MODULE -I

Network Theorems
Single Phase Circuit
Three Phase Circuit
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 KIRCHHOFF’S VOLTAGE LAW

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The summation of voltage rises and voltage drops around a closed loop
is equal to zero

Swagata Roy Chatterjee

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IDEAL VOLTAGE SOURCE
 VOLTAGE REGULATION
The V − I characteristic of the source is also called the source’s “regulation curve”
or “load line”.

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The open-circuit voltage is also called the “no-load” voltage, .
The maximum allowable load current (rated current) is known as full-load
current

Swagata Roy Chatterjee

the corresponding source or load terminal voltage is known as “full-load” voltage
.
The percentage change in source terminal voltage from no-load to full-load
current is termed the “voltage regulation” of the source.

For ideal voltage source, there should be no change in terminal voltage from
no-load to full-load and this corresponds to “zero voltage regulation”
Hence ideal voltage source has zero internal resistance

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Ideal and Practical Current Sources

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Swagata Roy Chatterjee
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Conversion of a Practical Voltage
Source to a Practical Current source

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Dependent Sources
Mesh Analysis

Mesh 1

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Mesh 2

2A
Mesh 3

ANSWER9
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ANSWER
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_
+

4-5(I1-I2)-3 I1=0
_
+
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What is Nodal Analysis?

Nodal analysis is a method that provides a general

procedure for analysing circuits using node voltages as
the circuit variables.

Also called Node-Voltage Method.

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 INTRODUCTION I1
Types of Nodes in Nodal
Analysis:
1 R2 2
▪ Non-Reference Node:
Node which has a definite
Node-Voltage. R1 R3
▪ Reference Node: It is the I1
node which acts as reference
0
point to all the other node.
Also called Node 0 is Reference node
Datum Node/Ground.
Nodes 1, 2 are Non-Reference node

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Reference node indication

Common symbols for

indicating reference node,
(a) common ground,
(b) (earth) ground,
(a) (b) (c) (c) chassis ground.

Chassis ground: used in devices where the case,

enclosure, or chassis acts as reference point for all circuits.
Earth ground: used when the potential of the earth
is used as reference.

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Steps to find Node Voltages:
1. Select a node as the reference (ground)
node. Assign voltages v1, v2, v3,……., vn-1 to the
remaining n-1 nodes.
2. Apply KCL to each of the n-1 non-reference nodes.
3. Use Ohm’s law to express the branch currents in terms
of node voltages.
4. Solve the resulting ‘n-1’ simultaneous
equations to obtain the unknown node voltages.
For ‘n’ nodes there will be ‘n-1’ simultaneous equations to
be solved to get all the node voltages.

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Example circuits depicting Nodes:

I1
R4

1 R2 2 1 R1 2 R3 3
ix
R1 R3 R2 2ix
I1 I1
0 0

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 NODE VOLTAGE ANALYSIS
• Dual of Mesh Analysis.
• Involves the application of KCL equations (KVL for mesh).
• One of the nodes is taken as reference or datum
node
(node with maximum number of branches connected preferred).
• The reference node is assumed to be at ground or zero potential.
• Potentials of all the other nodes are defined with
respect to reference node.
• KCL equations are written for each node except for
the reference node. On solving, node voltages are obtained.

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 SUPER NODE ANALYSIS
R Supernode
Definition of Super Node:
when a dependent or an i4
independent voltage source 1
R i 1 v2 5V
connected
is between two v1 3 v3

+
reference non- nodes,
then i3
to form 10V +
these
nodes can be combined i2 R 2 R
a generalised node which
0
is as Super Node.
known
Super node can be regarded as a
surface enclosing the voltage Node 2 and 3 can be
source and its two nodes. combined into a Super-Node.

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 SUPER NODE ANALYSIS
Properties of super-node:
• A super-node cannot be assigned a single voltage value
• A super-node requires application of both KCL and KVL to solve
it.
• Any element can be connected in parallel with the voltage source
forming the super-node without affecting the analysis.
• A super-node satisfies the KCL as like a simple node.

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 ADVANTAGES OF NODAL ANALYSIS
Nodal Analysis vs Mesh Analysis?
• Nodal analysis can be applied to both planar and nonplanar
circuits. But Mesh analysis can be applied to planar circuits only.

Planar circuit: a circuit that can be drawn on a plane with

no crossed wires.

• The choice between Node Voltage analysis and Mesh Current analysis
is usually unambiguous.
• To choose between methods, pick the one that involves
solving the fewest equations.

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 NODAL ANALYSIS VS MESH ANALYSIS
Nodal Analysis Mesh analysis
• The number of voltage variables, • The number of current variables,
and hence simultaneous and hence simultaneous
equations to solve, equals the equations to solve, equals the
number of nodes minus one. number of meshes.
• Every voltage source connected • Every current source in a mesh
to the reference node reduces the reduces the number of
number of unknowns by one. unknowns by one.
• Nodal analysis is thus best for • Mesh analysis is thus best for
circuits with voltage sources. circuits with current sources.

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 NUMERICAL
Q1. Calculate Node Voltages in the given
circuit.
5A
i1 2
1
4Ω i2 v2 4
v1
i3 i5
2Ω i 10A
6Ω
0

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 NUMERICAL

Soln: There are 3 nodes: 5A

0: ground, 1 & 2: non-reference i1
nodes. 1 2
4Ω i2 v2 i4
v1
Step1. Node voltages (v1 and v2) are i3 i5
assigned. Also direction of branch 2Ω 6Ω 10A
currents are marked. 0

Step2. Apply KCL to nodes 1 and

2

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 CONTD. 5A
KCL at Node 1: i1
1 2
i1 = i2 + i3 4Ω i2 v2 i4
v1
i3 i5
KCL at Node 2: 2Ω
6Ω 10A
i2 + i4 = i1 + 0
Step3. Apply Ohm’s Law to KCL
i5
equations. Ohm’s Law to KCL equation at
Node 1:
v − vv − 0
i1 = i2 + i3 ⇒ 5 1 2 ⇒ 3v1 v2 = …(1)
4 2 +
= − 20
1

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 CONTD.
Now, Ohm’s Law to KCL equation at Node
2:
v1 −2 v2 −
i2 + i4 = i1 + i5 +10 = 5 5A
v 4 06
⇒ +
⇒ −3v1 + 5v2= 60 …(2) i1 2
1
4Ω i2 v2 4
v1
i3 i5
On solving (1) and (2) we get, 2Ω i 10A
6Ω
0
v1 = 13.33V and v2 = 20
V

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 NUMERICAL
Q2.Calculate Node Voltages and Branch Currents in the
given circuit.

