1 Circuit Laws and Theorems-5

8/12/2019 1 Circuit Laws and Theorems-5

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Indian Institute of Technology Guwahati

Department of Electronics & Electrical Engineering

1

EE 101

Electrical Sciences


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Indian Institute of Technology Guwahati 2

Syllabus

Fundamental Laws of Electrical Engineering

Coulombs law

Ohms law

Kirchoffs law

Amperes law

Faradays law of electromagnetic induction

The circuit elements

Ideal independent current and voltage sources

Energy and power

Resistance, inductance and capacitance

Network Theory Series and parallel combination of circuit elements

Network analysis by mesh currents and node pair voltages

Thevenins theorem, Nortons theorem

Network reduction by star delta transformation


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Syllabus

Circuit differential equations

The differential operator General formulation of circuit differential equations

The forced solutions, and natural response

Circuit Dynamics and Forced Responses

First order circuits (R-L,R-Ccircuits)

Sinusoidal Steady State Response of Circuits

Average and effective values of a periodic function

Instantaneous and average power, power factor

Phasor representation of sinusoids

Application of network theorems to complex impedances, and

Balanced three phase circuits

Star and delta connected sources

Star and delta connected loads

Balanced 3-phase system analysis


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Syllabus

Magnetic Theory and circuits

Definition of Magnetic quantities Theory of magnetism The magnetic circuits Hysteresis and eddy current losses in magnetic materials

Electromechanical energy conversion

Basic analysis of electromagnetic torque Analysis of induced voltage Construction features of electric machines

Transformers Theory of operation and phasor diagrams Equivalent circuit and parameters

Three phase induction machines Theory of operation and phasor diagrams Equivalent circuit and parameters

Introduction to synchronous machines and DC machines


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Books

1. Vincent Del Toro, Electrical Engineering Fundamentals, SecondEdition, PHI Learning Private Limited, New Delhi, 2010.

2. Bhag S. Guru, Huseyin R. Hiziroglu, Electric Machinery and

Transformers, 3rdEdition, Oxford University Press, 2001.

3. G B Shrestha, M H Haque, AC Circuits and Machines, PearsonPrentice Hall, 2006

5


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Lectures 1-5

DC Circuits:Basic Concepts and Fundamental Laws

Network Techniques and Theorems

Circuit Differential Equations

Transient Behavior of RL, RC Circuits

G. B. Shrestha

Visiting Professor

Department of Electronics & Electrical Engineering

Room CET-102, Ph. x3475, Email: [email protected]

Thanks to

Dr. Praveen Kumar, Associate Professor,

for portions of his Lecture Notes on this topic


7/68Indian Institute of Technology Guwahati

Introduction

There are two broad branches of classical electrical engineering Electric circuit theory

Electromechanical energy conversion

Both these branches are based on of experimentally established

fundamental laws. The electric circuit theory mainly consists of :

Coulomb Law(1785), OhmsLaw (1827),

Faradayslaw (1831), and

Kirchhoff's law (1857)

The electromechanical energy conversion can be treated and analyzed byapplying just two of the fundamental laws, namely AmperesLaw (1825)

and Faraday's Law of Induction (1831).

7



Fig.1 shows two pointpositivecharges of values Q1and Q

2separated by a

distance of rmeters.

Charles A. Coulomb showed experimentally that a force of repulsion exists

between Q1and Q2 which is given by

Coulombs law for the force on Q2is also frequently written as

1 2

2

1 2

[N] (1)4

where, and are in coulombs

is the separation between the charges in metres

4 is a proportionality constant, and

is the permittivity of the medium

Q QF

r

Q Q

r

Coulombs Law

2

11 22

[N] 2

And, [N/C] is known as electric field intensity of at .4

F Q

QQ Q

r

E

E =

8

r Q2

Figure 1

Q1



Permittivity

In equation (1) all the quantities can be known through measurements with

the exception of the proportionality factor .

The factor is a property of the medium in which the experiment is

performed and this factor is known as permittivity. The permittivity of a

substance describes how easy/difficult it is to set up an electric field in it.

When the experiment is performed in a vacuum, the value of thepermittivity is: 0=8.85410

-12 [F/m]

Repeating the Coulombsexperiment in oil for the same values of Q1and

Q2 and r, it is found that the resulting force is only about half as much as

for air.

