fig. 2.9-(b). The two circuits in Figs. 2.9-(a) and 2.9-(b) are identical.
However, for the sake of clarity, nodes b and c are spread out with perfect conductors as in
Fig. 2.9-(a).
A branch represents a single element such as a voltage source or a resistor.
A node is the point of connection between two or more branches.
A loop is any closed path in a circuit.
Two or more elements are in series if they are cascaded or connected sequentially
and consequently carry the same current.
Two or more elements are in parallel if they are connected to the same two nodes
and consequently have the same voltage across them.

24
(a): Nodes, branches, and loops.
(b): The three-node circuit of Fig. 2.9-(a) is redrawn.
Figure 2. 8: An electric network showing nodes, branches, elements and loop.
A loop is a closed path formed by starting at a node, passing through a set of nodes,
and returning to the starting node without passing through any node more than once.
A loop is said to be independent if it contains a branch which is not in any other
loop. Independent loops or paths result in independent sets of equations. For example, the
closed path abca containing the 2- resistor in Fig. 2.9-(b) is a loop. Another loop is the
closed path bcb containing the 3- resistor and the current source. Although one can
identify six loops in Fig. 2.9-(b), only three of them are independent. A network with b
branches, n nodes, and l independent loops will satisfy the fundamental theorem of network
topology:
𝑏 = 𝑙 + 𝑛 − 1 (2. 29)

25
Example 2. 21:
Determine the number of branches and nodes in the circuit shown in Fig.2.10.
Identify which elements are in series and which are in parallel.
Solution:
Since there are four elements in the circuit, the circuit has four branches 10 V, 5,
6, and 2 A. The circuit has three nodes as identified in Fig. 2.11. The 5- resistor is in series
with the 10-V voltage source because the same current would flow in both. The 6- resistor
is in parallel with the 2-A current source because both are connected to the same nodes 2
and 3.
Figure 2. 10: The three nodes in the circuit of Fig. 2.10.
Example 2. 22:
How many branches and nodes does the circuit in Fig. 2.12 have? Identify the
elements that are in series and in parallel.

26
Solution:
Five branches and three nodes are identified in Fig. 2.13. The 1- and 2- resistors
are in parallel. The 4- resistor and 10-V source are also in parallel.
Figure 2. 12: Answer for example 2.22.
2.3.3 OHM’S LAW
See section ‎2.2.6.1.
2.3.4 KIRCHHOFF’S LAWS
Arguably the most common and useful set of laws for solving electric circuits are the
Kirchhoff’s voltage and current laws. Several other useful relationships can be derived based
on these laws. These laws are formally known as Kirchhoff’s current law (KCL) and
Kirchhoff’s voltage law (KVL).
Mathematically, KCL implies that:
𝑖 𝑛
𝑁
𝑛=1
= 0 (2. 30)
KCL is based on the law of conservation of charge, while KVL is based on the
principle of conservation of energy.
Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a
node (or a closed boundary) is zero. In other words; the sum of the currents
entering a node is equal to the sum of the currents leaving the node.
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a
closed path (or loop) is zero. In other words; sum of voltage drops = sum of voltage
rises

27
where N is the number of branches connected to the node and 𝑖 𝑛 is the nth current
entering (or leaving) the node. By this law, currents entering a node may be regarded as
positive, while currents leaving the node may be taken as negative or vice versa. From KCL
and (2. 30) we can write:
𝑖 𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 = 𝑖 𝑙𝑖𝑣𝑖𝑛𝑔 (2. 31)
For example consider the node in Fig. 2.14. Applying KCL gives
𝑖1 + −𝑖2 + 𝑖3 + 𝑖4 + −𝑖5 = 0
Thus,
𝑖1 + 𝑖3 + 𝑖4 = 𝑖2 + 𝑖5
Figure 2. 13: Currents at a node illustrating KCL.
Expressed mathematically, KVL states that:
𝑣 𝑚
𝑀
𝑚=1
= 0 (2. 32)
Where M is the number of voltages in the loop (or the number of branches in the
loop) and 𝑣 𝑚 is the 𝑚𝑡𝑕 voltage.
To illustrate KVL, consider the circuit in Fig. 2.15. The sign on each voltage is the
polarity of the terminal encountered first as we travel around the loop. We can start with
any branch and go around the loop either clockwise or counterclockwise. Thus, KVL yields:
−𝑣1 + 𝑣2 + 𝑣3 − 𝑣4 + 𝑣5 = 0
Rearranging terms gives:
𝑣1 + 𝑣4 = 𝑣2 + 𝑣3 + 𝑣5

28
Figure 2. 14: A single-loop circuit illustrating KVL.
Example 2. 23:
For the circuit in Fig. 2.16-(a), find voltages 𝑣1 and 𝑣2.
(a) (b)
Solution:
To find 𝑣1 and 𝑣2, we apply Ohm’s law and Kirchhoff’s voltage law. Assume that
current 𝑖 flows through the loop as shown in Fig. 2.16-(b).From Ohm’s law:
𝑣1 = 2𝑖 (1)
𝑣2 = −3𝑖 (2)
Applying KVL around the loop gives:
−20 + 𝑣1 − 𝑣2 = 0 (3)
Substituting Eq. (1) and Eq. (2) into Eq. (32), we obtain:
−20 + 2𝑖 − 3𝑖 = 0 → 𝑖 = 4 𝐴 (4)
Substituting 𝑖 in Eq. (1) and (2) finally gives:
𝑣1 = 8 𝑉 (5)
𝑣2 = −12 𝑉 (6)

29
Example 2. 24:
Find the currents and voltages in the circuit shown in Fig. 2.17(a).
(a)
(b)
Solution:
We apply Ohm’s law and Kirchhoff’s laws. By Ohm’s law :
𝑣1 = 8𝑖1 (1)
𝑣2 = 3𝑖2 (2)
𝑣3 = 6𝑖3 (3)
Since the voltage and current of each resistor are related by Ohm’s law as shown, we
are really looking for three things: 𝑣1, 𝑣2, 𝑣3 or 𝑖1, 𝑖2, 𝑖3 . At nod a, KCL gives:
𝑖1 − 𝑖2 − 𝑖3 = 0 (4)
Applying KVL to loop 1 as in Fig. 2.17(b):
−30 + 𝑣1 + 𝑣2 = 0 (5)
We express this in terms of i1 and i2:
−30 + 8𝑖1 + 3𝑖2 = 0 (6)
Or

30
𝑖1 =
30 + 3𝑖2
8
(7)
Applying KVL to loop 2:
−𝑣2 + 𝑣3 = 0 → 𝑣2 = 𝑣3 (8)
We can write:
6𝑖3 = 3𝑖2 → 𝑖3 =
𝑖2
2
(9)
Substituting Eqs. (7) and (9) into (4) gives :
30 − 3𝑖2
8
− 𝑖2 −
𝑖2
2
= 0 → 𝑖2 = 2 𝐴 (10)
Thus we obtain:
𝑖1 = 3𝐴, 𝑖3 = 1𝐴, 𝑣1 = 24𝑉, 𝑣2 = 6𝑉, 𝑣3 = 6𝑉 (11)
2.3.4.1 Series resistors and voltage division
Consider the single-loop circuit of Fig. 2.18. The two resistors are in series, since the
same current 𝑖 flows in both of them. Applying Ohm’s law to each of the resistors, we
obtain:
𝑣1 = 𝑖𝑅1 (2. 33)
𝑣2 = 𝑖𝑅2 (2. 34)
If we apply KVL to the loop (moving in the clockwise direction), we have
−𝑣 + 𝑣1 + 𝑣2 = 0 (2. 35)
Figure 2. 17: A single-loop circuit with two resistors in series.
Combining Eqs. (2. 33),(2. 34)and (2. 35), we get :
𝑣 = 𝑣1 + 𝑣2 = 𝑖 𝑅1 + 𝑅2 (2. 36)

