BEEE UNIT I

8/12/2019 BEEE UNIT I

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Basic Electrical & Electronics Engineering

Prepared By:Deependr Singh1

UNIT IELECTRICAL CIRCUIT ANALYSIS

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.

Basic Circuit Elements: Electric Circuits consist of two basic types of elements. These are the

active elements and the passive elements.

An Active element is capable of generating electrical energy. [In electrical engineering,

generating or producing electrical energy actually refers to conversion of electrical energy from anon-electrical form to electrical form. Similarly energy loss would mean that electrical energy is

converted to a non-useful form of energy and not actually lost. [Principle of Conservation of Mass

and Energy].

Examples of active elements are voltage source (such as a battery or generator) and current

source. Most sources are independent of other circuit variables, but some elements are dependant

(modeling elements such as transistors and operational amplifiers would require dependant sources).

Active elements may be ideal voltage sources or current sources. In such cases, the particular

generated voltage (or current) would be independent of the connected circuit.

A passive element is one which does not generate electricity but either consumes it or stores it.

Resistors, Inductors and Capacitors are simple passive elements. Diodes, transistors etc. are also

passive elements.

Also it can be said that 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.

Passive elements may either be linear or non-linear. Linear elements obey a straight line

law. For example, a linear resistor has a linear voltage vs current relationship which passes through

the origin (V = R.I). A linear inductor has a linear flux vs current relationship which passes through

the origin ( = k I) and a linear capacitor has a linear charge vs voltage relationship which passes

through the origin (q = CV). [R, k and C are constants].

Resistors, inductors and capacitors may be linear or non-linear, while diodes and transistors

are always non-linear.


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Branch: A branch represents a single element, such as a resistor or a battery. A branch is a 2-

terminal (end) element.

Node: A node is the point connecting two or more branches. The node is usually indicated by a dot

() in a circuit.

Loop or Mesh: A loop is any closed path in a circuit, formed by starting at a node, passing through anumber of branches and ending up once more at the original node.

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

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

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:

Linear Circuit: Roughly speaking, a linear circuit is one whose parameters do not change with

voltage or current.


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

voltage or current. More specifically, non-linear circuit does not obey the homogeneity and additive

properties. Volt-ampere characteristics of linear and non-linear elements are shown in figs. 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. Otherwise, it is a nonlinear system.

Kirchhoffs Current Law (KCL): KCL states that at any node

(junction) in a circuit the algebraic sum of currents entering and

leaving a node at any instant of time must be equal to zero. Here

currents entering (+ve sign) and currents leaving (-ve sign) the

node must be assigned opposite algebraic signs (see fig., I1-I2+I3-

I4+I5-I6= 0).

Kirchhoffs Voltage Law (KVL): It states that in a closed

circuit, the algebraic sum of all source voltages must be equal

to the algebraic sum of all the voltage drops. Voltage drop is

encountered when current flows in an element (resistance or

load) from the higher-potential terminal toward the lower

potential terminal. Voltage rise is encountered when current

flows in an element (voltage source) from lower potential

terminal (or negative terminal of voltage source) toward the

higher potential terminal (or positive terminal of voltage source). Kirchhoffs voltage law is

explained with the help of fig. KVL equation for the circuit shown in fig. is expressed as (we walk in

clockwise direction starting from the voltage source V1and return to the same point)

V1-IR1-IR2-V2-IR3-IR4+V3-IR5-V4= 0

V1-V2+V3-V4= IR1+IR2+IR3+IR4+IR5

Meaning of Circuit Ground and the Voltages referenced to Ground: In electric or electronic

circuits, usually maintain a reference voltage that is named ground voltage to which all voltages


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are referred. This reference voltage is thus at ground potential or zero potential and each other

terminal voltage is measured with respect to ground potential, some terminals in the circuit will have

voltages above it (positive) and some terminals in the circuit will have voltages below it (negative)

or in other words, some potential above or below ground potential or zero potential.

Consider the circuit as shown in fig. and the commonpoint of connection of elements V1 & V3 is selected as

ground (or reference) node. When the voltages at different

nodes are referred to this ground (or reference) point, we

denote them with double subscripted voltages VED, VAD, VBD

and VCD. Since the point D is selected as ground potential or

zero potential, we can write VEDas VE, VADas VAand so on.


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VOLTAGE AND CURRENT SOURCES

Ideal and Practical Voltage Sources: An ideal voltage source, is a device that produces a constant

voltage across its terminals (V = E) no matter what current is drawn from it (terminal voltage is

independent of load (resistance) connected across the terminals)

However, real or practical dc voltage sources do not exhibit such

characteristics (see fig. 3.14) in practice. We observed that as the load

resistance RL connected across the source is decreased, the

corresponding load current IL increases while the terminal voltage

across the source decreases. We can realize such voltage drop across

the terminals with increase in load current provided a resistance

element (RS) present inside the voltage source. Fig. 3.15 shows the

model of practical or real voltage source of value VS.

The V-I characteristic of the source is also called the sources regulation curve or load line.

