Module for download the PPT from -01.pptx

DC CIRCUITS
(Only Independent Sources)
Module- 01

Contents-
 Kirchhoff’s Laws
 Ideal and Practical Voltage and Current Sources
 Source Transformation
 Mesh and Nodal Analysis (no super node and super mesh)
 Star-Delta / Delta-Star Transformations
 Superposition Theorem
 Thevenin’s Theorem
 Norton’s Theorem and
 Maximum Power Transfer Theorem.

INTRODUCTION-
 When electrical circuit is driven by DC source then it is called as DC Circuit.
 Essential part of the circuits are source of power (battery), conductor through
which current flows and load (Lamp, Heater, Resistor and Motor).
 Source supplies electrical energy to load which convert into other form of energy.
 The direction of current is positive terminal of battery- bulb – Negative terminal
of battery.
 If the load in DC circuit is resistance then it can be connected in following ways-
 Series circuit
 Parallel circuit
 Series-Parallel circuit

SERIES CIRCUIT-
In an electric circuit, the different components are connected either in
series or in parallel to produce different resistive networks.
Resistors in Series-
Two or more resistors are said to be connected in series when the same
amount of current flows through all the resistors. In such circuits, the voltage
across each resistor is different.
For the above circuit, the total resistance is given as:
Rtotal = R1 + R2 + ….. + Rn
The total resistance of the system is just the total sum of individual resistances.

SERIES CIRCUIT-
Consider a circuit shown below.
The resistances R1, R2 and R3 are connected in series across voltage source
V, same current I will be flows through it.
According to ohm’s law
Voltage across R1 is VR1 = IR1
The total voltage (V) applied is sum of the voltages across R1, R2 and R3.
V = VR1 + VR2 + VR3 = IR1 + IR2 + IR3 = I(R1 + R2 + R3)
= IRtotal
R + R + R = V/I

SERIES CIRCUIT-
Voltage Division Rule-
We know the current through circuit is-
Similarly Voltage across R2 is VR2 = IR2
Note- The voltage drop across each resistor in series circuit is directly
proportional to the ratio of its resistance to total series resistance of the circuit
and applied voltage.

PARALLEL CIRCUIT-
Consider a circuit in which three resistors R1, R2 and R3 are connected in
parallel across supply voltage V. Due to voltage V current I flowing through
circuit divided into I1 through R1, I2 through R2 and I3 through R3. Voltage
across each resistor is same which is supply voltage V.
For the above circuit, the total resistance is given as:
By ohm’s law

PARALLEL CIRCUIT-
From the given circuit the total current is sum of branch currents.

PARALLEL CIRCUIT-
Current Division Rule-
Consider a circuit in which two resistors R1 and R2 are connected in
parallel across supply voltage V. Total current I is divided into two parts I1
through R1, I2 through R2.
Total Resistance Rtotal is
Voltage V is given as

PARALLEL CIRCUIT-
By ohm’s law
In a parallel circuit of two resistors, the current through one resistor is the
line current times the opposite resistance divided by sum of two resistances.

SERIES & PARALLEL CIRCUITS-
Examples- Determine the current through & voltage across three resistances of
ohmic values 5Ω, 7Ω and 8 Ω connected in series across 100 V source.
Examples- Determine the current through & voltage across three resistances of
ohmic values 5Ω, 7Ω and 8 Ω connected in parallel across 100 V source.
Examples- Two resistors R1 and R2 are connected in parallel to a certain
supply. The current taken from the supply is 5A. Calculate the value of R1 if R2
= 6Ω and current through R2 is 2 A. Also calculate total power absorbed by the
circuit.
Examples- Calculate effective resistance of the circuit and the current through
8 Ω resistance, when potential difference of 60 V is applied between point A &
B.
Examples- Calculate effective resistance between point A & B.

SHORT & OPEN CIRCUITS-
 Sort circuits- When two terminals of a circuit are connected by a wire, they
are said to be short circuited. A short circuit has following features:
 It has zero resistance.
 Current through it is very large.
 There is no voltage across it.
 Open circuits- When two terminals of a circuit have no direct connection
between them, they are said to be open circuited. An open circuit has the
following features:
 It has infinite resistance.
 Current through it is zero.
 The entire voltage appears across it.

