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READING MATERIAL FOR B.E STUDENTS

OF RGPV AFFILIATED ENGINEERING COLLEGES

SUBJECT BASIC ELECTRICAL AND ELECTRONICS

Professor MD Dutt

Addl General Manager (Retd)

BHARAT HEAVY ELECTRICALS LIMITED

Professor(Ex) of EX Department

Bansal Institute of Science and Technology

KOKTA ANANAD NAGAR BHOPAL

Presently Head of The Department ( EX)

Shri Ram College Of Technology

Thuakheda BHOPAL

Sub Code BE 104 Subject Basic Electrical & Electronics

UNIT I Electrical Circuit Analysis

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RGPV Syllabus

BE 104 BASIC ELECTRICAL & ELECTRONICS ENGINEERING

UNIT I

ELECTRICAL CIRCUIT ANALYSIS

Voltage and current source, Dependent and independent sources, source conversion,

DC circuit analysis using mesh and nodal method. Thevenin,s and superposition

theorem, Star delta transformation.

1 – Phase AC circuits under sinusoidal steady state , active reactive and apparent

power . Physical meaning of reactive power . Power Factor, 3 Phase balance and

unbalance supply. Star and Delta connections.

INDEX

S No Topic Page

1 Voltage and current source, Dependent and independent

sources, source conversion

3,4,5

2 DC circuit analysis using mesh and nodal method 6,7,8

3 Thevenin,s theorem, 9,10

4 Superposition theorem 11,12

5 Star delta transformation 13,14,15

6 1 – Phase AC circuits under sinusoidal steady state 15

7 Active reactive and apparent power . Physical meaning of

reactive power

16,17

8 Power Factor, 3 Phase balance and unbalance supply 17,18

9 Star and Delta connections 18,19,20

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VOLTAGE The rate at which energy is drawn from a source that produces a flow of

electricity in a circuit; expressed in volts.

Electric Current Flow of electrons from one place to another. Electric current means a

flow of charge in the same way that water current flows . A flow of electricity through a

conductor; measured in amperes

RESISTANCE A material’s opposition to the flow of electric current; measured in ohm.

VOLTAGE AND CURRENT SOURCE

Voltage source :- Most of the sources used in day to day life are constant voltage source,

such as batteries, dynamos and alternators.

Current Source :- Specifically the collector circuits of transistors, metadyne generators, are

considered as current sources.

VOLTAGE SOURCE CURRENT SOURCE

Independent Source:- A source may be said independent when it does not depend on any

other quantity of circuit. Figure below shows Independent DC source and Independent AC

Source varying with time.

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IDEAL CURRENT SOURCE:- In the figure below independent DC Current source and

independent time varying source.

DEPENDENT SOURCES:-

I) Voltage dependent voltage source

II) Current dependent voltage source

III) Voltage dependent current source

IV) Current dependent current source.

IDEAL VOLTAGE SOURCE:- A constant voltage source is an ideal source, almost

capable of supplying any current at a given voltage, considering internal resistance equal to

zero. The terminal voltage across the terminal remains constant. The ideal voltage can

never be short circuited. Practically an Ideal voltage source is not possible.

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IDEAL CURRENT SOURCE:- A source that supplies constant current to a load even its

impedance varies. The Ideal current source can never be open circuited.

SOURCE EQUIVALNCE OR TRANSFORMATION:- Consider a voltage source of Vs

and internal resistance Rin for conversion into an equivalent current source. The current

supplied by this voltage source, when short circuited is put across the terminal A and B will

be equal to Vs/Rin.

A current source supplying this current is Is = Vs/Rin and having the same resistance

across , It will represent the current source.

sss IRV

s

s

sR

VI

R

viRiv ssss or

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KIRCHOFF’s CURRENT LAW

At any junction point in an electrical circuit, the total current into the junction equals the

total current out of the junction.

(“What goes in must come out.”)

In the diagram at right,

I1 + I2 = I3

OR

Where N is the total number of branches connected to a

node.

Example 1 (KCL) Determine I, the current flowing out of the voltage source.

Use KCL

1.9 mA + 0.5 mA + I are entering the node.

I

3 mA is leaving the node.