31
Or
𝑖 =
𝑣
𝑅1 + 𝑅2
(2. 37)
Notice that Eq.(2. 36) can be written as:
𝑣 = 𝑖𝑅 𝑒𝑞 (2. 38)
Implying that the two resistors can be replaced by an equivalent resistor that is:
𝑅 𝑒𝑞 = 𝑅1 + 𝑅2 (2. 39)
Thus, Fig. 2.18 can be replaced by the equivalent circuit in Fig. 2.19. An equivalent circuit
such as the one in Fig. 2.19 is useful in simplifying the analysis of a circuit.
Figure 2. 18: Equivalent circuit of the Fig. 2.18 circuit.
For N resistors in series then:
𝑅 𝑒𝑞 = 𝑅1 + 𝑅2 + ⋯ + 𝑅 𝑁 = 𝑅 𝑛
𝑁
𝑛=1
(2. 40)
The voltage across each resistor of Fig. 2.18 is given by:
𝑣1 =
𝑅1
𝑅1 + 𝑅2
𝑣 (2. 41)
𝑣2 =
𝑅2
𝑅1 + 𝑅2
𝑣 (2. 42)
Notice that the source voltage v is divided among the resistors in direct proportion
to their resistances; the larger the resistance, the larger the voltage drop. This is called the
principle of voltage division, and the circuit in Fig. 2.29 is called a voltage divider. In general,
if a voltage divider has N resistors (R1,R2, . . . , RN) in series with the source voltage v, the
nth resistor (Rn) will have a voltage drop of.
𝑣 𝑛 =
𝑅 𝑛
𝑅1 + 𝑅2 + ⋯ + 𝑅 𝑛
𝑣 (2. 43)

32
2.3.4.2 Parallel resistors and current division
In the same manner, we can get the equivalent resistance of resistors in parallel by
applying KVL. Mathematically we can write:
1
𝑅 𝑒𝑞
=
1
𝑅1
+
1
𝑅2
+ ⋯ +
1
𝑅 𝑁
=
1
𝑅 𝑛
𝑁
𝑛=1
(2. 44)
𝑖 𝑛 =
𝐺 𝑛
𝐺1 + 𝐺2 + ⋯ + 𝐺 𝑛
𝑖 (2. 45)
This second equation introduces the principal of current division and the current
divider.
2.3.4.3 Series and parallel capacitors
The equivalent capacitance of series-connected capacitors is the reciprocal of the
sum of the reciprocals of the individual capacitances. Mathematically (Applying KCL) we can
write:
1
𝐶𝑒𝑞
=
1
𝐶1
+
1
𝐶2
+ ⋯ +
1
𝐶 𝑁
=
1
𝐶 𝑛
𝑁
𝑛=1
(2. 46)
The equivalent capacitance of N parallel-connected capacitors is the sum of the
individual capacitances. Mathematically (Applying KVL):
𝐶𝑒𝑞 = 𝐶1 + 𝐶2 + ⋯ + 𝐶 𝑁 = 𝐶 𝑛
𝑁
𝑛=1
(2. 47)
2.3.4.4 Series and parallel inductors
The equivalent inductance of series-connected inductors is the sum of the individual
inductances.
𝐿 𝑒𝑞 = 𝐿1 + 𝐿2 + ⋯ + 𝐿 𝑁 = 𝐿 𝑛
𝑁
𝑛=1
(2. 48)
The equivalent inductance of parallel inductors is the reciprocal of the sum of the
reciprocals of the individual inductances.
1
𝐿 𝑒𝑞
=
1
𝐿1
+
1
𝐿2
+ ⋯ +
1
𝐿 𝑁
=
1
𝐿 𝑛
𝑁
𝑛=1
(2. 49)
Example 2. 25:
Fine the equivalent resistance for the circuit given in figure 2.20.

33
Figure 2. 19: Circuit for this example.
Solution:
Figure 2. 20: Equivalent circuit for this example.
Hence, the equivalent resistance after steps (a) and (b) is given by:
𝑅 𝑒𝑞 = 4 + 2.4 + 8 = 14.4 Ω
Example 2. 26:
Fine the equivalent resistance for the circuit given in figure 2.22.
Solution:
𝑅 𝑒𝑞 = 6 Ω

34
Example 2. 27:
Find the equivalent capacitance seen between terminals a and b of the circuit in
Figure 2.23.
.
Solution:
The 20-μF and 5-μF capacitors are in series; their equivalent capacitance is
20 × 5
20 + 5
= 4 𝜇F
This 4-μF capacitor is in parallel with the 6-μF and 20-μF capacitors; their combined
capacitance is
4 + 6 + 20 = 30 𝜇F
This 30-μF capacitor is in series with the 60-μF capacitor. Hence, the equivalent
capacitance for the entire circuit is
𝐶𝑒𝑞 =
30 × 60
30 + 60
= 20 𝜇F
Example 2. 28:
Calculate the equivalent inductance for the inductive ladder network in Figure 2.24.
Solution:
25 mH.

35
2.3.5 ELECTRIC CIRCUITS ANALYSIS
The KVL, KCL and Ohm’s law are the primary tools to analyze DC electric circuits. The
term nodal analysis is generally used when analyzing an electric circuit with KCL whereas
loop or mesh analysis is designated for problem solving using KVL.
The mesh and nodal analysis methods outlined below are quite systematic and
usually provides the solution to a given problem. However, they are fairly computational and
an alternative straightforward solution may exist using circuit reduction techniques such as
series/parallel combination of resistors and/or VDR/CDR methods.
2.3.5.1 Mesh Analysis
The mesh analysis technique consists of the following steps
1. Transform all currents sources to voltage sources, if possible (see Section 3.8).
2. Identify and assign a current to each mesh of the network (preferably in the
same direction).
3. Write mesh equations using KVL and simplify them.
4. Solve the simultaneous system of equations.
5. Number of equations is equal to number of meshes in the network.
2.3.5.2 Nodal Analysis
The following steps describe the nodal analysis method
1. Transform all voltage sources to current sources, if possible (see Section 3.8).
2. Identify and assign an arbitrary voltage to each node including the reference
node in the network assuming all currents leaving the node. The reference node
is normally assumed to be at zero potential. Write nodal equations using KCL
and simplify them.
3. Solve the simultaneous system of equations.
4. Number of equations is N − 1 where N is the number of nodes in the network
including the reference node.
Example 2. 29:
Figure 2. 24: For this example.
Determine the voltages at the nodes in Figure. 2.25.

36
Solution:
At node 1,
3 = 𝑖1 + 𝑖 𝑥 ⟹ 3 =
𝑣1 − 𝑣3
4
+
𝑣1 − 𝑣2
2
Multiplying by 4 and rearranging terms, we get
3𝑣1 − 2𝑣2 − 𝑣3 = 12
At node 2,
𝑖 𝑥 = 𝑖2 + 𝑖3 ⟹
𝑣1 − 𝑣2
2
=
𝑣2 − 𝑣3
8
+
𝑣2 − 0
4
Multiplying by 8 and rearranging terms, we get
−4𝑣1 + 7𝑣2 − 𝑣3 = 0
At node 3,
2𝑖 𝑥 = 𝑖1 + 𝑖2 ⟹
𝑣1 − 𝑣3
4
+
𝑣2 − 𝑣3
8
=
2(𝑣3 − 𝑣2)
4
Multiplying by 8, rearranging terms, and dividing by 3, we get
2𝑣1 − 3𝑣2 + 𝑣3 = 0
Figure 2. 25: Circuit for analysis for this example
We have three simultaneous equations to solve to get the node voltages 𝑣1, 𝑣2, and
𝑣3. We shall solve the equations in two ways; using the elimination technique or using
Cramer’s rule.
Let’s use the second method
3 −2 −1
−4 7 −1
2 −3 1
𝑣1
𝑣2
𝑣3
=
12
0
0
From this, we obtain 𝑣1 =
∆1
∆
, 𝑣2 =
∆2
∆
, 𝑣3 =
∆3
∆

