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

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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.

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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 × 10−16𝐶

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Example 2. 3 :

The total charge entering a terminal is given by 𝑞 = 5𝑡 sin 4𝜋𝑡 𝑚𝐶. Calculate the

current at 𝑡 = 0.5 𝑠.

Solution:

𝒊 =𝒅𝒒

𝒅𝒕=

𝒅

𝒅𝒕 𝟓𝒕 𝐬𝐢𝐧 𝟒𝝅𝒕 = 𝟓 𝐬𝐢𝐧𝟒𝝅𝒕 + 𝟐𝟎𝝅𝒕 𝐜𝐨𝐬𝟒𝝅𝒕 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 = 𝒊𝒅𝒕𝟐

𝒕=𝟏

= 𝟑𝒕𝟐 − 𝒕 𝒅𝒕 = 𝒕𝟑 −𝒕𝟐

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:

𝑣𝑎𝑏 =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 𝑑𝑡𝑡

𝑡0

= 𝑣𝑖 𝑑𝑡𝑡

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 cos 60𝜋𝑡 𝐴

And the voltage is:

(a) 𝑣 = 3𝑖,

(b) 𝑣 = 3di

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 = 3𝑑𝑖

𝑑𝑡= 3 −60𝜋 5 sin 60𝜋𝑡 = −900𝜋 sin 60𝜋𝑡 𝑉

𝑝 = 𝑣𝑖 = −4500𝜋 sin 60𝜋𝑡 cos 60𝜋𝑡𝑊

At 𝑡 = 3 𝑚𝑠,

𝑝 = −4500𝜋 sin 0.18𝜋 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 = 𝜌𝑙

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 = 𝑖2𝑅 =𝑣2

𝑅 (2. 14)

The power dissipated by a resistor may also be expressed in terms of G as:

𝑝 = 𝑣𝑖 = 𝑣2𝐺 =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 = 𝐼2𝑅 = 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 = 𝐼2𝑅 = 6 × 10−3 25 × 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 sin𝜋𝑡 𝑉 is connected across a 5 𝑘Ω resistor. Find the current

through the resistor and the power dissipated.

Solution:

𝑖 =𝑣

𝑅=

20 sin𝜋𝑡

5 × 103= 4 sin𝜋𝑡 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 = 𝐶𝑑𝑣

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 = 𝑐𝑣𝑑𝑣

𝑑𝑡 (2. 19)


−∞

= 𝐶 𝑣𝑑𝑣

𝑑𝑡 𝑑𝑡

𝑡

−∞

= 𝐶 𝑣 𝑑𝑣𝑡

−∞

=1

2𝐶𝑣2

𝑡=−∞

𝑡

(2. 20)

We note that 𝑣 −∞ = 0, because the capacitor was unchanged at 𝑡 = −∞. Thus,

𝑤 =1

2𝐶𝑣2 (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

2𝐶𝑣2 =

1

2× 3 × 10−12 × 400 = 600 𝑝𝐽

Example 2. 16:

The voltage across a 5*F capacitor is:

19

𝑣 𝑡 = 10 cos 6000𝑡 𝑉

Calculate the current through it.

Solution:

By definition, the current is:

𝑖 𝑡 = 𝐶𝑑𝑣

𝑑𝑡= 5 × 10−6

𝑑

𝑑𝑡 10 cos 6000𝑡

𝑖 𝑡 = −5 × 10−6 × 6000 × 10 sin 6000𝑡 = −0.3 sin 6000𝑡 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−3000𝑡𝑑𝑡𝑡

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 < 1100 − 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 < 310𝑚𝐴 3 < 𝑡 < 40 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 = 𝐿𝑑𝑖

𝑑𝑡 (2. 23)

Or

𝑖 =1

𝐿 𝑣𝑑𝑡𝑡

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 = 𝐿𝑑𝑖

𝑑𝑡 𝑖 (2. 25)

The energy stored is:


−∞

= 𝐿𝑑𝑖

𝑑𝑡 𝑖 𝑑𝑡

𝑡

−∞

(2. 26)

𝑤 = 𝐿 𝑖 𝑑𝑡𝑡

−∞

=1

2𝐿𝑖2 𝑡 −

1

2𝐿𝑖2 −∞ (2. 27)

Since, 𝑖 −∞ = 0

𝑤 =1

2𝐿𝑖2 (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−5𝑡𝐴. Find the voltage across

the inductor and the energy stored in it.

