Circuit Theory Notes Studyhaunters

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EINSTEIN COLLEGE OF ENGINEERINGTIRUNELVELI

DEPARTMENT OF EEE

EE 25 CIRCUIT THEORYII SEM EEE

Prepared By,

SHIBU J.V.BRIGHT,

LECTURER/EEE


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EE25 CIRCUIT THEORY 3 1 0 100

(Common to EEE, EIE and ICE Branches)

UNIT I BASIC CIRCUITS ANALYSIS 12

Ohms Law Kirchoffs laws DC and AC Circuits Resistors in series and parallel circuits Mesh

current and node voltage method of analysis for D.C and A.C. circuits.

UNIT II NETWORK REDUCTION AND NETWORK THEOREMS FOR DC AND AC CIRCUITS: 12

Network reduction: voltage and current division, source transformation star delta conversion.

Thevenins and Novton & Theorem Superposition Theorem Maximum power transfer

theorem Reciprocity Theorem.

UNIT III RESONANCE AND COUPLED CIRCUITS 12

Series and paralled resonance their frequency response Quality factor and Bandwidth - Selfand mutual inductance Coefficient of coupling Tuned circuits Single tuned circuits.

UNIT IV TRANSIENT RESPONSE FOR DC CIRCUITS 12

Transient response of RL, RC and RLC Circuits using Laplace transform for DC input and A.C. with

sinusoidal input.

UNIT V ANALYSING THREE PHASE CIRCUITS 12

Three phase balanced / unbalanced voltage sources analysis of three phase 3-wire and 4-wire

circuits with star and delta connected loads, balanced & un balanced phasor diagram of

voltages and currents power and power factor measurements in three phase circuits.

TOTAL :60 PERIODS

TEXT BOOKS:

1. William H. Hayt Jr, Jack E. Kemmerly and Steven M. Durbin, Engineering Circuits

Analysis,Tata McGraw Hill publishers, 6th

edition, New Delhi, (2002).

2. Sudhakar A and Shyam Mohan SP, Circuits and Network Analysis and Synthesis,Tata

McGraw Hill, (2007).

REFERENCES:

1. Paranjothi SR, Electric Circuits Analysis, New Age International Ltd., New Delhi, (1996).

2. Joseph A. Edminister, Mahmood Nahri, Electric circuits, Schaums series, Tata

McGraw-Hill, New Delhi (2001).

3. Chakrabati A, Circuits Theory (Analysis and synthesis), Dhanpath Rai & Sons, New Delhi,

(1999).

4. Charles K. Alexander, Mathew N.O. Sadik, Fundamentals of Electric Circuits, Second

Edition, McGraw Hill, (2003).


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UNIT I

BASIC CIRCUITSANALYSIS


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INTRODUCTION

The interconnection of various electric elements in a prescribed manner

comprises as an electric circuit in order to perform a desired function. The electric

elements include controlled and uncontrolled source of energy, resistors, capacitors,inductors, etc. Analysis of electric circuits refers to computations required to determine

the unknown quantities such as voltage, current and power associated with one or more

elements in the circuit. To contribute to the solution of engineering problems one must

acquire the basic knowledge of electric circuit analysis and laws. Many other systems,

like mechanical, hydraulic, thermal, magnetic and power system are easy to analyze and

model by a circuit. To learn how to analyze the models of these systems, first one needs

to learn the techniques of circuit analysis. We shall discuss briefly some of the basic

circuit elements and the laws that will help us to develop the background of subject.

BASIC ELEMENTS & INTRODUCTORY CONCEPTS

Electrical Network: A combination of various electric elements (Resistor, Inductor,

Capacitor, Voltage source, Current source) connected in any manner what so ever is

called an electrical network. We may classify circuit elements in two categories, passive

and active elements.

Passive Element: The element which receives energy (or absorbs energy) and then

either converts it into heat (R) or stored it in an electric (C) or magnetic (L ) field is called

passive element.

Active Element: The elements that supply energy to the circuit is called active element.Examples of active elements include voltage and current sources, generators, and

electronic devices that require power supplies. A transistor is an active circuit element,

meaning that it can amplify power of a signal. On the other hand, transformer is not an

active element because it does not amplify the power level and power remains same

both in primary and secondary sides. Transformer is an example of passive element.

Bilateral Element: Conduction of current in both directions in an element (example: Resistance;

Inductance; Capacitance) with same magnitude is termed as bilateral element.


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Unilateral Element: Conduction of current in one direction is termed as

unilateral (example: Diode, Transistor) element.

