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EE2001

CIRCUIT ANALYSIS

(2014/2015 Session)

Lecture 1 to 6

Dr Er Meng JooProfessor

School of EEEOffice : S1-B1c-90Tel : 67904529

E-mail : [email protected]://ntulearn.ntu.edu.sg



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11-Aug-14 2

CONSULTATION HOURS

Academic Year 2014/2015 (Semester 1)

S/No Day of Week Time

1. Monday 3.30 pm – 4.30 pm

2. Thursday 3.30 pm – 4.30 pm

3. Friday 3.30 pm – 4.30 pm



11-Aug-

Learning Objective

• This courses focuses on the fundamental principles

of circuit theorems and circuit elements, DC/AC andthree-phase circuits, transient and steady-stateresponses, circuit analysis using Laplace Transforms

• In this course, we will learn various techniques(“tools”) to analyze the operation of real circuits.

• Our major concern is the analysis of circuits, i.e. thestudy of the behavior of the circuit, not the creationof circuits, i.e., the engineering design of the circuit.

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11-Aug-

Learning Outcome

• This course equips students with the knowledge for

the analysis of DC and AC linear circuits.

• Students would be able to set up independentequations of linear circuits and solve them using thetechniques and skills acquired in this course.

•

A sound knowledge of the analytical techniqueslearnt in this course can serve as good foundationof the study of linear control systems, powernetworks, electronics and communication systemsin later years.

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11-Aug-

11-Aug-14 6

New CA Weightage System

1. CA weightage is 40 % and final-examinationweightage is 60 %.

2. Two laboratory modules (L2001A and L2001B)are integrated in EE2001.

3. Summary:

Quiz Two take-

homeassignments (5% each)

Practical works

(L2001A andL2001B-5 %each)

Final

examination

20 % 10 % 10 % 60 %



11-Aug-

Quiz

• The Quiz mark contributes to 20 % of the final mark

• One quiz will be conducted after the recess week duringthe tutorial class i.e. Week 8 (6 to 10 Oct., 2014).

• Topics to be tested: Tutorials 1 to 6.

• Students can only take Quiz in their respective tutorialgroups.

• Students must present their ID with photo for takingattendance.

• Zero marks will be given for absentees without validreasons or MC’s.

• Absentees must inform the tutor through email withinthe same day or earlier of the Quiz to request for amake-up quiz.

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11-Aug-

Assignments

• Two take-home assignments.

• Assignment questions will be distributed by tutors during

the tutorial class.

• One from Part 1 (5%) and another from Part II (5%).

• Assignment 1 will be given out in Week 6 while Assignment 2 will be given out in Week 9.

• Students will be given two weeks to complete eachassignment

• Students have to submit solutions to their respectivetutors on or before the end of two weeks i.e. Week 8 and11 for Assignments 1 and 2 respectively.

• Late submissions will not be entertained.

• Marked assignments will be returned to students.

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11-Aug-

Laboratory Modules

• Two laboratory modules are integrated in thiscourse.

• L2001A: Circuit Theorems and Time Responses of Passive Networks (5%).

• L2001B: Two-port Network Parameters andTransient Response of a General Second-order

Circuit (5%).

• Each laboratory module contributes to 5 % of thefinal mark.

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11-Aug-

11-Aug-14 10

Text Book :

Charles K. Alexander and Matthew N.O. Sadiku, “Fundamentals of ElectricCircuits”, McGraw Hill, 5th Edition.

TK454.A375 2013 x5, Lee Wee Nam

Library, Reserves.



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11-Aug-14 11

References:

1. James W. Nilsson and Susan A. Riedel, “ElectricCircuits”, 9th Edition, Pearson/Prentice- Hall,2011. TK454.N712 2011 x7, LWNL, Reserves.

2. William H. Hart, Jr., Jack E. Kemmerly andSteven M. Durbin, “Engineering Circuit

Analysis”, 8th Edition, McGraw-Hill, 2012.

TK454.H426 2012 x2, LWNL, Reserves.

3. M. Nahvi and J.A. Edminister, “Schaum’s OutlinesElectric Circuits”, 5th Edition, McGraw-Hill, 2011 .TK454.E24 2011 x1, LWNL, Reserves.



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Outline

2. Basic Concepts

3. Basic Laws

4. Methods of Analysis

5. Circuit Theorems

6. Operational Amplifiers

1. Introduction



11-Aug-

Introduction

• Electric circuit theory is one of the fundamental

theories upon which all branches of electricalengineering are built.

• Many branches of electrical engineering, such aspower, electric machines, control, electronics,communications and instrumentation are basedon electric circuit theory.

• In electrical engineering, we are often interested

in transferring energy from one point to another.

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11-Aug-

Introduction• An interconnection of electrical devices is

required for transfer of energy.

• Such interconnection is referred to as an electric circuit and each component of the circuit isknown as an element .

• An electric circuit is an interconnection of electric elements.

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11-Aug-

A Simple Electric Circuit

It consists of three basic elements: abattery, a lamp and some connectingwires.

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11-Aug-

A Notch Filter with Op-amps

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11-Aug-

An Amplifier Circuit for a Microphone

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11-Aug-

11-Aug-14 18

Outline

2. Basic Concepts

3. Basic Laws


5. Circuit Theorems


1. Introduction



11-Aug-

Units

• When taking any

measurements, wemust use units toquantify values.

• We use theInternational Systems of Units (SI for short)

• Prefixes on SI unitsallow for easyrelationships betweenlarge and small values.

19



11-Aug-

Basic Concepts

• In carrying out circuit analysis, we often

deal with currents, voltages or power.We start with a brief description of these quantities.

Electric Charge and Current

• Electric charge and its movement arethe most basic items of interest inelectrical engineering.

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11-Aug-

• Charge : An electric property of matter,

measured in coulombs. Like charges repeland unlike charges attract each other.The magnitude of the electron’s charge is1.602 x 10-19 coulomb (unit is C).

• Current : Measures movement of charges. Current is measured in amperes(designated as A). One ampere is themovement of charges through a surfaceat the rate of 1 C/sec.11-Aug-14 21

Concept of Charge and Current



11-Aug-

Relationship Between Charge and Current

• The relationship between current i ,

charge q and time t is

• The direction of current flow is taken by convention as opposite to the directionof electron flow.