The permittivity of oil is greater than that of air. A convenient way ofexpressing this difference is to introduce a quantity called relative

permittivity defined as: so that0

/ ,r

1 2

2

0

[N]4 r

Q QF

r

9



Potential Difference In Fig. 2 the charge Q1is assumed to be located at some fixed point in space. The

charge Q2is assumed initially to be on a horizontal path from Q1and is infinitelyfar away. In this condition the force between Q1and Q2is zero.

If Q2is moved in the direction of,increasingly larger force is requiredto move closer to Q1.

The force exerted on Q2acts along the line Q1Q2 against the movement whilemoving from infinity up to r2, it follows that work is being done on Q2.

The work done on Q2is stored in the form ofpotential energyand is recovered inthe form of kinetic energyimmediately upon release of the force which keeps Q2fixed at a position r2meters from Q1.

The work done on Q2to bring it from infinity to r2can also be expressed in termsof the electric field intensityassociated with charge Q1(eqs. 1 and 2).

In moving Q2from infinity to r2, work must be done against this electric field.

Q1 Q2Figure 2a

r2

Q2Figure 2

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The quantity, work per unit chargeis also known as voltage,is defined as

work done per coulomb of charge while moving Q2from infinity to r2.

To obtain the voltageit is necessary to perform an integration of the electric

field intensity (eq. 2) with respect to the distance.

Mathematically voltage is defined as:

The minus sign is arbitrarily inserted to indicate that work is done on the

charge against the action of the electric field produced by Q1.

Potential Difference

2

22

2

1

22

workVoltage , or

charge

ie, [V] (3)4

r

WV

Q

Q

V Edr r

Q1 Q2Figure 2a

r2

Q2Figure 2


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To determine thepotential differencebetween any two point charges, another

charge Q3is considered and this charge is moved from infinity to r3.

From Fig. 3 it can be seen that r3is smaller than r2, hence more work per unit

charge must be done to place a charge at point 3.

Mathematically voltage or potential of Q3

is calculated as:

Since V3is greater than V2,

there exists a potential difference between these two points.

When a unit charge moves from a point of higher potential (eg, point 3) to one

of lower potential (eg, point 2) it gives away some energy.

Potential Difference

13

3

[V] (4)4

QV

r

r2Q1 Q2

Figure 3Q1

r3

Q3


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On the basis of experiments, Ohm showed that current flow in a circuit

composed of a battery and conductors can be expressed as

where, Iis the current,

Aand are the cross section and the resistivity of the conductor, and

is the rate of change of voltage along the conductor. In case of a uniform conductor, the current can be expressed as

where, V andLare total voltage and the length of the conductor.

Thus, Ohms law states that the current in a wire is proportional to the voltage

across its ends, which can be expressed as:

where, Ris called the resistance of the conductor.

13

Ohms Law

d [A] (5)

d

A vI

l

= [A] (6)/

A V VI

L L A

/ or where, / (7)I V R V IR R L A

d / dv l

Figure 4


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During Ohms time, experimental verification of his law was achieved

exclusively by using direct current sources, namely the voltaic cells.

Further experimentation has shown that Ohmslaw is also valid when the

potential difference across a linear resistor is time varying. In this case

Ohmslaw is expressed as

The lower case letters are used to indicate

the time varying, non constant natures of

the quantities vand i.

Ohms Law

(8)v iR i

v R

Figure 5


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A negative point charge Q1of magnitude 210-12C is placed in

vacuum.

a. Find the value of the electric field at a point 1 m from the

charge.

b. Compute the work done in bringing an electron from a point

very far removed from Q1to a point 0.5 m from the chargec. What force is acting on the electron in part (b)?

Example

Solutiona. 0.018 V/mb. 0.0576 10-19Joules

c. 115.2 10-22N


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Gustav Robert Kirchhoff published the first systematic formulation of the

principles governing the behavior of the electric circuits. His work is

embodied in two fundamental laws:

Kirchoffscurrent law, and

Kirchoffsvoltage law

Kirchhoff's current law states that the sumof the currents entering (or leaving) a node

at any instant is equal to zero. A node is a

junction in a circuit where two or more

circuit elements are joined together (Fig.6).