37
Where ∆, ∆1, ∆2 𝑎𝑛𝑑 ∆3 are the determinants to be calculated.
𝑣1 =
∆1
∆
=
48
10
= 4.8 V, 𝑣2 =
∆2
∆
=
24
10
= 2.4 V, 𝑣3 =
∆3
∆
=
−24
10
= −2.4 V
Example 2. 30:
For the circuit in Figure 2.27, find the branch currents 𝐼1, 𝐼2, and 𝐼3 using mesh
analysis.
Solution:
We first obtain the mesh currents using KVL. For mesh 1,
−15 + 5𝑖1 + 10 𝑖1 − 𝑖2 + 10 = 0
Or
3𝑖1 − 2𝑖2 = 1 (1)
For mesh 2,
6𝑖2 + 4𝑖2 + 10 𝑖1 − 𝑖2 − 10 = 0
Or
𝑖1 = 2𝑖2 − 1 (2)
We shall solve the equations in two ways; using the elimination technique or using
Cramer’s rule.
Let’s use the first method, we substitute equation (2) into equation (1), and write
6𝑖2 − 3 − 2𝑖2 = 1 ⟹ 𝑖2 = 1 A
From equation (2),
𝑖1 = 2𝑖2 − 1 ⟹ 𝑖1 = 1 A
Thus,
𝐼1 = 𝑖1 = 1 A, 𝐼2 = 𝑖2 = 1A, 𝐼3 = 𝑖1 − 𝑖1 = 0

38
2.3.6 SUPERPOSITION THEOREM
2.3.7 THVENIN’S THEOREM
2.3.8 NORTON’S THEOREM
2.3.9 SOURCE TRANSFORMATION
2.3.10 MAXIMUM POWER TRANSFER THEOREM
2.3.11 MESH AND NODAL ANALYSIS BY INSPECTION

39
2.4 SINUSOIDS AND PHASORS
2.4.1 INTRODUCTION
Thus far our analysis has been limited for the most part to dc circuits: those circuits
excited by constant or time-invariant sources. We now begin the analysis of circuits in which
the source voltage or current is time-varying.
2.4.2 SINUSOIDS
In this section, we are particularly interested in sinusoidally time-varying excitation
or simply, excitation by a sinusoid.
A sinusoidal current is usually referred to as alternating current (ac). Such a current
reverses at regular time intervals and has alternately positive and negative values. Circuits
driven by sinusoidal current or voltage sources are called ac circuits.
Consider the sinusoidal voltage:
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡
Where
𝑉𝑚 the amplitude of the sinusoid.
𝜔 the angular frequency in radians/s
𝜔𝑡 the argument of the sinusoid.
The sinusoid is shown in Figure 2.28(a) as a function of its argument and in
Figure 2.28(b) as a function of time. It is evident that the sinusoid repeats itself every T
seconds; thus, T is called the period of the sinusoid. From the two plots in Figure 2.28, we
observe that 𝜔T = 2𝜋,
𝑇 =
2𝜋
𝜔
(2. 50)
The fact that 𝑣(𝑡) repeats itself every 𝑇 seconds is shown by replacing 𝑡 by 𝑡 + 𝑇 in
Equation (2. 50). We get:
A sinusoid is a signal that has the form of the sine or cosine function.

40
𝑣 𝑡 + 𝑇 = 𝑉𝑚 sin 𝜔(𝑡 + 𝑇) = 𝑉𝑚 sin 𝜔 𝑡 +
2𝜋
𝜔
= 𝑉𝑚 sin 𝜔𝑡 + 2𝜋 = 𝑉𝑚 sin 𝜔𝑡
= 𝑣 𝑡
(2. 51)
Hence,
𝑣 𝑡 + 𝑇 = 𝑣 𝑡 (2. 52)
That is, v has the same value at 𝑡 + 𝑇 as it does at 𝑡 and 𝑣(𝑡) is said to be periodic.
In general,
.
Figure 2. 27: A sketch of Vm sin ωt: (a) as a function of ωt, (b) as a function of t.
As mentioned, the period T of the periodic function is the time of one complete
cycle or the number of seconds per cycle. The reciprocal of this quantity is the number of
cycles per second, known as the cyclic frequency 𝑓 of the sinusoid. Thus,
𝑓 =
1
𝑇
(2. 53)
A periodic function is one that satisfies 𝒇 (𝒕) = 𝒇 (𝒕 + 𝒏𝑻), for all 𝒕 and for all
integers 𝒏

41
And it is clear that
𝜔 = 2𝜋𝑓 (2. 54)
While 𝜔 is in radians per second (rad/s), 𝑓 in hertz (Hz).
Let us now consider a more general expression for sinusoid.
𝑣 𝑡 = 𝑉𝑚 sin(𝜔𝑡 + 𝜙) (2. 55)
Where 𝜔𝑡 + 𝜙 is the argument and 𝜙 is the phase. Both argument and phase can
be in radians or degrees.
Let us examine the two sinusoids
𝑣1 𝑡 = 𝑉𝑚 sin(𝜔𝑡) 𝑎𝑛𝑑 𝑣2 𝑡 = 𝑉𝑚 sin(𝜔𝑡 + 𝜙) (2. 56)
Figure 2. 28: Two sinusoids with different phases.
These two vectors are shown in Figure.2.29. The starting point of 𝑣2 in Figure 2.29
occurs first in time. Therefore, we say that 𝑣2 leads 𝑣1 by φ or that 𝑣1 lags 𝑣2 by 𝜙. If 𝜙 ≠ 0,
we also say that 𝑣1 and 𝑣2 are out of phase. If 𝜙 = 0, then 𝑣1 and 𝑣2 are said to be in phase;
they reach their minima and maxima at exactly the same time. We can compare 𝑣1 and 𝑣2 in
this manner because they operate at the same frequency; they do not need to have the
same amplitude.
Example 2. 31:
Find the amplitude, phase, period, and frequency of the sinusoid
𝑣 𝑡 = 12 cos 50𝑡 + 10°
Solution:
The amplitude is 𝑉𝑚 = 12 V

42
The pahse is 𝜙 = 10°
The angular frequency is 𝜔 = 50 𝑟𝑎𝑑/𝑠
The periode 𝑇 =
2𝜋
𝜔
=
2𝜋
50
𝑠
The frquency is 𝑓 =
1
𝑇
=
50
2𝜋
𝐻𝑧
Example 2. 32:
Calculate the phase angle between
𝑣1 = −10 cos 𝜔𝑡 + 50°
And
𝑣2 = 12 sin 𝜔𝑡 − 10°
State which sinusoide is leading.
Solution:
In order to compare 𝑣1 and 𝑣2, we must express them in the same form. If we
express them in cosine form with positive amplitudes,
𝑣1 = −10 cos 𝜔𝑡 + 50° = 10 cos 𝜔𝑡 + 50° − 180°
𝑣1 = 10 cos 𝜔𝑡 − 130°
Or
𝑣1 = 10 cos 𝜔𝑡 + 230°
and
𝑣2 = 12 sin 𝜔𝑡 − 10° = 12 cos 𝜔𝑡 − 10° − 90°
𝑣2 = 12 cos 𝜔𝑡 − 100°
Comparing both expressions of 𝑣1and 𝑣2 we can deduce that 𝑣2 leads 𝑣1 by 30°.
2.4.1 PHASORS
Sinusoids are easily expressed in terms of phasors, which are more convenient to
work with than sine and cosine functions.
A phasor is a complex number that represents the amplitude and phase of a
sinusoid.