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11-Aug-

Two Methods of Representing a Current

Conventional currentflow:

(a) + ve current flow

(b) – ve current flow

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(a) (b)



11-Aug-

Type of Current

• A Direct Current (dc) is a

current that remains constantwith time and is denoted byI .

• A common source of DC is abattery.

• An Alternating Current (AC) is

a current that variessinusoidally with time and isdenoted by i .

• Mains power is an example of AC

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11-Aug-

Concept of Voltage• We are interested in the potential difference (voltage )

between two points, not the absolute potential of a

point.

• The voltage between two points a and b in acircuit is the energy (or work) needed to move a unitcharge from a to b .

• The relationship between the energy w (in joules, J)

and the charge (in C) is

.

• Thus, voltage (measured in volts, V) is the energyrequired to move a unit charge through an elementand one V = one J/C.

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11-Aug-

• The + and – signs are used to designate which point is atthe assumed higher potential (the + point).

• The can be interpreted as follows: The potential atpoint a with respect to point b is .

• An arrow is used to point to the terminal of assumedhigher potential (the + point).

• Suppose 9 . We may say that there is a 9-Vvoltage rise from b to a or equivalently a 9-V voltage drop from a to b .

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Concept of Voltage



11-Aug-

Type of Voltage

• A constant voltage is called a dc voltage and isrepresented by V whereas a sinusoidally time-

varying voltage is called an ac voltage and isrepresented by v .

• A dc voltage is commonly produced by a battery.

• An AC voltage is produced by an electric generator.

• Note: Electric voltage is always across the element or between two points.

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11-Aug-

Concept of Power and Energy

• Although current and voltage are the two basic

variables in an electric circuit, they are notsufficient for circuit analysis. Power and energycalculations are also needed.

• Power: Power is the rate of absorbing orsupplying energy (measured in Watts (W)), i.e.

where p is the power, w is the energy(in Joules) and t is time (in sec).

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11-Aug-

• The relationship between power p andvoltage v and current i is given by

• The power absorbed or supplied byan element is the product of voltageacross the element and the currentpassing through it.

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Relationship Between Power, Voltage and Current



11-Aug-

Passive Sign Convention

• By convention, we say that anelement being supplied power has

positive power.

• A power source such as a batteryhas negative power.

• Passive sign convention is

satisfied if the direction of currentis selected such that currententers through the terminal that ismore positively biased.

30



11-Aug-


• Consider the element (represented by a block)

as shown:

• It has two terminals (also called nodes).

• It conducts current from one node to the otherand in the process voltage drop occurs acrossthe element in the direction of current flow(shown by arrow).

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11-Aug-

• The terminal at which the current enters acquires+ve polarity with respect to the terminal at

which the current exits.

• We assume that the current enters the terminalof higher potential.

• We use the passive sign convention. Theword “passive” means that the element isassumed to absorb power. The power assumes a+ve sign when the current enters the +vepolarity of the voltage across an element.

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11-Aug-

Element Absorbing Power

• The (actual ) current flows into the +veterminal of the element:

4 2 8

• The element is absorbing 8 W of power.

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11-Aug-

Basic Circuit Elements

• An element is the basic building block of a circuit.

• An electric circuit is simply an inter-connection of the elements.

• Circuit analysis is the process of determiningvoltages across (or the current through) the

elements of the circuit.

• There are two types of elements found in electriccircuits: passive and active elements.

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11-Aug-

Passive and Active Elements

• An active element is capable of generating energywhile a passive element is not.

• Passive elements: resistors, capacitors and inductors.

• Active elements: generators, batteries andoperational amplifiers.

•

Most important active elements: voltage or currentsources that deliver power to the circuit.

• Two kinds of sources: independent anddependent sources.

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11-Aug-

Independent Source

• An ideal independent source is anactive element that provides aspecified voltage or current that iscompletely independent of othercircuit elements.

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11-Aug-

Symbols for Independent Voltage Source

(a) Used for constant ortime-varying voltage

(b) Used for constantvoltage (dc).

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11-Aug-

Independent Current Source

• An ideal independent current source isan active element that provides aspecified current completelyindependent of the voltage across thesource.

• The current source delivers to thecircuit whatever voltage is necessary tomaintain the designated current.

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11-Aug-

• Note that the arrow

indicates the directionof current flow (acurrent source requiresthat a direction bespecified).

• The terminal voltage is

determined by thecondition of the circuitto which it is connected.

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Symbols for Independent Current Source



11-Aug-

Dependent Source

• An ideal dependent (or controlled)

source is an active element in whichthe source quantity is controlled byanother voltage or current elsewhere inthe circuit.

• Dependent sources are used a greatdeal in electronics to model both dc andac behavior of transistors, especially inamplifier circuits.

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11-Aug-

Type of Dependent Source

• A dependent source has its outputcontrolled by an input value.

• Symbolically represented as adiamond

• Four types:

A voltage-controlled voltagesource (VCVS).

A current-controlled voltage

source (CCVS). A voltage-controlled currentsource (VCCS).

A current-controlled currentsource (CCCS).

43



11-Aug-

Example 1

• The source on the right-hand side is a

current-controlled voltage source.• The value of the voltage supplied is 10i V

(not 10i A) as it depends on the current i through element C .

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11-Aug-

Power Absorbed or Delivered in a Circuit

The algebraic sum of power in a circuit is zero i.e.

Total power supplied + Total power absorbed = 0

Example 3

Calculate the power supplied or absorbed by eachelement in the following circuit.

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11-Aug-

20 5 100 W , power supplied.

12 5 60 W, power absorbed

8 6 48 W, power absorbed

For , note that the voltage is 8 V (+ve at the top), thesame as the voltage for since both the passiveelement and the dependent source are connected to thesame terminals. Since the current flows out of the +veterminal,

8 0.2 8 0.25 8 W

Note : Power supplied + Power absorbed

= -100-8+60+48 = 0.

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Outline

2. Basic Concepts

3. Basic Laws


5. Circuit Theorems


1. Introduction



11-Aug-

Basic Laws

• Ohm’s Law

• Kirchoff’s Current Law

•

Kirchoff’s Voltage Law

• Some commonly used techniques

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11-Aug-

Ohm’s Law

• A resistor is a device that resists theflow of current flow. It is the simplestpassive element.

• Resistors can be used to control currentflow in a circuit.