7 i v 1 i3
Ω 2 i1
+
60 4
1
Ω
V 12 0
Ω

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 NUMERICAL
7 i2 v 1 i3
Ω i11
+
Soln: There are 2 nodes: 60 4
12
0: ground, 1: non-reference node Ω
V Ω 0
Step1. Node voltage (v1) is
assigned. Also direction of
branch currents are marked.

Step2. Apply KCL to node 1

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 CONTD.
KCL at Node 1: 7 i2 v 1 i3
i1 + i2 + i3 = 0
Ω i11
+
60 4
12
Ω
Step3. Apply Ohm’s V Ω 0
Law to KCL
equations.
Ohm’s Law to KCL
equation at Node 1: ⇒ v1 − 0 + v1 − 60 + v1 − 0 =

0
12 7 4
⇒ v1 = 18V
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 CONTD.
Now, from v1 we can find the branch currents (i1, i2 and i3)

7Ω i v1 1 i3
v1 − 018 2
i1 = = = 1.5A i1
12 12 +
60 4
v1 − 18 − 12
Ω
i2
60 7
=
60 7
= V Ω 0
= −6A
v1 − 018
i3 = = = 4.5A
4 4
1

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 NUMERICAL
Q3. Calculate Node Voltages in the given
circuit.

2Ω

v1 4 v2

+
V
5 6Ω3Ω 2
A A

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 NUMERICAL
2Ω
Soln: Here 4V voltage source is
connected between Node-1 and Node-2
and it forms a Super-node with a 2Ω v1 4 v2

+
resistor in parallel. V
5A 6Ω 3Ω 2
A

v2 – v1 = 4V (always, whatever may be

the value of resistor in parallel). So 2Ω
resistor is irrelevant for applying KCL to
the super-node.

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 CONTD.
KCL in terms of nodes v1 v2
voltages: 5A i1 i2
v1 − 0 v2 − 0 3Ω
2A
5= + +
6Ω
2⇒ 10 = 2v + v
30 = ν1 + 2ν22 + 12
1
⇒ +12 6= −2v
vν2 + 3 − (1
1
2ν2 =1 18
2 )
v1 + 4 − v2 = 0 1 4V 2

+
⇒ v2 = v1 + 4

+
(2)
v1 v2
On solving (1) and (2) we get,
vν1 = 23.33
V and v2 ν= = 7.33 V
V and
1 2
6V
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 Super Node Analysis
If there is only voltage source between two nodes in the given network then it is difficult to
apply the nodal analysis. Because the voltage source has to be converted into a current source
in terms of the voltage source, write down the nodal equations and relate the node voltages to

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the voltage source. But this is a difficult approach .This difficulty can be avoided by creating
super node which encloses the two nodes that have common voltage source

Swagata Roy Chatterjee

Super Node: A super node is constituted by two adjacent nodes that have a common voltage
source

Find the node voltage V3 and current through the 5 Ω resistor

V1/3+(V1-V2)/2=10 (1)

(V2-V1)/2+V2/1+(V3-10)/5+V3/2=0 (2)

V2-V3=20 (3)

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 NUMERICAL
Q4. Calculate current through the 4Ω resistor in the given circuit.

1 6V 4Ω
3
2A 3Ω + 2
5Ω 7A
0

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 NUMERICAL
1 6V 3
4Ω

+
2 7A
2A 3Ω 5Ω
0

Soln. Here 6V voltage source is connected between Node-1 and

Node-2 and it forms a Super node with a 4Ω resistor in series.
The two node voltages are related as:
v1 – v2 = 6 …(1)

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 CONTD.
KCL for this super-node 1 and
2:
v1 − 0 v2 − v3
+ =2
3 4 1 6V 4Ω
⇒ 4v1 + 3v2 − 3v3 = 24 (2) 3

+
…(2) 2 7A
2A 3Ω 5Ω
And applying KCL at node 3: 0
v3 v3 − v2
+ = −7
5 4
…(3) (3
⇒ 9v3 − 5v2 =
−140 )
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 CONTD.
On solving (1), (2) and (3) we get,
v1 = −2.77 V,v2 = −8.77V and v3 =
−20.43V 6V 4Ω
1
3

+
2 7A
2A 3Ω 5Ω
Now, current
0
through 4Ω resistor
is:
v2 − v3 −8.77 − (−20.42)
= =
2.9125A
4 4 39
 NUMERICAL
Q5. Solve for voltage V across the 3Ω resistor using node voltage
analysis for the given circuit.
1 2v1 2Ω
4A

+
1Ω 5A 3Ω v
+ 4
3 6V

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 NUMERICAL
Soln: Here 6V voltage source is
connected between Node-3 and Node-4,
so node 3 and 4 are constrained to one
1 v1 2Ω 2
another.
4A

+
1Ω 5A 3Ω
v 4

+
To find the number of independent
nodes, the circuit is redrawn as shown
3 6V
below by turning off the sources.

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 CONTD.
There are three nodes, in which two of them
are independent. 1 2
2Ω
However, if we add the two series resistors
(2Ω and 3Ω) then we will have only one 3Ω
1Ω
independent node i.e. node 1 only.
And hence we will have to solve only one
equation. 0
Then using voltage divider rule we can find
the unknown voltage across 3Ω resistor.

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 CONTD.
Applying KCL ate node 1, Node 4 is taken as ground (0
V)
v1 − 6 v1 − 0 1 v1 2Ω 2
4= + +5
15
4A

+
⇒ 6v1 = 25 3Ω v
1Ω 5A
⇒ v1 = 4.1667V 4

+
Now, voltage across 3Ω resistor is
3 3 6V
v = 4.1667 =
× 2 + 2.5V
3

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 NUMERICAL
Q6. Find Is for the circuit shown. Take V0 =
16V.
1 4Ω 2
+

+
Is V 1
V 8Ω V
0
6Ω 14
0

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 NUMERICAL
Soln. Applying KCL at node 1: KCL at node 2
V1 V1 0 gives:
Is + V1 V0 −V1 V0
−V = +
= 6 4
4 4 8
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⇒ Is − (1
V V1 16 −V1
= 12 4 ) ⇒ = +
2 16
1 4Ω 2V
⇒ 44 14= 8
+