Solution:

Since,𝑣 = 𝐿𝑑𝑖

𝑠𝑡 and 𝐿 = 0.1𝐻

𝑣 = 0.1𝑑

𝑑𝑡 10𝑡𝑒−5𝑡 = 𝑒−5𝑡 + 𝑡 −5 𝑒−5𝑡 = 𝑒−5𝑡 1 − 5𝑡 𝑉

The energy stored is:

𝑤 =1

2𝐿𝑖2 =

1

2 0.1 100𝑡2𝑒−10𝑡 = 5𝑡2𝑒−10𝑡 𝐽

Example 2. 20:

Find the current through a 5 H inductor if the voltage across it is:

𝑣 𝑡 = 30𝑡2 𝑡 > 00 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 30𝑡2𝑑𝑡 + 0 = 2𝑡3 𝐴𝑡

0

The power is:

𝑝 = 𝑣𝑖 = 60𝑡5

The energy stored is then:

𝑤 = 𝑝 𝑑𝑡 = 60𝑡5𝑑𝑡5

0

= 156.25 𝑘𝐽

Alternatively, we can obtain the energy stored by writing:

𝑤05 =

1

2𝐿𝑖2 5 −

1

2𝐿𝑖 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:

𝑖𝑛

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 𝑖𝑛 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:

𝑣𝑚

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 𝑣𝑚 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 =𝑖22

(9)

Substituting Eqs. (7) and (9) into (4) gives :

30 − 3𝑖28

− 𝑖2 −𝑖22

= 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 = 𝑖𝑅𝑒𝑞 (2. 38)

Implying that the two resistors can be replaced by an equivalent resistor that is:

𝑅𝑒𝑞 = 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:

𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + ⋯+ 𝑅𝑁 = 𝑅𝑛

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.

𝑣𝑛 =𝑅𝑛

𝑅1 + 𝑅2 + ⋯+ 𝑅𝑛 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

𝑅𝑒𝑞=

1

𝑅1+

1

𝑅2+ ⋯+

1

𝑅𝑁=

1

𝑅𝑛

𝑁

𝑛=1

(2. 44)

𝑖𝑛 =𝐺𝑛

𝐺1 + 𝐺2 + ⋯+ 𝐺𝑛 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

𝐶𝑁=

1

𝐶𝑛

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 + ⋯+ 𝐶𝑁 = 𝐶𝑛

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.

𝐿𝑒𝑞 = 𝐿1 + 𝐿2 + ⋯+ 𝐿𝑁 = 𝐿𝑛

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

𝐿𝑒𝑞=

1

𝐿1+

1

𝐿2+ ⋯+

1

𝐿𝑁=

1

𝐿𝑛

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:

𝑅𝑒𝑞 = 4 + 2.4 + 8 = 14.4 Ω

Example 2. 26:

Fine the equivalent resistance for the circuit given in figure 2.22.


Solution:

𝑅𝑒𝑞 = 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 + 𝑖𝑥 ⟹ 3 =𝑣1 − 𝑣3

4+𝑣1 − 𝑣2

2

Multiplying by 4 and rearranging terms, we get

3𝑣1 − 2𝑣2 − 𝑣3 = 12

At node 2,

𝑖𝑥 = 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,

2𝑖𝑥 = 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 −12 −3 1

𝑣1

𝑣2

𝑣3

= 1200

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 sin𝜔𝑡

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 sin𝜔(𝑡 + 𝑇) = 𝑉𝑚 sin𝜔 𝑡 +2𝜋

𝜔 = 𝑉𝑚 sin 𝜔𝑡 + 2𝜋 = 𝑉𝑚 sin𝜔𝑡

= 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

50𝑠

The frquency is 𝑓 =1

𝑇=

50

2𝜋𝐻𝑧

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 = 𝑟𝑒𝑗𝜙 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 = cos𝜙 ± 𝑗 sin𝜙 (2. 61)

We may write

cos𝜙 = Re 𝑒𝑗𝜙 (2. 62)

sin𝜙 = Im 𝑒𝑗𝜙 (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 𝑉𝑚𝑒𝑗 𝜔𝑡+𝜙 (2. 64)

or

𝑣 𝑡 = Re 𝑉𝑚𝑒𝑗𝜙 𝑒𝑗𝜔𝑡 (2. 65)

Thus,

𝑣 𝑡 = Re 𝐕𝑒𝑗𝜔𝑡 (2. 66)

Where

𝐕 = 𝑉𝑚𝑒𝑗𝜙 = 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 𝑒𝑗𝜔𝑡 , 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 = 𝑉𝑚𝑒𝑗𝜙 = 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 = 𝐼𝑚 cos 𝜔𝑡 + 𝜙 ⟹ 𝐈 = 𝐼𝑚 𝜙 (2. 71)

𝑖 𝑡 = 𝐼𝑚 sin 𝜔𝑡 + 𝜙 ⟹ 𝐈 = 𝐼𝑚 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 cos 50° + 𝑗 sin 50° = 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 − sin𝐴 = 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 − 3𝑑𝑖