• The physical property or ability to resistelectric current is known as resistance (represented by the symbol R ) and ismeasured in ohms (designated as Ω).

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11-Aug-

Resistor and its circuit symbol

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Circuit Symbol of Resistor



11-Aug-

Resistivity

• Materials tend to resist the flow of

electricity through them.• This property is called “resistance”.

• The resistance of an object is a functionof its length, l , cross sectional area, A and the material’s resistivity, :

52

l R



11-Aug-

Resistivity of Common Materials

53



11-Aug-

Ohm’s Law

• Ohm’s law states that the voltage v across a

resistor is directly proportional to the current i flowing through the resistor i.e. ∝ .

• Ohm defines the constant of proportionality fora resistor to be the resistance, R .

• The mathematical form of Ohm’s law is

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11-Aug-

Application of Ohm’s law

• To apply Ohm’s law, we must pay careful

attention to the current direction and voltagepolarity.

• The direction of current i and the polarity of voltage must conform with the passive signconvention i.e. the current flows from a higherpotential to a lower potential so that .

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11-Aug-

Short Circuit and Open Circuit

• Since the value of R can range from zero to

infinity, it is important that we consider thetwo extreme possible values of R.

A short circuit is a circuit element with resistance approaching zero, R = 0.

An open-circuit is a circuit element with resistance approaching infinity, R = .

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11-Aug-

Short Circuit and Open Circuit

For a short circuit,

0 , i.e. thevoltage is zero butthe current could beanything.

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For an open circuit,

⟶

0 , i.e. the

current is zero thoughthe voltage could beanything.



11-Aug-

Conductance

• A useful quantity in circuit analysis is

the reciprocal of resistance R , known asconductance,

.

• Conductance is the ability of an elementto conduct electric current and is

measured in Siemens (S).

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11-Aug-

Power Dissipated by a Resistor

• The power dissipated by a resistor can

be expressed as

• The power dissipated in a resistor isalways +ve i.e. a resistor alwaysabsorbs power from the circuit.

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11-Aug-

Example 1

In the following circuit, calculate the current

i , the conductance G and the power p .

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11-Aug-

Solution:

6 10 6

1

1

510 0.210 0.2

180

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11-Aug-

Branches, Nodes and Loops• Branch

A branch represents a single two-terminal element

like a voltage source or a resistor.

• Node

A node is the point of connection between two or morebranches. A node is indicated by a dot in a circuit.

• Loop

A loop is a closed path formed by starting at a node,passing through a set of nodes and returning to thestarting node without passing through any node morethan once.

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11-Aug-

Example 2

Consider the following circuit:

We can redraw the circuit as shown:

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11-Aug-

• The circuit has 5 branches: the 10-V voltagesource, the 2-A current source and the 3

resistors.• The circuit has 3 nodes: a, b and c .

• A node is indicated by a dot in a circuit.

• Note that the 3 points that form node b areconnected by wires and therefore constitute asingle point. The same is true for the 4 pointsthat form node c .

• abca with the 2 Ω resistor is a loop.

• bcb with the 2 Ω and 3 Ω resistors is a loop.

• bcb with the 3 Ω resistor and the 2- A current

source is a loop.

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11-Aug-

• Mesh

A mesh is a loop which does not contain any

other loops within it (it is an independent loop). For the above example: abca with the 2 Ω resistor

is a mesh; bcb with the 2 Ω and 3 Ω resistors is amesh; bcb with the 3 Ω resistor and the 2- A current source is a mesh.

In a circuit with b branches and n nodes, the

number of meshes is 1. For the above example, 5 3 1 3.

Independent loops result in independent sets of equations (to be used later).

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11-Aug-

Series and Parallel Connection

• Two or more elements are in series if

they exclusively share a single node andconsequently carry the same current.

• Two or more elements are in parallel if they are connected to the same twonodes and consequently have the samevoltage across them.

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11-Aug-

• The 10-V source and the 5 Ω resistor are in series asthe same current flows through them.

• The 2 Ω and 3 Ω resistors and the 2-A current sourceare in parallel because they are connected to the sametwo nodes b and c and consequently have the samevoltage across them.

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Series and Parallel Connection



11-Aug-

Example 3

How many branches and nodes does thethe following circuit have ? Identify theelements that are in series and parallel.

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11-Aug-

Solution:

• There are 5 branches and 3 nodes in thecircuit.

• The 1 Ω and 2 Ω resistors are in parallel.

• The 4 Ω resistor and the 10-V source arealso in parallel.

11-Aug-14 69



11-Aug-

Kirchoff’s Current Law

• Kirchoff’s Current Law (KCL) states thatthe algebraic sum of the currents

entering any node is zero.

• Consider the node in the figure. Applying KCLyields

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11-Aug-

11-Aug-14 71

• An alternative form of KCL: The sum of

the currents entering a node is equal to the sum of the currents leaving the node .

• KCL is based on conservation of charge.



11-Aug-

• Kirchoff’s Voltage Law (KVL) states that thealgebraic sum of all voltages around a loop

in a specified direction is zero.

•

Express the loop current in the clockwise (CW)direction (preferred direction).

• We take voltage rise as –ve and voltage drop as+ve.

• KVL is based on conservation of energy.

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11-Aug-

• Begin at the -source and go CW

around the loop applying KVL:

• An alternative form of KVL: Around a loop in a clockwise direction, the sum of voltage rises equals to the sum of voltage drops.

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11-Aug-

Example 5

Consider the voltage sources as shown.

Applying KVL (in the CW direction) 0

⟹

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11-Aug-

• To avoid violating KVL, a circuit cannot contain two different voltages sources in parallel unless their terminal voltages are the same.

Valid Invalid Invalid

11-Aug-14 77



11-Aug-

Example 6

• Determine and in the following

circuit.

• Express the current in the CW direction.

11-Aug-14 78



11-Aug-

Series Resistor and Voltage Divider

• For 2 resistors connected in series, the

equivalent resistance of these two resistors is the sum of the individual resistances , i.e., .

11-Aug-14 80



11-Aug-

• The above two circuits are equivalent asthey have the same v-i relationship at the

terminals a-b i.e.

because

.

• An equivalent circuit is useful in simplifyingthe analysis of a circuit.

• For N resistors connected in series, we

have ⋯ .