Open Circuits and Short Circuits in a Series Circuit-
When an open circuit appears in a series circuit, the equivalent resistance becomes
infinite and no current flows through the circuit.
Current through circuit is
Voltage across open circuit is
When a short circuit appears in a series circuit, as shown, the resistance R2
becomes zero.

Open Circuits and Short Circuits in a Parallel Circuit-
When an open circuit appears in a parallel circuit, no current flows through that
branch. The other branch currents are not affected by the open circuit.
When a short circuit appears in a parallel circuit, the equivalent resistance
becomes zero.

Open Circuits and Short Circuits-
Examples- Find an equivalent resistance between terminal A & B.
Examples- Determine the current delivered by the source in the circuit shown
below-
Examples- Determine the current delivered
by the source in the circuit shown below-

KIRCHHOFF’S LAWS-
The entire electric circuit analysis is based mainly on Kirchhoff’s laws. Some of the
term related to Kirchhoff’s laws are as follows-
Kirchhoff’s law are classified as-
 Kirchhoff’s Current law/Kirchhoff’s First Law (KCL)
 Kirchhoff’s Voltage law/Kirchhoff’s Second Law (KVL)

KIRCHHOFF’S LAWS-
Kirchhoff’s Current law/Kirchhoff’s First Law (KCL)-
It is related to current at a junction point. It states that- “ The algebraic
sum of currents meeting at a point/junction/node is zero.”
Consider a conductor, carrying currents I1, I2, I3, I4 and I5 at point o.
According to Kirchhoff’s law-
I = 0
Ʃ
Assuming incoming current positive and outgoing
current negative. Above equation becomes-
+I1 +I2 +I3+(-I4) +(-I5) = 0
I1 +I2 +I3 -I4 -I5 = 0
It also state that, “Sum of the currents flowing towards any junction in an electric

KIRCHHOFF’S LAWS-
Example- Find the currents in all branches
Example- Find VXY as shown in figure

KIRCHHOFF’S LAWS-
Kirchhoff’s Voltage law/Kirchhoff’s Second Law (KVL)-
It is related to electromotive force and voltage drop in a circuit and stated
that- “ The algebraic sum of all the voltages in a closed circuit/mesh/loop is
zero.”
If we start from any point in a closed circuit and go back to that point,
after going around the circuit, there is no increase or decrease in potential at
that point. This means that sum of emf’s and sum of voltage drops or rise meet
on the way is zero.
According to Kirchhoff’s law-
V = 0
Ʃ
In a given diagram V is supply voltage and V1, V2
and V3 are voltage drop across R1, R2 and R3.
V = V1+V2+V3 = IR1+IR2+IR3

KIRCHHOFF’S LAWS-
Sign Convention-
While applying KVL to closed loop circuit, algebraic sum is considered.
It is important to assign proper sign to emf and voltage drops in a closed loop
circuits.
Assumptions – Consider rise in voltage as positive and fall in voltage as
negative.
 If we go from positive terminal of battery/source to the negative terminal
there is fall in potential and so the emf should be assigned a negative sign,
While if we move negative to positive terminal of battery then there is rise if
potential so emf should be assigned positive sign.

KIRCHHOFF’S LAWS-
 When current flows through the resistor there is voltage drop across it. If we
go from resistor in the same direction as current there is fall in potential and
sign of voltage drop is negative. If we go to opposite to the direction of
direction flow, there is rise in potential and voltage drop is trated as positive.

KIRCHHOFF’S LAWS-
While Solving problems on Kirchhoff’s Law-
 Assume unknown current and show directions.
 Write KVL for all loops.
 In a solution if the value of any unknown current come out to be negative, it
means that actual direction of current is opposite to that of assumed
direction.

KIRCHHOFF’S LAWS-
Example- By using Kirchhoff’s law calculate the branch current shown in
figure,
I1 = 5 A( ) I2 = 5A( )
Example- By using Kirchhoff’s law calculate the branch current shown in
figure,
Example- Find the current through 2 Ω resistor,
I = 0.3 A( )
I2Ω = 1.66 A

KIRCHHOFF’S LAWS-
Example- Determine the current through 20 Ω resistor shown in figure,
Example- For the network shown in figure, Determine a) I1, I2 and I3 b)
resistance R c) value of emf E.
Example- Determine potential difference VAB for
given network.

