Electric circuits and network theorems

Department of Electrical and Electronics Engineering
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ELECTRIC CIRCUITS AND NETWORK THEOREMS

Prepared by
Mrs.K.Kalpana.,M.E.,(Ph.D).,
Assistant Professor
Department of Electrical and Electronics
Engineering
Arasu Engineering College
Kumbakonam
Department of Electrical and Electronics
Engineering
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INTRODUCTION
 The interconnection of various electric elements in a
prescribed manner comprises as an electric circuit in
order to perform a desired function.
 An electrical network is an interconnection
of electrical components (e.g., batteries, resistors,
inductors, capacitors, switches, transistors) or a model of
such an interconnection, consisting of electrical elements
(e.g., voltage sources, current sources, resistances,
inductances, capacitances).
 The electric elements include controlled and uncontrolled
source of energy, resistors, capacitors, inductors, etc.
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INTRODUCTION
 Analysis of electric circuits refers to computations
required to determine the unknown quantities such as
voltage, current and power associated with one or more
elements in the circuit.
 Many other systems, like mechanical, hydraulic, thermal,
magnetic and power system are easy to analyze and
model by a circuit.
 To learn how to analyze the models of these systems,
first one needs to learn the techniques of circuit analysis.
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BASIC ELEMENTS & INTRODUCTORY CONCEPTS
• Electrical Network: A combination of various
electric elements (Resistor, Inductor, Capacitor,
Voltage source, Current source) connected in any
manner what so ever is called an electrical network.
We may classify circuit elements in two categories,
passive and active elements.
• Passive Element: The element which receives
energy (or absorbs energy) and then either converts
it into heat (R) or stored it in an electric (C) or magnetic
(L ) field is called passive element.
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Active Element: The elements that supply energy to
the circuit is called active element.
 Examples of active elements include voltage and
current sources, generators and electronic devices that
require power supplies.
 A transistor is an active circuit element, meaning that it
can amplify power of a signal.
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 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.
 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
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Potential Energy Difference: The voltage or
potential energy difference between two points in an
electric circuit is the amount of energy required to
move a unit charge between the two points.
 Resistor opposes the flow of current through it and it
dissipates energy in the form of heat. (Ohms (Ω)).
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 Inductor stores energy in its
magnetic field.
 Unit of inductance is Henry (H).
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 Capacitor stores energy in its electric field.
 Unit of capacitance is Farad (F).
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Classification of sources( active elements)
Sources (active elements) are classified as
1. Independent sources
2. Dependent sources
1. Independent sources:
Batteries and generators are called independent sources
which can directly generate electrical energy. Independent
sources do not depend on other electrical sources.
2. Dependent sources:
Transistors and Op-amps are called dependent sources
whose output energy depends on other independent sources.
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EE8251- CIRCUIT THEORY

V-I characteristics of ideal and practical sources:
Ideal voltage sources:
A voltage source is a two-terminal device whose voltage at
any instant of time is constant and is independent of
the current drawn from it. Such a voltage source is called
an Ideal Voltage Source and have zero internal resistance.
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Practical Voltage sources:
Sources having some amount of internal resistances are known
as Practical Voltage Source. Due to this internal resistance,
voltage drop takes place, and it causes the terminal voltage to
reduce.
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Ideal Current sources:
• An ideal current source is a two-terminal circuit
element which supplies the same current to any load
resistance connected across its terminals. Here the
current supplied by the current source is independent of
the voltage of source terminals. It has infinite resistance.

• Practical Current sources
• A practical current source is a two terminal device having
some resistance connected across its terminals. Unlike
ideal current source, the output current of practical
source depends on the voltage of the source.
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Active Network:
A network that contains at least one active
element, such as an independent voltage or
current source, is an active network.
Passive Network:
A network that does not contain any active
elements is a passive network.
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Lumped elements :
Physically separated elements such as resistor,
capacitor and inductors are called as lumped elements.
Distributed element :
It is not separable for electrical purposes. A
transmission line has distributed resistance, inductance
& capacitance along its length.
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Bilateral element :
 Conduction of current in both directions in an
element with same magnitude. (Ex: R,L,C)
In a bilateral element, the voltage and current
relation is the same for current flowing in either
direction.
Unilateral elements-:
Conduction of current in one direction.(Ex:
Diode, Transistor).
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D.C circuits:
The direction of current flow, at any point in the circuit,
does not change with time.
A.C circuits:
The direction of current flow (as well as its magnitude)
changes with time.
CHARGE
Charge is an electrical property of the atomic particles of
which matter consists, measured in coulombs (C).
Charge, positive or negative, is denoted by the letter q
or Q.
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EE8251- CIRCUIT THEORY

CURRENT
 Current can be defined as the rate of flow of electrons
in a conductive or semi conductive material, measured in
Ampere (A).
 Electric current, is denoted by the letter i or I.
 The unit of current is the ampere abbreviated as (A) and
corresponds to the quantity of total charge that passes
through an arbitrary cross section of a conducting
material per unit second.
 Mathematically,
• Where Q is the symbol of charge measured in Coulombs
(C),
• I is the current in amperes (A) and t is the time in
second (s).
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 The current can also be defined as the rate of charge
passing through a point in an electric circuit.
 Mathematically, i= dq/dt
Two types of currents:
1) A direct current (DC) is a current that remains constant with
time.
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2) An alternating current (AC) is a current that varies with time.
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• The difference in potential energy of the charges is
called potential difference.
• It is also known as voltage.
• It is denoted by V, unit Volts
• V=W/Q or dw/dQ
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VOLTAGE (or) POTENTIAL DIFFERENCE :

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POWER
Power is the rate of change of energy, measured in
watts (W). Power, is denoted by the letter p or P.
Mathematically, P=dw/dt
Where P is power in watts (W),
w is energy in joules (J), and
t is time in seconds (s).
From voltage and current equations, it follows that;
P=dw/dt= dw/dq X dq/dt
= V X I
Thus, if the magnitude of current I and voltage are given,
then power can be evaluated as the product of the two
quantities and is measured in watts (W).