Meaning of Response: An application of input signal to the system will

produce an output signal, the behavior of output signal with time is known as

the response of the system

Linear and Nonlinear Circuits

Non-Linear Circuit: Roughly speaking, a non-linear system is that whose

parameters change with voltage or current. More specifically, non-linear circuit


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does not obey the homogeneity and additive properties. Volt-ampere

characteristics of linear and non-linear elements are shown in figs. 3.2 - 3.3. In

fact, a circuit is linear if and only if its input and output can be related by a

straight line passing through the origin as shown in fig.3.2. Otherwise, it is a

nonlinear system.

Potential Energy Difference: The voltage or potential energy difference

between two points in an electric circuit is the amount of energy required to

move a unit charge between the two points.

KIRCHHOFFS LAWS

Kirchhoffs laws are basic analytical tools in order to obtain the solutions

of currents and voltages for any electric circuit; whether it is supplied from a

direct-current system or an alternating current system. But with complex

circuits the equations connecting the currents and voltages may become so

numerous that much tedious algebraic work is involve in their solutions.

Elements that generally encounter in an electric circuit can be

interconnected in various possible ways. Before discussing the basic analytical

tools that determine the currents and voltages at different parts of the circuit,

some basic definition of the following terms are considered.


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Meaning of Circuit Ground and the Voltages referenced to Ground


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In many cases, such as in electronic circuits, the chassis is shorted to the earth

itself for safety reasons.

Voltage Divider


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Current divider


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Potentiometer and its function


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Ideal and Practical Voltage Sources


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Ideal and Practical Current Sources

Another two-terminal element of common use in circuit modeling is

current source` as depicted in fig.3.17. An ideal current source, which is

represented by a model in fig. 3.17(a), is a device that delivers a constant

current to any load resistance connected across it, no matter what the terminal

voltage is developed across the load (i.e., independent of the voltage across its

terminals across the terminals).


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Conversion of a Practical Voltage Source to a Practical Current source and vise-

versa


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Current source to Voltage Source


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Solution of Electric Circuit Based on Mesh (Loop) Current Method

Let us consider a simple dc network as shown in Figure 4.1 to find the currents

through different branches using Mesh (Loop) current method.


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Solution of Electric Circuit Based on Node Voltage Method


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UNIT II

NETWORK REDUCTION

AND NETWORK

THEOREMS FOR DC

AND AC CIRCUITS


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DeltaStar conversion


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Conversion from Delta to Star


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Conversion from Star to Delta


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SOLUTION:


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SUPERPOSITION THEOREM

INTRODUCTION

Statement of superposition theorem

In any linear bilateral network containing two or more independent sources

(voltage or current sources or combination of voltage and current sources), the

resultant current / voltage in any branch is the algebraic sum of currents / voltages

caused by each independent sources acting along, with all other independent sources

being replaced meanwhile by their respective internal resistances.

Superposition theorem can be explained through a simple resistive network as

shown in fig.7.1 and it has two independent practical voltage sources and one

practical current source.


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Limitations of superposition Theorem

Superposition theorem doesnt work for power calculation. Because power calculations

involve either the product of voltage and current, the square of current or the square of the

voltage, they are not linear operations. This statement can be explained with a simple exampleas given below.


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Thevenins and Nortons theorems


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The procedure for applying Thevenins theorem


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Maximum Power Transfer Theorem


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Remarks: The Thevenin equivalent circuit is useful in finding the maximum power that a linear circuit

can deliver to a load.


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Proof of Thevenin Theorem

The basic concept of this theorem and its proof are based on the principle of superposition theorem. Let

us consider a linear system in fig.L.8.8(a).


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Norton's Theorem

Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex,

to an equivalent circuit with just a single current source and parallel resistance connected to a

load. Just as with Thevenin's Theorem, the qualification of linear is identical to that found in the

Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Contrasting our original example circuit against the Norton equivalent: it looks something like

this:

. . . after Norton conversion . . .

Remember that a current source is a component whose job is to provide a constant amount of

current, outputting as much or as little voltage necessary to maintain that constant current.


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As with Thevenin's Theorem, everything in the original circuit except the load resistance has

been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's

Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton)

and Norton resistance (RNorton).