The open-circuit voltage is also called the no-load voltage, VOC. The maximum allowable load

current (rated current) is known as full-load current IFL and the corresponding source or load

terminal voltage is known as full-load voltage VFL. We know that the source terminal voltage

varies as the load is varied and this is due to internal voltage drop inside the source. The percentage

change in source terminal voltage from no-load to full-load current is termed the voltage regulation

of the source. It is defined as

Voltage Regulation (%) = 100

FL

FLOC

V

VV

For ideal voltage source, there should be no change in terminal voltage from no-load to full-load and

this corresponds to zero voltage regulation. For best possible performance, the voltage source

should have the lowest possible regulation and this indicates a smallest possible internal voltage drop

and the smallest possible internal resistance.

Ideal and Practical Current Sources:An ideal current source is a device that delivers a constant

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


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across the load (i.e., independent of the voltage across its terminals across the terminals); while a

practical current source depends on the voltage across its terminals.


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Dependent and Independent Sources

Independent and Dependent Sources that encountered in electric circuits:

Independent Sources:The voltage and current sources (whether ideal or practical) that havebeen discussed till now are known as independent sources and these sources play an

important role to drive the circuit in order to perform a specific job. The internal values of

these sources (either voltage source or current source) that is, the generated voltage VSor

the generated current ISare not affected by the load connected across the source terminals or

across any other element that exists elsewhere in the circuit or external to the source.

Dependent Sources:Another class of electrical sources is characterized by dependent sourceor controlled source. In fact the source voltage or current depends on a voltage across or a

current through some other element elsewhere in the circuit. Sources, which exhibit this

dependency, are called dependent sources. Both voltage and current types of sources may be

dependent, and either may be controlled by a voltage or a current. In general, dependent

source is represented by a diamond ()-shaped symbol as not to confuse it with an

independent source. One can classify dependent voltage and current sources into four types of

sources as shown in fig.3.21. These are listed below:

(i) Voltage controlled voltage source (VCVS) (ii)Current controlled voltage source (ICVS)(iii)Voltage controlled current source (VCIS) (iv)Current controlled current source (ICIS)

Note: When the value of the source (either voltage or current) is controlled by a voltage (vs)

somewhere else in the circuit, the source is said to be voltage-controlled source. On the other hand,

when the value of the source (either voltage or current) is controlled by a current (ix) somewhere else


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in the circuit, the source is said to be current-controlled source. KVL and KCL laws can be applied to

networks containing such dependent sources. Source conversions, from dependent voltage source

models to dependent current source models, or visa-versa, can be employed as needed to simplify the

network. One may come across with the dependent sources in many equivalent-circuit models of

electronic devices (transistor, BJT (bipolar junction transistor), FET (field-effect transistor) etc.) andtransducers.

Source Conversion

A practical voltage source can be transformed to an equivalent practical current source.

Similarly, a practical current source can be transformed to an equivalent practical voltage source.

Replacing one source by an equivalent source is called as Source transformation or conversion. It

serves as a powerful tool to simplify the analysis of circuits especially with mixed sources.

WhereR

VI SS

Where RIV SS

Volta e Source to Current Source Conversion

Current Source to Voltage Source Conversion


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Introduction: The Series-parallel reduction technique that we learned for analyzing DC

circuits simplifies every step logically from the preceding step and leads on logically to the next step.

Unfortunately, if the circuit is complicated, this method becomes mathematically laborious, time

consuming and likely to produce mistake in calculations. In fact, to elevate these difficulties, some

methods are available which do not require much thought at all and we need only to follow a well-defined faithful procedure. One most popular technique will be discussed in this lesson is known as

mesh or loop analysis method that based on the fundamental principles of circuits laws, namely,

Ohms law and Kirchhoffs voltage law. Some simple circuit problems will be analyzed by hand

calculation to understand the procedure that involve in mesh or loop current analysis.

MEANING OF CIRCUIT ANALYSIS: The method by which one can determine a variable (either

a voltage or a current) of a circuit is called analysis.

DC CIRCUITS ANALYSIS USING MESH METHOD

MESH ANALYSIS:A mesh is any closed path in a given circuit that does not have any element

(or branch) inside it. A mesh has the properties that (i) every node in the closed path is exactly

formed with two branches (ii) no other branches are enclosed by the closed path. Meshes can be

thought of a resembling window partitions. On the other hand, loop is also a closed path but inside

the closed path there may be one or more than one branches or elements.

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.

Applying KVL around mesh (loop)-1: (note in mesh-1, I1is known as local current and other mesh

currentsI2&I3are known as foreign currents)

0)()( 421231 RIIRIIVV CA

0)( 3224142 CA VVIRIRIRR


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0)( 1313212111 VIRIRIR (1)

Applying KVL around mesh (loop)-2: (similarly in mesh-2,I2is local current andI1&I3are known

as foreign currents)

0)()( 332412 RIIRIIVB

0)( 3324314 BVIRIRRIR

02323222121 VIRIRIR (2)

Applying KVL around mesh (loop)-3:

0)()( 21332313 RIIRIIRIVC

0)( 33212312 CVIRRRIRIR

03333232131 VIRIRIR (3)

NOTE: EQUATION (1), (2) & (3) WILL BE SOLVED USING CALCULATOR


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DC CIRCUITS ANALYSIS USING NODAL METHOD

NODE VOLTAGE ANALYSIS: In the previous section, it has been discussed in detail the analysis

of a dc network by writing a set of simultaneous algebraic equations (based on KVL only) in which

the variables are currents, known as mesh analysis or loop analysis. On the other hand, the node

voltage analysis (Nodal analysis) is another form of circuit or network analysis technique, which will

solve almost any linear circuit. In a way, this method completely analogous to mesh analysis method,

writes KCL equations instead of KVL equations, and solves them simultaneously.