Mathematically the law is expressed as:

Kirchhoff's Laws

1

0 (9)

where, number of circuit elements connected to a node.

k

jj

I

k

Figure 6

Node 1

16


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Applying the Kirchoffscurrent law to node 1 in Fig.6 gives

The Kirchhoff's law is the restatement of

theprinciple of conservation of charge.

It may be noted that if the current entering

a node is taken as +ve, then the currentgoing out of the node will beve, and

vice versa.

For two resistances in parallel, KCL can be

used to show that:

Current division in parallel resistances:

Parallel combination of resistances:

Figure. 6: Kirchhoff's current law

Node 1

Kirchoffs Current Law (KCL)

1 2 3 4 5 6 0 (10)I I I I I I

17

2 11 2

1 2 1 2

, (11)R R

I I I IR R R R

1 2

1 2 1 2

1 1 1, or (12)eq

eq

R RR

R R R R R

I

1

I

2

R

1

R

2

I

I

R

eq

Figure. 7: Parallel Resistances


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Kirchoffs Voltage Law (KVL)

Kirchoffsvoltage law states that at

any instant of time, the sum of the

voltages in a closed circuit is zero.

Application of Kirchhoffs voltage law

to the circuit shown inn Figure.7 gives:

It should be noted that the polarity of

various voltages should be properly taken care of.

For two resistances in series, KVL can be used to show that:

Voltage division in series resistances:

Series combination of resistances:

1 2 3 0 (13)V V V E Figure.8: Kirchhoff's voltage law

18

1 21 2

1 2 1 2

, V (14)R R

V V VR R R R

1 2 (15)eqR R R

Figure. 9: Series Resistances

I

V

1

-

R

1

V

2

-

R

2

V

IR

eq

V -


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For an independent voltage source, the magnitude of the voltage is not

affected by either the magnitude or direction of the current.

The characteristics of an independent voltage source is shown in Fig. 8.

A notable point about the ideal voltage source is its zero internal

resistance. This is readily demonstrated by invoking Ohms law in

incremental form:

It may be noted that an ideal voltage source is capable of supplying any amount

of power, which is not true with practical voltage sources.

Ideal Independent Voltage Source

2 1

From Fig. 8,

0

VR

I

V

R I I

I1 I2

Voltage [V]

Current [A]

Figfure. 8: Characteristics of ideal voltage source

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The ideal independent current source is a two terminal element which

supplies its specified current to the circuit it is connected to, independent

of the value and direction of the voltage appearing across its terminals.

The voltage vs. current characteristics of an ideal current source is shown

In Fig. 9.

It may also be noted that an

ideal current source is capable of

supplying any amount of power,

which is not true with practical

current sources.

Ideal Independent Current Source

V1

V2

Voltage [V]

Current [A]

Figure. 9: Characteristics of ideal current source

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In case of ideal dependent or controlled source, the source quantity is

determined by a voltage or current existing at some other location in thesystem or circuit being analyzed.

The dependent sources appear in the equivalent electrical models for

many electronic devices such as transistors, operational amplifiers and

integrated circuits.

The dependent sources are represented as diamonds to distinguish them

from independent sources (Fig. 10).

Ideal Dependent Sources

Figure 10: Controlled or dependent sources

kix gvx

Controlled current sources

kvx+

- rix +-

Controlled voltage sources

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Determine vxin the circuit shown in Fig. 11(a)

Example 1

22

Fig.11a: Circuit for the example 1


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First the given network is properly labeled (Fig.11b)

Using KVL in loop I,

Using KCL at node X

Using KVL in loop II,

Solution

10 4 0xv v v

10

10 10

20 V

And, / 10 2 A

v

i v

104 10

4

5 5 3 A10

4 3 12 V

vi i

v

23

8 5 8 40 Vv

8 1060 0v v

10 4 20 12 8 Vxv v v

Fig.11(b): Labeled circuit

I

IIX


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Star and Delta Networks

Consider the circuit shown, which is not easily

amenable to series parallel circuit reductiontechniques.

A or Star network is a three-terminal three-

branch network with a common point as shown.

The resistance connected between any terminal

i(1, 2 or 3) and the common point in the Ynetwork is denotedRi.

A or Delta network is a three-terminal three-

branch network as shown.