As before, the first step is to identify the load resistance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a

direct wire (short) connection between the load points and determine the resultant current. Note

that this step is exactly opposite the respective step in Thevenin's Theorem, where we replaced

the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R1 is

strictly a function of B1's voltage and R1's resistance: 7 amps (I=E/R). Likewise, the current

through R3 is now strictly a function of B2's voltage and R3's resistance: 7 amps (I=E/R). Thetotal current through the short between the load connection points is the sum of these two

currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source

current (INorton) in our equivalent circuit:


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Remember, the arrow notation for a current source points in the direction opposite that of

electron flow. Again, apologies for the confusion. For better or for worse, this is standard

electronic symbol notation. Blame Mr. Franklin again!

To calculate the Norton resistance (RNorton), we do the exact same thing as we did for calculating

Thevenin resistance (RThevenin): take the original circuit (with the load resistor still removed),

remove the power sources (in the same style as we did with the Superposition Theorem: voltage

sources replaced with wires and current sources replaced with breaks), and figure total

resistance from one load connection point to the other:

Now our Norton equivalent circuit looks like this:


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If we re-connect our original load resistance of 2 , we can analyze the Norton circuit as a

simple parallel arrangement:

As with the Thevenin equivalent circuit, the only useful information from this analysis is the

voltage and current values for R2; the rest of the information is irrelevant to the original circuit.However, the same advantages seen with Thevenin's Theorem apply to Norton's as well: if we

wish to analyze load resistor voltage and current over several different values of load resistance,

we can use the Norton equivalent circuit again and again, applying nothing more complex than

simple parallel circuit analysis to determine what's happening with each trial load.

REVIEW:

Norton's Theorem is a way to reduce a network to an equivalent circuit composed of asingle current source, parallel resistance, and parallel load.

Steps to follow for Norton's Theorem:

(1) Find the Norton source current by removing the load resistor from the originalcircuit and calculating current through a short (wire) jumping across the open connection

points where the load resistor used to be. (2) Find the Norton resistance by removing all power sources in the original circuit

(voltage sources shorted and current sources open) and calculating total resistance

between the open connection points.

(3) Draw the Norton equivalent circuit, with the Norton current source in parallel withthe Norton resistance. The load resistor re-attaches between the two open points of theequivalent circuit.

(4) Analyze voltage and current for the load resistor following the rules for parallelcircuits


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Nortons Theorem

In some ways Norton's Theorem can be thought of as the opposite to "Thevenins Theorem", in that Thevenin

reduces his circuit down to a single resistance in series with a single voltage. Norton on the other hand reduces his

circuit down to a single resistance in parallel with a constant current source. Nortons Theorem states that "Any linearcircuit containing several energy sources and resistances can be replaced by a single Constant Current generator in

parallel with a Single Resistor". As far as the load resistance, RL is concerned this single resistance, RS is the value

of the resistance looking back into the network with all the current sources open circuited and IS is the short circuit

current at the output terminals as shown below.

Nortons equivalent circuit.

The value of this "constant current" is one which would flow if the two output terminals where shorted together while

the source resistance would be measured looking back into the terminals, (the same as Thevenin).

For example, consider our now familiar circuit from the previous section.

To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40 load resistor and short

out the terminalsA and B to give us the following circuit.


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When the terminalsA and B are shorted together the two resistors are connected in parallel across their two

respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can

now be calculated as:

with A-B Shorted Out

If we short-out the two voltage sources and open circuit terminalsA and B, the two resistors are now effectively

connected together in parallel. The value of the internal resistorRs is found by calculating the total resistance at the

terminalsA and B giving us the following circuit.


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Find the Equivalent Resistance (Rs)

Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following



Ok, so far so good, but we now have to solve with the original 40 load resistor connected across terminalsA and B

as shown below.

Again, the two resistors are connected in parallel across the terminalsA and B which gives us a total resistance of:


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The voltage across the terminalsA and B with the load resistor connected is given as:

Then the current flowing in the 40 load resistor can be found as:

which again, is the same value of0.286 amps, we found usingKirchoffscircuit law in the previous tutorials.

Nortons Theorem Summary

The basic procedure for solving a circuit using Nortons Theorem is as follows:

1. Remove the load resistorRL or component concerned.

2. Find RS by shorting all voltage sources or by open circuiting all the current sources. 3. Find IS by placing a shorting link on the output terminalsA and B.

4. Find the current flowing through the load resistorRL.

In a circuit, power supplied to the load is at its maximum when the load resistance is equal to the source resistance.

In the next tutorial we will look atMaximum Power Transfer. The application of the maximum power transfer

theorem can be applied to either simple and complicated linear circuits having a variable load and is used to find the

load resistance that leads to transfer of maximum power to the load.