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ENERGY
Energy is the capacity for doing work, and is
measured in joules (J).
The energy absorbed or supplied by an element from
time 0 to t is given by,

 Node: A node is a point in a network, where two or
more elements are joined.
 Principal node: Junction of three or more elements.
(Current is divided)
 Simple node: Junction of any two elements. (Current is
not divided)
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 Branch: Branch represents a single element connected
between two nodes such as a voltage source or resistor.
In the above circuit, the branches are: af, ab, bc, cd, be.
 Loop: Loop is an any closed path in the circuit. In the
above circuit, the loops are: abefa, bcdeb, abcdfa.
 Mesh: Mesh is a closed path that does not contain any
other loops .In the above circuit, the meshes are: abefa,
bcdeb.
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Ohm’s Law:
 Ohm's law states that the current through a conductor
is directly proportional to the potential difference or
voltage across the two points, that the temperature of
the conductor remains constant
V ∝ I (Or) V = IR
 The mathematical equation that describes this
relationship is: V / I= R
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 where I is the current through the resistance in units of
amperes,
 V is the potential difference measured across the
resistance in units of volts, and
 where R is the proportionality constant.
R is called Resistance of the conductor and is measured
in ‘Ohms’ (Ω).
 The power absorbed by the resistance is calculated as :
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Since I=V/R

• The V-I relation for resistor according to Ohm’s law is
depicted in Fig.
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Limitations of Ohm’s Law:
1. Ohm’s law is not applicable to non-linear elements like
diode, transistor etc.
2. Ohm’s law is not applicable for non-metallic conductors
like silicon carbide.
3. Ohm’s law holds good only for constant temperature. If
the temperature change, this law cannot be applied.

KIRCHOFF'S LAW
Kirchoff's First Law - The Current Law, (KCL)
 It states that “the algebraic sum of currents meeting at a
junction(node) is equal to zero".
 In other words the algebraic sum of ALL the currents
entering and leaving a node must be equal to zero,
 I(exiting) + I(entering) = 0.
 In case of AC circuits, Kirchoff’s current law
states that phasor sum of incoming current is
equal to the phasor sum of outgoing current.
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I1 + I2 + I3 = I4 + I5
 Here, the 3 currents entering the node, I1, I2, I3 are all positive
in value and the 2 currents leaving the node, I4 and I5 are
negative in value.
 Then the equation can also rewrite as; I1 + I2 + I3 - I4 - I5 = 0
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Kirchhoff’s Voltage Law (KVL):
 The algebraic sum of electromotive forces plus the
algebraic sum of voltages across the impedances
(resistances) in any closed electrical circuit is equal to
zero.
 For any closed path in a circuit, the algebraic sum of
the voltage is zero.
 In other words, for any closed path in a circuit, the
sum of voltage rise is equal to sum of voltage drop.
 Mathematically, ∑emf + ∑ IR =0 in any closed
electrical circuit.
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Consider the above fig., Applying KVL: V1- V2 + V3-V4 = 0
It can be also written as, V1 +V3 = V2 + V4

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Consider the circuit shown in figure.
Apply KVL to the circuit,
-E1+ E2-E3+E4+I1R1-I2R2-I3R3-I4R4=0

Problem :1
A current of 0.5 A is flowing through the resistance of
10Ω. Find the potential difference between its ends.
Solution:
Current I = 0.5A.
Resistance R = 10Ω
Potential difference V =?
V = IR
= 0.5 × 10
= 5V.
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Problem: 2
A supply voltage of 220V is applied to a resistor100Ω.
Find the current flowing through it.
Solution:
Voltage V = 220V
Resistance R = 100Ω
Current I = V/ R
= 2 2 0 /100
= 2.2 A.

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Problem: 3
Calculate the resistance of the conductor if a current of 2A
flows through it when the potential difference across its
ends is 6V.
Solution:
Current I = 2A.
Potential difference V = 6.
Resistance R = V/I
= 6 /2
R= 3 ohm.

Problem: 4
Calculate the current and resistance of a 100 W, 200V electric
bulb.
Solution:
Power, P = 100W
Voltage, V = 200V
Power P= VI
Current I = P/V
= 100/200
= 0.5A
Resistance R = V /I
= 200/0.5
= 400Ω
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Resistors In Series
When resistors are connected as shown in figure such that
the same current passes through all of them, they are said
to be in series.
Each resistor has a voltage drop across it given by Ohm’s
law. Thus,
V1=I1R1
V2=I2R2 and
V3=I3R3
The total drop in the three resistors put together is
V = V1+V2+ V3
V =I(R1+R2+R3)

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Power dissipated in R1 is given by

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Resistors In Parallel
When resistors are connected across one another such
that the same voltage is applied to each, then they are
said to be in Parallel. Such an arrangement is shown in
fig.
The current in each resistor is different and the current
I taken from the supply divides among all the three
resistors.

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If we replace the three resistors in parallel by a single
equivalent resistor R, as shown in figure , it will draw the
same current, I.
therefore V= I Reff (by Ohm’s law)

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If there were only two resistors in parallel, then the effective
resistance is given by

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Difference between Series and Parallel Circuit

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1.Find the voltage across the resistor R in figure.

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2.Find the current through 12 Ωresistor in figure.

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Voltage division rule:
Current division rule:

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Problem:
Find the current flowing through each branch for the
circuit shown in figure.
Solution:
Here two resistor are connected in parallel.
Therefore Req is

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Req = 2.4Ω
Input current I= V/R = 230 / 2.4 =95.83A
Using current division rule, the current through 4 Ω resistor is

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Using current division rule, the current through 6Ω resistor is
Problem:
Three loads A, B and C are connected in parallel across a
250V source. Loads A takes 50 A. Load B is a resistor of 10
ohms and load C takes 6.25 kW. Calculate(I) RA and RC (ii)
Currents lB and IC (iii) Power in loads A ad B (v) Total Power
and (vi) Total effective resistance.

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Solution:

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Problem:
Determine the total current taken from the source.