3ma .5ma

1.9ma

I

Kirchhoff’s Voltage Law

In any complete path in an electrical circuit, the sum of the potential increases equals the

sum of the potential drops.

(“What goes up must come down.”)

nodenode

1

0

leaveenter

N

n

n

ii

i

mAI

mAmAmAI

mAImAmA

6.0

)5.09.1(3

35.09.1

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Example 2 (KVL)

Find the voltage across R1. Note that the polarity of the voltage has been assigned in the

circuit schematic.

First, define a loop that include R1.

If the outer loop is used:

Follow the loop clockwise.

rises drops

M

1m

v v

0 v

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By convention, voltage drops are added and voltage rises are subtracted in KVL.

Summary

The currents at a node can be calculated using Kirchhoff’s Current Law (KCL).

The voltage dropped across components can be calculated using Kirchhoff’s Voltage Law

(KVL).

Ohm’s Law is used to find some of the needed currents and voltages to solve the problems.

VV

VVV

R

R

2

035

1

1

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THEVNIN’S THEOREM :- This theorem provides a mathematical technique for replacing

a two terminal network by a voltage source Vt, and resistance Rt connected in series. The

voltage source Vt ( called Thevenin’s equivalent voltage), this is the open circuit voltage

that appears across the load terminals when the load is removed or disconnected and

resistance Rt, called the Thevenin’s equivalent resistance, is equal to the resistance of

network looking back into the load terminals.

The steady state current will be I = Vt/(Rt +RL) This is in the case of DC network

Let us consider a circuit shown here below, which consists of a source of EMF

E volts and internal resistance r ohms connected to an external circuit consists of resistance

R1 and R2 Ohms in series. AB across which a resistance R2 ohms is connected the network

acts as source of open circuit voltage Voc ( also called Thevenin’s equivalent voltage Vt

and internal resistance Rin ( also called Thevenin’s resistance Rt)

For determination of open circuit voltage Voc ( Vt) Disconnect the load resistance RL from

the terminals A B to provide open circuit

Now current through resistance R2,

I = E

R1 + R2 +r

For determination of internal resistance Rin or Rt of the network under consideration,

remove the voltage source from the circuit, leaving behind only its internal resistance r , as

shown in in figure. Now view the circuit inwards from open terminal A and B. It is found

that in circuit below now consists resistance R2 only, and the other consists of resistance R1

and r in series.

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Thus the equivalent resistance Rt as viewed from open terminals A and B

Rt = R2(R1+r)

R2 +R1+r

When load resistance Rl is connected across terminal A and B, the network behaves as a

source of Voltage Vt and terminal resistance Rt and the current flowing through the load

resistance Rl is given as

I1 = Vt = ER2 ( R1+R2+r)

Rt+Rl R2 (R1 +r) + Rl

R1 +r +R2

= ER2

R2 (R1+r) + Rl ( R1+R2 +r)

The current in any passive circuit element ( which may be called Rl ) in a network is the

same as would be obtained if Rl were supplied with a voltage source Voc or Vt in series

with an equivalent resistance Rin or Rt . Voc being the open circuit voltage at terminal

from which Rl has been removed and Rin or Rt being the resistance that would be measured

at these terminals, after all sources have been removed and each have been replaced with its

internal resistance.

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SUPERPOSITION THEOREM

The superposition principle states that the voltage across (or current through) an element in

a linear circuit is the algebraic sum of the voltages across (or currents through) that element

due to each independent source acting alone.

Steps to apply superposition principle

1. Turn off all independent sources except one source. Find the output (voltage or

current) due to that active source using nodal or mesh analysis.

Turn off voltages sources = short voltage sources; make it equal to zero voltage

Turn off current sources = open current sources; make it equal to zero current

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to

the independent sources.

Dependent sources are left intact.

2mA Source Contribution

I’0 = -4/3 Ma

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4 ma Source Contribution

I’’0 = 0

12 voltage source contribution

I’’’0 = -4 mA

Final result

I’0 = -4/3 mA

I’’0 = 0

I’’’0 = -4 mA

I0 = I’0+ I’’0+ I’’’0 = -16/3 mA

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STAR DELTA TRANSFORMATION:- Three resistances connected nose to tail as

shown in figure are said to be Delta connected or mesh connected.