SOLUTION OF ELECTRIC CIRCUIT BASED ON NODE VOLTAGE METHOD: In the node

voltage method, we identify all the nodes on the circuit. Choosing one of them as the reference

voltage (i.e., zero potential) and subsequentlyassign other node voltages (unknown) with

respect to a reference voltage (usually ground

voltage taken as zero (0) potential and denoted

by ( ). If the circuit has n nodes there are

n-1 node voltages are unknown (since we

are always free to assign one node to zero or

ground potential). At each of these n-1

nodes, we can apply KCL equation. The

unknown node voltages become the

independent variables of the problem and the

solution of node voltages can be obtained by

solving a set of simultaneous equations. Let us consider a simple dc network as shown in Figure 5.1

to find the currents through different branches using Node voltage method.

KCL equation at Node-1:

;02

31

4

2131

R

VV

R

VVII SS 0

11113

2

2

4

1

42

31

V

RV

RV

RRII SS

31321211131 VGVGVGII SS (4)


;023

32

4

21

SI

R

VV

R

VV3

3

2

43

1

4

2

1111V

RV

RRV

RIS

3232221212 VGVGVGIS (5)


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;01

3

2

31

3

323

R

V

R

VV

R

VVIS 3

321

2

3

1

2

3

11111V

RRRV

RV

RIS

3332321313 VGVGVGIS (6)

NOTE: EQUATION (4), (5) & (6) WILL BE SOLVED USING CALCULATOR


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SUPERPOSITION THEOREM

If the circuit has more than one independent (voltage and/or current) sources, one way to

determine the value of variable (voltage across the resistance or current through a resistance) is to

use nodal or mesh current methods. Alternative method for any linear network, to determine the

effect of each independent source (whether voltage or current) to the value of variable (voltage

across the resistance or current through a resistance) and then the total effects simple added. This

approach is known as the superposition.

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.

One may consider the resistances R1 and

R3 are the internal resistances of the

voltage sources whereas the resistance R4

is considered as internal resistance of the

current source. The problem is to

determine the response I in the in the

resistor R2. The current Ican be obtained

from

According to the application of the superposition theorem, it may be noted that each independent

source is considered at a time while all other sources are turned off or killed. To kill a voltage source

means the voltage source is replaced by its internal resistance (i.e.R1orR3; in other wordsE1or E2

should be replaced temporarily by a short circuit) whereas to kill a current source means to replace

the current source by its internal resistance (i.e.R4; in other wordsISshould be replaced temporarily

by an open circuit).


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Remarks:Superposition theorem is most often used when it is necessary to determine the individual

contribution of each source to a particular response.

PROCEDURE FOR USING THE SUPERPOSITION THEOREM:

Step-1:Retain one source at a time in the circuit and replace all other sources with their internalresistances.

Step-2: Determine the output (current or voltage) due to the single source acting alone using the

Nodal or Mesh techniques.

Step-3:Repeat steps 1 and 2 for each of the other independent sources.

Step-4: Find the total contribution by adding algebraically all the contributions due to the

independent sources.


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THEVENINS THEOREM

A simple circuit as shown in fig.8.1 is considered to illustrate the concept of equivalent

circuit and it is always possible to view even a very complicated circuit in terms of much simpler

equivalent source and load circuits. Subsequently the reduction of computational complexity that

involves in solving the current through a branch for different values of load resistance (RL) is also

discussed. In many applications, a network may contain a variable component or element while other

elements in the circuit are kept constant. If the solution for current (I) or voltage (V) or power (P) in

any component of network is desired, in such cases the whole circuit need to be analyzed each time

with the change in component value. In order to avoid such repeated computation, it is desirable to

introduce a method that will not have to be repeated for each value of variable component. Such

tedious computation burden can be avoided provided the fixed part of such networks could be

converted into a very simple equivalent circuit that represents either in the form of practical voltagesource known as Thevenins voltage source (VTh = magnitude of voltage source, RTh = internal

resistance of the source). In true sense, this conversion will considerably simplify the analysis while

the load resistance changes. Although the conversion technique accomplishes the same goal, it has

certain advantages over the techniques that we have learnt in earlier sections.

Let us consider the circuit

shown in fig. (a). Our problem is to

find a current through RL using

different techniques; the following

observations are made

Find

Mesh current method needs 3 equations to be solved Node voltage method requires 2 equations to be solved Superposition method requires a complete solution through load resistance (RL) by

considering each independent source at a time and replacing other sources by their internal

source resistances.

Suppose, if the value ofRLis changed then the three (mesh current method) or two equations

(node voltage method) need to be solved again to find the new current in RL. Similarly, in case of

superposition theorem each time the load resistance RL is changed, the entire circuit has to be

analyzed all over again. Much of the tedious mathematical work can be avoided if the fixed part of

circuit (fig. 8.1(a)) or in other words, the circuit contained inside the imaginary fence or black box


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with two terminals A & B, is replaced by the simple equivalent voltage source (as shown in fig.