11-Aug-14 81

Equivalent Resistance



11-Aug-

• For the above circuit with 2 resistors connected

in series, we have

• The method is known as the voltage divider. Voltages across the two resistors can be easilydeduced by this method.

11-Aug-14 82

Voltage Divider



11-Aug-

Example 7

Using the voltage divider as shownabove, calculate so that 0.75

when 100 Ω.

We have

0.75.

So,

0.75 ⟹ 300 Ω.

11-Aug-14 83



11-Aug-

Parallel Resistor and Current Divider

• For 2 resistors connected in parallel, the

equivalent resistance of these two resistors is the product of their resistances divided by their sum, i.e.

.

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11-Aug-

• For the circuit with 2 resistors connected in parallel, wehave

• The circuit is known as the current divider.

• Note that when 0 (a short circuit has occurred), 0 and i.e. the input current will bypass

and flow through the short-circuited path.

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Current Divider



11-Aug-

Example 8

• Calculate the equivalent resistance R ab in thefollowing circuit:

• The 3 Ω and 6 Ω resistors are in parallel asthey are connected to the same two nodes c and b . Their combined resistance is

3 Ω||6 Ω

2 Ω

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11-Aug-

• Also, the 12 Ω and 4 Ω resistors are in

parallel as they are connected to the same

two nodes d and b . Hence, 12 Ω||4 Ω

3 Ω

• The 1 Ω and 5 Ω resistors are in series;hence, their equivalent resistance is 1 + 5 =6 Ω. With these combinations, we have thefollowing circuit.

11-Aug-14 88



11-Aug-

• The 3 Ω and 6 Ω resistors in parallel gives 2 Ω.This 2 Ω equivalent resistance is now in series

with the 1 Ω resistance to give a combinedresistance of 1 + 2 = 3 Ω. With these, we

have the following circuit.

• The 2 Ω and 3 Ω resistors in parallel gives1.2 Ω. This 1.2 Ω resistor is in series with the10 Ω resistor so that 10 1.2 11.2 Ω.

11-Aug-14 89



11-Aug-

Example 9

Find and in the following circuit.Calculate the power dissipated in the 3 Ω

resistor.

11-Aug-14 90



11-Aug-

• First, we need to find the total current i . The6 Ω and 3 Ω resistors in parallel gives 2 Ω. The

circuit can be simplified to

•

Using Ohm’s law,

2 A and 2 4 V . Also,

. The power

dissipated by the 3Ω resistor is

4

5.33 W.

11-Aug-14 91



11-Aug-

Y- Δ (Wye-Delta) Transformation

• There are cases in circuit analysis where theresistors are neither in parallel nor in series(see the following bridge circuit).

• Many circuits of the type shown can besimplified to a three-terminal equivalentcircuit shown next.

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11-Aug-

• These are the Y or Tee (T) network and ∆ orpi (Π) network as shown below.

Y or T network

∆ or Π network

11-Aug-14 93

Y- Δ (Wye-Delta) Transformation



11-Aug-

Δ to Y conversion ( Δ known)

• If it is more convenient to work with a Y network in a place where the circuit contains a ∆

configuration, then we superimpose a Y network on the existing ∆ network and find the equivalentresistances in the Y network.

• To obtain the equivalent resistances in the Y network, we compare the two networks and makesure that the resistance between each pair of nodes in the ∆ network is the same as theresistance between the same pair of nodes in the Y network. For example, for terminals 1 and 2, Y and Δ || .Then, we set Y Δ .

11-Aug-14 94



11-Aug-

• The conversion formula for a delta to wye

transformation is:

95

1

2

3

b c

a b c

c a

a b c

a b

a b c

R R R

R R R

R R R

R R R

R R R

R R R

Δ to Y conversion ( Δ known)



11-Aug-

• The conversion formula for a wye to

delta transformation is:

96

1 2 2 3 3 1

1

1 2 2 3 3 1

2

1 2 2 3 3 1

3

a

b

c

R R R R R R R

R

R R R R R R R

R

R R R R R R R

R

Y to Δ Conversion (Y known)



11-Aug-

Example 10

Obtain the equivalent resistance forthe circuit shown and use it to findcurrent i .

11-Aug-14 97



11-Aug-

Note that there are 2 Y networks (one at n andone at c ) and 3 delta networks (can , abn , cnb ).

We only need to transform one Y network comprising the 5Ω, 10 Ω and 20 Ω resistors to a ∆ network. We let 10Ω , 20Ω , 5Ω and we obtain the following using theconversion formulae:

35 Ω

17.5 Ω

70 Ω

11-Aug-14 98



11-Aug-

With the Y converted to ∆ , the equivalentcircuit (with the voltages source removedfor now) is shown as follows:

11-Aug-14 99



11-Aug-

1

Combining the 3 pairs of resistors in parallel, weobtain

70||30 = 21 Ω, 12.5||17.5 = 7.292 Ω,

15||35 = 10.5 Ω

so that the equivalent circuit is as shown.

Hence , 7.292 10.5 ||21 9.632 Ω.

Then,

. 12.458 A.

11-Aug-14 100



11-Aug-

1

Methods of Analysis

• Nodal Analysis

• Mesh Analysis

11-Aug-14 102



11-Aug-

1

Overview

• With Ohm’s and Kirchoff’s law established, theymay now be applied to circuit analysis.

• Two techniques will be presented in this chapter:

Nodal analysis, which is based on Kirchoff Current Law (KCL)

Mesh analysis, which is based on Kirchoff Voltage Law (KVL)

• Any linear circuit can be analyzed using these two

techniques.• The analysis will result in a set of simultaneous

equations which may be solved by Cramer’s rule orcommercial software such as MATLAB.

103



11-Aug-

1

Nodal Analysis

• It is based on the application of KCL.

• It uses node voltages as the circuitvariables.

• In nodal analysis, we are interested infinding the node voltages.

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11-Aug-

1

Nodal Analysis Without Voltage Sources

• To simplify matters, we begin with a circuit with n nodes without voltage sources. The nodal analysisof the circuit is as follows:

Select a node as the reference node. Assignvoltages , , … , to the remaining n- 1

nodes. The voltages are referenced withrespect to the reference node .

Apply KCL to each of the (n –1) non-reference

nodes. Use Ohm’s law to express the branchcurrents in terms of the node voltages.

Solve the resulting simultaneous equations toobtain the unknown node voltages.