The resistance connected between any two

terminals iandjin the network is denotedRij.

Star and Delta networks are commonly found inmany circuits and are basic configurations in 3-phase circuits.

24

1

1

32

2

1 3

R12 R23

R13

Y or Star

or Delta


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Delta Star Transformation

Every (Delta) network has an equivalent (Star)

network and vice versa.

If the two circuits shown are equivalent, then their

impedances should match under all circumstances.

If terminal 1 is open, then,

If terminal 2 is open,

And, if terminal 3 open,

Adding all three,

Then subtracting each from the total:

25

1

32

2

1 3

R12 R23

R13

231312

23131232

)(

RRR

RRRRR

231312

13231231 )(RRRRRRRR

231312

12231321

)(

RRR

RRRRR

231312

323123211312321

)(

RRR

RRRRRRRRR

,231312

13121

RRR

RRR

,

231312

23212

RRR

RRR

231312

32313

RRR

RRR


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1

32

Star Delta Transformation

Every Y (Star) network has an equivalent (Delta)

network.

Using to Y conversion relations:

We get:

Taking the ratios:

It should be noted that above conversion equations

are equally valid for AC circuits with complex

impedances.

26

2

1 3

R12 R23

R13

)()(

)(

231312

3123122

231312

231312312312133221

RRR

RRR

RRR

RRRRRRRRRRRR

231312

13121

RRR

RRR

,

231312

23212

RRR

RRR

231312

32313

RRR

RRR

,1

323223R

RRRRR ,2

131331R

RRRRR 3

212112 R

RRRRR


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Star Delta Transformation - Example

Example:

Find the current supplied by E, when

E = 10 V,

R1= 3 , R2= 10 , R3= 5 ,

RL

=10 , R4

= 2 , and R5

= 4 .

Ans. 1.25 A

27

Exercise:

Del Toro Problems: 1-10, 3-22, 3-27


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Network Techniques and Theorems

There are several systematic techniques and useful theorems which arebeneficially employed to solve network problems.

Nodal Analysis

Mesh Analysis

Superposition Theorem TheveninsTheorem

Nortons Theorem

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Nodal Analysis

To illustrate the basic mechanics of the technique, consider the three node

circuit shown in Fig.a. In this network there are three nodes.

A three node circuit should have two unknown voltages and two equations;

a 10-node circuit will have nine unknown voltages and nine equations;

an N-node circuit will need (N-1) voltages and (N-1) equations.

In the Fig. b, it is shown that the network has three unique nodes. One node isdesignated as a reference node(Fig.c) and it is the negative terminal of the

nodal voltages.

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(a) (b)


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In the Fig. b, one node isdesignated as a reference node

(also in Fig. c) and it is the

negative terminal of the nodal

voltages.

Nodal Analysis

--

+ +

V1 V2

1 2

3(c)

30

As a general rule, a node having maximum number of branches is chosen

as the reference node. By doing so the number of equations defining the

network is reduced.

In Fig. c, the voltage of node 1 relative to the reference node is defined asV1and V2is defined as the voltage of node 2 with respect to the reference

node. The voltage between any other pair of nodes may be found in terms

of V1and V2.


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Nodal AnalysisGeneral Procedure

The node analysis is a general method and

can always be applied to any electrical

network. Consider the circuit shown in the

Fig.

Identify the goal of the problem: There are

four nodes in this network. The bottom mostnode is selected as the reference node and all

the other three nodes are labeled as shown in

Fig.

Collect the information: There are threeunknown voltages V1, V2and V3. All current

sources and resistors have designated values

which are marked on the schematic.

31

A 4 node network

Redrawn network

1 2

3


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Nodal AnalysisGeneral Procedure

Devise a plan: This problem is well suited for

nodal analysis and three independent KCLequations may be written.

Construct an appropriate set of equati ons:

There are three unknowns and three equations.

Solve for voltages.

1 31 2

2 32 12

3 3 2 3 1

8 3 at node 13 4

3 at node 23 7

1.4 at node 3

5 7 4

V VV V

V VV VV

V V V V V

32

1 2

3


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The Supernode

Consider the network shown in the fig. If

KCL is applied as in the previous example

then we will run into difficulty at nodes 2

and 3 because the current in the branch 23

cannot be expressed in terms of V23.