8.1(b)).

Thevenins Theorem: Thevenins theorem states that any two output terminals (A &B, shown in fig.

8.2.(a)) of an active linear network containing independent sources (it includes voltage and

current sources) can be replaced by a simple voltage source of magnitude VThin series with a

single resistor RTh (see fig. 8.2(d)) where RTh is the equivalent resistance of the network

when looking from the output terminals A & B with all sources (voltage and current)

removed and replaced by their internal resistances (see fig. 8.2(c)) and the magnitude of VTh

is equal to the open circuit voltage across the A & B terminals.

THE PROCEDURE FOR APPLYING THEVENINS THEOREM:

To find a currentILthrough the load resistanceRL(as shown in fig. 8.2(a)) using Thevenins

theorem, the following steps are followed:

Step-1:Disconnect the load resistance (RL) from the circuit, as indicated in fig. 8.2(b).


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Step-2:Calculate the open-circuit voltage VTh(shown in fig. 8.2(b)) at the load terminals (A & B)

after disconnecting the load resistance (RL). In general, one can apply any of the techniques (mesh-

current, node-voltage and superposition method) learnt in earlier sections to compute VTh.

Step-3:Redraw the circuit (fig. 8.2(b)) with each practical source replaced by its internal resistance

as shown in fig.8.2(c). (NOTE: voltage sources should be short-circuited (just remove them andreplace with plain wire) and current sources should be open-circuited (just removed).

Step-4:Look backward into the resulting circuit from the load terminals (A & B), as suggested by

the eye in fig. 8.2(c). Calculate the resistance that would exist between the load terminals. The

resistanceRThis described in the statement of Thevenins theorem.

Note:Calculating this resistance may be a difficult task but one can try to use the standard circuit

reduction technique or YorYtransformation techniques.

Step-5: Place RTh in series with VTh to form the Thevenins equivalent circuit (replacing the

imaginary fencing portion or fixed part of the circuit with an equivalent practical voltage source) as

shown in fig. 8.2(d).

Step-6:Reconnect the original load to the Thevenin voltage circuit as shown in fig. 8.2(e); the loads

voltage, current and power may be calculated by a simple arithmetic operation only.

Load Current,LTh

ThL

RR

VI

(7)


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Voltage across the load, LLLLTh

ThL RIR

RR

VV

(8)

Power absorbed by the load, LLL RIP 2 (9)

REMARKS:

(i) One great advantage of Thevenins theorem over the normal circuit reductiontechnique or any other technique is this: once the Thevenin equivalent circuit has been

formed, it can be reused in calculating load current (IL), load voltage (VL) and load

power (PL) for different loads using the equations (7)-(9).

(ii) With help of this theorem one can find the choice of load resistance RL that results inthe maximum power transfer to the load. On the other hand, the effort necessary to

solve this problem-using node or mesh analysis methods can be quite complex andtedious from computational point of view.


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Star-Delta Transformation

One of the most basic three-terminal network equivalents is that of three resistors connected in

Delta () and in Wye (Y). These two circuits identified in fig. 6.1(e) and Fig. 6.1(f) are

sometimes part of a larger circuit and obtained their names from their configurations. These three

terminal networks can be redrawn as four-terminal networks as shown in fig. 6.1(c) and fig. 6.1(d).

We can obtain useful expression for direct transformation or conversion from to Y or Y to by

considering that for equivalence the two networks have the same resistance when looked at the

similar pairs of terminals.

CONVERSION FROM DELTA () TO STAR OR WYE (Y):

Let us consider the network shown in fig. 6.1(e) and assumed the resistances (RAB,RBCand

RCA) in network are known. Our problem is to find the values of RA, RB and RC in Wye (Y)

network (see fig. 6.1(e)) that will produce the same resistance when measured between similar pairs

of terminals. We can write the equivalence resistance between any two terminals in the following

form.

BETWEEN A & C TERMINALS:

CABCAB

BCABCACA

RRR

RRRRR

)( (10)

BETWEEN C & B TERMINALS:

CABCAB

CAABCBCB

RRR

RRRRR

)( (11)

BETWEEN B & A TERMINALS:

CABCAB

CABCAB

AB RRR

RRR

RR

)(

(12)

By combining above three equations, we have


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CABCAB

ABCACABCBCABCBA

RRR

RRRRRRRRR

(13)

Subtracting equations (10), (11), and (12) from (13) we write the expressions for for unknown

resistances of Wye (Y) network as

CABCAB

CAABA

RRRRRR

(14)

CABCAB

BCABB

RRR

RRR

(15)

CABCAB

CABCC

RRR

RRR

(16)

CONVERSION FROM STAR OR WYE (Y) TO DELTA ():

To convert a Wye (Y) to a Delta (), the relationshipsRAB,RBC&RCAmust be obtained in

terms of the Wye (Y) resistancesRA,RB&RC(referring to fig.6.1 (f)). Considering the Y-connected

network, we can write the current expression throughRA resistor as

A

NAA

R

VVI

)( (For Ynetwork) (17)