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Solution:
Here, 50Ω and 50Ω are connected in series. So, 50+50=100Ω
100Ω and 100Ω are connected in series. So, 100+100=200Ω

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100Ω and 100Ω are connected in parallel. So,

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Total current I= V/R = 100/50= 2A
I= 2A

Problem : For the circuit shown in figure, determine (a) the
battery voltage V, (b) the total resistance of the circuit,
and (c) the values of resistance of resistors R1, R2 and
R3, given that the p.d.’s R1, R2 across and R3 are 5V, 2V
and 6V respectively.
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EE8251-CIRCUIT THEORY

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(a) Battery voltage V =V1 + V2 + V3 =5 + 2 + 6=13V
(b)Total circuit resistance R= V/ I
= 13/4=3.25Ω
(c) Resistance R1 = V1/ I
= 5/4 =1.25 Ω
Resistance R2 = V2/ I
= 2/4 =0.5 Ω
Resistance R3 = V3/ I = 6/4 =1.5 Ω

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Mesh Analysis
In mesh method, Kirchhoff's Voltage Law(KVL) is applied
to a network. For this network, we have to write mesh
equations in terms of mesh currents. Here, we are using
mesh current instead of branch currents. Here each mesh
is assigned a separate mesh current. Assume that the
mesh current direction is clockwise. Then, KVL is applied
to the network, in order to write equations in terms of
unknown mesh currents.

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The branch Current can be found out by taking the
algebraic sum of the mesh currents which are common to
that branch.
The following steps should be implemented to find out the
mesh currents and branch currents.
Consider a simple network as shown in figure.
Step 1: First, each mesh is assigned a separate mesh
current. Assume all mesh currents directions are
clockwise. It consists of two meshes
(PQSP and QRSQ) and two mesh currents (I1 and I2).

Step 2: If two mesh currents (I1 and I2) are flowing
through a network elements, the actual current in the
circuit element is the algebraic sum of the two mesh
currents (I1 and I2). In this network, mesh currents, I1
and I2 are flowing through R2. First we consider one
direction i.e., Q to S, current is I1-I2 and in for another
direction. i.e., S to Q, current is I2-I1
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Step 3: Write equation for each mesh in terms of mesh
currents by applying KVL. When writing mesh equations,
we assign rise in potential as positive (+) sign and fall in
potential as negative (—) sign.
Step 4: Suppose any value of mesh current becomes
negative in the solution, the actual or true direction of
the mesh current is anticlockwise, i.e., opposite to the
clockwise direction.
By applying KVL to the circuit , we get two equations,
Mesh PQSP
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MESH QRSQ
The mesh currents I1 and 12 can be found out by solving
the above equations
Step-5 : The branch currents can be easily found out by
using the mesh currents.
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Mesh Analysis by Inspection Method
Step 1: Convert the current source into equivalent voltage
source by source transformation.
Step 2: All the resistance through which the loop current I1
flows are summed up and denoted by R11. It is called self
resistance of loop 1.
Step 3: All the resistance through which loop currents I1
in the first loop, and I2 in the second loop flow are
summed up. This is denoted by R12. The sign of the term
R12 is negative, if the two currents I1 and I2 through R12
are in opposite directions; otherwise the sign is positive.

Step 4: Let V1 be the effective voltage on the first loop
through which the loop current I1 flows. The sign of V1 is
positive if the direction of V1 is same as that of I1 (i.e.
aiding the current I1, otherwise the sign of V1 is
negative). V1 is written on the right hand side of the
equation.
Now, zero is written on the right hand side if there is no
source in the first loop through which I1 flows.
The general matrix form of mesh equation is
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(The minus sign indicates the loop current I2 flows from the
voltage terminal + to --)
By using Cramer’s rule, we can find mesh or loop currents
I1 and I2
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Nodal Analysis
 The nodal method is used to analyze multisource
circuits. In this method, we solve the simultaneous
equations using Kirchhoff’s current law (KCL) applied at
various nodes in an electric circuit. Node defined as
junction or joining point of two or more component
terminals.
 In nodal method, we select one node as a reference
node, with respect to the voltages at all other nodes are
measured. Thus, the reference node acts as a ground or
common for the circuit.
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The first step in nodal method is to convert all voltage
sources to equivalent current sources.
The second step is to identify all the principal nodes in
the circuit and to choose a reference node.
The choice is arbitrary, but it is usually convenient to
select the reference node as the one having the most of
the components connected to it.
All the nodes except the reference node are then
numbered and their corresponding voltages are designated
as V1, V2….etc.
The reference node and other nodes are indicated as
shown in figure .

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Steps
1. Select one major node as the reference node and assign
each of the (n -1) remaining nodes with its own unknown
potential (with respect to the reference node).
2. Assign branch current to all branches.
The arrow is drawn from V1 towards V2 because current
always flows from higher potential to the lower potential.

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3. Express the branch currents in terms of the node
potentials.
4.Write the current equation at each of the (n-1) unknown
nodes.
5. Substitute the current expressions (Step-3) into the
current equations (Step-4), which then become a set of
simultaneous equations of unknown node voltages.
6. Solve for the unknown voltages and the branch
currents.

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In many circuits the reference node is most conveniently
selected as a common terminal or the ground terminal. The
above circuit diagram consists of three nodes. It is possible to
write (n-1) equations for the circuit having ‘n’ nodes. Applying
KCL at node 1 gives,

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The generalized node equations can be written as
[G] [V]= [I] ……………..3
where the square matrix G is called the node
conductance matrix, V is the column matrix of the node
voltages with respect to the reference node and I is the
column matrix of input currents.

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Nodal Analysis by Inspection Method:
Step 1: First, convert all the voltage sources to equivalent
current sources.
Step 2: The conductances of all branches connected to
node 1 are added and denoted by G11 . G11 is called the
self conductance of node 1.
Step 3: All the conductances connected to nodes 1 and 2
are added and denoted by G12. G12 is called mutual
conductance of nodes 1 and 2. This G12 is written with
negative sign. If no conductance is connected between
nodes 1 and 2 then G12 = 0, G12=G21.

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Step 4: I1 denotes the value of the current source to node
1 and is written on the right hand side of the equation.
The sign of I1 is positive if it is flowing towards node 1,
otherwise it is negative.
If no current source is connected to node 1 , then I1 = 0.
For example, consider the circuit given below,
First convert voltage sources into equivalent current
sources.

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The circuit consists of two nodes 1 and 2 and common (ref)
node.

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By Solving this matrix, we can find out nodal voltages V1 and
V2.