𝑑𝑡= 50 cos 2𝑡 + 75°

Solution:

We transform each term in the equation from time domain to phasor domain

4𝐈 + 88𝐈

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 = 𝐼𝑚 cos(𝜔𝑡 + 𝜙), the voltage across it is

given by Ohm’s law as

47

𝐕 = 𝑖𝑅 = 𝑅𝐼𝑚 cos(𝜔𝑡 + 𝜙) (2. 75)

The phasor form of this voltage is

𝐕 = 𝑅𝐼𝑚 𝜙° (2. 76)

But the phasor representation of the current is

𝐈 = 𝐼𝑚 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 = 𝐼𝑚 cos(𝜔𝑡 + 𝜙). The

voltage across the inductor is

𝑣 = 𝐿𝑑𝑖

𝑑𝑡= −𝜔𝐿𝐼𝑚 sin(𝜔𝑡 + 𝜙) (2. 79)

We know that

− sin𝐴 = cos 𝐴 + 90° (2. 80)

We can write the voltage as

𝑣 = 𝐿𝑑𝑖

𝑑𝑡= 𝜔𝐿𝐼𝑚 cos(𝜔𝑡 + 𝜙 + 90°) (2. 81)

Which transforms to the pasor

𝑉 = 𝜔𝐿𝐼𝑚𝑒𝑗 𝜙+90° = 𝜔𝐿𝐼𝑚𝑒

𝑗𝜙 𝑒𝑗90° = 𝜔𝐿 𝐼𝑚 𝜙𝑒𝑗90° (2. 82)

But the phasor representation of the current is

𝐈 = 𝐼𝑚 𝜙 (2. 83)

And

𝑒𝑗90° = 𝑗 (2. 84)

Thus

𝐕 = 𝑗𝜔𝐿𝐈 (2. 85)

Showing that the voltage has a magnitude of 𝜔𝐿𝐼𝑚 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 = 𝐶𝑑𝑣

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 = 𝐿𝑑𝑖

𝑑𝑡 𝐕 = 𝑗𝜔𝐿𝐈

C 𝑖 = 𝐶𝑑𝑣

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 ,… , 𝑣𝑛 be the voltages around a closed loop.

Then

𝑣1 + 𝑣2 + ⋯+ 𝑣𝑛 = 0 (2. 94)

In the sinusoidal steady state, each voltage may be written in cosine form, so that

𝑉𝑚1 cos 𝜔𝑡 + 𝜃1 + 𝑉𝑚2 cos 𝜔𝑡 + 𝜃2 + ⋯+ 𝑉𝑚𝑛 cos 𝜔𝑡 + 𝜃𝑛 = 0 (2. 95)

This can be written as

Re(𝑉𝑚1𝑒𝑗𝜃1𝑒𝑗𝜔𝑡 ) + Re(𝑉𝑚2𝑒

𝑗𝜃2𝑒𝑗𝜔𝑡 ) + ⋯Re(𝑉𝑚𝑛 𝑒𝑗𝜃𝑛 𝑒𝑗𝜔𝑡 ) = 0 (2. 96)

If we let 𝐕𝑘 = 𝑉𝑚𝑘 𝑒𝑗𝜃𝑘

𝑅𝑒[ 𝐕1 + 𝐕2 + ⋯+ 𝐕𝑛 𝑒𝑗𝜔𝑡 ] = 0 (2. 97)

Since 𝑒𝑗𝜔𝑡 ≠ 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 𝑑𝑡𝑇

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 𝑑𝑡𝑇

0

(2. 100)

Mean Square =1

𝑇 𝑣 𝑡 2𝑑𝑡𝑇

0

(2. 101)

Root Mean Square 𝑉𝑅𝑀𝑆 = 1

𝑇 𝑣 𝑡 2𝑑𝑡𝑇

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) = 𝐼𝑚 cos(𝜔𝑡 + 𝜃𝑖) (2. 105)

The instantaneous power absorbed by the circuit is

𝑝 𝑡 = 𝑣 𝑡 𝑖 𝑡 = 𝑉𝑚 𝐼𝑚 cos 𝜔𝑡 + 𝜃𝑣 cos(𝜔𝑡 + 𝜃𝑖) (2. 106)

Using the trigonometric identity

cos𝐴 cos𝐵 =1

2 cos 𝐴 − 𝐵 + cos 𝐴 + 𝐵

𝑝 𝑡 = 𝑉𝑚 𝐼𝑚 cos 𝜔𝑡 + 𝜃𝑣 cos(𝜔𝑡 + 𝜃𝑖)

=1

2𝑉𝑚 𝐼𝑚 cos 𝜔𝑡 + 𝜃𝑣 −𝜔𝑡 − 𝜃𝑖 + cos 𝜔𝑡 + 𝜃𝑣 + 𝜔𝑡 + 𝜃𝑖

=1

2𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 + cos 2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖

(2. 107)

𝑝 𝑡 =1

2𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 +

1

2𝑉𝑚 𝐼𝑚 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.