Appling KCL at N for Y connected network (assume A, B, C terminals having higher potential

than the terminalN) we have,

C

C

B

B

A

A

CBA

N

C

NC

B

NB

A

NA

R

V

R

V

R

V

RRR

V

R

VV

R

VV

R

VV 1110

)()()(

Or,

CBA

C

C

B

B

A

A

N

RRR

R

V

R

V

R

V

V111

(18)

For -network (see fig. 6.1(f)),

Current entering at terminal A = Current leaving the terminal A

AC

AC

AB

ABAR

V

R

VI (for network) (19)

From equations (17) and (18),


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AC

AC

AB

AB

A

NA

R

V

R

V

R

VV

)(

Using the VN expression in the above equation, we get

AC

AC

AB

AB

A

CBA

C

CA

B

BA

AC

AC

AB

AB

A

CBA

C

C

B

B

A

A

A

R

V

R

V

R

RRR

R

VV

R

VV

R

V

R

V

R

RRR

R

V

R

V

R

V

V

111111

OrAC

AC

AB

AB

A

CBA

C

AC

B

AB

R

V

R

V

R

RRR

R

V

R

V

111

(19)

Equating the coefficients of VABand VAC and in both sides of eq. (19), we obtained the following

relationship.

C

BABAAB

CBA

BA

AB R

RRRRR

RRRRR

R

111

11 (20)

B

CACAAC

CBA

CAAC R

RRRRR

RRRRRR

111

11 (21)

Similarly,IBfor both the networks (see fig. 61(f)) are given by

B

NBB

R

VVI

)( (For Y network)

BA

BA

BC

BCB

R

V

R

VI (For network)

Equating the above two equations and using the value of VN(see eq (18)), we get the final expression

as

BA

BA

BC

BC

B

CBA

A

BA

C

BC

R

V

R

V

R

RRR

R

V

R

V

111

Equating the coefficients of VBCin both sides of the above equations we obtain the following relation


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A

CBCBBC

CBA

CB

BC R

RRRRR

RRRRR

R

111

11 (22)

NOTE: When we need to transform a Delta () network to an equivalent Wye (Y) network, the

equations (14)-(16) are the useful expressions. On the other hand, the equations (20)(22) are used

for Wye (Y) to Delta () conversion.

REMARKS:

In order to note the symmetry of the transformation equations, the Wye (Y) and Delta ()

networks have been superimposed on each other as shown in fig. 6.2.

The equivalent star (Wye) resistance connected to a given terminalis equal to the product of the two Delta () resistances connected

to the same terminal divided by the sum of the Delta ()

resistances (see fig. 6.2).

The equivalent Delta () resistance between two-terminals is thesum of the two star (Wye) resistances connected to those terminals

plus the product of the same two star (Wye) resistances divided by

the third star (Wye (Y)) resistance (see fig.6.2).


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1-Phase AC circuits

A multi-turn coil is placed inside a magnet with an air gap as shown in Fig. 12.1. The flux

lines are from North Pole to South Pole. The coil is rotated at an angular speed, = 2n (rad/s).

Instantaneous value of e.m.f. induced in the coil

dt

dNe (-ve sign indicates that in effect the induced e.m.f. is opposite

to the very cause which produces it)

tdt

dNe m cos ( tm cos )

tNe m sin

tNe m sin . (i)

The value of induced e.m.f will be maximum when angle or t = 900(i.e. sint = 1)

mm NE

Putting this value in equation (i), we get,

e = Emsint = Emsin


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From the above equation it is clear that the magnitude of the induced e.m.f. varies according

to the sine of angle. The wave shape of the induced e.m.f. is shown in fig. below. This is called

sinusoidal wave

If this voltage is applied across a resistor, an alternating current will flow through it varying

sinusoidally. This alternating current is given by the equation:

i = Imsint = ImsinIMPORTANT TERMS:

Waveform: The shape of the curve obtained by plotting the instantaneous values ofalternating quantity is called a Waveform.

Instantaneous Value: The value of an alternating quantity at any instant is calledinstantaneous value and is represented by eor i.

Cycle: When an alternating quantity goes through a complete set of +ve and ve values, it issaid to complete one cycle.

Time Period:The time taken in seconds to complete one cycle by an alternating quantity iscalled time period.

Frequency:The no. of cycles made per second by an alternating quantity is called frequency. Amplitude:The maximum value (positive or negative) attained by an alternating quantity in

one cycle is called itsamplitudeorpeak valueor maximum value.


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1-PHASE AC CIRCUITS UNDER SINUSOIDAL STEADY STATE

CASE- 1: PURELY RESISTIVE CIRCUIT (R ONLY)

If we were to plot the current and voltage for a very simple AC circuit consisting of a source and a

resistor (Figure 3.1), it would look something like this: (Figure 3.2)

Because the resistor simply and directly resists the flow of electrons at all periods of time, the

waveform for the voltage drop across the resistor is exactly in phase with the waveform for the

current through it. We can also calculate the power dissipated by this resistor, and plot those valueson the same graph: (Figure 3.3)

Note that the power is never a negative value. When the current is positive (above the line),

the voltage is also positive, resulting in a power (p=ie) of a positive value. Conversely, when the

current is negative (below the line), the voltage is also negative, which results in a positive value for

power. This consistent polarity of power shows that the resistor is always dissipating power, taking

it from the source and releasing it in the form of heat energy. Whether the current is positive or

negative, a resistor still dissipates energy.