IDEAL & PRACTICAL VOLTAGE AND CURRENT SOURCE-
There are two types of sources available-
 Voltage source
 Current source
Ideal & Practical Voltage source-
Ideal voltage source is the type of source whose output remains constant whatever
changes in load current. It has zero internal resistance. In practice ideal voltage source is
not possible because it has some internal resistance. Smaller is the value of internal
resistance, more it will approaches to ideal voltage source.

IDEAL & PRACTICAL VOLTAGE AND CURRENT SOURCE-
Ideal & Practical Current source- Ideal current source is the type of source whose output
remains constant whatever changes in load resistance. It’s internal resistance is infinity. At
any load resistance it supplies constant current. In practice ideal current source has very
large internal resistance. Higher is the value of internal resistance, more it will approaches
to ideal current source.

SOURCE TRANSFORMATION-
As we know, we have two sources i.e. current source and voltage source. As per the
requirement for circuit simplification of complex network one type of source can be
converted to other type of source. This process is called as source transformation.
There are different types of source transformation can be used which are-
 Voltage to current transformation
 Current to voltage transformation
 Voltage series source
 Current parallel source
Voltage to Current (V-I) Transformation- A voltage source in series with resistance can be
transfer/replace/converted to current source in parallel with resistance.
Example-

Current to Voltage (I-V) Transformation- A current source in parallel with resistance can be
transfer/replace/converted to voltage source in series with resistance.
Example-
Series Voltage Source- Voltage sources are connected in series with additive polarity, the
resultant voltage source will be the addition of voltage source. For subtraction property it
will be subtracted.
Example-

Parallel Current Source- Current sources connected in parallel can be added or clubbed
together as per direction of currents.
Example-

Example- Using source transformation find the current through 4 Ω resistor.
I4Ω = 1.33 A (↓)
I10Ω = 0.0516 A (↓)
Example- Find the voltage across 4 Ω resistor.
V4Ω = 7.12 V
I3Ω = 2 A (↓)

MESH & NODALANALYSIS-
Mesh Analysis(Maxwell’s Mesh current Method)- In this method, Kirchhoff’s voltage law
is applied to each mesh in terms of mesh current instead of branch current. Each mesh
assigned a mesh current. The mesh current is assumed to be flow in clockwise direction.
Once mesh current is known we can determine branch current.
Steps to follow in Mesh Analysis-
Consider the network as shown in figure.
 Identify the mesh and assign the unknown current in each mesh in clockwise direction
only.

 Assign polarity of voltage across the branches.
 Apply KVL to each mesh around the mesh and use ohm’s law to express in terms of
unknown current & resistance.
Apply KVL to Mesh 1
………………………… (1)
Apply KVL to Mesh 2
………………………… (2)
Apply KVL to Mesh 3
………………………… (3)
 Solve the above simultaneous equations to find branch currents.

Example- Find the value of current flowing through 1 Ω resistor.
I1Ω = 0.5 A
I5Ω = 0.78 A
I2Ω = 5.71 A

Example- Find the current supplied by battery.
I = 0.66 A
I10Ω = 0.78 A
I5Ω = 2 A

MESH AND NODALANALYSIS-
Super Mesh-
When a current source is present in branch and are common to two
meshes then it is known as super mesh.
In this cases we need some manipulation and concept of super mesh to
solve the problem of super mesh.
Example- By using Mesh analysis find the current through 10 Ω resistor.

Nodal Analysis (Node Voltage Method)- This method is based on Kirchhoff’s
current law to determine voltage at different nodes with respect to reference
node. After determination of node voltage, current in all branches can be
determine.
Step to solve the Node Analysis problem-
 Mark all nodes.
 Assign any one node as reference node.
 Assign unknown potential to all nodes with respect to reference node.
 Assume unknown current and directions.
 Apply KCL for simultaneous node voltage and solve it.
 Using these voltage find any branch current which is required.

Example- Find the current through 2 Ω resistor.
VA = 11 V, VB = 12 V I2 Ω = 6A
VA = 5.5 V, VB = 8.5 V I10 Ω = 0.75A
VA = 67.25 V, VB = 48 V I100 Ω = 0.48A

VA = 15 V, VB = -26.25 V I15 Ω = 2.75A
Example- Find the voltage at node A and B.
VA = 8 V, VB = 8 V
Example- Find the current through 4 Ω & 3 Ω resistor.
V1 = 10 V, V2 = 14.09 V, V3 = 11.16 V I4 Ω = 0.29A, I3 Ω = 0.98A

Super Node-
When a voltage source is present between two nodes make them to be
super node.
In this cases separate equation is obtained between two nodes and these
two nodes are combines during nodal analysis.
Example- Determine the current in 5 Ω resistor for the network shown.