Problem 1: Use mesh-current analysis to determine the
current flowing in the 1Ω resistance of the d.c. circuit
shown in fig.
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The mesh currents I1, I2 and I3 are shown in Figure
Using Kirchhoff’s voltage law:
For loop 1, (3 + 5) I1 −5I2 = 4 ………………………………(1)
For loop 2, (4 + 1 + 6 + 5) I2 −(5) I1 −(1) I3 = 0……..(2)
For loop 3, (1 + 8) I3 −(1) I2 =- 5 …………………………..(3)
Thus
8I1 −5I2 =4
−5I1 + 16I2 −I3 =0
−I2 + 9I3 =-5
By solving the above equations by Cramer’s rule , we get
I1=0.595A ; I2=0.152A ; I3= -0.539A

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Problem 2: Write the mesh equations and solve for the
currents I1 , and I2 .
Mesh 1 4I1 + 6(I1 – I2) = 10 – 2 ................(1)
Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20 .............(2)

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The previous equations can be written in matrix form as:
Solving the above equation using Cramer’s rule, we get
I1 = 2.2105A
I2 = 2.3509A

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Problem: 3
Find the current in a circuit using Kirchhoff's voltage
law.
80= 20(I) + 10(I)
80 = 30 (I)
I= 80/30=2.66 A

Example 5
Write the mesh equations for the circuit shown in the
figure and solve for the current in the 12 Ω resistor.
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Solution: The loop currents are assumed as shown. It is
required to find i3. By inspection we can write,

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(—) indicates that the assumed direction of I3 is not the
actual direction. The actual direction of I3 is anti-
clockwise. i. e., the current through 12 Ω is 1O A flowing
from the right to the left terminal of the resistor.

Example 6
Write and solve the equations for the mesh currents in
the network shown in figure.
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• For solving the problems on the mesh current (loop
current) method, the preferable method is to convert the
practical current sources into practical voltage sources.
The original circuit is redrawn as in the figure below after
transforming the current sources into voltage sources.
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Example 7
Apply mesh current method and determine currents
through the resistors of the network shown in fig.
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Example 8
Frame the nodal equations of the network of figure and
hence find the difference of potential between nodes 2
and 4.

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Let the node 4 be the reference i.e. zero potential node. Let
V1,V2,V3 be the node voltages at the nodes 1, 2 and 3 respectively.
By inspection,

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It is required to find the difference of potential between
nodes 2 and 4. i.e., V2 — V4 = V2 —0 = V2. So, it is enough if
we know the value of V2.

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Example 9
Compute the voltage at nodes A and B in the circuit of
figure shown below.

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First converting the voltage source into its equivalent current source
and re-drawing the circuit the following figure is obtained.

There are two nodes A and B other than the reference node.
Let VA and VB be the voltages of the nodes A and B respectively
with respect to reference node. By inspection,
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EE8251&CIRCUIT THEORY

 Example 10
 Use nodal voltage method and hence find the power
dissipated in the 10 ohms resistor on the circuit shown in
figure.
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• Taking the node 4 as reference, and converting the voltage
source into current source, the above network is re-drawn as
below:
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Example 11
In the network shown in figure, find the node voltages
V1 and V2. Find also the current supplied by the source.

• The above circuit is re-drawn as below after converting the
voltage source into its equivalent current source.
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• LOOP ANALYSIS (Dependent Sources or Controlled
Sources)
Example 1
Determine the current through 4Ω resistor of the
network shown.

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Example
Determine the voltage across each conductance
of the circuit shown in the figure.
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Nodal analysis (Dependent Sources or Controlled
Sources)

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DC CIRCUITS:
A DC circuit (Direct Current circuit) is an electrical
circuit that consists of any combination of constant
voltage sources, constant current sources, and resistors.
 In this case, the circuit voltages and currents are
constant, i.e., independent of time.
 More technically, a DC circuit has no memory.
 That is, a particular circuit voltage or current does not
depend on the past value of any circuit voltage or
current.
This implies that the system of equations that
represent a DC circuit do not involve integrals or
derivatives.

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AC CIRCUITS:
Fundamentals of AC:
An alternating current (AC) is an electrical current, where the
magnitude of the current varies in a cyclical form, as opposed
to direct current, where the polarity of the current stays
constant.
The usual waveform of an AC circuit is generally that of a
sine wave, as these results in the most efficient transmission
of energy. However in certain applications different waveforms
are used, such as triangular or square waves.
Used generically, AC refers to the form in which electricity is
delivered to businesses and residences. However, audio and
radio signals carried on electrical wire are also examples of
alternating current.

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Sinusoidal alternating quantity
An alternating quantity which varies according to the sine of
the angle θ is known as sinusoidal alternating quantity.
All over the world sinusoidal voltages and currents are
selected for generation of electric power.
The following are the reasons:
a)The sinusoidal voltage and currents produce low iron and
copper losses in AC rotating machines and transformers.
b)Sinusoidal voltage and current will offer less interference to
nearby telephone lines.
c)They produce less disturbance in the electrical circuit.

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Cycle:
One complete set of positive and negative values of an
alternating quantity is called cycle.
Time period:
The time taken to complete one complete cycle.
Frequency:
The number of cycles made by an alternating quantity per
second is called frequency. The unit of frequency is Hertz (Hz)
Amplitude or Peak value:
The maximum positive or negative value of an alternating
quantity is called amplitude or peak value. In the voltage
waveform the peak value is Em. It occurs when the angle is
π/2 (positive cycle) and 3π/2 (negative cycle).
Average value:
This is the average of instantaneous values of an
alternating quantity over one complete cycle of the wave.

Sinusoidal Voltage and current equations
A sinusoidal voltage and current can be represented as
E=Em Sinωt or V=Vm Sinωt and
I=Im Sinωt
• Where E or V and I are instantaneous values and Em or Vm
and Im are the peak values.
• ω = angular frequency in radians per second.
V=Vm Sin (2πf)t and
I=Im Sin(2πf)t
Where, ω = 2πf
f= ω / 2 π = 1/ T
Where, f is the frequency in cycles per second(Hz)
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Waveform:
It is the graph between the alternating quantity(only
instantaneous value) as ordinate and time.
The alternating quantity may be either voltage or current or
flux.

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R.M.S (Root mean square) value or effective value:
The steady current which when flowing through a
given resistor for a given time produces the same amount of
heat as produced by the alternating current when flowing
through the same resistor for the same time is called R.M.S or
effective value of the alternating current.