Three resistances connected together at a common point O are said to be connected in

star.

The replacement of delta or mesh by the equivalent star system is known as

DELTA STAR TRANSFORMATION

The two system will be equivalent or identical if the resistance measured between any

pair of lines is same in both the system when the third line is open the resistance

between point B and C

RBC = R3║ (R1 + R2)

= R3 (R1 + R2)

R3+ R1+ R2

Where as In Star

RBC = RB + RC

Since two system are identical, resistance between terminals B and C in both system

must be equal

RB + RC = R3 (R1 + R2) equation(i)

R3+ R1+ R2

Similarly between A & B

RA + RB = R2 (R1 + R3) equation(ii)

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R3+ R1+ R2

Similarly between C & A

RA + Rc = R1 (R2+ R3) equation(iii)

R3+ R1+ R2

Adding equations i ,ii and iii we get

2(RA + Rc+ RB) = 2(R1 R3 + R2 R3+ R1 R2)

R3+ R1+ R2

(RA + Rc+ RB) = (R1 R3 + R2 R3+ R1 R2) equation (iv)

R3+ R1+ R2

Subtracting equations i,ii,iii from iv we get respectively

RA = R1 R2 equation (v)

R3+ R1+ R2

RB = R2 R3 equation (vi)

R3+ R1+ R2

Rc = R1 R3 equation (vii)

R3+ R1+ R2

STAR TO DELTA TRANSFORMATION

Multiply equations v with vi , vi with vii and vii with v

(RARB + RB Rc+ RcRA) = (R1R2²R3 + R1R2R3² + R1²R2R3)

(R3+ R1+ R2 )²

(RARB + RB Rc+ RcRA) = R1R2R3(R3 + R1 + R2)

(R3+ R1+ R2 )²

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(RARB + RB Rc+ RcRA) = R1R2R3 Equation (viii)

(R3+ R1+ R2 )

Dividing equation viii with v , vi and vii we get

R3 = (RARB + RB Rc+ RcRA)

RA

R3 = RB + Rc+ RcRB

RA

R1 = RA + Rc+ RcRA

RB

R2 = RA + RB+ RBRA

RA

SINGLE PHASE A.C. CIRCUITS UNDER STEADY STATE POWER IN

RESISTIVE CIRCUIT

An alternating current or voltage is one which changes continuously in magnitude

and alternate in direction at regular interval of time. It rises from zero to maximum

positive value, falls to zero, increases to maximum in reverse direction. The EMF

voltage and current repeats the procedure.

WAVE FORM:- The shape plotted against time for increase and decrease of voltage,

current is called wave form.

INSTANTANEOUS VALUES:- The value of alternating quantity ( voltage or

current) at particular instant is called instantaneous values.

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TIME PERIOD AND FREQUENCY:- Time taken for alternating quantity to

complete one cycle is called time period.

The number of cycles completed per second by an alternating quantity is called

frequency.

POWER in AC CIRCUITS

In AC circuits there are three type of powers:

i) Apparent power (S)

ii) Active power (P)

iii) Reactive power (Q)

APPARENT POWER :- It is the product of R.M.S value of applied voltage and circuit

current. It is also called idle or wattles power.

Apparent power S= VI, V=IZ so S = I²Z

Normally this power is represented when system power factor is not known.

POWER FACTOR :- The cosine of angle between Voltage and current vector’s of A.C

system is called power factor. The power factor can be lagging , leading or unity.

Cos ϕ = P/S = R/Z

ACTIVE POWER :- (P) It is the power which is actually dissipated in circuit resistance. It

is seen that power is consumed in resistance only. A pure inductor and a pure capacitor do

not consume any power. Since in half cycle power is received from source by these

components and next half cycle the same is returned to source.

Active Power = VI Cos ϕ = I²R

REACTIVE POWER:- Is the power developed in the reactance of the circuit.