CASE- 2: PURELY INDUCTIVE CIRCUIT (L ONLY)

Inductors oppose changes in current through them, by dropping a voltage directly

proportional to the rate of changeof current. In accordance with Lenzs Law, this induced voltage is

always of such a polarity as to try to maintain current at its present value. That is, if current is


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increasing in magnitude, the induced voltage will push against the electron flow; if current is

decreasing, the polarity will reverse and push with the electron flow to oppose the decrease. This

opposition to current change is called reactance, rather than resistance. Expressed mathematically,

the relationship between the voltage dropped across the inductor and rate of current change through

the inductor is as such:

dt

diLe

The expression di/dt is one from calculus, meaning the rate of change of instantaneous current (i)

over time, in amps per second. The inductance (L)is in Henry, and the instantaneous voltage (e), of

course, is in volts. To show what happens with alternating current, lets analyze a simple inductor

circuit: (Figure 3.4)

If we were to plot the current and voltage for this very simple circuit, it would look

something like this: (Figure 3.5)

The voltage dropped across an inductor is a reaction against the changein current through it.

Therefore, the instantaneous voltage is

zero whenever the instantaneous current is

at a peak, and the instantaneous voltage is

at a peak wherever the instantaneous

current is at maximum. This results in a

voltage wave that is 90oout of phase with

the current wave. Looking at the graph, the

voltage wave seems to have a head start

on the current wave; the voltage leads

the current, and the current lags behind


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CASE- 3: PURELY CAPACITIVE CIRCUIT (C ONLY):

Capacitors oppose changes in voltage by drawing or supplying current as they charge or

discharge to the new voltage level. The flow of electrons through a capacitor is directly

proportional to the rate of changeof voltage across the capacitor. This opposition to voltage change

is another form of reactance, but one that is precisely opposite to the kind exhibited by inductors.

Expressed mathematically, the relationship between the current through the capacitor and rate of

voltage change across the capacitor is as such:

dt

deCi

Or idtCe

1

To show what happens with alternating current, lets analyze a simple capacitor circuit: (Figure 4.4)

If we were to plot the current and voltage for this very simple circuit, it would look

something like this: (Figure 4.5)

Remember, the current through a capacitor is a reaction against the change in voltage across

it. Therefore, the instantaneous current is zero whenever the instantaneous voltage is at a peak (zero


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change, or level slope, on the voltage sine wave), and the instantaneous current is at a peak wherever

the instantaneous voltage is at maximum change (the points of steepest slope on the voltage wave,

where it crosses the zero line). This results in a voltage wave that is -90oout of phase with the current

wave. Looking at the graph, the current wave seems to have a head start on the voltage wave; the

current leads the voltage, and the voltage lags behind the current. (Figure 4.6)

As you might have guessed, the same unusual power wave that we saw with the simple

inductor circuit is present in the simple capacitor circuit, too: (Figure 4.7)

As with the simple inductor circuit, the 90 degree phase shift between voltage and current results in a

power wave that alternates equally between positive and negative. This means that a capacitor does

not dissipate power as it reacts against changes in voltage; it merely absorbs and releases power.

Since capacitors conduct current in proportion to the rate of voltage change, they will pass

more current for faster-changing voltages (as they charge and discharge to the same voltage peaks in

less time), and less current for slower-changing voltages. This means that reactance in ohms for any

capacitor is inverselyproportional to the frequency of the alternating current.

fCXC

2

1


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From these equations, the magnitude and phase angle of the current, Iare derived. From the second

one, )/(tan RL

So phase angle, )/(tan 1 RL

Two relations, )/(cos ZR , and )/(sin ZL , are derived with the term impedance,

22 )( LRZ

If these two expressions are substituted in the first one, it can be shown that the magnitude of the

current isI=V/Z, with both VandZin magnitude only.

The steps required to find the r.m.s. value of the current I, using complex form of impedance, are

given here.

The impedance (Fig. 14.5) of the inductive (R-L) circuit is,

LjRjXRZ L

Where,

2222 )()( LRXRZL

and

R

L

R

XL

11

tantan

LjR

jV

jXR

jV

Z

VI

L

0000

2222 )( LR

V

XR

V

Z

VI

L

Note that the current lags the voltage by the angle , value as given above. In this case, the voltage

phasor has been taken as reference phase, with the current phasor lagging the voltage phasor by the

angle, . But normally, in the case of the series circuit, the current phasor is taken as reference phase,

with the voltage phasor leading the current phasor by . This can be observed both from phasor

diagram (Fig. 14.4b), and waveforms (Fig. 14.4c). The inductive reactance XL is positive. In the

phasor diagram, as one move from voltage phasor to current phasor, one has to go in the clockwise

direction, which means that phase angle; is taken as positive, though both phasors are assumed to

move in anticlockwise direction as shown in the previous lesson.

The complete phasor diagram is shown in Fig. 14.4b, with the voltage drops across the two

components and input (supply) voltage (OA), and also current (OB). The voltage phasor is taken as

reference. It may be observed that


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VOC+ VCA= VOA

=> IR + I(jXL) = IZ

CASE- 5: SERIES R-C CIRCUIT:

This part is discussed in brief. The voltage balance equation for the R-C series circuit (Fig. 14.6a) is,

tVidtC

Riv sin21

The current is

tIi sin2

The reasons for the above choice of the current, i, and the steps needed for the derivation ofthe above expression, have been described in detail, in the case of the earlier example of inductive

(R-L) circuit. The same set of steps has to be followed to derive the current, iin this case.