43
A complex number can be written as
𝑧 = 𝑥 + 𝑗𝑦 (2. 57)
Where 𝑗 = −1, x is the real part of z and y is the imaginary par of z.
The complex number z can also be written in polar or exponential form as
𝑧 = 𝑥 + 𝑗𝑦 Rectangular form (2. 58)
𝑧 = 𝑟 𝜙 Polar from (2. 59)
𝑧 = 𝑟𝑒 𝑗𝜙 Exponential form (2. 60)
Where 𝑟 = 𝑥2 + 𝑦2 is the magnitude of z, and 𝜙 = tan−1 𝑦
𝑥
is the phase of z.
The idea of phasor representation is based on Euler’s identity. In general,
𝑒±𝑗𝜙
= cos 𝜙 ± 𝑗 sin 𝜙 (2. 61)
We may write
cos 𝜙 = Re 𝑒 𝑗𝜙 (2. 62)
sin 𝜙 = Im 𝑒 𝑗𝜙 (2. 63)
where Re and Im stand for the real part of and the imaginary part of.
Given a sinusoid
𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙 = Re 𝑉𝑚 𝑒 𝑗 𝜔𝑡+𝜙 (2. 64)
or
𝑣 𝑡 = Re 𝑉𝑚 𝑒 𝑗𝜙
𝑒 𝑗𝜔𝑡 (2. 65)
Thus,
𝑣 𝑡 = Re 𝐕𝑒 𝑗𝜔𝑡 (2. 66)
Where
𝐕 = 𝑉𝑚 𝑒 𝑗𝜙
= 𝑉𝑚 𝜙 (2. 67)
𝐕 is the phasor representation of the sinusoid 𝑣 𝑡 = as we said earlier.
To get the phasor corresponding to a sinusoid, we first express the sinusoid in the
cosine form so that the sinusoid can be written as the real part of a complex number.
Then we take out the time factor 𝑒 𝑗𝜔𝑡
, and whatever is left is the phasor
corresponding to the sinusoid. By suppressing the time factor, we transform the sinusoid
from the time domain to the phasor domain. This transformation is summarized as follows:

44
Time domain representation Phasor domain representation
𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙 ⟹ 𝐕 = 𝑉𝑚 𝑒 𝑗𝜙
= 𝑉𝑚 𝜙 (2. 68)
Given a sinusoid 𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙 ,we obtain the corresponding phasor as
𝐕 = 𝑉𝑚 𝜙. Equation (2.68) is also demonstrated in Table 2.2, where the sine function is
considered in addition to the cosine function.
From Eq. (6.68), we see that to get the phasor representation of a sinusoid, we
express it in cosine form and take the magnitude and phase. Given a phasor, we obtain the
time-domain representation as the cosine function with the same magnitude as the phasor
and the argument as 𝜔𝑡 plus the phase of the phasor. The idea of expressing information in
alternate domains is fundamental to all areas of engineering.
𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜙 ⟹ 𝐕 = 𝑉𝑚 𝜙 (2. 69)
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡 + 𝜙 ⟹ 𝐕 = 𝑉𝑚 𝜙 (2. 70)
𝑖 𝑡 = 𝐼 𝑚 cos 𝜔𝑡 + 𝜙 ⟹ 𝐈 = 𝐼 𝑚 𝜙 (2. 71)
𝑖 𝑡 = 𝐼 𝑚 sin 𝜔𝑡 + 𝜙 ⟹ 𝐈 = 𝐼 𝑚 𝜙 (2. 72)
We can write some useful relation between time representation and phasor
representation
𝑑𝑣
𝑑𝑡
𝑗𝜔𝐕 (2. 73)
𝑣 𝑑𝑡
𝐕
𝑗𝜔
(2. 74)
Example 2. 33:
Evaluate these complex numbers:
1.
40 50° + 20 −30°
2.
10 −30° + (3 − 𝑗4)
2 + 𝑗4 (3 − 𝑗5)∗

45
Solution :
1. Using polar to rectangular transformation
40 50° = 40 cos50° + 𝑗 sin50° = 25.71 + 𝑗30.64
20 −30° = 20 cos −30° + 𝑗 sin −30° = 17.32 − 𝑗10
Adding them up gives
40 50° + 20 −30° = 43.03 + 𝑗20.64 = 47.72 25.63°
Taking the square root of this
40 50° + 20 −30° = 6.91 12.81°
2. Using polar-rectangular transformation, addition, multiplication and division
10 −30° + (3 − 𝑗4)
2 + 𝑗4 (3 − 𝑗5)∗
=
8.66 − 𝑗5 + (3 − 𝑗4)
2 + 𝑗4 (3 − 𝑗5)∗
=
11.66 − 𝑗9
−14 + 𝑗22
=
14.73 −37.66°
20.08 122.47°
= 0.565 −160.31°
Example 2. 34:
1. Transform these sinusoids to phasors:
1.1 𝑣 = −4 sin 30𝑡 + 50°
1.2 𝑖 = 6 cos 50𝑡 − 40°
2. Find the sinusoids represented by these phasors:
2.1 𝐕 = 𝑗8𝑒−𝑗20
2.2 𝐈 = −3 + 𝑗4
Solution:
1.
1.1 Since − sin 𝐴 = cos 𝐴 + 90°
𝑣 = −4 sin 30𝑡 + 50° = 4 cos 30𝑡 + 50° + 90°
= 4 cos 30𝑡 + 140°
The phasor from of 𝑣 is
𝐕 = 4 140°
1.2 𝑖 = 6 cos 50𝑡 − 40° has the phasor
𝐈 = 6 −40°
2.

46
2.1 Since 𝑗 = 1 −90°
𝐕 = 𝑗8 −20° = 1 90° 8 −20°
= 1 90° − 20° = 8 70° V
Converting this to time domain gives
𝑣 𝑡 = 8 cos 𝜔𝑡 + 70° V
2.2 𝐈 = −3 + 𝑗4 = 5 126.87°. Transforming this to time domain gives
𝑖 𝑡 = 5 cos 𝜔𝑡 + 126.87° A
Example 2. 35:
Using the phasor approach, determine the current i(t) in a circuit described by the
integrodifferential equation.
4𝑖 + 8 𝑖 𝑑𝑡 − 3
𝑑𝑖
𝑑𝑡
= 50 cos 2𝑡 + 75°
Solution:
We transform each term in the equation from time domain to phasor domain
4𝐈 + 8
8𝐈
𝑗𝜔
− 3𝑗𝜔𝐈 = 50 75°
But 𝜔 = 2
𝐈 4 − 𝑗4 − 6𝑗 = 50 75°
𝐈 =
50 75°
4 − 𝑗10
=
50 75°
10.77 68.2°
= 4.642 143.2° A
Converting this to the time domain
𝑖 𝑡 = 4.624 cos 2𝑡 + 143.2° A
Keep in mind that this is only the steady-state solution, and it does not require
knowing the initial values.
2.4.2 PHASOR RELATIONSHIP FOR CIRCUIT ELEMENT
Now that we know how to represent a voltage or current in the phasor or frequency
domain, one may legitimately ask how we apply this to circuits involving the passive
elements R, L, and C. What we need to do is to transform the voltage-current relationship
from the time domain to the frequency domain for each element. Again, we will assume the
passive sign convention.
2.4.2.1 Resistor
If the current through a resistor R is 𝑖 = 𝐼 𝑚 cos(𝜔𝑡 + 𝜙), the voltage across it is
given by Ohm’s law as

47
𝐕 = 𝑖𝑅 = 𝑅𝐼 𝑚 cos(𝜔𝑡 + 𝜙) (2. 75)
The phasor form of this voltage is
𝐕 = 𝑅𝐼 𝑚 𝜙° (2. 76)
But the phasor representation of the current is
𝐈 = 𝐼 𝑚 𝜙 (2. 77)
Hence,
𝐕 = 𝑅𝐈 (2. 78)
This equation shows that the voltage-current relation for the resistor in the phasor
domain continues to be Ohm’s law, as in the time domain. Figure 2.30 illustrates the
voltage-current relations of a resistor. We should note from this equation to, that voltage
and current are in phase, as illustrated in the phasor diagram in Figure 2.31.
Figure 2. 29: Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.
Figure 2. 30: Phasor diagram for the resistor.