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To find the R.M.S. value of a Sinusoidal alternating
Quantity

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For a Sinusoidal quantity, R.M.S value = 0.707 x maximum
value

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To find the average
value of sinusoidally
varying quantity.
i=Im Sinθ
It is a symmetrical wave
form.
To calculate the
average value, only half
cycle must be
considered.

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For a sinusoidally varying quantity, average value
= 0.637 x maximum value

• Form factor (Kf) : It is the ratio of rms value to average value.
Mathematically,
Form factor Kf = RMS value / Average value
• Peak factor (Ka) : It is the ratio of peak or maximum value to
r.m.s value.
• Mathematically
Peak factor Ka = Peak value / RMS Value
Peak factor is also known as amplitude factor or crest factor.
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ANALYSIS OF AC CIRCUIT
It is assumed that the sinusoidally varying voltage is connected
to
(a) purely resistive
(b) purely inductive and
(c) purely capacitive circuits.
In each case it is required to find the following quantities
1. The expression for the instantaneous current.
2. The polar form of voltage and current
3. The phasor diagram.
4. Ratio of voltage to current
5. Average power
6. Phase angle between voltage and current and hence
power factor.

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Purely Resistive Circuit Excited by a Sinusoidally Varying Voltage

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Circuit Consisting of Pure Inductance Excited by a
Sinusoidally Varying Voltage
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i

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Purely Capacitive Circuit Excited by Sinusoidally Varying
Voltage

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Voltage division

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CURRENT DIVISION
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Source Transformation
Voltage Source to Current Source
If a voltage source has a resistance connected
series, it can be transformed into an equivalent current
source with a resistance connected in parallel

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Source Transformation
Current Source to Voltage Source
If a current source has a resistance connected in parallel, it can
be transformed into an equivalent voltage source with a
resistance connected in series.

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1.Convert the given current source into a voltage source

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2. Convert the given voltage source into a current
source for the circuit given below.

Star(Y) and Delta (Δ) transformation or T and π:
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Between three terminals A, B and
C, we can connect three resistances
(impedances) in two fashions.
One is called star connection (Y),
the other is Delta connection (Δ).
 In star Connection, one end of
each resistance is connected at a
point called star point and the
remaining three terminals are
connected to A, B and C.
It is called star connection Since it
appears like a star.

 In Delta connection, the three
resistances (impedances) are connected
end to end, so as to form Delta shape.
 In some electrical circuits, calculations
become simpler if it is star connected.
In some other problems, Delta
connection may enable us to solve it
easily.
 Hence it is necessary to convert given
star connection resistances into its
equivalent Delta and vice-versa.
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Given Resistances in Delta- To Find Resistance in
Equivalent Star
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Given Resistances in Star To Find Resistance in
Equivalent Delta

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Problem: For the passive circuit consisting of resistances
(in ohms) as shown, calculate the equivalent resistance
between the terminals A and D.

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Solution: Between the terminals A and D in the circuit
given, the combination is neither series nor parallel.
Hence simplification is not possible as it is. So. we can
convert the delta between A, B and C into equivalent star.
As a result the circuit becomes as drawn below:

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Now, it is series parallel combination of resistances.
Hence the total resistance between A and D is equal to,

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Problem: For the network shown in the figure below, find the
equivalent resistance between the terminals B and C.

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Solution: The given combination of resistances between the
terminals B and C is neither series combination nor parallel
combination.
If the star Connection between A, B and C is converted
into equivalent delta, we will have a known combination
which can be simplified by Series parallel simplification.

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The given figure is re-drawn after replacing the star by its
equivalent Delta.
The resistance between B and C terminals is equivalent to
the parallel combination of 3 and 1.5 Ohms.

Super position theorem
 With the help of this theorem, we can find the current
through or the voltage across a given element in a linear
circuit consisting of two or more sources.
 The statement is as follows:
“In a linear circuit containing more than one source, the
current that flows at any point or the voltage that exists
between any two points is the algebraic sum of the
currents or the voltages that would have been produced
by each source taken separately with all other sources
removed.”
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Note:
1. Removal of an ideal voltage source means short
circuiting.
2. Removal of an ideal current source means replacing it
by an open circuit.
3. Removal of practical voltage source means replacing it
by an internal resistance.
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Example
Compute the current through 23 ohm resistor of the
figure below by using superposition theorem.
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Solution: Step 1 : Allow only the voltage source to act. The
corresponding circuit is as below:
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Step 2: Allow only the current source to act. The circuit becomes
as
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Example
Using superposition theorem or otherwise, obtain the
current in EA in figure below:

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• Step 1 : First allow only 1V source to act. Let the current in EA
be I’. The relevant circuit diagram is as below:

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• I’ can be found by applying mesh current method or
converting delta to star and then applying Ohm’s law.
The above circuit becomes as drawn below after
converting delta connection between A, B and C into its
equivalent star.

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• Step 2: When only 2V source is acting and the
other source short circuited, the circuit
becomes as shown in fig. Now let the current
in EA be I”.

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Thevenin’s theorem:
A linear network consisting of a number of voltage
sources and resistances can be replaced by an
equivalent network having a single voltage source called
Thevenin’s voltage (VTh ) and a single resistance called
Thevenin’s resistance (RTh )
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Explanation:

To find VTh :
The load resistance RL is removed.
The current I in the circuit is I= E / (R1 + R2)
. The voltage across AB = Thevenin’s voltage VTh .
VTh = I R2 VTh = E R2 / (R1 + R2)
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• To find RTh :
• The load resistance RL is removed. The cell is
disconnected and the wires are short as shown.
• The effective resistance across AB = Thevenin’s
resistance RTh .
• R1 is parallel to R2 and this combination in series with R3 ]

 The equivalent circuit is given by
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Problem : Use Thevenin’s theorem to find the current
flowing in the 10 Ω resistor for the circuit shown in
figure.
To find VTh :
 The load resistance RL is
removed.
 The 10 Ω resistance is removed
from the circuit as shown in
figure
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• There is no current flowing in the 5 Ω resistor and
current I1 is given by:
I1 = 10/(R1 + R2 )
= 10/(2 + 8)
= 1A
VTh = IR2 = 1 X 8 = 8V
To find RTh :
The load resistance RL is removed. The cell is
disconnected and the wires are short as shown.
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Resistance, RTh = R3 + (R1R2 / R1 + R2 )
=5+ (2×8/2+8)
= 5 + 1.6 = 6.6 Ω
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Thevenin’s equivalent circuit is shown in fig.
Load Current IL= E / (R+ RTh ) = 8/ (10+6.6)
= 8/16.6=0.482A

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• Problem
• Find the Thevenin ‘s equivalent for the network of the
figure between a and b.