+
V 8Ω ⇒ V1 =
8 (2)
Is V1 6Ω =16 V
V0 12V
0 14 5×12
−4=5−4=
Is = 12
1A
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 NUMERICAL
Q7. Solve the following network using the nodal analysis and determine
the current through 2S resistor. (S indicates Siemens i.e. conductance)
-3A
-8A
1 3S 2 2S v3 3
v1 v2
4S -25A
1S 5S
-8A
0

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 NUMERICAL
Soln: Here there are 4 nodes, node 0 is reference node. So we
have to write to 3 KCL equations.
-3A
-8A
1 3S 2 2S v3 3
v1 v2
4S -25A
1S 5S
-8A
0

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 CONTD.
We can write nodal voltage equation in matrix form, as
below v+ (−8)
(4 + 3) −3 −4 (−3)
⎤ ⎡
⎡ ⎡ ⎤
⎥⎤ ⎢ 2 ⎥ ⎢ −(−3) ⎥
1
⎢ −3 (3 + 2 +1) −2 ⎥ ⎢v ⎥ = ⎢ ⎥
⎢ (4 + 2 + 5) ⎢ −(−25) ⎥
−4 −2 ⎥⎦ ⎣ -3A
⎣ v3
-8A⎦
⎢⎣ ⎥⎦ 2
7 −3 −4 v1 −11
1 3S 2S v3 3
⎡⎢ ⎤⎡ ⎤ ⎡ ⎤ v1 v2
⇒ −3 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎢ ⎥⎥
6 −2 v2 = 3
⎢ 11 v3 25
4S
1S
-25A 5S
⎥⎦ ⎢⎣ ⎢⎣ -8A
−4 −2 0
⎢⎣
⎥⎦ ⎥⎦

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 CONTD.
On solving ,we get -3A
-8A
v1 = 1V, v2 = 2V and v3 1 2 2S
3S v3 3
= 3V v1 v2
4S -25A
1S 5S
-8A
0

Putting the value of v2 to find current through 2S resistor we have

v2 − v3
I = 3
2S 2
R = G(v − v ) = 2(2 − 3) =
−2A
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 NUMERICAL
Q8. Determine the voltages at the nodes of the given circuit.
4Ω

1 2 2Ω 8Ω
3
ix
4Ω 2ix
3A
0

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 NUMERICAL
Soln: Here there are 4
nodes, node-0 is reference node. 4Ω

There are 3 non-reference nodes in 1 2Ω 2

8Ω
this network. So we have to write 3
three KCL equations. ix
4Ω 2ix
3A
Assign voltages to the three nodes
0
as shown below and label the
currents.

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 CONTD.
Applying KCL at node 1:
3 = i1 + ix 4Ω
v −v v −v i1
⇒ 3 = 1 3+ 1 2
42 1 2Ω 2 8Ω
v2 v3
⇒ 3v1 − 2v2 − v3 = 12 (1)
ix i3
v1 i2 3
Applying KCL at node 2: 4Ω 2ix
3A
ix = i2 + i3 0
v −v v −v v −
⇒ 1 2 = 2 3+ 2
0
(2)
28 4 42

⇒ −4v1 + 7v2 − v3 = 0
 CONTD.
Applying KCL at node 3: 4Ω
i1 + i2 =
i1
2ix
v −v v −v 1 2Ω 2 8Ω
⇒ 1 3+ 2 3 = v2 v3
2(v1 − v2 )
ix i3
48 2 v1 i2 3
4Ω 2ix
⇒ 2v1 − 3v2 + v3 = 0 3A
(3) 0
On solving (1), (2) and (3) we get,
v1 = 4.8V, v2 = 2.4V and v3 =
−2.4V

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 UNSOLVED NUMERICAL.
Q1.Determine the node voltages and the current ix for the given
circuit having 4 nodes with the following values, R1 = 2Ω, R2 = 1Ω, R3 = 5Ω,
R4 =and I1 = 6A.
6Ω
R4

1 R1 2 R3 3
ix
R2 2ix
I1
0

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Calculate node voltages using super node analysis

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Swagata Roy Chatterjee
Answer:

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Find the current Ix in the circuit using supernode concept

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The presence of a voltage source between
designated nodes 1 and 2 makes the combination

Swagata Roy Chatterjee

of nodes 1 and 2 a supernode.

At node 3

the auxiliary equation is

Answer
:
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 Super Mesh Analysis
If there is only current source between two meshes in the given network then it is difficult
to apply the mesh analysis. Because the current source has to be converted into a voltage

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source in terms of the current source, write down the mesh equations and relate the mesh
currents to the current source. But this is a difficult approach .This difficulty can be
avoided by creating super mesh which encloses the two meshes that have common
current source .

Swagata Roy Chatterjee

Super Mesh: A super mesh is constituted by two adjacent meshes that have a common
current source.

10( I2– I1) + 2I2 + I3 + 5( I3 – I1) =

0…..(1)

10( I1 – I2) + 5( I1 – I3) =

50……….(2)

I2 – I3 = 2
...................(3) 57
Answer
I1 = 19.99 A I2 = 17.33 A and I3 = 15.33
A
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Answer
 SUPERPOSITION THEOREM

APPLICATION

Circuit with more than one energy/power supply units

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 SUPERPOSITION THEOREM

APPLICATION

System with more than one energy sources

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 SUPERPOSITION
PRINCIPLE
• Helps us to analyze a linear circuit with more than
one independent source.
• It is used to determine the value of some circuit
variable (voltage across or current through a particular
impedence)
• It is applied by calculating the contribution of each
independent source separately.
• The output of a circuit is determined by summing
the individual responses of each independent source.
• The idea of superposition rests on the linearity
property
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(specifically, additivity)
 SUPERPOSITION
LINEARITY – ADDITIVE PROPERTY
• The response to a sum of inputs is the sum of the responses to
each input applied separately.

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 SUPERPOSITION
LINEARITY – ADDITIVE PROPERTY
• Example: Ohm’s Law
Voltage (output) developed across a resistor in response
to applied current (input)
v1 = i1R (for applied current i1)
and
v2 = i2 R (for applied current i2)
Then applying current (i1 + i2) gives
v = (i1 + i2 )R = i1R + i2 R = v1 + v2

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 SUPERPOSITION

STATEMENT
•In any linear bilateral network containing two
or more independent sources (voltage and/or
current sources), the resultant current / voltage
in any branch is the algebraic sum of currents /
voltages caused by each independent source
(with all other independent sources turned off).