Alternatively, the steps required to find the rms value of the current I, using complex form of

impedance, are given here.

The impedance of the capacitive (R-C) circuit is,

CjRjXRZ C

1

Where

2

222 1

CRXRZ C

And

CRCRR

XC

1tan

1tantan 111

CjRjV

jXR

jV

Z

VI

C

1

0000

2222 )1( CR

V

XR

V

Z

VI

C


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CASE- 6: SERIES R-L-C CIRCUIT:

The voltage balance equation for the circuit with R, L and C in series (Fig. 15.1a), is

tVidtCdt

diLRiv sin2

1

The current,Iis of the form,

)sin(2 tIi

As described in the previous topic on series (R-L) circuit, the current in steady state is sinusoidal innature. The procedure given here, in brief, is followed to determine the form of current. If the

expression for )sin(2 tIi is substituted in the voltage equation, the equation shown here is

obtained, with the sides (LHS & RHS) interchanged.

tVtIC

tILtIR

sin2)cos(21

)cos(2)sin(2

Or

tVtIC

LtIR

sin2)cos(21

)sin(2

The steps to be followed to find the magnitude and phase angle of the current i, are same as

described for R-L circuit.

So, the phase angle is

R

CL

R

XX CL

1

tantan 11

And the magnitude of the current isI = V/Z

Where the impedance of the series circuit is 22 )]1([ CLRZ

Alternatively, the steps to find the rms value of the currentI, using complex form of impedance, are

given here.

CLjRXXjRZ CL

1)(

Where,

22 )]1([ CLRZ , and


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R

CL

R

XX CL

1

tantan 11

22

22 1)(

CLR

V

XXR

V

Z

VI

CL

Two cases are: (a)

CL

1, and (b)

CL

1

ACTIVE, REACTIVE AND APPARENT POWER

APPARENT POWER, VOLT-AMPERES (VA) AND REACTIVE POWER:

TheApparent Poweris the product of the voltage and complex conjugate of the current, both

in phasor form. For the inductive circuit, described earlier, the voltage ( 00V ) is taken as reference

and the current ( sincos jIII ) is lagging the voltage by an angle, . The complex power

is jQPjVIVIIVIVIVS sincos)(00

The Volt-Amperes (S), a scalar quantity, is the product of the magnitudes the voltage and the

current. So,

22 QPIVS . It is expressed in VA.

TheActive or Real Power (P)is

cos)Re()Re( VIIVSP , it is expressed in W.

TheReactive Power (Q)is given by

sin)Im()Im( VIIVSP , it is

expressed in VAr

As the phase angle, is taken as positive ininductive circuits, the reactive power is positive. The real

part, (Icos) is in phase with the voltage V, whereas the imaginary part,Isinis in quadrature (-900)


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with the voltage V. But in capacitive circuits, the current ( I ) leads the voltage by an angle ,

which is taken as negative. So, it can be stated that the reactive power is negative here, which can

easily be derived.

PHYSICAL MEANING OF REACTIVE POWER

EXPLANATION FOR REACTIVE POWER:Explanation for reactive power says that in

an alternating current system, when the voltage and current go up and down at the same time, only

real power is transmitted and when there is a time shift between voltage and current both active and

reactive power are transmitted. But, when the average in time is calculated, the average active power

exists causing a net flow of energy from one point to another, whereas average reactive power is

zero, irrespective of the network or state of the system. In the case of reactive power, the amount of

energy flowing in one direction is equal to the amount of energy flowing in the opposite direction (or

different parts -capacitors, inductors, etc of a network, exchange the reactive power). That means

reactive power is neither produced nor consumed. But, in reality we measure reactive power losses,

all the methods we introduce to reduce electricity consumption are actually employed to reduce

reactive power so as to reduce total electricity consumption cost.

POWER FACTORPower factor = (Average power/Apparent Power)

=222

2 1coscos

XR

R

CLR

R

Z

R

VI

VI

The power factor in this circuit is less than 1 (one), as 00 900, being positive as given

above.

For the resistive (R) circuit, the power factor is 1 (one), as = 0 0, and the average power is

VI.

For the circuits with only inductance, L or capacitance, C as described earlier, the power

factor is 0 (zero), as = 900. For inductance, the phase angle, or the angle of the impedance, =

+900(lagging), and for capacitance, = -900(leading). It may be noted that in both cases, the average

power is zero (0), which means that no power is consumed in the elements, L and C.