To overcome the dilemma, node 2, node 3together with the voltage source is

considered a supernodeand KCL is then

applied.

The above super node is shown as a region

enclosed by the broken lines. A supernode

may be formed covering any region of a

circuit.

33

V1 V2 V3

supernode

1 2

3


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Supernode Example

The super node is shown as a region

enclosed by the broken lines.

The system of equations are:

Then solve for the three unknown voltages.

1 31 2

3 1 32 12

2 3

at node 1

8 33 4

at supernode

3 1.43 4 5

the third equation is22

V VV V

V V VV VV

V V

34

V1 V2 V3

Supernode (nodes 2 & 3

1

2

3


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Mesh Analysis

The node analysis is a general method and can always be applied to anyelectrical network.

Another convenient technique to analyze the networks is mesh analysis.

This technique is not general and can be applied to only planar networks.

If it is possible to draw the diagram of the network on a plane surface in

such a way that no branch passes over or under any other branch, then the

network is said to be aplanar network.

If a network is planar, mesh analysis can be used to accomplish the

analysis. This concept involves the concept of a mesh current.

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Planar and Non Planar Networks

Fig: Planar and non planar networks

a. Planar network

b. Non-planar network

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Consider the two mesh circuit shown in

the Fig. We define a mesh currentas acurrent that flows only around the

perimeter of a mesh.

The left hand mesh has a current i1

flowing in a clockwise direction and a

mesh current of i2is established in theremaining mesh.

The system of equations for the network

in terms of mesh currents are as follows:

Solving the above two equations, the

mesh currents i1 and i2are obtained as

Mesh Analysis

1 1 2

2 1 2

6 3( ) 42, for mesh 13( ) 4 10, for mesh 2

i i ii i i

1 26 A, and 4 Ai i

37

Fig.: Mesh Analysis

i1 i2


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According to superposition theorem, in a linear network containing two or

more sources, the response in any element is equal to the algebraic sum ofthe responses caused by individual sources acting alone while the other

sources are non-operative.

This theorem is useful for analyzing networks that have large number of

independent sources as it makes it possible to consider the effects of eachsource separately.

Superposition theorem is applicable to any linear circuit having time varying

or time invariant elements.

The limitations of superposition theorem are

It is not applicable to the networks consisting of non linear elements like

transistors, diodes, etc., in general (unless linearized).

It is not applicable to the networks consisting of any dependent sources,

in general.

Superposition Theorem

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Superposition Theorem - Example

Compute the current in the 23 Ohm resistor using superposition theorem.

The circuit contains 2 independent sources

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I23a

Superposition Theorem - Solution

With 200V source acting alone

and the 20A current open

circuited:

23

27 (4 23)47 60.5

54

200 3.31 A60.5

273.31 1.65 A

54

eq

total

a

R

I

I

40

With 20A source acting alone and

the 200V source short circuited:

23

27 474 21.15

74

21.1520 9.58 A

21.15 23

eq

b

R

I

I23b

23 23 23Curent through the 23 is: 11.23 Aa bI I I


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Thevenins theorem states that any two-terminal linear network having anumber of sources and resistances can be replaced by a simple equivalent

circuit consisting of a single voltage source in series with a resistance.

The value of the voltage source is equal to the open circuit voltage across

the two terminals of the network.

The resistance is equal to the equivalent resistance measured between the

terminals with all the energy sources replaced by their internal

resistances.

Thevenins Theorem

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The steps involved in determining the Theveninsequivalent circuit are:

Temporarily remove the load resistance where the current is required

Find the open circuit voltage VOCacross the two terminals from where the

load resistance has been removed. This is known the Theveninsvoltage VTH.

Calculate the resistance of the whole network as seen from these two

terminals, after all voltage sources are replaced by short circuit and allcurrent sources are replaced by open circuit leaving internal resistance (if

any). This is called Theveninsresistance, RTH.

Replace the entire network by a single Theveninsvoltage source VTHand

resistance RTH.

Connect the resistance (RL), across which the current value is desired, back

to its terminals from where it was previously removed.

Finally, calculate the current flowing through RLusing the following

expression:

Procedure to Obtain Thevenins Equivalent Circuit

THL

TH L

VI

R R

42


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Nortons theorem states that any two-terminal linear network with

current sources, voltage sources and resistances can be replaced by anequivalent circuit consisting of a current source in parallel with a

resistance.