11-Aug-14 105



11-Aug-

1

Example 1

Calculate the node voltages in the circuit asshown. Node 0 is the reference node ( 0),while node 1 and 2 are assigned voltages and respectively. These node voltages are

defined with respect to the reference node.

11-Aug-14 108



11-Aug-

1

Redraw the circuit by adding , and as the currentsthrough resistors , and respectively.

Applying the KCL to each non-reference node:Node 1: 0 (1)

Node 2: 0 (2)

11-Aug-14 109



11-Aug-

1

Now apply Ohm’s law to express the unknown currents, and in terms of the node voltages:

,

,

,

Using these in (1) and (2) gives

0

0

Rearranging these yields

11-Aug-14 110



11-Aug-

1

We can now solve for the node voltages and

using the elimination method or Cramer’s rule.

The equations can be rewritten into the matrix form:

A X B

Here, is a 2x2 matrix and and are 2x1 columnvectors. We can then use the Cramer’s rule to solve forthe node voltages and .

11-Aug-14 111



11-Aug-

1

Redraw the circuit as shown. Note that the labeling ofcurrents is arbitrary.

Applying the KCL and Ohm’s law:Node 1: 0 ⟹ 5

0

Node 2: 0 ⇒

10 5

11-Aug-14 113



11-Aug-

1

The two equations can be written as follows:

5 (1)

5 (2)

=

1 5

5

5

5

=

2 5

5

5 = 5

⟹ =

=

=13.33 ; =

=20

11-Aug-14 114



11-Aug-

1

Example 3

Determine the nodal voltages of the

following circuit.

11-Aug-14 115



11-Aug-

1

Redraw the circuit as follows:

Node 1:

3 0 ⇒ 3

0 (1)

Node 2: 0,

⇒

0 (2)

Node 3: 2 0,

⇒

0 (3)

11-Aug-14 116



11-Aug-

1

Equations (1) – (3) can be written as follows:

3

0

0

In matrix form, we have

3

0

0

11-Aug-14 117



11-Aug-

1

Here, is a 3x3 matrix, and are 3x1column vectors.

Using Cramer’s rule, we have

5

128

1

0

0

11-Aug-14 118



11-Aug-

1

2

3

0

0

3

32

3

3

0

0

3

32

⇒

4.8 ,

2.4

2.4 .

11-Aug-14 119



11-Aug-

1

We now consider how voltage sources

affect nodal analysis.Consider the following circuit.

11-Aug-14 120

Nodal Analysis Without Voltage Sources



11-Aug-

1

• If a voltage source is connected between thereference node and a non-reference node ,

the voltage at the non-reference node isequal to that of the voltage source, i.e., 10 V.

• If the voltage source (dependent orindependent) is connected between 2 non-

reference nodes , the 2 non-reference nodesform a supernode.

• We apply both KCL and KVL to determine thenode voltages.

11-Aug-14 121



11-Aug-

1

• In the circuit shown above, nodes 2 and 3form a supernode (indicated by the region

enclosed by the broken line).• We analyze the circuit with a supernode using

the same 3 steps mentioned before exceptthat the supernode is treated differently inthat we apply KCL to both the nodes bynoting that all currents flowing into the regionsum to zero.

• At the supernode of the circuit shown, 0

⟹

0 (1)

11-Aug-14 122



11-Aug-

1

• The voltage source inside the supernodeprovides a constraint equation , i.e. it

constrains the difference between the nodevoltages at these two nodes to be equal tothe voltage of the source i.e.

5 (2)

• The constraint equation is needed to solve for

the unknown node voltages.

11-Aug-14 123



11-Aug-

1

Equation (1) gives

(3)

Using (2) and 10 V gives

5

⟹ 9.2 V and 5 4.2 V.

11-Aug-14 124



11-Aug-

1

Example 4

Consider the following circuit, find the

node voltages.

11-Aug-14 125



11-Aug-

1

• The supernode contains the 2-V source,

nodes 1 and 2 and the 10Ω resistor asshown in the following circuit.

11-Aug-14 126

supernode



11-Aug-

1

Applying KCL to the supernode gives

2 7 0

Expressing and in terms of the nodevoltages gives

2

7 0 ⟹

5

or 20 2 (1)

The constraint equation provided by the voltage

source in the supernode is 2 (2)

Using (2) in (1) gives

3 22 ⇒ 7.33

and 2 5.33 . 127



11-Aug-

1

11-Aug-14 128

Note that the 10 Ω resistor does not make anydifference because it is connected across the

supernode.

Note: Nodal analysis is a straightforward analysis technique when only current sources are present and voltage sources are easily accommodated with the supernode concept.



11-Aug-

1

Mesh Analysis

• Recall that a mesh is a loop that does not

contain any other loops within it .• The current through the mesh is known

as the mesh current .

• Mesh analysis provides another generalprocedure for analyzing circuits usingmesh currents as the circuit variables.

• Mesh analysis applies KVL to findunknown currents and is only applicableto planar circuits .

11-Aug-14 129



11-Aug-

1

• A planar circuit is one that can be

drawn in a plane with no branchescrossing one another; otherwise, it isnonplanar .

A nonplanar circuit

11-Aug-14 130



11-Aug-

1

• A circuit that is drawn with crossingbranches still is considered planar if itcan be redrawn with no crossoverbranches.

11-Aug-14 131

⟹



11-Aug-

1

Comments

• The direction of the mesh currents isarbitrary; it can be Clockwise (CW) orCounter Clockwise (CCW) and thechoice will not affect the validity of thesolution.

• We assume the mesh current flows CW.

11-Aug-14 133



11-Aug-

1

Example 5


11-Aug-14 134



11-Aug-

1

Applying KVL to each mesh:

Mesh 1: 0 (1)

Mesh 2: 0 (2)

⟹

The branch currents can be calculated asfollows:

, ,

11-Aug-14 136



11-Aug-

1

Example 6

Use mesh analysis to find the current

in the following circuit:

11-Aug-14 137



11-Aug-

1

Need to find , first.

Applying KVL to each mesh:

Mesh 1: 24 10 12 0⟹ 11 5 6 12

Mesh 2: 24 4 10 0

⟹ 5 19 2 0

Mesh 3: 4 12 4 0

⟹ 2 0

Solving for the mesh currents gives 2.25 , 0.75 and 1.5 . Then,

2.25 0.75 1.5 .11-Aug-14 138



11-Aug-

1

• Case 1: When a current source exists

only in one mesh.