3-PHASE BALANCED AND UNBALANCED SUPPLY

Generation of Three-phase Balanced Voltages

In the previous section, the generation of single-phase voltage, using a multi-turn coil placed

inside a magnet, was described. It may be noted that, the scheme shown was a schematic one,


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whereas in a machine, the windings are distributed in number of slots. Same would be the case with a

normal three-phase generator. Three windings, with equal no. of turns in each one, are used, so as to

obtain equal voltage in magnitude in all three phases. Also to obtain a balanced three-phase voltage,

the windings are to be placed at an electrical angle of 1200with each other, such that the voltages in

each phase are also at an angle of 120

0

with each other, which will be described in the next section.The schematic diagram with multi-turn coils, as was shown earlier in Fig. 12.1 for a single-phase

one, placed at angle of 1200with each other, in a 2-pole configuration, is shown in Fig. 18.1a. The

waveforms in each of the three windings (R, Y & B), are also shown in Fig. 18.1b. The windings are

in the stator, with the poles shown in the rotor, which is rotating at a synchronous speed of NS(r/min,

or rpm), to obtain a frequency off = ((p.NS)/120)(Hz), p being no. of poles [p = 2].


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STAR AND DELTA CONNECTIONS

THREE-PHASE VOLTAGES FOR STAR CONNECTION:

The connection diagram of a star (Y)-connected three-phase system is shown in Fig. The

connection diagram of a star (Y)-connected three-phase system is shown in Fig. 18.2a, along with

phasor representation of the voltages (Fig. 18.2b). These are in continuation of the figures 18.1a-b.

Three windings for three phases are R (+) & R(),Y (+) & Y(), and B (+) & Y(). Taking the

winding of one phase, say phase R as an example, then R with sign (+) is taken as start, and R with

sign () is taken as finish. Same is the case with two other phases. for making star (Y)-connection,

R, Y & B are connected together, and the point is taken as neutral, N. Three phase voltages are:

eRN=Emsin ; eYN=Emsin( -1200);eBN=Emsin( - 2400) =Emsin( + 1200)

It may be noted that, if the voltage in phase R (eRN) is taken as reference as stated earlier, then the

voltage in phase Y (eYN) lags eRNby 1200, and the voltage in phase B(eBN) lags eYNby 120

0, or leads

eRNby 1200. The phasors are given as:

The phase voltages are all equal in magnitude, but only differ in phase. This is also shown in

Fig. 18.2b. The relationship between EandEmis 2mEE . The phase sequence is R-Y-B. It can

be observed from Fig. 18.1b that the voltage in phase Y attains the maximum value, after = t =

1200 from the time or angle, after the voltage in phase R attains the maximum value, and then the

voltage in phase B attains the maximum value. The angle of lag or lead from the reference phase, R

has been stated earlier.

RELATION BETWEEN THE PHASE AND LINE VOLTAGES FOR STAR CONNECTION:


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Three line voltages (Fig. 18.4) are obtained by the following procedure. The line voltage,ERY

is

866.05.0011200 00 jjEEEEEE YNRNRY

0303)866.05.1( EjEERY

The magnitude of the line voltage,ERYis 3 times the magnitude of the phase voltageERN, andERY

leadsERNby 300. Same is the case with other two line voltages as shown in brief (the steps can easily

be derived by the procedure given earlier).

000 903120120 EEEEEE BNYNYB

000 15030120 EEEEEE RNBNBR

So, the three line voltages are balanced, with their magnitudes being equal, and the phase angle being

displaced from each other in sequence by 1200

. Also, the line voltage, say ERY, leads the

corresponding phase voltage,ERNby 300

33 LPLP VVEE

Also, LP II

RELATION BETWEEN PHASE AND LINE QUANTITIES FOR DELTA CONNECTION

The connection diagram of a delta ()-connected three-phase system is shown in Fig. 18.4a,

along with phasor representation of the voltages (Fig. 18.4b). For making delta ()-connection, the

start of one winding is connected to the finish of the next one in sequence, for instance, starting from

phase R, R is to connected to Y, and then Y to B, and so on (Fig. 18.4a). The line and phase

voltages are the same in this case, and are given as

00EERY ;0120EEYB ;

0120EEBR


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If the phasor sum of the above three phase (or line) voltages are taken, the result is zero (0).

The phase or line voltages form a balanced one, with their magnitudes being equal, and the phase

being displaced from each other in sequence by 1200.

3LP II

Also, LPLP VVEE

TOTAL POWER CONSUMED IN THE CIRCUIT

STAR-CONNECTED CIRCUIT

In the previous section, the power consumed in a circuit fed from a single-phase supply was

presented. Using the same expression for the above star-connected balanced circuit, fed from three-

phase supply (Fig. 18.4a-b), the power consumed per phase is given by:

PPPP IVW cos

Hence, the power consumed in 3-phase is:

PPP IVW cos3

It has been shown that magnitude of phase voltage is given by 3LP VV , where magnitude

of line voltage is VL. The magnitudes of line and phase currents are same LP II . Substituting the

two expressions, the total power consumed by 3-phase supply is obtained as

PLLPLL IVIVW cos3cos33

NOTE: The phase angle P is the angle between phase voltage VPand phase current,IP.

DELTA-CONNECTED CIRCUIT

The power consumed per phase is given by:

PPPP IVW cos

Hence, the power consumed in 3-phase is:

PPP IVW cos3

It has been shown that magnitude of phase voltage is given by LP VV , where magnitude of

line voltage is VL. The magnitudes of line and phase currents are 3LP II . Substituting the two

expressions, the total power consumed by 3-phase supply is obtained as

PLLPLL IVIVW cos3cos33

NOTE: The phase angle P is the angle between phase voltage VPand phase current,IP.

BEEE UNIT I

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