The value of the current source is equal to the short circuit current

between the two terminals of the network and the resistance is

equivalent resistance measured between the terminals of the network

with all the energy sources replace by their internal resistance.

Nortons Theorem

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The steps involved in obtaining the Nortons equivalent circuit are:

Temporarily remove the load resistance across the two terminals and

short circuit these terminals.

Calculate the short circuit current IN.

Calculate the resistance of the whole network as seen at these two

terminals, after all voltage sources are replaced by short circuit andcurrent sources are replaced by open circuit leaving internal

resistances (if any). This is Nortons resistance RN.

Replace the entire network by a single Nortons current source whose

short circuit current is INand parallel with Nortons resistance RN.

Connect the load resistance (RL) back to its terminals from where it

was previously removed.

Finally calculate the current flowing through RL.

Procedure to Obtain Nortons Equivalent Circuit

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Obtain the Thevenins and Nortons Equivalent Circuit for the followingnetwork:

Thevenins and Nortons Equivalent Circuits - Example

45


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With the terminals aand bopen, the two

voltage sources drive a clockwise currentthrough the 3 and 6 resistors

Since no current passes through the upper

right 3 resistor, the Thevenins voltage

can be taken from either active branch:

The equivalent resistance as seen from

terminals aand bis

The Thevenins equivalent circuits is:

Thevenins Equivalent Circuit

20 10 30A

3 6 9I

3020 3 10 V

9THV

46

3 63 5

9eqR

Thevenins eq. circuit


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When a short circuit is applied to theterminals, current Iscresults from the two

sources. By superposition theorem, this

current is given by

The Nortons equivalent circuit is:

Nortons Equivalent Circuit

6 20

6 3 3 3 6 / 9

3 10 2 A

3 3 6 3 3 / 6

scI

47

Nortons eq. circuit


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Circuit Components R, L, C

When a (time varying) voltage eis applied to a simple resistorR, the

current is given by Ohms law as:

If the voltage is applied to a simple inductor, the relation between eand iis

given by Faradays law as

When voltage eis applied to a simple capacitor the pertinent relationship is

(1)e

iR

or (2)

where, the differential operator /

di ee L e pLi i

dt pL

p d dt

(3)1/

de eC i Cpe i i

dt pC

48


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Operational Impedance

The use of the differential operator permits the differential equation to be

written as an algebraic expression, wherepLplays a role for inductance that

is the same asRfor resistance.

The inductor is said to posses an operational impedanceofpL.

From the analogy

the capacitor can be said to present to the forcing function ean operationalimpedance of 1/pC

The term operational impedance can be applied to single elements or to

several interconnected elements.

49

Since,e

ipL

Since,1 /

ei

pC


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Consider the network shown in Fig. 1. By KVL, the describing differential

equation for the circuit is:

In terms of the poperator, or the operational impedances, the expression

becomes:

Such representation can be very useful in many situations.

Fig.1: Circuit to illustrate equivalency between p operator

algebraic equation and differential equation

Circuit Differential Equations - Operational Impedance

1(4)

die Ri L idt

dt C

1

1( ) (5)

1

where, ( )

e Ri pLi ipC

e R pL i Z p ipC

Z p R pL pC

50


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There are situations when it is required to relate an electrical quantity in one

part of the circuit to an appropriate electrical quantity in another part of thecircuit.

In the network of Fig. 1, it may be required to express the ratio of voltage

appearing across the inductor to the source voltage.

If vLis used to denote the voltage developed across

the inductor, the desired result can then be written as

The expression (6) is commonly called

the transfer function. The transfer function relates two quantities that are

identified at different terminal pairs.

The transfer functions may be used to relate ratios of voltages, ratios of

currents, ratios of current to voltage, ratios of voltage to current provided that

the quantities are not identified at the same terminal pairs.

Circuit Equations using Operational Impedances

(6)1 /

Lv pL

e R pL pC

51


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General Formulation of Circuit Differential Equation

Consider the network shown in Fig. 2, where we like to relate voto e2.