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• Solution: The network given is a combination of
voltage and current sources. By converting voltage
source into current source and vice versa, wherever
necessary and simplifying we can obtain the required
network.
• Step 1: Converting or transforming the voltage source
of 1OV in series with resistance 3Ω into equivalent
current source, the following circuit is obtained.

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Step 2: Replacing the 2 current sources in parallel by its
equivalent current source, we get the following circuit.

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Step 3: Transforming the 2 current sources in series by
their equivalent voltage sources we get the following
network.

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Step 4: Transforming the voltage sources which are
parallel in the above circuit, into their equivalent current
sources we get the following network.

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Step 5: Converting the current source into equivalent
voltage source, we get the Thevenin’s equivalent circuit
as shown below:

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Maximum Power Transfer Theorem
 This theorem describes the condition for maximum
power transfer from an active network to an external
load resistance.
 It states that in a linear, active, bilateral DC network,
the maximum power will be transferred from source to
the load when the external load resistance equals to
the internal resistance of the source.
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Explanation of Maximum Power Transfer Theorem:
 Basically, the condition at which maximum power
transfer can be obtained by deriving an expression of
power absorbed by the load using mesh or nodal current
techniques and then finding its derivative with respect to
the load resistance.

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From the above circuit, the current flowing through the
load, ‘I’ is given as

 In the above equation RL is a variable, therefore the
condition for maximum power delivered to the load is
determined by differentiating load power with respect
to the load resistance and equating it to zero.
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This is the condition for maximum power transfer, which
states that power delivered to the load is maximum,
when the load resistance RL matches with Thevenin’s
resistance RTH of the network.
Under this condition, power transfer to the load is

 For a given values the Thevenin’s voltage and Thevenin’s
resistance, the variation of power delivered to the load with
varying load resistance is shown in below figure.
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Problem: For the circuit shown below determine the
value of load resistance, RL for which maximum power
will transfer from source to load.

 Now, the given circuit can be further simplified by
converting the current source into equivalent voltage
source as follows.
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• We need to find the Thevenin’s equivalent voltage Vth
and Thevenin’s equivalent resistance Rth across the load
terminals in order to get the condition for maximum
power transfer. By disconnecting the load resistance, the
open-circuit voltage across the load terminals can be
calculated as;

By applying Kirchhoff’s voltage law, we get
12 – 6I – 2I – 16 = 0
– 8I = 4
I = –0.5 A
The open-circuit voltage across the terminals A and B,
VAB = 16 – 2 ×0.5
= 15 V
Thevenin’s equivalent resistance across the terminals A and
B is obtained by short-circuiting the voltage sources as
shown in the figure.
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Req = (6 × 2) / (6 + 2)
= 1.5 Ω
So the maximum power will transferred to the load when
RL = 1.5 ohm.
Current through the circuit, I = 15 / (1.5 + 1.5)
= 5 A
Therefore, the maximum power = 52 × 1.5 = 37.5 W

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Norton’s Theorem
Norton’s theorem is the dual of the Thevenin’s Theorem:
Statement: “Any linear active network with output
terminals A, B as shown in the figure can be
replaced by a single current source. ISC(IN) in parallel with a
single impedance ZTh (Zn) = RTh (RN)
Isc is the current through the terminals AB of the active
network when shorted.
ZTh is the Thevenin’s impedance.
The current through an impedance connected to the
terminals of the Norton’s equivalent circuit must have the
same direction as the current through the same impedance
connected to the original active network.

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After evaluating the values of ISc and RTh, the current
through RL connected across A and B can be calculated by
applying division of current formula which is expressed
below:

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Example
Determine the voltage across 200Ω resistor in circuit by
Norton ‘s theorem.

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Solution :
Step 1: To find the short circuit current ISC .Replace the
200 Ω by short circuit. The current through short
circuited AB is ISC. Refer the following figure.

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The voltage source will drive a current of 10/50= 0.2 A,
which flows through short circuited AB only. Similarly the
current of 1A flows through 20 and then through short-
circuited AB only. 0.2 A will not flow through 2OΩ and 1A will
not flow through 50Ω .It is because of short-circuit. So, Isc=
0.2+1=1.2A.
Step 2: To find RTh, : From the given circuit, disconnect RL
= 200 Ω between A and B and also kill the sources. The
resultant circuit becomes as below:

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Step 3: Drawing Norton’s equivalent circuit and showing
the load resistance RL, we get the following circuit:

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Reciprocity Theorem
Statement
In a linear, bilateral network a voltage source V volts in
a branch gives rise to a current I, in another branch. If
V is applied in the second branch, the current in the
first branch will be I.‘This V/I is called transfer
impedance or resistance.

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On changing the voltage source from branch 1 to
branch 2, the current I in the branch 2 appears in
branch 1.

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Example
For the circuit shown in the figure below, find I3 and
verify reciprocity theorem.

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Solution : Step 1: By inspection putting in the matrix
form, we get

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Step 2: Transferring the battery, to the branch with
1.375 Ω resistor, the following circuit is obtained. The
current through 1Ω is taken as I3. According to
reciprocity theorem this I3= 2A.
Let us calculate the value of I3 in figure below:

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Thus reciprocity theorem is verified.

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GENERATION OF 3-PHASE VOLTAGE
 The 3-phase voltage can be produced in a
stationary armature with rotating field or in a rotating
armature with a stationary field.
 3-phase voltages are generated in three separate but
identical sets of windings or coils which are displaced by
120 electrical degrees in the armature.
 Hence, the voltages generated in them are 120 degree
apart in time phase. This arrangement is shown in Fig.
 Here, RR’ constitutes one coil (R-phase) ; YY’ another
coil (Y-phase) and BB’ constitutes the third coil (B phase).
The field magnets are assumed to be rotating in clock-
wise direction.