48
2.4.2.2 Inductor
For the inductor L, assume the current through it is 𝑖 = 𝐼 𝑚 cos(𝜔𝑡 + 𝜙). The
voltage across the inductor is
𝑣 = 𝐿
𝑑𝑖
𝑑𝑡
= −𝜔𝐿𝐼 𝑚 sin(𝜔𝑡 + 𝜙) (2. 79)
We know that
− sin 𝐴 = cos 𝐴 + 90° (2. 80)
We can write the voltage as
𝑣 = 𝐿
𝑑𝑖
𝑑𝑡
= 𝜔𝐿𝐼 𝑚 cos(𝜔𝑡 + 𝜙 + 90°) (2. 81)
Which transforms to the pasor
𝑉 = 𝜔𝐿𝐼 𝑚 𝑒 𝑗 𝜙+90°
= 𝜔𝐿𝐼 𝑚 𝑒 𝑗𝜙
𝑒 𝑗90°
= 𝜔𝐿 𝐼 𝑚 𝜙𝑒 𝑗90°
(2. 82)
But the phasor representation of the current is
𝐈 = 𝐼 𝑚 𝜙 (2. 83)
And
𝑒 𝑗90°
= 𝑗 (2. 84)
Thus
𝐕 = 𝑗𝜔𝐿𝐈 (2. 85)
Showing that the voltage has a magnitude of 𝜔𝐿𝐼 𝑚 and a phase of 𝜙 + 90°.The
voltage and current are 90° out of phase. Specifically, the current lags the voltage by 90°.
Figure 2.32 shows the voltage-current relations for the inductor. Figure 2.33 shows the
phasor diagram.
Figure 2. 31: Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.

49
Figure 2. 32: Phasor diagram for the inductor; I lags V.
2.4.2.3 Capacitor
For the capacitor C, assume the voltage across it is 𝑣 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙). The
current through the capacitor is
𝑖 = 𝐶
𝑑𝑣
𝑑𝑡
= −𝜔𝐿𝑉𝑚 sin(𝜔𝑡 + 𝜙) (2. 86)
By following the same steps as we took for the inductor we obtain:
𝐈 = 𝑗𝜔𝐶𝐕 ⟹ 𝐕 =
𝐈
𝑗𝜔𝐶
(2. 87)
Showing that the current and voltage are 90° out of phase. To be specific, the
current leads the voltage by 90°. Figure 2.34 shows the voltage-current relations for the
capacitor; Figure 2.35 gives the phasor diagram.
Figure 2. 33: Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.

50
Figure 2. 34: Phasor diagram for the capacitor; I leads V.
Following is the summary of the time-domain and phasor-domain representations of
the circuit elements.
Element Time Domain Frequency Domain
R 𝐕 = 𝑅𝑖 𝐕 = 𝑅𝐈
L 𝑣 = 𝐿
𝑑𝑖
𝑑𝑡
𝐕 = 𝑗𝜔𝐿𝐈
C 𝑖 = 𝐶
𝑑𝑣
𝑑𝑡
𝐕 =
𝐈
𝑗𝜔𝐶
Example 2. 36:
The voltage 𝑣 = 12 cos(60𝑡 + 45 ◦) is applied to a 0.1-H inductor. Find the steady-
state current through the inductor.
Solution :
For the inductor, 𝐕 = 𝑗𝜔𝐿𝐈 where 𝜔 = 60 𝑟𝑎𝑑/𝑠 and 𝐕 = 12 45° V. Hence,
𝐈 = 2 −45° A
Converting this to time domain
𝑣 = 2 cos 60𝑡 − 45° ◦ A
2.4.3 IMPEDANCE AND ADMITTANCE
In the preceding section, we obtained the voltage-current relations for the three
passive elements as
𝐕 = 𝑅𝐈 𝐕 = 𝑗𝜔𝐿𝐈 𝐕 =
𝐈
𝑗𝜔𝐶
(2. 88)
These equations may be written in terms of the ratio of the phasor voltage to the
phasor current as

51
𝐕
𝐈
= 𝑅
𝐕
𝐈
= 𝑗𝜔𝐿
𝐕
𝐈
=
1
𝑗𝜔𝐶
(2. 89)
From these three expressions, we obtain Ohm’s law in phasor form for any type of
element as
𝐙 =
𝐕
𝐈
or 𝐕 = 𝐙𝐈 (2. 90)
where Z is a frequency-dependent quantity known as impedance, measured in
ohms.
Sinusoids are easily expressed in terms of phasors, which are more convenient to
work with than sine and cosine functions.
The impedance represents the opposition which the circuit exhibits to the flow of
sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor,
because it does not correspond to a sinusoidally varying quantity.
The admittance Y of an element (or a circuit) is the ratio of the phasor current
through it to the phasor voltage across it, or
𝐘 =
𝐈
𝐕
or 𝐘 =
1
𝐙
(2. 91)
As a complex quantity, the impedance and admittance may be expressed in
rectangular form as
𝐙 = 𝑅 + 𝑗𝑋 (2. 92)
𝐘 = 𝐺 + 𝑗𝐵 (2. 93)
Where
R: is the resistance.
The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor
current I, measured in ohms ().
The admittance Y is the reciprocal of impedance, measured in siemens (S).

52
X: is the reactance.
G: is the conductance.
B: is the susceptance..
2.4.4 KIRCHHOFF LAWS IN THE FREQUENCY DOMAIN
We cannot do circuit analysis in the frequency domain without Kirchhoff’s current
and voltage laws. Therefore, we need to express them in the frequency domain.
For KVL, let 𝑣1, 𝑣2, … , 𝑣 𝑛 be the voltages around a closed loop.
Then
𝑣1 + 𝑣2 + ⋯ + 𝑣 𝑛 = 0 (2. 94)
In the sinusoidal steady state, each voltage may be written in cosine form, so that
𝑉 𝑚1 cos 𝜔𝑡 + 𝜃1 + 𝑉 𝑚2 cos 𝜔𝑡 + 𝜃2 + ⋯ + 𝑉𝑚𝑛 cos 𝜔𝑡 + 𝜃 𝑛 = 0 (2. 95)
This can be written as
Re(𝑉 𝑚1 𝑒 𝑗 𝜃1 𝑒 𝑗𝜔𝑡
) + Re(𝑉 𝑚2 𝑒 𝑗 𝜃2 𝑒 𝑗𝜔𝑡
) + ⋯ Re(𝑉𝑚𝑛 𝑒 𝑗 𝜃 𝑛 𝑒 𝑗𝜔𝑡
) = 0 (2. 96)
If we let 𝐕𝑘 = 𝑉 𝑚𝑘 𝑒 𝑗 𝜃 𝑘
𝑅𝑒[ 𝐕1 + 𝐕2 + ⋯ + 𝐕𝑛 𝑒 𝑗𝜔𝑡
] = 0 (2. 97)
Since 𝑒 𝑗𝜔𝑡
≠ 0
𝐕1 + 𝐕2 + ⋯ + 𝐕𝑛 = 0 (2. 98)
Indicating that Kirchhoff’s voltage law holds for phasors.
By following a similar procedure, we can show that Kirchhoff’s current law holds for
phasors.
Thus, it is easy to do many things, such as impedance combination, nodal and mesh
analyses, superposition, and source transformation.
2.4.5 OTHER SINUSOIDAL PARAMETERS
2.4.5.1 Mean or Average Value
For a continuous periodic waveform such as a sinusoid, the mean value can be found
by averaging all the instantaneous values during one cycle. This is given by