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Thus, the average power is given by

𝑃 =1

𝑇 𝑝 𝑡 𝑑𝑡𝑇

0

(2. 109)

After some development we find the average power as:

𝑃 =1

2𝑉𝑚 𝐼𝑚 cos 𝜙 𝜙 = 𝜃𝑣 − 𝜃𝑖 (2. 110)

In terms of RMS voltage and current, the average power is given by

𝑃 = 𝑉𝑅𝑀𝑆𝐼𝑅𝑀𝑆 cos𝜙 (2. 111)

2.4.7 POWER FACTOR

The term cos𝜙 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

𝐼𝑅𝑀𝑆 =𝑃𝑅𝑀𝑆

𝑉𝑅𝑀𝑆 cos𝜙 (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 𝐼2𝑅 losses in the

transmission lines. A purely reactive load where 𝜙 = 90°and cos𝜙 = 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

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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

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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.

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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.

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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.

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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.

𝑣𝐴 𝑡 = 𝑉 cos 𝜔𝑡 (2. 113)

𝑣𝐵 𝑡 = 𝑉 cos 𝜔𝑡 − 120° 𝑜𝑟 𝑣𝐵 𝑡 = 𝑉 cos 𝜔𝑡 −2𝜋

3 (2. 114)

𝑣𝐶 𝑡 = 𝑉 cos 𝜔𝑡 − 240° 𝑜𝑟 𝑣𝐶 𝑡 = 𝑉 cos 𝜔𝑡 −4𝜋

3

𝑜𝑟 𝑣𝐶 𝑡 = 𝑉 cos 𝜔𝑡 + 240° 𝑜𝑟 𝑣𝐶 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.

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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 = 𝐼𝑃𝑕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.

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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.

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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 = 𝐼𝑃𝑕𝑎𝑠𝑒 × 3 (2. 120)

And

𝐼𝑃𝑕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).

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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 = 𝐼𝑃𝑕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 × 𝐸𝑃𝑕𝑎𝑠𝑒 × 𝐼𝑃𝑕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.

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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.

𝐸𝑃(𝐿𝑜𝑎𝑑 ) : Phase voltage of the load.

𝐼𝑃(𝐿𝑜𝑎𝑑 ) : Phase current of the load.

𝐼𝐿(𝐿𝑜𝑎𝑑 ) : Line current to the load

𝐼L(Alt ) : Line current delivered by the alternator

𝐼𝑃(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.

𝐸𝑃(𝐿𝑜𝑎𝑑 ) = 𝐸𝐿(𝐿𝑜𝑎𝑑 )

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𝐸𝑃(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.

𝐼𝑃(𝐿𝑜𝑎𝑑 ) = 𝐸𝑃(𝐿𝑜𝑎𝑑 )

𝑍

𝐼𝑃(𝐿𝑜𝑎𝑑 ) = 480

8

𝐼𝑃(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 ) = 𝐼𝑃(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.

𝐼𝑃(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:

𝐸𝑃(𝐴𝑙𝑡 ) =𝐸𝐿(𝐴𝑙𝑡 )

1.732

𝐸𝑃(𝐴𝑙𝑡 ) =480

1.732

𝐸𝑃(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

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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.

𝐸𝑃(𝐿𝑜𝑎𝑑 ) : Phase voltage of the load.

𝐼𝑃(𝐿𝑜𝑎𝑑 ) : Phase current of the load.

𝐼𝐿(𝐿𝑜𝑎𝑑 ) : Line current to the load

𝐼L(Alt ) : Line current delivered by the alternator

𝐼𝑃(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.

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𝐸𝑃 𝐿𝑜𝑎𝑑 =240

1.732

𝐸𝑃(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.

𝐼𝑃(𝐿𝑜𝑎𝑑 ) =𝐼𝑃(𝐿𝑜𝑎𝑑 )

𝑍

𝐼𝑃(𝐿𝑜𝑎𝑑 ) =138.57

6

𝐼𝑃(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.

𝐼𝑃(𝐴𝑙𝑡 ) = 𝐼𝐿(𝐴𝑙𝑡 )

1.732

𝐼𝑃(𝐴𝑙𝑡 ) = 23.1

1.732

𝐼𝑃(𝐴𝑙𝑡 ) = 13.34 𝐴

The phase voltage of a delta is the same as the line voltage.

𝐸𝑃(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

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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.

fundamentals of electrical engineering

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