11-Aug-14 139

Mesh Analysis With Current Sources



11-Aug-

1

• It is clear that 5 A (thus we eliminatemesh 2 from consideration).

• The mesh equation for mesh 1 is

10 4 6 0 ⟹ 2 .

Note: We could have calculated the currenti

using nodal analysis, but we have to find the

node voltage v first (using

5

0) and then find the current using i 10

v/4 . In this case, mesh analysis is simpler.

11-Aug-14 140



11-Aug-

1

• We create a supermesh by excluding thecurrent source and any elements connected

in series with it as shown. A supermesh results when two meshes have a (dependent or independent ) current source in common.The current source and the element connected in series with it is in the interior of the supermesh.

11-Aug-14 142



11-Aug-

1

Applying KVL to the supermesh gives20 6 10 4 0

⟹ 6 14 20 (1)

11-Aug-14 143

Note: The voltages around the supermeshare in terms of the original mesh currents .



11-Aug-

1

Applying KCL to a node in the branch where thetwo meshes intersect. At node 0 (or at the top

node), 6 (2)

The current source in the supermesh providesthe constraint equation (2). Using (2) in (1)gives 3.2 , 2.8 .

Note: Again, we could have used nodal analysisto find the currents i and i.

11-Aug-14 144



11-Aug-

1

Selecting an Appropriate Approach

• In principle, both the nodal analysis andmesh analysis are useful for any givencircuit.

• What is the more efficient method forsolving a circuit problem?

• There are two factors that dictate the bestchoice:

The nature of the particular network is thefirst factor

The second factor is the informationrequired

145



11-Aug-

1

• If the network contains:

Many parallel connected elementsCurrent sources

Supernodes

Circuits with fewer nodes than meshes

• If node voltages are what are being solved for.

• Non-planar circuits can only be solved using

nodal analysis.• This format is easier to solve by a computer.

147

When to Use Nodal Analysis ?



11-Aug-

1

Circuit Theorems

• Linearity Property

• Superposition

• Source Transformation

• Thevenin’s Theorem

• Norton’s Theorem

• Maximum Power Transfer

11-Aug-14 149



11-Aug-

1

• The resistor is a linear element because

it satisfies both the scaling and theadditivity properties.

• For , if is plotted as a functionof , the result is a straight line i.e. thev-i relationship is linear.

11-Aug-14 151

Linearity Property



11-Aug-

1

• In general, a circuit is linear if it has both theadditivity and scaling properties.

• A linear circuit is one whose output is directlyproportional to its input.

• A linear circuit consists of linearelements, linear dependent sources andindependent sources.

• A linear dependent source is a source

whose output current or voltage isproportional only to the first power of aspecified current or voltage variable in thecircuit (i.e. ∝ or ∝ .

11-Aug-14 152

Linearity Property



11-Aug-

1

Example 1

For the circuit shown, find when 15 and

30 A.

11-Aug-14 153



11-Aug-

1

Superposition

• The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

• The superposition principle helps us to

analyze a linear circuit with more than oneindependent source by calculating thecontribution of each independent sourceseparately.

11-Aug-14 155



11-Aug-

1

Steps to Apply Superposition Principle

1. Turn off all independent sources except onesource. Find the output (voltage or current)due to that active source using thetechniques learned previously.

2. Repeat step 1 for each of the otherindependent sources.

3. Find the total contribution by addingalgebraically all the contributions due to theindependent sources.

11-Aug-14 156



11-Aug-

1

Points to Note:

• We consider one independent source at atime while all other independent sources are

turned off. This implies that we replace everyvoltage source by 0 V (or a short circuit), andevery current source by 0 A (or an opencircuit) i.e.

voltage source short circuit

current source open circuit

• One major disadvantage of usingsuperposition is that it involves more worksalthough it helps to reduce a complex circuitto simpler circuits.

11-Aug-14 157



11-Aug-

1

Example 2

Use the superposition to find in the

following circuit.

11-Aug-14 158



11-Aug-

1

Since there are two independent sources, let where and are the

contributions due to the 6-V voltage source andthe 3-A current source, respectively.

To obtain , we set the current source to zeroas shown.

11-Aug-14 159



11-Aug-

1

Source Transformation

• A source transformation is the processof replacing a voltage source

inseries with a resistor by a currentsource in parallel with a resistor , orvice-versa.

Transformation of independent sources

11-Aug-14 162



11-Aug-

1

• Thus,

in order for the two

circuits to be equivalent.

• Source transformation requires that

.

• Note: The arrow of the current sourceis directed toward the positive terminalof the voltage source.

11-Aug-14 164



11-Aug-

1

• Source transformation also applies todependent sources, provided wecarefully handle the dependent variable,

i.e.

.

Transformation of dependent sources11-Aug-14 165



11-Aug-

1

Example 3

Use source transformation to find in

the following circuit.

11-Aug-14 166



11-Aug-

1

• First transform the current and voltagesources to obtain the following circuit.

• Combining the 4Ω and 2Ω resistors in seriesand transforming the 12-V voltage sourcegives

11-Aug-14 167



11-Aug-

1

Combine the 2-A and 4-A current sources to geta 2-A source.

Using current division,

2 0.4

and

8 8 0.4 3.2 11-Aug-14 168



11-Aug-

1

Example 4

Find in the circuit shown using source

transformation.

11-Aug-14 169



11-Aug-

1

Transform the dependent current source andthe 6-V independent source as shown. Note

that the 18-V voltage source is not transformedbecause it is not connected in series with anyresistors.

11-Aug-14 170



11-Aug-

1

Applying KVL around the loop gives 3 1 4 18 0 (1)

Applying KVL to the loop containing only the 3- V voltage source, the 1Ω resistor, and yields

3 1 0 ⟹ 3 (2)

Using (2) and (1) gives

15 5 3 0 ⟹ 4.5

11-Aug-14 172



11-Aug-

1

11-Aug-14 174

Thevenin’s Theorem



11-Aug-

1

• Suppose Figs. 1(a) and 1(b) are equivalenti.e. they have the same v-i relationship attheir terminals.