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 SUPERPOSITION

Superposition theorem for two

independent sources
(either voltage or
current).
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 SUPERPOSITION

THINGS TO KEEP IN MIND WHILE APPLYING SUPERPOSITION:

•To turn off a voltage source: Replace by its internal
resistance (for non-ideal source) or short circuit (for ideal
source).
•To turn off a current source: Replace by its internal
resistance (for non-ideal source) or open circuit (for ideal
source).
•The dependent sources should not be zeroed. They remain the
same for every particular solution with each independent source.

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 SUPERPOSITION
THINGS TO KEEP IN MIND WHILE APPLYING SUPERPOSITION:

• To turn off a voltage source,

• To turn off a current

source,

67
 SUPERPOSITION
STEPS TO APPLY
• Step-1: Retain one source at a time in the circuit and replace all
other sources with their internal resistances.
• Step-2: Determine the output (current or voltage) due to the
single source acting alone using any circuit analysis techniques
(mesh, node, transformations etc.).
• Step-3: Repeat steps 1 and 2 for every independent source.
• Step-4: Find the total contribution by adding algebraically all the
contributions due to all the independent sources.

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Q)Calculate Iab and Vcg using Superposition theorem

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Voltage Source Only (retain one source at a time)

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Current source only (retain one source at a time)

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Current through 3 ohm

Current through 1 ohm

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 SUPERPOSITION

Example 2: Calculate the current Iab flowingthrough the 3Ω

resistor using the superposition theorem.

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 SUPERPOSITION
Solution:
Step 1: Consider only the 3A current source on the
left
2 6
I1(dueto3
= 3×
source)
Acurrent= A(atob)
7 7

Step 2: Consider only the 2V voltage source on the left

I = 0A(a tob)
2(dueto 2V votagesource)

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 SUPERPOSITION
Step 3: Consider only the 1V voltage source
1
= − votagesource)
I 3( dueto1V A(atob) 7

Step 4: Consider only the 3A current source on the

right
2 6
I 4(dueto3
= 3× =
Acurrent source) A(atob) 7 7

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 SUPERPOSITION

Step 5: Consider only the 2V voltage source

on the right
2
= − votagesource)
I 5( dueto1V A(atob) 7

Step 6: Resultant current Iab flowingthrough3Ω resistor due to

the combination of all sources
I =I +I +I
ab 1( dueto 3 Avotage source ) 2( dueto 2V votage source ) 3(dueto1V votage source)

+I 4( dueto 3 Avotage source )

+I
5( dueto 2V votage source )

7 6 7 71 76 27 9
= +0− + − = = 1.285 A(a to b)

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 SUPERPOSITION
LIMITATIONS
1. It doesn’t work for calculation of power.
• Power calculation is not linear operation.

Example:
• When current i1 flows through resistor R, the power is P1 = Ri12,
When current i2 flows through R, the power is P2 = Ri22

• If current i1 + i2 flows through R, the power absorbed is

P3 = R (i1 + i2)2 = Ri12 + Ri2 2 + 2Ri1i2 ≠ P1 + P2.

77
 SUPERPOSITION
LIMITATIONS
EXAMPLE :

Using superposition power consumed bythe individual sources are

P P = 12watts; = 12watts;
W 1(dueto12V sourceleft ) W 2(dueto12V sourceleft )
Total power consumed = 24watts

But current flowingthroughthe resistor is zero, so total

power consumed is actually zero!

78
 SUPERPOSITION

LIMITATIONS
2. Superposition theorem cannot be applied for circuits with
nonlinear elements (eg. Diodes and
Transistors).
3. In order to calculate load current or voltage for several choices of load
resistance, one needs to solve for every voltage and current source in
the network several times. With a simple circuit this is fairly easy, but
in a large circuit with many sources this method becomes a painful
experience! Thevenin/Norton equivalent with Mesh/Node analysis is
a better choice in that case.

79
 THEVENIN’S THEOREM

Thevenin’s Theorem Application

• It often occurs in practice that a particular element in a circuit is
variable (usually called the load) while other elements are fixed.
• As a typical example, a household outlet terminal may be connected
to different appliances constituting a variable load.
• Each time the variable element is changed, the entire circuit has to be
analyzed all over again.
• To avoid this problem, Thevenin’s theorem provides a technique by
which the entire fixed part of the circuit is replaced by a very simple
equivalent of a voltage source in series with an impedence.

80
 INTRODUCTION

Thevenin’s Theorem
R2 Vth= Thevenin’s equivalent voltage
voltage R5 Rth= Thevenin’s equivalent resistance
A
R1

V1 I1 R3 R4 R6 Load Terminals
+-

+-
B
V2 Rth
A

Thevenin equivalent circuit

+ -
V Load Terminals
th

81
Thevenin’s Theorem application areas

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 THEVENIN’S THEOREM
Statement: A linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source VTh in series with a
resistor RTh, where VTh is the open-circuit voltage at the terminals and
R is the input or equivalent resistance at the terminals
Th
when the
independent sources are turned off.

Linear Circuit: A linear circuit is one whose output is linearly related

(or directly proportional) to its input i.e. containing only linear
elements eg. R, L, C, transformer

83
 THEVENIN’S THEOREM

How to find Thevenin equivalent voltage VTh and resistance RTh ?

• The two circuits given below are said to be equivalent if they have
the same voltage-current relation at their terminals.

(a) Original Circuit (b) Thevenin equivalent circuit

84
 THEVENIN’S THEOREM

• If the terminals a-b are made open-circuited (by removing the load), no
current flows, then the open circuit voltage across the terminals a-b is
equal to the voltage source VTh.
• Thus, VTh is the open-circuit voltage across the terminal i.e. VTh = voc
• The input resistance (or equivalent resistance) of the dead circuit (all
independent sources turned off) at the terminals a-b in the Fig (a) must
be equal to RTh in Fig. (b) (input resistance with VTh turned off).
• Thus, Ris the input resistance at the terminals when the
Th
independent sources are turned off, i.e. RTh = Rin

85
 THEVENIN’S THEOREM

Finding the Thevenin equivalent resistance RTh:

Case 1: When the networks has no Dependent Sources

• Turn off all the independent sources, then RTh is the input
resistance of the network looking between terminals a and b.

Case 2: When the networks has Dependent Sources

• Turn off all independent sources, apply voltage source vo across
terminals a-b and determine the resulting current io. Then RTh = vo ⁄ io.
• Alternatively, insert a current source io across terminals a-b and find
the terminal voltage vo. Again RTh = vo ⁄ io.