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Three phase wave form

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Representation of three phase voltage in polar and rectangular form

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There are six types of wires
Vulcanized Indian rubber wire(VIR)
Tough rubber sheathed wire(TRS)
Poly vinyl chloride wire(PVC)
Lead alloy sheathed wire
Weather proof wires
Flexible wires

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General construction of cable

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Cable Structure

Distribution of Electric Energy to
Domestic Consumer

•Types of wiring
•Suitability of a particular wiring system for a
given installation
•Corridor and staircase lighting
•Necessity of earthing
•Different types of earthing
•Fuses
•Different types of fuses
•Imp VTU Questions
Wiring

Introduction
•A network of wires drawn from the meter board to
the various energy consuming points (lamps,
fans, motors etc) through control and protective
devices for efficient distribution of power is known as
electrical wiring.
•Electrical wiring done in residential and commercial
buildings to provide power for lights, fans, pumps and
other domestic appliances is known as domestic
wiring. There are several wiring systems in practice.
They can be classified into different Types

Types of
Wiring:
1. Cleat wiring
2. CTS wiring or TRS wiring or batten wiring
3. Metal sheathed wiring or lead sheathed
wiring
4. Casing and capping
5. Conduit wiring

Cleat
wiring
•: In this type of wiring, insulated conductors (usually
VIR, Vulcanized Indian Rubber) are supported on
porcelain or wooden cleats.
•The cleats have two halves one base and the other
cap.
•The cables are placed in the grooves provided in the
base and then the cap is placed. Both are fixed
securely on the walls by 40mm long screws.
•The cleats are easy to erect and are fixed 4.5 – 15
cms apart.
•This wiring is suitable for temporary installations
where cost is the mPravaeeinnkumcarr.Cit, Deeprt.iaof EEbE,uSKtIT,not
Bangalore

Advantages and
Disadvantages:
Advantages:
1.Easy installation
2.Materials can be retrieved for
reuse
3.Flexibility provided for
inspection, modifications and
expansion.
4.Relatively economical
5.Skilled manpower not
required.
Disadvantages:
1.Appearance is not good

CTS ( Cable Tyre Sheathed) / TRS ( Tough
Rubber Sheathed ) / Batten wiring:
• In this wiring system, wires sheathed in tough
rubber are used which are quite flexible. They are
clipped on wooden battens with brass clips (link
or joint) and fixed on to the walls or ceilings by
flat head screws.
• These cables are moisture and chemical proof.
• They are suitable for damp climate but not
suitable for outdoor use in sunlight.
• TRS wiring is suitable for lighting in low voltage
Bangalore
installations. Praveen kumar .C, Dept. of EEE, SKIT,

CTS ( Cable Tyre Sheathed) / TRS ( Tough
Rubber Sheathed ) / Batten wiring:

Advantages and
Disadvantages:
Advantages:
1.Easy installation and is
durable
2.Lower risk of short circuit.
3.Cheaper than casing and
capping system of wiring
4.Gives a good appearance
if properly erected.
Disadvantages:
1.Danger of mechanical
injury.
2.Danger of fire hazard.

Metal Sheathed or Lead Sheathed wiring
•The wiring is similar to that of CTS but the
conductors (two or three) are individually
insulated and covered with a common outer
lead-aluminum alloy sheath.
•The sheath protects the cable against
dampness, atmospheric extremities and
mechanical damages.

•The sheath is earthed at every junction to
provide a path to ground for the leakage
current.
•They are fixed by means of metal clips on
wooden battens.
•The wiring system is very expensive. It is
suitable for low voltage installations.

Precautions to be taken during
installation
•The clips used to fix the cables on battens
should not react with the sheath
2.Lead sheath should be properly earthed to
prevent shocks due to leakage currents.
3.Cables should not be run in damp places
and in areas where chemicals (may react with
the lead) are used.

Advantages and
Disadvantages:
Advantages:
1.Easy installation and
is aesthetic in
appearance.
2.Highly durable
3.Suitable in adverse
climatic conditions
provided the joints are
not exposed
Disadvantages:
1.Requires skilled
labor

Casing and Capping
•It consists of insulated conductors laid inside
rectangular, teakwood or PVC boxes having
grooves inside it.
•A rectangular strip of wood called capping having
same width as that of casing is fixed over it.
•Both the casing and the capping are screwed
together at every 15 cms.
•Casing is attached to the wall. Two or more wires
of same polarity are drawn through different
grooves.
•The system is suitable for indoor and domestic
Bangalore
installations. Praveen kumar .C, Dept. of EEE, SKIT,

Advantages and
Disadvantages:
Advantages:
1.Cheaper than lead sheathed
and conduit wiring.
2.Provides good isolation as
the conductors are placed
apart reducing the risk of
short circuit.
3.Easily accessible for
inspection and repairs.
4.Since the wires are not
exposed to atmosphere,
insulation is less affected by
dust, dirt and climatic
Bangalore

Conduit wiring
Praveen kumar .C, Dept. of EEE, SKIT,
•In this system PVC (polyvinyl chloride) or VIR cables
are run through metallic or PVC pipes providing good
protection against mechanical injury and fire due to
short circuit.
•They are either embedded inside the walls or
supported over the walls, and are known as concealed
wiring or surface conduit wiring (open conduit)
respectively.
•The conduits are buried inside the walls on wooden
gutties and the wires are drawn through them with fish
(steel) wires.
•The system is best suited for public buildings,
industries and workshopBsan.galore

Advantages and
Disadvantages:
Advantages:
1.No risk of fire and good protection
against mechanical injury.
2.The lead and return wires can be
carried in the same tube.
3.Earthing and continuity is assured.
4.Waterproof and trouble shooting is
easy.
5.Shock- proof with proper earthing
and bonding
6.Durable and maintenance free
7.Aesthetic in appearance
Disadvantages:
1.Very expensive system of wiring.
2.Requires good skilled workmanship.

FACTORS AFFECTING THE CHOICE
OF WIRING SYSTEM
The choice of wiring system for a particular installation
depends on technical factors and economic viability.
1.Durability: Type of wiring selected should conform to
standard specifications, so that it is durable i.e. without
being affected by the weather conditions, fumes etc.
2.Safety: The wiring must provide safety against
leakage, shock and fire hazards for the operating
personnel.
3.Appearance: Electrical wiring should give an
aesthetic appeal to the interiors.
4.Cost: It should not be prohibitively expensive.