• When the terminals a-b are made open-circuited (by removing the load), no currentflows so that the open-circuit voltage acrossthe terminals a-b in Fig. 1(a) must be equal

to the voltage source in Fig. 1(b) since thetwo circuits are equivalent. Thus, is the

open-circuit voltage across the terminals asshown i.e. .

11-Aug-14 175

How to Find V and R?



11-Aug-

1

• Again, with the load removed and terminalsa-b open-circuited, we turn off allindependent sources. The input resistance (orequivalent resistance) of the dead circuit atthe terminals a-b of Fig. 1(a) must be equalto in Fig. 1(b) because the circuits areequivalent. Thus, is the input resistance

at the terminals when the independent

sources are turned off as shown i.e. .

11-Aug-14 176

How to Find V and R?



11-Aug-

1

• To find the Thevenin resistance , we

need to consider two cases.

• Case 1: If the circuit has no dependent sources , we turn off all independentsources. is the input resistance of the circuit seen between terminals a -b

as shown in Fig. 2(b).

11-Aug-14 178

How to Find R?



11-Aug-

1

• Case 2: If the circuit has dependent sources ,we turn off all independent sources.Dependent sources are not turned off because they are controlled by circuitvariables. We apply a known voltage source(say, 1 V) at terminals a -b and determinethe resulting current . Then / as

shown.

• Alternatively, we may insert a known currentsource (say, 1 A) at terminals a-b asshown and find the terminal voltage and /.

11-Aug-14 179

How to Find R?



11-Aug-

1

11-Aug-14 180

How to R?



11-Aug-

1

• Consider a linear circuit terminated by a load as shown.

• The current through the load and thevoltage

across the load are easily

determined once the Thevenin equivalentcircuit at the load’s terminals is obtained asfollows:

,

11-Aug-14 181

Use of Thevenin Equivalent Circuit



11-Aug-

1

Example 5

Find the Thevenin equivalent circuit of the

following circuit seem by the load i.e. tothe left of the terminals a-b . Find for 6 Ω and 26 Ω.

11-Aug-14 182



11-Aug-

1

To find , remove the load. Turn off the 32-V

voltage source (replacing it with a short circuit)

and the 2-A current source (replacing it with anopen circuit). The circuit becomes

Thus,

4||12 1

1 4 Ω

11-Aug-14 1-183



11-Aug-

1

To find , consider the circuit as shown.

Note that no current flows through the 1 Ω

resistor (since the load is removed).

We can use mesh analysis to find .Mesh 1: 32 4 12 0

Mesh 2: 2 A ⟹ 0.5 A

12 12 0.5 2 30 V

11-Aug-14 184



11-Aug-

1

The Thevenin equivalent circuit is as shown.

The current through is

For 6Ω,

3 .

For 26Ω,

1 .

11-Aug-14 185



11-Aug-

1

The circuit contains a dependent voltagesource. To find , set the independent

current source equal to zero but leave thedependent voltage source alone. Furthermore,we excite the circuit with a voltage source 1 V across the terminals a-b as shown.

11-Aug-14 187



11-Aug-

1

The aim is to find the current through theterminals a-b and then obtain /.

Applying mesh analysis to the circuit:

Mesh 1: 2 2 0 or

But 4 ⟹ 3

Mesh 2: 4 2 6 0

Mesh 3: 6 2 1 0

11-Aug-14 188



11-Aug-

1

These equations can be written as follows:

3 0

2 12 6 0

6 8 1

But, we only need to solve for andsolving it gives

A ⟹

A

Hence,

/ 6Ω.

11-Aug-14 189



11-Aug-

1

To get , we find in the circuit as shown.

Note that no current flows through the 2 Ω

resistor on the right (with load removed).

11-Aug-14 190



11-Aug-

1

Applying mesh analysis:

Mesh 1: 5 A (1)

Mesh 2: 4 2 6 0 (2)

Mesh 3: 2 2 0 , ⟹

But 4 . (3)

From (1) – (3), we get12 2 20

3 20

11-Aug-14 191



11-Aug-

1

Norton’s Theorem

• Norton’s theorem states that a linear

two-terminal circuit can be replaced by an equivalent circuit consisting of a current source in parallel with a resistor , where is the short-circuit current through the terminals and is the input or equivalent resistance at the

terminals when the independent sources are turned off

11-Aug-14 193



11-Aug-

1

• Thus, as shown.

• Dependent and independent sourcesare treated the same way as inThevenin’s Theorem.

11-Aug-14 196

How to Find and R?



11-Aug-

1

• The Thevenin and Norton equivalent

circuits are related by a sourcetransformation.

• , ,

11-Aug-14 197

Thevenin vs Norton Equivalent



11-Aug-

1

Example 7

Find the Norton equivalent circuit of the

circuit as shown at the terminals a-b .

11-Aug-14 198



11-Aug-

1

We may find the same way as we find

in the Thevenin equivalent circuit. Setting the

independent sources to zero leads to thefollowing circuit:

5 8 4 8 5 20

4 Ω

11-Aug-14 199



11-Aug-

2

To find , we short circuit terminals a and b as

shown.

The 5Ω resistor can be ignored because it has

been short-circuited. Applying mesh analysis:Mesh 1: 2 A

Mesh 2: 12 4 88 0,

⇒ 1 ⇒ 1

11-Aug-14 200



11-Aug-

2

The Norton’s equivalent circuit is as shown.

11-Aug-14 201



11-Aug-

2

Open-circuit and Short-circuit Tests

• The open-circuit and short-circuit tests

are sufficient to find any Thevenin or Norton equivalent circuit of a given circuit which contains at least one independent source.

• We have , , /.

11-Aug-14 202



11-Aug-

2

Example 8

Use the open-circuit and short-circuit

tests to find the Thevenin equivalentcircuit of the following circuit acrossterminals a-b .

11-Aug-14 203



11-Aug-

2

Find as the open-circuit voltage across

terminals a and b as shown.

11-Aug-14 204

Open-circuit Test



11-Aug-

2

Applying mesh analysis:

Mesh 1:

5 A (1)

Mesh 2: 4 2 6 0 (2)

Mesh 3: 2 2 0 ⟹

But 4 (3)

From (1) – (3), we obtain

12 2 20

3 20

Solving for gives

A . We only need as

6. Hence, 6 20 V.

11-Aug-14 205



11-Aug-

2

To find , we short circuit terminals a and b as

shown.