86
 THEVENIN’S THEOREM

Steps to determine Thevenin’sEquivalent Resistance (RTH)

and Voltage (VTH):

•Remove load resistor RL or any component connected across the

terminals a-b through which Thevenin equivalent is required.
•DeterminebyR shorting all voltage sources and open-
TH
circuiting all current sources, and then calculating the circuit’s total
resistance as seen from the open terminals a-b.
•DeterminebyV calculating the voltage between open
TH
terminals a-b (by usual circuit analysis methods).

87
 Thevenin’s theorem
RTh with all sources (voltage and current) removed and
replaced by their internal resistances (fig. 3) and the Vth is

8/30/2024
equal to the open circuit voltage across the A & B terminals.

Swagata Roy Chatterjee

88
Example: To find Thevenin equivalent between terminals ‘a’ and ‘b’
Case 1: No dependent source

89
 Step1: Remove all the independent sources.
a) Replace voltage source by short circuit
b) Replace current source by open circuit
20Ω

a 10Ω 20Ω
b

10Ω 10Ω 10Ω

90
 Step2: Finding RTh
20Ω

a 10Ω 20Ω a 10Ω 10Ω

b b

10Ω 10Ω 10Ω 10Ω 10Ω 10Ω

91
 Step2: Finding RTh

a 10Ω 10Ω a 30
b Ω b

30 30
10Ω 10Ω 10Ω Ω Ω 10
10 Ω
Ω

Star to Delta transformation

92
 Step2: Finding RTh
a 30 a 30Ω b
Ω b

30 30 7.5Ω
10 7.5Ω
10 Ω Ω
Ω
Ω

a 30Ω b

RTh = 30||15 = 10Ω 15Ω

93
 THEVENIN’S THEOREM
Step3: Finding VTh

a 10Ω 10Ω
b
10V
10Ω 10Ω 10Ω
30V i1 i2
50V

Source Transformations

94
Step3: Finding VTh
a 10Ω 10Ω
For loop 1: b
−30 + 50 + 30i1 −10i2 = 10V
10Ω 10Ω
⇒
0 −2 = 3i − i
1 2
(1) 10Ω

For loop 2: 30V i1 i2

50V
−50 −10 + 30i2 −10i1 =
⇒
0 6 = −i + 3i
1 2
(2)
On solving (1) and (2), i1 = 0, i2 = 2A

95
Step3: Finding VTh

Applying KVL to the output loop,

−vab −10i1 + 30 −10i2= 0
⇒ vab = 10 V

Therefore,
VTh = vab = Thevenin Equivalent Circuit
10 V

96
Example: To find Thevenin equivalent between terminal ‘a’ and ‘b’
Case 2: With dependent source
Turn off all
independent sources,
apply voltage source
v0 at terminals a and b
and determine the
resulting current i0.
Then RTh = v0 ∕i0.

97
Step 1. Remove dependent sources
5A current source is replaced by open circuit and set v0 = 1V.

98
Step 2. Find RTh

KVL for loop 1:

But,

Therefore,

99
Mesh analysis for loop 2 and 3,

On solving we get, i0 = −i3 = 1/6 A.

Therefore,

10
0
Step 3. Find VTh
We have to find voc for this circuit.
Applying mesh analysis we get,

and

10
1
On solving (1),(2),(3) and (4), we get, i2 = 10/3 =>

The Thevenin equivalent circuit is shown below.

10
2
Find Thevenin voltage and resistance across load resistance(a-b)

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Swagata Roy Chatterjee
103
8/30/2024 Swagata Roy Chatterjee
104
KCL at node
C
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105
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106
 Maximum Power Transfer Theorem
In an electric circuit, the load receives electric energy via the supply sources and converts that

8/30/2024
energy into a useful form. The maximum allowable power receives by the load is always
limited either by the heating effect (incase of resistive load) or by the other power conversion
taking place in the load. The Thevenin and Norton models imply that the internal circuits

Swagata Roy Chatterjee

within the source will necessarily dissipate some of power generated by the source.

A logical question will arise in mind, how much power can be transferred to the load
from the source under the most practical conditions? In other words, what is the value of
load resistance that will absorbs the maximum power from the source?

107
 maximum power dissipated to

8/30/2024
the load resistance is given by

Swagata Roy Chatterjee

This result is known as
“Matching the load” or
maximum power transfer
occurs when the load
resistance RL matches the
Thevenin’s resistance of a
given systems.
108
 8/30/2024 Swagata Roy Chatterjee
109
Efficiency
 Norton’s Theorem
Norton’s theorem states that any two terminals A & B of a network composed of
linear resistances and independent sources (voltage or current, combination of voltage

8/30/2024
and current sources) may be replaced by an equivalent current source and a parallel
resistance. The magnitude of current source is the current measured in the short
circuit placed across the terminal pair A & B . The parallel resistance is the equivalent

Swagata Roy Chatterjee

resistance looking into the terminal pair A & B with all independent sources has been
replaced by their internal resistances. Norton’s theorem is a dual of Thevenin’s theorem.

110
8/30/2024 Swagata Roy Chatterjee
111
8/30/2024 Swagata Roy Chatterjee
112
8/30/2024 Swagata Roy Chatterjee
113
Q) Find the value of RL such that the power transferred to RL is maximum

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Swagata Roy Chatterjee
114
Q) For the circuit shown in fig find the value of RL that absorbs maximum power from the
circuit and the corresponding power under this condition.

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Swagata Roy Chatterjee
115
applying ‘Super position theorem

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Swagata Roy Chatterjee
the current through ‘b-c’ branch 20/20=1A

116
Vab = − 10
volt
Considering only 10v source only

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Swagata Roy Chatterjee
No current is flowing through ‘cb’-branch. ∴ Vab = 5v (‘a’ is higher
potential than ‘b’)
117
 Consider only 2 A current source only

8/30/2024
Swagata Roy Chatterjee
Note that the current flowing the ‘c-a’ branch is zero ∴ Vab =10 v (‘a’ is higher potential
than ‘b’ point).

The voltage across the ‘a’ and ‘b’ terminals due to the all sources = VTh = Vab (due to
20v) + Vab (due to 10v) + Vab (due to 2A source) = - 10 + 5 + 10 = 5v (a is higher
potential than the point ‘b’). 118
 8/30/2024
Swagata Roy Chatterjee
Under this condition, the maximum power
Thevenin equivalent circuit
dissipated to RL is

119
Q) Applying Norton’s theorem, calculate the value of R that results in maximum power
transfer to the 6.2 Ω resistor in fig.. Find the maximum power dissipated by the resistor
6.2Ω under that situation.