Writing the node equation at node ausing operational impedances:

Then,

This implies that, in the form of differential equation, the voltage vois related

to e2by:

2

0 22

1 2 2 5 2

3 2 4 11 8a

p p pv v e

p p p

52

Fig. 2: An RLC network

a2

22

01 2 / 2 1 2 1

(2 )(3 2 )

4 11 8

a a a

a

v v v e

p p

p pv e

p p

2 20 0 2 2

0 22 24 11 8 2 5 2 (7)

d v dv d e dev e

dt dt dt dt


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Eq. 7 is often referred to as an equilibrium equation. It relates a linear

combinations of the response function and its derivatives to linear

combinations of the source function and its derivatives.

The operator version of eq. 7 is more generally written as:

Or,

2

0 222 5 2 (8)4 11 8p pv ep p

0 2( ) (9)

where,

( ) operational network function.

v G p e

G p

Fig. 2: An RLC network

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For any network, the following generalization holds true

Eq. (10) is generally written as

( ) ( ) ( ) ( ) (11)D p y t N p f t

( ) ( )

( )( ) ( ) ( ) ( ) (10)

( )

( ) response function such as current or voltage in a particular part of the circuit

( )= total networ

response function network function applied source function

N py t G p f t f t

D p

where

y t

G p

k function

( )=numerator polynomial of network function G(p)

( )= denominator polynomial of the network function G(p)

( )= applied voltage source function

N p

D p

f t

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Write the expression for the operational impedance for each branch of the

network shown in Fig.3 and determine the expression for the driving pointimpedance appearing at the terminal a-d.

Example 1

a b c

d

Fig.3: Circuit for the example 1

1 1

2 2

2

1

1

Driving point impedance

Z

ab

bd

bcd

bd bcd ad ab

bd bcd

Z RpC

Z R pL

Z R pLpC

Z ZZ

Z Z

Solution 1

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Solution of Circuit Differential Equation

Consider the response function,

If f(t) is an applied source function, then y(t) is said to be a solution to the

homogeneous linear differential equation.

The completeor fullsolution will consist of two components:

1. The SteadyStateor the Particularor the Forcedsolution.

This component is primarily dependent on the source.

2. The Transientor the Generalor the Naturalsolution

This component is primarily dependent on the network or the system.

The complete solution will be the sum of the two solutions.

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( ) ( ) ( ) ( ) (11)D p y t N p f t


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The Steady State (Forced or Particular) Solution

Consider, the first part of the solution to the network equation

1. The solutiony(t) will exhibit the same nature as the inputf(t). For

example, iff(t)is a sinusoidal function, theny(t)will also be sinusoidal,

althoughy(t)will differ from the source function in amplitude and

phase, in general. If the source function is exponential, the solution willbe exponential. This is why this part of the solution is termed the

particular solution, - the form ofy(t) is particularized to the nature of

the source function.

2. y(t) will continue to exist in the circuit so long as the source function

exists. This is the reason whyf(t)is referred to as theforced solution.

3. y(t) is also referred to as thesteady state solutionbecause this is the

component which remains in the circuit after the transient terms disappear.

( ) ( ) ( ) ( ) (11)D p y t N p f t

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The response to constant sources

The response to a constant source is developed for the network shown in Fig.

4. It is required to find the battery current if(t) due to the source voltage E.

The general procedure to obtain the solution is:

Obtain the equilibrium equation by applyingcircuit theory to the configuration of circuitelements described, in terms of their

Operational impedances.

Put the resulting expression in the formof eq.11. Then examine the right side ofthis equation to determine the exact formneeded by the forced solution to qualify as

a solution to the equilibrium equation. Finally, solve for the unknown quantities.

To find the expression for the current if(t) in the circuit shown in Fig.4, it isrequired to find the impedance Zgd.

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Fig. 4: The network 1

a b c

g

d


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The operation impedance across the terminals g-d is:

The current if(t) is given by

The corresponding differential equation is

2

2

( )

2(3 2 )

4 11 82 1 (12)2 2 4 2

3 2

ab bcd gd bg

ab bcd

Z ZZ p Z

Z Z

pp

p p

p p pp

p

2

24 11 8 2 (14)

f f

f

d i dii E

dt dt

2

2

2 2

2 4 2( )

4 11 8

(4 11 8) ( ) (2 4 2) (13)

f

gd

f

E p pi t

Z p p

p p i t p p E

a b c

g

d

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Examination of the equilibrium equation (eq. 13) reveals that because there are

no derivative terms on the right side of the equation, the derivative terms of theresponse function must be identically equal to zero.

