RGPV BE Ist SEM BEE 104 UNIT I

1. 1
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

2. 2
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

3. 3
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.

4. 4
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.

5. 5
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
s
R
V
I 
R
v
iRiv s
sss  or

6. 6
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
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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.
 




risesdrops
M
1m
vv
0v

8. 8
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



9. 9
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

12. 12
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.

16. 16
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.

17. 17
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 by120 ͦ 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.

19. 19
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

20. 20
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 phasorIYR , (-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 CURRENTIR = IY= IB = √ I²P + I²P + 2 IP IP/2
= √3 IP