53
𝑉𝑎𝑣 =
1
𝑇
𝑣 𝑡 𝑑𝑡
𝑇
0
(2. 99)
Clearly, the average value of a complete sine wave is 0 because of equal positive and
negative half cycles. This is regardless of the peak amplitude.
2.4.5.2 Effective or RMS Value
The effective or root mean square (RMS) value of a periodic signal is equal to the
magnitude of a DC signal which produces the same heating effect as the periodic signal
when applied across a load resistance.
Consider a periodic signal, 𝑣(𝑡), then
Mean =
1
𝑇
𝑣 𝑡 𝑑𝑡
𝑇
0
(2. 100)
Mean Square =
1
𝑇
𝑣 𝑡 2
𝑑𝑡
𝑇
0
(2. 101)
Root Mean Square 𝑉𝑅𝑀𝑆 =
1
𝑇
𝑣 𝑡 2 𝑑𝑡
𝑇
0
(2. 102)
All the above expressions are independent of the phase angle 𝜙.
2.4.6 POWER IN AC CIRCUITS
Power analysis is of paramount importance. Power is the most important quantity in
electric utilities, electronic, and communication systems, because such systems involve
transmission of power from one point to another. Also, every industrial and household
electrical device— every fan, motor, lamp, pressing iron, TV, personal computer—has a
power rating that indicates how much power the equipment requires; exceeding the power
rating can do permanent damage to an appliance.
The most common form of electric power is 50- or 60-Hz ac power. The choice of ac
over dc allowed high-voltage power transmission from the power generating plant to the
consumer.
The instantaneous power 𝑝(𝑡) absorbed by an element is the product of the
instantaneous voltage 𝑣(𝑡) across the element and the instantaneous current 𝑖(𝑡) through
it. Assuming the passive sign convention,
𝑝 𝑡 = 𝑣 𝑡 𝑖 𝑡 (2. 103)
The instantaneous power is the power at any instant of time. It is the rate at which
an element absorbs energy.
Let the voltage and current at the terminals of the circuit be

54
𝑣 𝑡 = 𝑉𝑚 cos 𝜔𝑡 + 𝜃𝑣 (2. 104)
𝑖 (𝑡) = 𝐼 𝑚 cos(𝜔𝑡 + 𝜃𝑖) (2. 105)
The instantaneous power absorbed by the circuit is
𝑝 𝑡 = 𝑣 𝑡 𝑖 𝑡 = 𝑉𝑚 𝐼 𝑚 cos 𝜔𝑡 + 𝜃𝑣 cos(𝜔𝑡 + 𝜃𝑖) (2. 106)
Using the trigonometric identity
cos 𝐴 cos 𝐵 =
1
2
cos 𝐴 − 𝐵 + cos 𝐴 + 𝐵
𝑝 𝑡 = 𝑉𝑚 𝐼 𝑚 cos 𝜔𝑡 + 𝜃𝑣 cos(𝜔𝑡 + 𝜃𝑖)
=
1
2
𝑉𝑚 𝐼 𝑚 cos 𝜔𝑡 + 𝜃𝑣 − 𝜔𝑡 − 𝜃𝑖 + cos 𝜔𝑡 + 𝜃𝑣 + 𝜔𝑡 + 𝜃𝑖
=
1
2
𝑉𝑚 𝐼 𝑚 cos 𝜃𝑣 − 𝜃𝑖 + cos 2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖
(2. 107)
𝑝 𝑡 =
1
2
𝑉𝑚 𝐼 𝑚 cos 𝜃𝑣 − 𝜃𝑖 +
1
2
𝑉𝑚 𝐼 𝑚 cos 2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖 (2. 108)
This shows us that the instantaneous power has two parts. The first part is constant
or time independent. Its value depends on the phase difference between the voltage and
the current. The second part is a sinusoidal function whose frequency is 2ω, which is twice
the angular frequency of the voltage or current.
Figure 2. 35: The instantaneous power p(t) entering a circuit.
We also observe that 𝑝(𝑡) is positive for some part of each cycle and negative for
the rest of the cycle. When 𝑝(𝑡) is positive, power is absorbed by the circuit. When 𝑝(𝑡) is
negative, power is absorbed by the source; that is, power is transferred from the circuit to
the source. This is possible because of the storage elements (capacitors and inductors) in the
circuit.
The instantaneous power changes with time and is therefore difficult to measure.
The average power is more convenient to measure. In fact, the wattmeter, the instrument
for measuring power, responds to average power.

55
Thus, the average power is given by
𝑃 =
1
𝑇
𝑝 𝑡 𝑑𝑡
𝑇
0
(2. 109)
After some development we find the average power as:
𝑃 =
1
2
𝑉𝑚 𝐼 𝑚 cos 𝜙 𝜙 = 𝜃𝑣 − 𝜃𝑖
(2. 110)
In terms of RMS voltage and current, the average power is given by
𝑃 = 𝑉𝑅𝑀𝑆 𝐼 𝑅𝑀𝑆 cos 𝜙 (2. 111)
2.4.7 POWER FACTOR
The term cos 𝜙 in Equation 2.110 is called the power factor and is an important
parameter in determining the amount of actual power dissipated in the load. In practice,
power factor is used to specify the characteristics of a load.
1. For a purely resistive load 𝜙 = 0°, hence Unity Power Factor.
2. For a capacitive type load I leads V , hence Leading power factor
3. For an inductive type load I lags V , hence Lagging power factor
From Equation 2.111, the current can be specified as
𝐼 𝑅𝑀𝑆 =
𝑃𝑅𝑀𝑆
𝑉𝑅𝑀𝑆 cos 𝜙
(2. 112)
Clearly, for a fixed amount of demanded power, P, at a constant load voltage, V , a
higher power factor draws less amount of current and hence low 𝐼2
𝑅 losses in the
transmission lines. A purely reactive load where 𝜙 = 90°and cos 𝜙 = 0will draw an
excessively large amount of current and a power factor correction is required.
Example 2. 37:
Given that
𝑣(𝑡) = 120 cos(377𝑡 + 45 ◦) V and 𝑖(𝑡) = 10 cos(377𝑡 − 10 ◦) A
Find the instantaneous power and the average power.
Solution:
The instantaneous power is given by
𝑝(𝑡) = 344.2 + 600 cos(754𝑡 + 35 ◦) W.
The average power is given by
P = 344.2 W

56
Example 2. 38:
Calculate the instantaneous power and average power if
𝑣(𝑡) = 80 cos(10𝑡 + 20 ◦) V and 𝑖(𝑡) = 15 cos(10𝑡 − 60 ◦) A
Solution:
The instantaneous power is given by
𝑝(𝑡) = 385.7 + 600 cos(20𝑡 − 10 ◦) W.
The average power is given by
P = 385.7 W
Example 2. 39:
For the circuit shown in Fig. 2.37, find the average power supplied by the source and
the average power absorbed by the resistor.
Figure 2. 36: For this example.
Solution:
The current I is given by
The average power supplied by the voltage source is
The current through the resistor is
and the voltage across it is
The average power absorbed by the resistor is

57
which is the same as the average power supplied. Zero average power is absorbed
by the capacitor.
2.4.8 POWER FACTOR CORRECTION
To be done as a homework.