Example-L.8.5 For the circuit shown in fig.8.10(a), find the current through resistor ( branch) using

Nortons theorem & hence calculate the voltage across the current source (). 21LRR==abIcgV

Solution:

Step-1: Remove the resistor through which the current is to be found and short the terminals a andb (see fig.8.10(b)).
http://www.electronics-tutorials.ws/dccircuits/dcp_4.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_4.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_4.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_8.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_8.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_8.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_8.htmlhttp://www.electronics-tutorials.ws/dccircuits/dcp_4.html


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Step-2: Any method can be adopted to compute the current flowing through the a-b branch. Here, we

apply mesh current method.Loop-13R

4(I

1I

2) = 0, where I

2= - 2A

R4I1= 3 + R

4I2= 32 2 = - 1 I

1= - 0.5A

Loop-3

133323333- RI-R(I-I)=0- 3I-4(I+2)=0- 7I-8=08 I=-=7

N138-7+1I=(I-I)=-0.5+=7149A14=(current is flowing from a to b)Step-3: To compute R

N, all sources are replaced with their internal resistances. The equivalent

resistance between a and b terminals is same as the value of Thevenins resistance of the circuit

shown in fig.8.3(d).


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Step-4: Replace the original circuit with an equivalent Nortons circuit as shown in fig.8.10(d).

NLNNLR1.555I=I=0.643=0.39A (a to b)R+R1.555+1

In order to calculate the voltage across the current source the following procedures are adopted.Redraw the original circuit indicating the current direction in the load.

bgbgcbcg V=3-10.39=2.61volt2.61 I==1.305A2 I=1.305-0.39=0.915A ('c' to 'b')

V=21.305+4.915=6.26volt ('c' is higher potential than 'g')

L.8.8 Test Your Understanding [Marks: 60]T.1 When a complicated dc circuit is replaced by a Thevenin equivalent circuit, it consists of one ----

--- in series with one --------- . [2]


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T.2 When a complicated dc circuit is replaced by a Norton equivalent circuit, it consists of ------ in --

--- with one -------. [2]T.3 The dual of a voltage source is a -----------. [1]T.4 When a Thevenin theorem is applied to a network containing a current source; the current source

is eliminated by --------- it. [1]T.5 When applying Nortons theorem, the Norton current is determined with the output terminals ----

----------, but the Norton resistance is found with the output terminals ---------.and subsequently allthe independent sources are replaced -----------. [3]T.6 For a complicated circuit, the Thevenin resistance is found by the ratio of -------- voltage and -----

------- current. [2]T.7 A network delivers maximum power to the load when its -------- is equal to the -------- of circuitat the output terminals. [2]

T.8 The maximum power transfer condition is meaningful in ------------ and --------- systems. [2]T.9 Under maximum power transfer conditions, the efficiency of the system is only --------- %. [1]


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UNIT III

RESONANCE AND

COUPLED CIRCUITS


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Ideal Transformer

In this lesson, we shall study two winding ideal transformer, its properties and workingprinciple under no load condition as well as under load condition. Induced voltages in primary and

secondary are obtained, clearly identifying the factors on which they depend upon. The ratio between

the primary and secondary voltages are shown to depend on ratio of turns of the two windings. At theend, how to draw phasor diagram under no load and load conditions, are explained. Importance ofstudying such a transformer will be highlighted. At the end, several objective type and numerical

problems have been given for solving.Key Words: Magnetising current, HV & LV windings, no load phasor diagram, reflected current,

equivalent circuit

23.2 Introduction

Transformers are one of the most important components of any power system. It basicallychanges the level of voltages from one value to the other at constant frequency. Being a staticmachine the efficiency of a transformer could be as high as 99%.

Big generating stations are located at hundreds or more km away from the load center (where the

power will be actually consumed). Long transmission lines carry the power to the load centre from

the generating stations. Generator is a rotating machines and the level of voltage at which it generates

power is limited to several kilo volts only


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UNIT IV

TRANSIENT RESPONSE

FOR DC CIRCUITS


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UNIT V

ANALYSING THREE

PHASE CIRCUITS


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Example


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Solution of Current in AC Parallel and Series-parallel Circuits


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Generation of Sinusoidal Voltage Waveform (AC) and Some Fundamental

Concepts


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As shown earlier, normally the voltage generated, which is also transmitted and then distributed to

the consumer, is the sinusoidal waveform with a frequency of 50 Hz in


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Three-phase Balanced Supply


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Three-phase Delta-Connected Balanced Load


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