STAR (Υ) TO DELTA (∆) TRANSFORMATION-
When a circuit cannot be simplified by normal series–parallel reduction
technique, the star-delta transformation can be used. Consider delta and star
network as shown.
Delta Connection
Star Connection
Resistors RA, RB and RC forms delta network and resistors R1, R2 and R3
forms star network. These two networks will be electrically equivalent if the
resistance as measured between any pair of terminals is the same in both the
arrangements.

STAR (Υ) TO DELTA (∆) TRANSFORMATION-
In figure (a) of delta network, Resistors between point 1 and 2 is
…………….. (1)
In figure (b) of star network, Resistors between point 1 and 2 is
…………….. (2)
Since the two networks are electrically equivalent, if equation (1) and (2)
are equal.
…………….. (3)
Similarly, resistors between point 2 and 3 is
Delta to Star Transformation-
…………….. (4)

STAR DELTA TRANSFORMATION-
Also resistors between point 3 and 1 is
Subtracting equation (4) from (3)
Adding equation (5) and (6), we get
Similarly
…………….. (5)
…………….. (6)
…………….. (7)
…………….. (8)
…………….. (9)

Thus, star resistor connected to a terminal is equal to the product of the
two delta resistors connected to the same terminal divided by the sum of the
delta resistors.
Star to Delta Transformation-
Multiply equation (7) & (8), (8) & (9) and (9)& (7)
…………….. (10)
…………….. (11)
…………….. (12)

Adding equations (10), (11) & (12)
Hence

Thus, delta resistor between the two terminals is the sum of two star
resistors connected to the same terminals plus the product of the two resistors
divided by the remaining third star resistor.
Note- If all resistances are of equal values-

Example- Find the equivalent resistance between point X and Y
RXY = 8.67 Ω
Example- Find the equivalent resistance between point A and B
RAB = 7.45 Ω
RAB = 32.36 Ω

Example- Find the equivalent resistance between terminal A and B
RAB = 10 Ω
Example- Calculate RAB for the circuit shown in figure
RAB = 1.64 Ω
RAB = 23 Ω

RAB = 23.52 Ω
Example- Calculate RAB for the circuit shown in figure
RAB = 4.23 Ω
Example- Find the current flowing through 8 Ω resistor
I8Ω = 1.18 A

SUPERPOSITION THEOREM-
This theorem is used when source transformation is not possible and the
network have number of sources acting on it.
Statement- It states that ‘In a linear network containing more than one
independent sources, the resultant current in any element is the algebraic sum
of the currents that would be produced by each independent source acting
alone, all the other independent sources being represented meanwhile by their
respective internal resistances.’
The independent voltage sources are represented by their internal
resistances if given or simply with zero resistances, i.e., short circuits if internal
resistances are not mentioned.
The independent current sources are represented by infinite resistances,
i.e., open circuits. A linear network is one whose parameters are constant, i.e.,
they do not change with voltage and current.

Steps for superposition theorem-
Consider the network as shown in figure-
Calculate current flowing through terminal A and B.
Case (i)- According to superposition theorem, each source act independently.
Consider only source V1 is acting independently while replacing V2 by its
internal resistance if not short it.
Apply KVL to each mesh and find current through AB (IAB’)
Case (ii)- Consider only voltage source V2 is act independently while replacing
V1 by its internal resistance, if not then short it.
Apply KVL to each mesh and find current through AB (IAB’’)

Case (iii)- Total current through terminal A & B is-
IAB = Current through AB because of only V1 + Current through AB because of only V1
IAB = IAB’ + IAB’’
Example- Using superposition theorem find the value of current flowing through 1 Ω resistor.
I3Ω = 5 A, I3Ω = 10 A, I4Ω = 5 A
I1Ω = 4 A

Example- Using superposition theorem find the value of current flowing through 10 Ω
resistor.
I8Ω = 0.316 A
I10Ω = 0.316 A
I4Ω = 4.29 A

THEVENIN’S THEOREM-
This theorem is used to find the current through complex network.
Statement- It states that ‘Any two terminals complex network can be replaced
by an equivalent voltage source (EMF/Thevenin;s voltage/VTH) in series with
equivalent resistance (Thevenin’s resistance).
The voltage source is the voltage across the two terminals with load, if any,
removed. The series resistance is the resistance of the network measured between
two terminals with load removed.
Constant voltage source being replaced by its internal resistance (or if it is not
given with zero resistance, i.e., short circuit) and constant current source replaced by
infinite resistance, i.e., open circuit.