58
2.5 THREE PHASE CIRCUITS
2.5.1 OBJECTIVES
After studying this unit, you should be able to:
1. Discuss the differences between three-phase and single-phase voltages.
2. Discuss the characteristics of delta and wye connections.
3. Compute voltage and current values for delta and wye circuits.
Most of the electrical power generated in the world today is three-phase. Three-
phase power was first conceived by Nikola Tesla. In the early days of electric power
generation, Tesla not only led the battle concerning whether the nation should be powered
with low-voltage direct current or high-voltage alternating current, but he also proved that
three-phase power was the most efficient way that electricity could be produced,
transmitted, and consumed.
2.5.2 THREE-PHASE CIRCUITS OVERVIEW
There are several reasons why three-phase power is superior to single phase power.
1. The horsepower rating of three-phase motors and the KVA (kilo-voltamp) rating of
three-phase transformers is about 150% greater than for single-phase motors or
transformers with a similar frame size.
2. The power delivered by a single-phase system pulsates, Figure 2.38. The power
falls to zero three times during each cycle. The power delivered by a three-phase circuit
pulsates also, but it never falls to zero, Figure 2.38. In a three-phase system, the power
delivered to the load is the same at any instant. This produces superior operating
characteristics for three-phase motors.
3. In a balanced three-phase system, the conductors need be only about 75% the
size of conductors for a single-phase two-wire system of the same KVA rating. This helps
offset the cost of supplying the third conductor required by three-phase systems.
Figure 2. 37: (a) Single-phase power falls to zero three times each cycle, (b) Three-phase power never
falls to zero.

59
A single-phase alternating voltage can be produced by rotating a magnetic field
through the conductors of a stationary coil, as shown in Figure 2.40.
Figure 2. 38: Producing a single-phase voltage
Since alternate polarities of the magnetic field cut through the conductors of the
stationary coil, the induced voltage will change polarity at the same speed as the rotation of
the magnetic field. The alternator shown in Figure 2.39 is single phase because it produces
only one AC voltage.
Figure 2. 39: The voltages of a three-phase system are 120° out of phase with each other.

60
If three separate coils are spaced 120° apart, as shown in Figure 2.40, three voltages
120° out of phase with each other will be produced when the magnetic field cuts through
the coils. This is the manner in which a three-phase voltage is produced. There are two basic
three-phase connections, the wye or star connection and the delta connection.
𝑣 𝐴 𝑡 = 𝑉 cos 𝜔𝑡 (2. 113)
𝑣 𝐵 𝑡 = 𝑉 cos 𝜔𝑡 − 120° 𝑜𝑟 𝑣 𝐵 𝑡 = 𝑉 cos 𝜔𝑡 −
2𝜋
3
(2. 114)
𝑣 𝐶 𝑡 = 𝑉 cos 𝜔𝑡 − 240° 𝑜𝑟 𝑣 𝐶 𝑡 = 𝑉 cos 𝜔𝑡 −
4𝜋
3
𝑜𝑟 𝑣 𝐶 𝑡 = 𝑉 cos 𝜔𝑡 + 240° 𝑜𝑟 𝑣 𝐶 𝑡 = 𝑉 cos 𝜔𝑡 +
2𝜋
3
(2. 115)
2.5.3 WYE CONNECTION
The wye or star connection is made by connecting one end of each of the three-
phase windings together as shown in Figure 2-41.
Figure 2. 40: A wye connection is formed by joining one end of each of the windings together.
The voltage measured across a single winding or phase is known as the phase
voltage, as shown in Figure 2.42. The voltage measured between the lines is known as the
line-to-line voltage or simply as the line voltage.
Figure 2. 41: Line and phase voltages are different in a wye connection.

61
In Figure 2.43, ammeters have been placed in the phase winding of a wye-connected
load and in the line supplying power to the load. Voltmeters have been connected across the
input to the load and across the phase. A line voltage of 208 V has been applied to the load.
Notice that the voltmeter connected across the lines indicates a value of 208 V, but the
voltmeter connected across the phase indicates a value of 120 V.
Two formulas used to compute the voltage in a wye connected system are:
𝑉𝐿𝑖𝑛𝑒 = 𝑉𝑃𝑕𝑎𝑠𝑒 × 3 (2. 116)
And
𝑉𝑃𝑕𝑎𝑠𝑒 =
𝑉𝐿𝑖𝑛𝑒
3
(2. 117)
.
𝐼𝐿𝑖𝑛𝑒 = 𝐼 𝑃𝑕𝑎𝑠𝑒 (2. 118)
Notice in Figure 2.43 that 10 A of current flow in both the phase and the line. In a
wye-connected system, phase current and line current are the same.
Figure 2. 42: Line current and phase current are the same in a wye connection.
In a Wye connected system, the line voltage is higher than the phase voltage by a
factor of the square root of 3.
In a wye connected system, phase current and line current are the same.

62
Figure 2. 43: Adding voltage vectors of two-phase voltage values.
An illustration of vector addition is shown in Figure 2.44. In this illustration two-
phase voltage vectors are added and the resultant is drawn from the starting point of one
vector to the end point of the other. The parallelogram method of vector addition for the
voltages in a wye-connected three-phase system is shown in Figure 2.45.
Figure 2. 44: The parallelogram method of adding three-phase vectors.
2.5.4 DELTA CONNECTIONS
In Figure 2.46, three separate inductive loads have been connected to form a delta
connection. This connection receives its name from the fact that a schematic diagram of this
connection resembles the Greek letter delta (Δ).
Figure 2. 45: Three-phase delta connection.

63
In Figure 2.47, voltmeters have been connected across the lines and across the
phase. Ammeters have been connected in the line and in the phase. In the delta connection,
line voltage and phase voltage are the same. Notice that both voltmeters indicate a value of
480 V.
.
𝑉𝐿𝑖𝑛𝑒 = 𝑉𝑃𝑕𝑎𝑠𝑒 (2. 119)
Figure 2. 46: Voltage and current relationships in a delta connection.
Formulas for determining the current in a delta connection are:
𝐼𝐿𝑖𝑛𝑒 = 𝐼 𝑃𝑕𝑎𝑠𝑒 × 3 (2. 120)
And
𝐼 𝑃𝑕𝑎𝑠𝑒 =
𝐼𝐿𝑖𝑛𝑒
3
(2. 121)
Notice that the line current and phase current are different, however. The line
current of a delta connection is higher than the phase current by a factor of the square root
of 3 (1.732). In the example shown, it is assumed that each of the phase windings has a
current flow of 10 A. The current in each of the lines, however, is 17.32 A. The reason for this
difference in current is that current flows through different windings at different times in a
In a delta connection, line voltage and phase voltage are the same.
The line current of a delta connection is higher than the phase current by a factor
of the square root of 3 (1.732).

64
three-phase circuit. During some periods of time, current will flow between two lines only.
At other times, current will flow from two lines to the third, Figure 2.48.
𝐼𝐿𝑖𝑛𝑒 = 𝐼 𝑃𝑕𝑎𝑠𝑒 × 3 (2. 122)
The delta connection is similar to a parallel connection because there is always more
than one path for current flow. Since these currents are 120° out of phase with each other,
vector addition must be used when finding the sum of the currents (Figure 2.48).
(a) (b)
Figure 2. 47: (a) Division of currents in a delta connection, (b) Vector addition is used to compute the sum of
the currents in a delta connection.
2.5.5 THREE-PHASE POWER
Students sometimes become confused when computing power in threephase
circuits. One reason for this confusion is that there are actually two formulas that can be
used. If line values of voltage and current are known, the power (watts) of a pure resistive
load can be computed using the formula:
VA = 3 × 𝐸𝐿𝑖𝑛𝑒 × 𝐼𝐿𝑖𝑛𝑒
(2. 123)
If the phase values of voltage and current are known, the apparent power can be
computed using the formula:
VA = 3 × 𝐸 𝑃𝑕𝑎𝑠𝑒 × 𝐼 𝑃𝑕𝑎𝑠𝑒 (2. 124)
Notice that in the first formula, the line values of voltage and current are multiplied
by the square root of 3. In the second formula, the phase values of voltage and current are
multiplied by 3. The first formula is used more often because it is generally more convenient
to obtain line values of voltage and current, which can be measured with a voltmeter and
clamp-on ammeter.