Hence, the expected form of the solution for if(t) is

Substituting the solution in eq. (15) into eq. (13) gives the forced or the steadystate solution.

The examination of the network given in Fig. 4 reveals that this result is

entirely expected. The current if(t) finds a closed path only in loopgbcdg. The

left hand part of the loop is open circuit in the steady state due to presence ofthe capacitor.

The voltage across the 2-H inductor is zero in steady state. Hence, the current

is just a d-c current of magnitude E/4.

0( ) (15)fi t I

0 (16)4

EI

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Fig. 5: Series R-L Circuit

The forced solution is the response which has settled down solely to the effects

of the source function. It is also known as thesteady state solution. The forced or the steady state solution always satisfies the defining differential

equation but in general it is not the complete solution over the entire time

domain.

To illustrate the point consider the R-L circuit shown in Fig. 5. In this network

it is desired to find the complete solution for the voltage vappearing across the

resistor for all time tafter the switch is closed.

The equation that relates vto the source voltageEis:

The forced or the steady state solution of eq. 18 is:

1(17)

1 ( / )

(18)

v R

E R pL p L R

L dvv E

R dt

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The Steady State Response of R-L Circuit

(19)fv E


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The Natural Response (Transient Solution)

Solution (19) satisfies eq. 18, however this solution does

not satisfy the circuit behavior in the period immediatelyfollowing the closing of the switch at time t=0+

At time t=0+the current is zero because of the presence of the inductor. Hencev, which is iR, is also zero. Hence, the solution in eq. 19 cannot be a completedescription at t=0+.

Therefore, in the period, t=0+, immediately after the switching of the sourcefunction, it needs to add a complementary functionto the forced solution.

This complementary function will be significant only immediately afterswitching but will disappear as steady state is reached. It is the purpose of thecomplementary function to provide a smooth transition from the initial state of

the response to the final state, in the presence of energy storing elements. The complementary function (natural response) is obtained by solving

equation 18 with no inputs,

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0 (20)L dv

vR dt


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The Natural Response (Transient Solution) of RL Circuit

The solution to the eq. 20 should satisfy that function, i.e, vand its derivative,

dv/dt, must be of the same form to make the left hand side equal to the righthand side.

The plausible solution for eq. 20 is

Substitution of vfrom eq. 21 into eq. 20 gives

Hence, the complementary solution becomes

It is clearly seen that this solution will have a value of K at t=0, but will become 0

when t.

(21)stv Ke

( / ) 1 0 (22)

/ ) (23)

stL R s Ke

s R L

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( / ) (24)R L tv Ke


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The Complete Response of RL Circuit

The complete solution of the original governing differential equation (18) is

The quantityKis found from the initial condition which requires that vto bezero at t=0+

Hence, the complete solution is

The nature of the complimentarySolution is shown in the figure.

It begins with an initial value ofE and exponentially decays to a value ofzero.

0 (24)E K K E

( / )(1 ) (25)R L tv E e

( / ) (25)R L tv E Ke

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The Complete (Transient+ Steady State) Solution

The behavior of the complete solution (eq. 25) is shown in Fig. 6.

Note that the total response now starts with a value of zero and

exponentially rises to the value ofE.

The rate at which this

transition takes place is

entirely dependent uponthe value of (R/L). WhenRis

large (orLis small) the

transition occurs quickly

because the transient term

dies out in little time.

65

( / )(1 )R L tv E e

Fig. 6: Complete solution of voltage across the resistor

in RL circuit of Figure 5.


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The Response to Different Kinds of Source

Similar analyses can be conducted for

different types of inputs, say sinusoidal

source as shown in Fig.7.

These cases are not covered in this course.

Similarly, the behavior of various circuit

components can be investigated.

Homework: (RC Circuit)

Investigate the behavior of the voltage across the

resistor in a series RC circuit after switching a dc

source.

Fig.7: Network with sinusoidal source

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Thank you!

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Exercise:

Del Toro Problems: 3-23, 3-32, 3-33, 3-34

1 Circuit Laws and Theorems-5

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