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Swagata Roy Chatterjee
120
 Complex Numbers using the Complex or
s-plane:

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A complex number is represented by a real part and an imaginary part that takes the
generalized form of:
Z=a+jb

Swagata Roy Chatterjee

Where Z - is the Complex Number representing the Vector
a - is the Real part or the Active component
b - is the Imaginary part or the Reactive component
j - is defined by √-1

for example, Z = 6 +
j4

z = a + bj
z* = a - bj
121
COMPLEX NUMBERS FOR AC CIRCUITS
The magnitude of z

Magn z = z = z* z= a - bj
( ) ( ) (
a + bj
)( )
y
z = a2 + abj - abj - j 2b2
b -1 b
tanθ = 2 ;θ =
z = a a+ b 2 tan
( )
a z
(a,b)

z = z cosθ + jsinθ = z e jθ
(
The exponential )
representation of a
θ
complex number will prove useful in
x
solving the RLC differenial eqn.

122
 Complex Numbers using Polar Form
Polar Form of a complex number is written in terms of its magnitude and angle. Thus, a polar
form vector is presented as: Z = A ∠±θ, where: Z is the complex number in polar form, A is

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the magnitude or modulo of the vector and θ is its angle or argument of A which can be either
positive or negative. The magnitude and angle of the point still remains the same as for the
rectangular form above, this time in polar form the location of the point is represented in a

Swagata Roy Chatterjee

“triangular form” as shown below.

123
COMPLEX NUMBERS FOR AC CIRCUITS

j= -1
j 2 = jj = -1 -1 = -1

j3 = jj 2 = j -1 = -j
( )
j4 = j 2 j 2 = -1 -1 =
( )( )
+1 j5 = jj4 = j +1 =
( )
j 124
 Rotation of complex numbery
z =z cosθ + jsinθ = z e jθ
( )
e jθ can be viewed as a rotation operator in a complex space

8/30/2024
Swagata Roy Chatterjee
Complex numbers simplify the solution of the integral- differential equations
in series RLC AC circuits.The use of complex numbers simplifies the lead-lag 125
nature of the voltage and current in AC circuits.
 RLC CIRCUIT SOLUTION

These are complex

variables

The quantity Z is called the impedance

and it is a complex variable
E = I Z is a complex version of
Ohm’s Law

126
 COMPLEX IMPEDANCE

1
-X ωLωC
-
XL =
tanθ =
R R

127
R= 250Ω, L = 1.20mH,
C = 1.80µF, Vp = 120v, f = 60Hz

Determine the following:

(d.) θ - Phase angle

θ =-80.4

128
8/30/2024 Swagata Roy Chatterjee
129
 R= 250Ω, L = 1.20MH,
C = 1.80µF, Vp = 120v, f = 60Hz
Determine the following:

8/30/2024
(a.) XL - Inductive reactance
(b.) XC - Capacitive reactance

Swagata Roy Chatterjee

(c.) Z - Impedance

First calc: ω = 2πf = 2(3.14)60 = 377

rad/s XL = ωL= 377(1.20x10-3) = 0.452Ω

XC = 1/ωC= 1/((377)(1.80x10-6)) = 1474Ω

Z= R2 +(X - X )2 = 2502 +(0.452 - 1474)2

L C

Z = 1495 Ω 130
 Single-phase series a.c. circuits
Purely resistive a.c. circuit Purely inductive a.c.
I circuit

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In a purely inductive a.c. circuit, the current IL
lags the applied voltage VL by 90° . In a purely
inductive circuit the opposition to the flow of

Swagata Roy Chatterjee

alternating current is called the inductive
reactance, XL

131
 Purely capacitive a.c. circuit
In a purely capacitive a.c. circuit, the current IC leads the applied voltage VC by 90°. In a
purely capacitive circuit the opposition to the flow of alternating current is called the

8/30/2024
capacitive reactance, XC

Swagata Roy Chatterjee

where C is the
capacitance in farads.

132
 R–L series a.c. circuit
In an a.c. circuit containing inductance L and resistance R, the applied voltage V is the
phasor sum of VR and VL , and thus the current I lags the applied voltage V by an angle

8/30/2024
lying between 0° and 90° (depending on the values of VR and VL). In any a.c. series
circuit the current is common to each component and is thus taken as the reference
phasor.

Swagata Roy Chatterjee

133
 R–C series a.c. circuit
In an a.c. series circuit containing capacitance C and resistance R, the applied voltage V is
the phasor sum of VR and VC and thus the current I leads the applied voltage V by an angle

8/30/2024
lying between 0° and 90° (depending on the values of VR and VC), shown as angle ˛ From
the phasor diagram , the ‘voltage triangle’ is derived.

Swagata Roy Chatterjee

134
 R–L–C series a.c. circuit
In an a.c. series circuit containing resistance R, inductance L and capacitance C, the applied
voltage V is the phasor sum of VR, VL and VC (see Figure 15.12). VL and VC are anti-phase,

8/30/2024
i.e. displaced by 180°, and there are three phasor diagrams possible — each depending on the
relative values of VL and VC

Swagata Roy Chatterjee

135
 8/30/2024 Swagata Roy Chatterjee
136
Series resonance
 Q-factor and Bandwidth
At resonance, if R is small compared with XL and XC, it is possible for VL and VC to have
voltages many times greater than the supply voltage the circuit is I2R. At I D 0.707Ir, the
power is 0.707Ir 2R D 0.5I2 rR, i.e., half the power that occurs at frequency fr. The points

8/30/2024
corresponding to f1 and f2 are called the half-power points. The distance between these
points, i.e. (f2 f1), is called the bandwidth.

Swagata Roy Chatterjee

Selectivity is the ability of a circuit to respond more readily to signals of a particular frequency
to which it is tuned than to signals of other frequencies. The response becomes progressively
weaker as the frequency departs from the resonant frequency. The higher the Q-factor, the
narrower the bandwidth and the more selective is the circuit. Circuits having high Qfactors
137
(say, in the order of 100 to 300) are therefore useful in communications engineering. A high
Q-factor in a series power circuit has disadvantages in that it can lead to dangerously high
voltages across the insulation and may result in electrical breakdown.
 Power in a.c. circuits
The value of power at any instant is given by the product of the voltage and current at that
instant, i.e. the instantaneous power, p=VI, (V and I being rms values). as shown by the
broken lines in the figures.