Cnt.
.
5.Accessibility: The switches and plug points
provided should be easily accessible. There must be
provision for further extension of the wiring system, if
necessary.
6.Maintenance Cost: The maintenance cost should
be a minimum
7.Mechanical safety: The wiring must be protected
against any mechanical damage

Two- way and Three- way Control
of Lamps:
•The domestic lighting circuits are quite simple
and they are usually controlled from one point.
But in certain cases it might be necessary to
control a single lamp from more than one point
(Two or Three different points).
•For example: staircases, long corridors, large
halls etc.

1 Two-way Control of lamp:
•Two-way control is usually used for staircase
lighting.
•The lamp can be controlled from two different
points: one at the top and the other at the bottom -
using two- way switches which strap wires
interconnect.
•They are also used in bedrooms, big halls and large
corridors.
•The circuit is shown in the following figure.

1 Two-way Control of lamp:
Explanation: -Switches S1 and S2 are two-way
switches with a pair of terminals 1&2, and 3&4
respectively. When the switch S1 is in position1 and
switch S2 is in position 4, the circuit does not form a
closed loop and there is no path for the current to flow
and hence the lamp will be OFF. When S1 is changed to
position 2 the circuit gets completed and hence the lamp
glows or is ON. Now if S2 is changed to position 3 with
S1 at position 2 the circuit continuity is broken and the
lamp is off. Thus the lamp can be controlled from two
different points.

2. Three- way Control of
lamp:
•In case of very long corridors it may be necessary to
control the lamp from 3 different points. In such cases,
the circuit connection requires two; two-way switches
S1and S2 and an intermediate switch S3.
•An intermediate switch is a combination of two, two
way switches coupled together. It has 4 terminals ABCD.
It can be connected in two ways.

a) Straight connection
b) Cross connection
• In case of straight connection, the terminals or points
AB and CD are connected as shown in figure 1(a)
while in case of cross connection, the terminals AB
and C D is connected as shown in figure 1(b). As
explained in two – way control the lamp is ON if the
circuit is complete and is OFF if the circuit does not
form a closed loop.
lamp:

lamp:

Truth table :
2. Three- way Control of lamp:

1.To protect the operating personnel from danger
of shock in case they come in contact with the
charged frame due to defective insulation.
2.To maintain the line voltage constant under
unbalanced load condition.
3.Protection of the equipments
4.Protection of large buildings and all machines
fed from overhead lines against lightning.
Necessity of Earthing:

1 ohm.
•The important methods of earthing are the plate
earthing and the pipe earthing.
•The earth resistance for copper wire is 1 ohm
and that of G I wire less than 3 ohms.
•The earth resistance should be kept as low as
possible so that the neutral of any electrical
system, which is earthed, is maintained almost at
the earth potential.
•The typical value of the earth resistance at
powerhouse is 0. 5 ohm and that at substation is
Methods of Earthing:

.
•In this method a copper plate of 60cm x 60cm x 3.18cm
or a GI plate of the size 60cm x 60cm x 6.35cm is used
for earthing.
•The plate is placed vertically down inside the ground at
a depth of 3m and is embedded in alternate layers of
coal and salt for a thickness of 15 cm.
•In addition, water is poured for keeping the earth
electrode resistance value well below a maximum of 5
ohms.
•The earth wire is securely bolted to the earth plate.
•A cement masonry chamber is built with a cast iron
cover for easy regular maintenance.
1. Plate Earthing:

•Earth electrode made of a GI (galvanized) iron pipe of
38mm in diameter and length of 2m (depending on the
current) with 12mm holes on the surface is placed
upright at a depth of 4.75m in a permanently wet ground.
•To keep the value of the earth resistance at the desired
level, the area (15 cms) surrounding the GI pipe is filled
with a mixture of salt and coal..
•the efficiency of the earthing system is improved by
pouring water through the funnel periodically. The GI
earth wires of sufficient cross- sectional area are run
through a 12.7mm diameter pipe (at 60cms below) from
the 19mm diameter pipe and secured tightly at the top
as shown in the figure.
2. Pipe Earthing

•Protection for electrical installation must be provided in
the event of faults such as short circuit, overload and
earth faults.
•The protective circuit or device must be fast acting and
isolate the faulty part of the circuit immediately.
•It also helps in isolating only required part of the circuit
without affecting the remaining circuit during
maintenance.
•The following devices are usually used to provide the
necessary protection:
PROTECTIVE DEVICES

1. Fuses
2. Relays
3. Miniature circuit breakers (MCB)
4. Earth leakage circuit breakers (ELCB)
Types

1. Re-wirable or kit -kat fuses: These fuses are simple in
construction, cheap and available up-to a current
rating of 200A. They are erratic in operation and their
performance deteriorates with time.
2. Plug fuse: The fuse carrier is provided with a glass
window for visual inspection of the fuse wire.
3. Cartridge fuse: Fuse wire usually an alloy of lead is
enclosed in a strong fiber casing. The fuse element is
fastened to copper caps at the ends of the casing. They
are available up-to a voltage rating of 25kV. They are
used for protection in lighting installations and power
lines.

circuits.
4. Miniature Cartridge fuses: These are the miniature version of the
higher rating cartridge fuses, which are extensively used in
automobiles, TV sets, and other electronic equipment‟s.
5. Transformer fuse blocks: These porcelain housed fuses are placed
on secondary of the distribution transformers for protection
against short circuits and overloads.
6. Expulsion fuses: These consist of fuse wire placed in hollow tube
of fiber lined with asbestos. These are suited only for out door use
for example, protection of high voltage circuits.
7. Semi-enclosed re-wirable fuses: These have limited use because of
low breaking capacity.
8. Time-delay fuse: These are specially designed to withstand a
current overload for a limited time and find application in motor

Under abnormal conditions such as short circuit, overload or
any fault the current raises above this value, damaging the
equipment and sometimes resulting in fire hazard. Fuses are
pressed into operation under such situations.
Fuse is a safety device used in any electrical installation,
which forms the weakest link between the supply and the
load. It is a short length of wire made of lead / tin /alloy of
lead and tin/ zinc having a low melting point and low ohmic
losses. Under normal operating conditions it is designed to
carry the full load current. If the current increases beyond
this designed value due any of the reasons mentioned
above, the fuse melts (said to be blown) isolating the power
supply from the load as shown in the following figures.
FUSE

Electric circuits and network theorems

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