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Thus reactive power Q = VI Sin ϕ = I²X Volts Ampere Reactance VAR

The above triangle is known as POWER TRIANGLE , AB is equal to Q, and BC is equal to

and AC is equal to S

Cos ϕ= Active Power/ Apparent Power = P/S = R/Z

Power Triangle is the multiplication of voltage to current element . This right angle triangle

gives relationship between Active Power, Reactive Power and Apparent Power.

i) When a active component of current is multiplied with circuit voltage it gives active

or true power. This is the power which produces torque in the motor.

ii) When the reactive component of current is multiplied with circuit voltage it results in

reactive power. It is the power which flows back without doing any work.

iii) When the circuit current is multiplied with voltage it gives apparent power. It does

not give real power , To avoid confusion it is measured in Volt Amperes.

THREE PHASE BALANCED AND UNBALANCED SUPPLY

Three phase balanced supply means the system which consists of three phases windings

or circuits.

A supply system is said to be symmetrical when the three voltages of same frequency

have equal magnitude and displaced from one another by equal time angle, 3 phase 3

wire or 3phase 4 wire supply system will be symmetrical when the line to line voltages

are equal in magnitude and displaced in phase angle by 120 ͦ electrical degrees with

respect to each other. Further in four wire 3 phase voltage with respect to neutral of the

three phases are displaced at 120 ͦ electrical degrees and equal in magnitude.

UNBALANCED SUPPLY:- A three phase supply will be unbalanced when either of

three phase voltages are unequal in magnitude or phase angle between three phases are

not displaced at 120 ͦ electrical degrees.

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A load circuit is balanced when the connected in various phases are same in magnitude

as well in phase.

Any three phase load in which the impedances in one or more phases differ from

impedances of other phase is called unbalanced three phase load.

COMPARISON OF POLYPHASE AND SINGLE PHASE SUPPLY

i) In a single phase circuit the power delivered is pulsating.

ii) The rating of given machine increases 1.5 times the output of 1 ϕ phase.

iii) Power factor of single phase motor is poor compare to 3 phase motors.

iv) Single phase induction motor’s have no starting torque. In case of three phase motors

( induction) the starting torque is produced in the machine.

v) Rotating magnetic flux can be set up by polyphase currents.

vi) Polyphase systems are more reliable than single phase.

vii) Parallel operation of polyphase alternators is simple as compare to that of single

phase alternators, because of pulsating reaction in single phase alternator.

However three phase operation is not a practical for domestic application where

motor’s are smaller than I Kw and where lighting circuits supplies most of the

load.

The demand for two phase system has almost disappeared because that does not

have any advantage over three phase system. The three phase system is used

widely for transmission and distribution of electric power.

STAR DELTA CONNECTIONS

STAR CONNECTION:- this system is obtained by joining together similar ends,

either the start or finish, the other are joined to the line wire. The common point N

at which similar ( start or finish) ends are connected is called the neutral or star

point. Mostly 3 wire star connected system is used but some times 4 wire system

is also used. The neutral wire is carried to the external load circuit.

The voltage between any line and neutral point is called phase voltage, where as

the voltages between two outer is called line voltage.

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The EMF’s induced in 3 phase are shown by phasor, In star connection there are

two windings between each pair of outer line and due to joining of similar ends

together. EMF’s induced in them are in opposition.

For example the potential difference between R and Y or line voltage ERY

Is the phasor difference of phase EMF’s ER and EY or phasor sum of ER and ( - EY)

ERY =ER – Ey or ER + (-Ey).Since phase angle between phasor ER and is (-Ey) 60ͦ

MESH OR DELTA CONNECTED SYSTEM

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When the starting end of one coil is connected to the finishing end of another coil, the

delta or mesh connection is obtained. The direction of e.m.f’s taken as positive from

start end to finish end.

It is obvious that line current is phasor difference of phase currents of two phases

concerned . Line current in R (-IR) will be equal to the phasor difference of phase

currents IYR and IRB

The line current IR = IYR - IRB = IYR+(-IRB)

Since phase angle between phase current phasor IYR , (-IRB) is 60ͦ

IR = √ I²YR + I²RB + 2 IYR IRB COSϕ

Assuming balanced load, the phase current in each winding is equal and let be each

equal to IP

Therefore LINE CURRENT IR = IY= IB = √ I²P + I²P + 2 IP IP/2

= √3 IP