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(a) For a purely resistive a.c. circuit, the (b)For a purely inductive a.c. circuit, the
average power dissipated, P, is given by: P = average power is zero.
VI = I2R = V2/R watts

Swagata Roy Chatterjee

(c) For a purely capacitive a.c. circuit, the average power is zero.

138
8/30/2024 Swagata Roy Chatterjee
139
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140
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141
Using the superposition theorem, find the current I
through the 4Ω reactance

8/30/2024
Swagata Roy Chatterjee
Considering the effects of the voltage source E1

142
Considering the effects of the voltage source E2

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Swagata Roy Chatterjee
143
Determine the voltage V3 using superposition theorem.

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Swagata Roy Chatterjee
For the dc analysis, the capacitor can be replaced by an open circuit equivalent, and the
inductor by a short-circuit equivalent.

144
8/30/2024 Swagata Roy Chatterjee
145
Find the Thévenin equivalent circuit for the network external to branch a-a

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Swagata Roy Chatterjee
146
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147
 A.C voltage generator

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The voltage induced by a single coil when rotated in a uniform
magnetic field

Swagata Roy Chatterjee

148
 Three-phase supply

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A three-phase supply is generated when three
coils are placed 120° apart and the whole rotated
in a uniform magnetic field

Swagata Roy Chatterjee

149
 8/30/2024
Swagata Roy Chatterjee
Star Connection Delta Connection

150
STAR CONNECTIONS IS GENERALLY USED IN LONG DISTANCE TRANSMISSION
LINES AS INSULATION REQUIREMENT IS LESS IN STAR CONNECTION.

Transmission Network
Delta connections are generally used in distribution networks for short
distances.

Distribution Network
Alternators and generators are usually star connected.

600 MW Turbo-Generator at power

plant
A Three Phase Transformer with Name
Plate
GENERATING TRANSFORMER NEAR TO POWER PLANT GENERATOR ARE
CONNECTED IN STAR CONNECTION TO PROVIDE GROUNDING PROTECTION.

Generating
Transformer
AC MOTORS WINDING ARE CONNECTED IN STAR/DELTA CONNECTION DEPENDING
ON REQUIREMENT AND APPLICATION.

Three Phase Induction Motor

Winding
STAR AND DELTA CONNECTIONS ARE USED IN STARTING OF THREE PHASE
INDUCTION MOTORS USING STAR-DELTA STARTER.

Star Delta Starter for Three phase Induction Motor

Delta-Star Starters are installed in cement industries for high inertial load
applications.

Cement Industry
POWER CAPACITORS IN 3 PHASE CAPACITOR BANK CONNECTIONS ARE EITHER
DELTA CONNECTED OR STAR (WYE) CONNECTED.

Delta connected capacitor bank

THE APPLICATION OF SUCH CONNECTION IS ALSO USED IN HIGH VOLTAGE
DIRECT CURRENT (HVDC) SYSTEMS.

Generating
Transformer
The application of such connection is also used in Wheatstone bridge
resistance measurement device.

Wheatstone Bridge
STAR/DELTA TRANSFORMATIONS AND EQUIVALENT CIRCUIT CALCULATIONS
HELP IN SIMPLIFICATION AND UNDERSTANDING OF COMPLEX ELECTRICAL
CIRCUITS.

Complex Electric Circuits

 Delta-Wye
Conversion
We calculate the equivalent resistance between terminals a and c while terminal b is open
in both cases:

8/30/2024
Swagata Roy Chatterjee
Solving for R1 , R2 , and R3 we
have:

163
Each resistor in the Y network is the product if the resistors in the two adjacent Delta
branches, divided by the sum of the three Delta resistors.
 Wye-Delta Conversion
From the previous slide

8/30/2024
Swagata Roy Chatterjee
Dividing this equation by each of the previous slide
equations

Each resistor in the Delta network is the sum of all the possible products of Y resistors taken
two at a time, divided by the opposite Y resistor 164
8/30/2024 Swagata Roy Chatterjee
165
Q. Convert the Y network to an equivalent Δ
network.

166
Soln:

7.5× 5
Ra = 7.5 + 5 + 3 = 25
ohms
7.5× 3
Rb = 7.5 + 3 + 5 = 15
ohms
5× 3
Rc = 5 + 3 + 7.5 = 10 ohms
Q. Convert the Δ network to an equivalent Y
network.

168
Soln:
Rb 10×
R1 = =5
R 25
Ra + cRb + 15 +10 + ohms
= Rc 25
R2 25×15
R R
Ra + cRba + = 50 = 7.5
=
Rc ohms
R3 15×10
R
Ra + R R
a b+b Rc
= 50 =
=
3ohms
EXERCISE/NUMERICAL
ANALYSIS
Q. Calculate equivalent resistance across terminals A and
B.

170
 EXERCISE/NUMERICAL ANALYSIS
Soln: Converting inner STAR (3 ohms, 3 ohms and 1 ohms ) into
Delta.
3× 3
R = 3 + 3 +1
1 = 15
ohms
3×1
R2 = 3 +1+ 3 = 5
ohms
1× 3
R = 1+ 3 +3
3 =5
ohms
 EXERCISE/NUMERICAL
ANALYSIS
Circuit now becomes,

172
 EXERCISE/NUMERICAL ANALYSIS

Delta-connected resistances 1 Ω, 5 Ω and 8 are converted in star,

'
1× 4
R1 = ohms
8
1+ 5 +
= 7
8
R2' =
5 + 8= 5
1+5×1
14 ohms
8×
R3' 20
= 1+5 5 + 8 7 ohms
=
 EXERCISE/NUMERICAL ANALYSIS

Circuit now becomes,

4 ⎡⎛ ⎞ || 20 20 ⎞⎤
R AB = + + 2.5 ⎟ ⎜ + + 7.6 = 10 ohms
5
⎝ ⎠ 9 ⎠
⎣⎢⎜ ⎛ ⎦
7 14 7 ⎟
⎝ ⎥
 EXERCISE/NUMERICAL
ANALYSIS
Q. Using delta/star transformation, find equivalent resistance across AC.