Steps for Thevenin’s Theorem-
 Remove the load resistance RL.
 Find open circuited voltage VTH across the point A & B. (remove the resistance
across which we have to calculate current through it).
 Find resistance RTH as seen from point A and B with voltage source and current
source replace by it’s internal resistance.
 Replace the circuit with Thevenin’s circuits i.e Voltage source VTH in series with
equivalent resistance RTH.
 Find the current through RL using Ohm’s law.

VTH = 23.33 V RTH = 3.33 Ω IL = 4.38 A
VTH = 50 V RTH = 12.5 Ω IL = 2.44 A
Example- Find the value of current
flowing through 2 Ω resistor
connected between terminal A & B.
VTH = 2.8 V RTH = 1.43 Ω IL = 0.82 A

VTH = 10 V RTH = 17 Ω IL = 0.37 A
VTH = 4.2 V RTH = 8.33 Ω IL = 0.09 A
flowing through 30 Ω resistor.
VTH = 70 V RTH = 26.09 Ω IL = 1.25 A

NORTON’S THEOREM-
Norton’s theorem is converse of Thevenin’s theorem. It uses current source instead of
voltage source and a resistance RN (which is same as RTH) in parallel with the source.
Statements- It states that, Any two terminal network can be replaced by a
single current source with magnitude IN (called Norton’s current/Short circuit
current) in parallel with single resistance RN (called Norton’s resistance).

NORTON’S THEOREM-
Steps for Norton’s Theorem-
 Remove the load resistance RL and put a short circuit across the terminals.
 Find the short-circuit current ISC or IN across the terminals.
 Find the resistance RN as seen from points A and B by replacing the voltage
sources and current sources by internal resistances.
 Replace the network by a current source IN in parallel with resistance RN.
 Find current through RN by current–division rule,

NORTON’S THEOREM-
IN = 5 A RTH = 0.95 Ω IL = 0.43 A
Example- Find the value of current flowing through 15 Ω resistor in given ckt.
IN = 1.4 A RTH = 10 Ω IL = 0.56 A
IN = -0.67 A RTH = 9.43 Ω IL = 0.33 A

NORTON’S THEOREM-
IN = 0.81 A RTH = 12.3 Ω IL = 0.45 A
Example- Find the value of current flowing through 08 Ω resistor in given ckt.
IN = 6.58 A RTH = 3 Ω IL = 1.79 A
IN = -13.17 A RTH = 1.46 Ω IL = 1.68 A

MAXIMUM POWER THEOREM-
This theorem deals with transfer of maximum power from source to load.
It states that ‘the maximum power is delivered from a source to a load when the
load resistance is equal to the source resistance.’
Consider a complex circuit supplying power to load RL. The complex circuit is
replaced by Thevenin’s equivalent circuit which consist of single source of emf
VTH(Thevenin’s voltage) in series with single resistance RTH (Thevenin’s resistance), RL is
load resistance.
The load current IL through RL is given as-

Proof-
Power delivered to the load RL
………….. (1)
For a given circuit RTH and VTH are constants, therefore power delivered to load
depends on RL. In order to find value of RL for which the value of P is maximum, it is
necessary to differentiate equation (1) with respect to RL and set the result equal to zero.
For Pmax, Differentiate equation (1) w. r. to RL, we get
Hence Maximum power Transfer theorem proved.
RL = RTH
Maximum power transfer from source to load
when load resistance equal to source resistance

Power delivered to load RL is given as-
When RTH = RL and P = Pmax
Steps for Maximum Power Transfer Function-
 Remove the variable load resistor RL.
 Find the open circuit voltage VTH across points A and B.
 Find the resistance RTH as seen from points A and B with voltage sources and current
sources replaced by internal resistances.
 Find the resistance RL for maximum power transfer. RL = RTH
 Find the maximum power.

Example- Find the value of RL for maximum power transfer. Calculate maximum power.

Module for download the PPT from -01.pptx

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