65
2.5.6 THREE-PHASE CIRCUIT CALCULATIONS
In the following examples, values of line and phase voltage, line and phase current,
and power will be computed for different types of threephase connections.
Example 2. 40:
A wye-connected three-phase alternator supplies power to a delta-connected
resistive load, Figure 2.49. The alternator has a line voltage of 480 V. Each resistor of the
delta load has 8  of resistance. Find the following values:
𝐸𝐿(𝐿𝑜𝑎𝑑 ) : Line voltage of the load.
𝐸 𝑃(𝐿𝑜𝑎𝑑 ) : Phase voltage of the load.
𝐼 𝑃(𝐿𝑜𝑎𝑑 ) : Phase current of the load.
𝐼𝐿(𝐿𝑜𝑎𝑑 ) : Line current to the load
𝐼L(Alt ) : Line current delivered by the alternator
𝐼 𝑃(𝐴𝑙𝑡 ) : Phase current of the alternator
𝐸P(Alt ) : Phase voltage of the alternator
P : True power
Figure 2. 48: Computing three-phase values using a wye-connected power source and a delta
connected load.
Solution:
The load is connected directly to the alternator. Therefore, the line voltage supplied
by the alternator is the line voltage of the load.
𝐸𝐿(𝐿𝑜𝑎𝑑 ) = 480 𝑉
The three resistors of the load are connected in a delta connection. In a delta
connection, the phase voltage is the same as the line voltage.
𝐸 𝑃(𝐿𝑜𝑎𝑑 ) = 𝐸𝐿(𝐿𝑜𝑎𝑑 )

66
𝐸 𝑃(𝐿𝑜𝑎𝑑 ) = 480 𝑉
Each of the three resistors in the load is one phase of the load. Now that the phase
voltage is known (480 V), the amount of phase current can be computed using Ohm’s Law.
𝐼 𝑃(𝐿𝑜𝑎𝑑 ) =
𝐸 𝑃(𝐿𝑜𝑎𝑑 )
𝑍
480
8
𝐼 𝑃(𝐿𝑜𝑎𝑑 ) = 60 𝐴
The three load resistors are connected as a delta with 60 A of current flow in each
phase. The line current supplying a delta connection must be 1.732 times greater than the
phase current.
𝐼𝐿(𝐿𝑜𝑎𝑑 ) = 𝐼 𝑃(𝐿𝑜𝑎𝑑 ) × 1.732
𝐼𝐿(𝐿𝑜𝑎𝑑 ) = 60 × 1.732
𝐼𝐿(𝐿𝑜𝑎𝑑 ) = 103.92 𝐴
The alternator must supply the line current to the load or loads to which it is
connected. In this example, only one load is connected to the alternator. Therefore, the line
current of the load will be the same as the line current of the alternator.
𝐼𝐿(𝐴𝑙𝑡) = 103.92 𝐴
The phase windings of the alternator are connected in a wye connection. In a wye
connection, the phase current and line current are equal. The phase current of the
alternator will, therefore, be the same as the alternator line current.
𝐼 𝑃(𝐴𝑙𝑡) = 103.92 𝐴
The phase voltage of a wye connection is less than the line voltage by a factor of the
square root of 3. The phase voltage of the alternator will be:
𝐸 𝑃(𝐴𝑙𝑡) =
𝐸𝐿(𝐴𝑙𝑡)
1.732
𝐸 𝑃(𝐴𝑙𝑡 ) =
480
1.732
𝐸 𝑃(𝐴𝑙𝑡) = 277.13 𝑉
In this circuit, the load is pure resistive. The voltage and current are in phase with
each other, which produces a unity power factor of 1. The true power in this circuit will be
computed using the formula:
𝑃 = 1.732 × 𝐸𝐿(𝐴𝑙𝑡) × 𝐼𝐿(𝐴𝑙𝑡) × 𝑃𝐹

67
𝑃 = 1.732 × 480 × 103.92 × 1
𝑃 = 86,394.93 𝑊
Example 2. 41:
A delta-connected alternator is connected to a wye-connected resistive load, Figure
2.50. The alternator produces a line voltage of 240 V and the resistors have a value of 6 
each. The following values will be found:
𝐸𝐿(𝐿𝑜𝑎𝑑 ) : Line voltage of the load.
𝐸 𝑃(𝐿𝑜𝑎𝑑 ) : Phase voltage of the load.
𝐼 𝑃(𝐿𝑜𝑎𝑑 ) : Phase current of the load.
𝐼𝐿(𝐿𝑜𝑎𝑑 ) : Line current to the load
𝐼L(Alt ) : Line current delivered by the alternator
𝐼 𝑃(𝐴𝑙𝑡 ) : Phase current of the alternator
𝐸P(Alt ) : Phase voltage of the alternator
P : True power
Figure 2. 49: Computing three-phase values using a delta-connected source and a wye-connected
load.
Solution:
As was the case in Example 1, the load is connected directly to the output of the
alternator. The line voltage of the load must, therefore, be the same as the line voltage of
the alternator.
𝐸𝐿(𝐿𝑜𝑎𝑑 ) = 240 𝑉
The phase voltage of a wye connection is less than the line voltage by a factor of
1.732.

68
𝐸 𝑃 𝐿𝑜𝑎𝑑 =
240
1.732
𝐸 𝑃(𝐿𝑜𝑎𝑑 ) = 138.57 𝑉
Each of the three 6 resistors is one phase of the wye-connected load. Since the
phase voltage is 138.57 V, this voltage is applied to each of the three resistors. The amount
of phase current can now be determined using Ohm’s Law.
𝐼 𝑃(𝐿𝑜𝑎𝑑 )
𝑍
138.57
6
𝐼 𝑃(𝐿𝑜𝑎𝑑 ) = 23.1 𝐴
The amount of line current needed to supply a wye-connected load is the same as
the phase current of the load.
𝐼𝐿(𝐿𝑜𝑎𝑑 ) = 23.1 𝐴
Only one load is connected to the alternator. The line current supplied to the load is
the same as the line current of the alternator.
𝐼𝐿(𝐴𝑙𝑡 ) = 23.1 𝐴
The phase windings of the alternator are connected in delta. In a delta connection
the phase current is less than the line current by a factor of 1.732.
𝐼 𝑃(𝐴𝑙𝑡) =
𝐼𝐿(𝐴𝑙𝑡 )
1.732
𝐼 𝑃(𝐴𝑙𝑡) =
23.1
1.732
𝐼 𝑃(𝐴𝑙𝑡 ) = 13.34 𝐴
The phase voltage of a delta is the same as the line voltage.
𝐸 𝑃(𝐴𝑙𝑡) = 240 𝑉
Since the load in this example is pure resistive, the power factor has a value of unity,
or 1. Power will be computed by using the line values of voltage and current.
𝑃 = 1.732 × 𝐸𝐿 × 𝐼𝐿 × 𝑃𝐹
𝑃 = 1.732 × 240 × 23.1 × 1
𝑃 = 9,602.21 𝑊

69
2.5.7 SUMMARY
1. The voltages of a three-phase system are 120° out of phase with each other.
2. The two types of three-phase connections are wye and delta. 3. Wye connections
are characterized by the fact that one terminal of each device is connected together.
4. In a wye connection, the phase voltage is less than the line voltage by a factor of
1.732. The phase current and line current are the same.
5. In a delta connection, the phase voltage is the same as the line voltage. The phase
current is less than the line current by a factor of 1.732.

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