11-Aug-14 206

Short-circuit Test



11-Aug-

2

The supernode contains the top two nodes, thedependent source and the 2 Ω resistor as shown.

At the supernode, applying KCL gives

5

0 (1)

The constraint equation is

2 (2)

Solving for from (1) and (2) gives

V and

A.

Thus,

6 Ω .

11-Aug-14 207



11-Aug-

2

The Thevenin equivalent circuit is as shown:

11-Aug-14 208



11-Aug-

2

Maximum Power Transfer

• Often we deal with small amount of

power in electronics and we want tomake full use of the power available.Obtaining the maximum power out of acircuit is very important.

• The Thevenin equivalent circuit is usefulin finding the maximum power a linear

circuit can deliver to a load.• We assume that we can adjust the load

resistance .

11-Aug-14 209



11-Aug-

2

Condition for Maximum Power Transfer

• Suppose that the entire circuit is

replaced by its Thevenin equivalentcircuit except for the load as shown.

• The power delivered to the load is

(1)

11-Aug-14 210



11-Aug-

2

• For a given circuit, and are

fixed. By varying the load resistance ,the power delivered to the load variesas shown.

11-Aug-14 211

Condition for Maximum Power Transfer



11-Aug-

2

• The maximum power delivered by a source tothe load occurs when is equal to i.e.

the Thevenin resistance at the terminals of theload.

• Using (2) in (1) yields the maximum power transferred (for :

• When , the power delivered to the loadis given by (1). i.e.,

11-Aug-14 213

Maximum Power Transfer



11-Aug-

2

Using the techniques of finding the Theveninresistance and the Thevenin voltage, we find

that 9 Ω and 22 V (exercise).

For maximum power transfer, 9 Ω

and the maximum power is

13.44 W

11-Aug-14 215



11-Aug-

2

Operational Amplifiers (Op-amps)

• Operational amplifiers are commonly used in

a large variety of electronic applications (acand dc signal amplification, active filters,oscillators, comparators and regulators).

• An op-amp is an active element of an electriccircuit.

• It acts like a voltage controlled voltagesource.

• Our focus is on the terminal behavior of theop-amp i.e. we are not interested in theinternal structure of the op-amp nor thecurrents and voltages that exist in thisstructure.11-Aug-14 217



11-Aug-

2

• Op-amps are commercially available in

IC packages in several forms. Thefollowing figure shows a typical op-amppackage which is an eight-pin dual in-line package (DIP) as shown.

Op-amp package Pin configuration

11-Aug-14 218

Operational Amplifiers (Op-amps)



11-Aug-

2

• Pin or terminal 8 is unused and terminals

1 and 5 are of little concern to us.

• The 5 important terminals are:

The inverting input, pin 2.

The non-inverting input, pin 3.

The output, pin 6.The positive power supply V+, pin 7.

The negative power supply V–, pin 4.

11-Aug-14 219

Pin Configuration of an Op-amp



11-Aug-

2

• The op-amp has two inputs and one

output.• The inputs are marked with minus (–)

and plus (+) to specify inverting andnon-inverting inputs respectively.

• An input applied to the non-invertingterminal will appear with the same

polarity at the output while an inputapplied to the inverting terminal willappear inverted at the output.

11-Aug-14 221

Circuit Symbol of an Op-amp



11-Aug-

2

• The op-amp is typically powered by a

voltage supply as shown.

11-Aug-14 222

Operation of an Op-amp



11-Aug-

2

• The power supplies are often ignored in

an op-amp circuit diagram for the sakeof simplicity. However, note that .

• The output voltage is limited by thevalues of the voltage sources i.e.

.

11-Aug-14 223

Operation of an Op-amp



11-Aug-

2

Ideal Op-amps

• To facilitate the understanding and

analysis of op-amp circuits, we consideronly ideal op amps.

• An ideal op-amp is an amplifier with infinite open-loop gain, infinite input resistance and zero output resistance,i.e., it has the following characteristics:

Infinite open-loop gain, ∞.

Infinite input resistance, ∞.

Zero output resistance, 0.

11-Aug-14 224



11-Aug-

2

• Because an op-amp has a very high open-

loop voltage gain, , negative feedback isusually considered to control the outputvoltage and to limit the voltage gain.

• A negative feedback is achieved when theoutput is fed back to the inverting terminalof the op-amp as shown.

• The voltage gain of the op-amp withnegative feedback is called closed-loopgain.

11-Aug-14 225

Ideal Op-amps



11-Aug-

2

For circuit analysis, the ideal op-amp is as

shown.

11-Aug-14 226

Analysis of the Ideal Op-amp



11-Aug-

2

• Two important characteristics of the ideal op-amp:

The currents into both the input terminals are zero, i.e. 0 , 0. This is due to the infinite input resistance.This implies that there is an open circuit between theinput terminals and current cannot enter the op amp.But, the output current is not necessarily zero, accordingto .

The voltage across the input terminals is equal to zero,i.e. 0 ⟹ .

• These two equations: (1) 0 & 0 and (2) are extremely important and should be

regarded as the key information to analyzing op-amp circuits.

11-Aug-14 227

Characteristics of the Ideal Op-amp



11-Aug-

2

Example 1

For the following circuit, find the voltage

gain /.

11-Aug-14 228



11-Aug-

2

• The non-inverting input is grounded,

is connected to the inverting inputthrough and the feedback resistor

is connected between the invertinginput and output.

• Applying KCL to node 1:

0 0 ⟹

0

11-Aug-14 229



11-Aug-

2

But, 0 for an ideal op-amp since

the non-inverting terminal is grounded.Hence,

⟹

The circuit is an inverting amplifier whichreverses the polarity of the input signalwhile amplifying it.

11-Aug-14 230



11-Aug-

2

Example 2

For the following circuit, find the voltage

gain /.

11-Aug-14 231



11-Aug-

2

The input voltage is applied directly at

the non-inverting input terminal andresistor is connected between theground and the inverting terminal.

Applying KCL to the inverting terminalgives

0 0

⟹

0

11-Aug-14 232



11-Aug-

2

The figure is redrawn as shown.

Note that .

Since 0, the 40 kΩ and 5 kΩ resistors are inseries, the same current flows through them.

Since is the voltage across the 5 kΩ resistor,

using the voltage division principle,

11-Aug-14 235



11-Aug-

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