175
 EXERCISE/NUMERICAL ANALYSIS
Soln: Delta can be replaced by equivalent star-connected resistances,

R 10×
R1 AB = = 2.86 ohms
RA + RDA+ RBD 1020+ 40 +
R
=
B DA 20
R2 R = = 5.72
AB
RA + RDA+ RBD 10× 40
10 + 40 + ohms
= R
B BD 20
R3 R = = 11.4
DA
= RA + RDA+ RBD 1010× 40+
+ 40 ohms
R
B BD 20
 EXERCISE/NUMERICAL ANALYSIS
FIGURE NOW BECOMES,

RAC = 2.86 + (30 + 5.72) × (15 +11.4)

= 18.04
30 + 5.72 + 15
ohms ( ) (
+11.4
)
 EXERCISE/NUMERICAL
ANALYSIS
Q. Calculate equivalent resistance across terminals A and
B.

178
 EXERCISE/NUMERICAL
ANALYSIS
Soln: Replacing inner STAR into
DELTA.

179
 EXERCISE/NUMERICAL
ANALYSIS
15.8 ohm is in parallel with 5 ohm and 26.3 ohm is in parallel with
4 ohm, circuit becomes

180
EXERCISE/NUMERICAL
ANALYSIS
Converting upper delta into star,

181
EXERCISE/NUMERICAL ANALYSIS
Now equivalent resistance can be calculated as,

Req = (3·8 + 2·98) || (1·99 +

3·5) + 1·2
= 4.23 ohms
 EXERCISE/NUMERICAL
ANALYSIS
Q. Obtain the equivalent resistance Rab for the circuit and use it to
find current i.

183
 EXERCISE/NUMERICAL
Soln:
ANALYSIS
In this circuit, there are two Y networks and three Δ
networks. Transforming just one of these will simplify the circuit.

184
 EXERCISE/NUMERICAL ANALYSIS
We convert the Y-network comprising the 5-Ω, 10-Ω, and 20-Ω resistors
into delta.
5×10
R1 = 5 +10 = 17.5
20 ohms
+ (comes in parallel with 12.5
Ω)

5× 20
R2 = 5 + 20 +10 = 35
(comes in parallel with 15
ohms
Ω)
10 × 20
R3 = 10 + 20 + 5 = 70
ohms
(comes in parallel with 30
Ω)
 EXERCISE/NUMERICAL ANALYSIS

Combining the three pairs of resistors in parallel, we obtain.

(12.5 || 17.5 Ω) R = (7.292 +10.5)||21 =

ab

17.792 × 21
17.792 + 21 =9.632
(30 || 70 Ω) ohms
(15 || 35 Ω)
v
i = Rs = 9.63 =
ab
120 12.458A
2
 EXERCISE/NUMERICAL
ANALYSIS
Q. Determine the load current in branch EF in the circuit shown.

187
 EXERCISE/NUMERICAL ANALYSIS
Sol. ACGA forms delta, Converting it to equivalent
star.

200 × 500 500 × 200 200 × 200

RA = 90 = RG = 90 = RC = 90 =
N N N
111.11ohms
0 111.11ohms
0 44.44ohms
0
EXERCISE/NUMERICAL ANALYSIS
Circuit can be redrawn as

R = 111.11+ 600 =
NE
R 711.11ohms
F ND = 600 + 44.44 =
644.44ohms
 EXERCISE/NUMERICAL ANALYSIS
Branches NCD and NEF are in parallel, 711.11 || 644.44=338
ohms.

V 100
Total current I in the circuit = I = = 0.222
R 111.11+ 338 A
eq
=
 EXERCISE/NUMERICAL ANALYSIS
To obtain current in branch EF, we apply current division
formula.
I NEF = I × R
NCD
RNCD+ R NEF

644.44
= 0.222
× 711.11+ 644.44

= 0.1055A
 EXERCISE
Q. A square and its diagonals are made of a uniform covered wire. The
resistance of each side is 1 Ω and that of each diagonal is 1·414 Ω.
Determine the resistance between two opposite corners of the square.

192
 EXERCISE
Q. Determine the resistance between the terminals A and B of the
network.

193
 EXERCISE
Q. Find the current in 10 Ω resistor in the network shown by
star-delta transformation.

194
 EXERCISE
Q. Using star/delta transformation, determine the value of R for
the network shown such that 4Ω resistor consumes the maximum
power.

195
Star connection (three phase supply)

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Swagata Roy Chatterjee
196
A star-connected load is shown in Figure where the
three line conductors are each connected to a load
and the outlets from the loads are joined together at

8/30/2024
N to form what is termed the neutral point or the
star point.
(ii) The voltages, VR, VY and VB are called phase

Swagata Roy Chatterjee

voltages or line to neutral voltages. Phase voltages
are generally denoted by Vp
(iii) The voltages, VRY, VYB and VBR are called line
voltages
(iv) From Figure, it can be seen that the phase
currents (generally denoted by Ip) are equal to their
respective line currents IR, IY and IB, i.e. for a star
connection: I = Ip
L
The line voltage, VRY, is given by VRY = VR − VY By
trigonometry, or by measurement, VRY = √3VR, i.e. for a
balanced star connection:
VL = √3 Vp 197
Q) A balanced, three-wire, star-connected, 3-phase load has a phase voltage of 240 V, a line
current of 5 A and a lagging power factor of 0.966. Draw the complete phasor diagram.

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Swagata Roy Chatterjee
198
 Delta connection

(i)A delta (or mesh) connected load is

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shown in Figure where the end of one
load is connected to the start of the
next load.

Swagata Roy Chatterjee

(ii) From Figure it can be seen that
the line voltages VRY, VYB and VBR
are the respective phase voltages, i.e.
for a delta connection: VL = Vp

(ii)Using Kirchhoff’s current law , IR

= IRY - IBR. From the phasor diagram
, by trigonometry or by measurement,
IR = √3IRY, i.e. for a delta connection:
IL = √3Ip

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 Power in three-phase systems

The power dissipated in a three-phase load is given by the sum of the power dissipated in
each phase. If a load is balanced then the total power P is given by: P = 3 p power

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consumed by one phase.

The power consumed in one phase p=VpIp cosφ (where φ is the phase angle between

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Vp and Ip) For a star connection, Vp = VL/ √3 and Ip = IL hence total power
P = 3( VL /√3) IL

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 Measurement of power in three-phase systems

Power in three-phase loads may be measured by

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the following methods:
(i) One-wattmeter method for a balanced load
Wattmeter connections for star topology

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shown in the figure
Total power = 3 × wattmeter reading
(i) Two-wattmeter method for balanced or
unbalanced loads
A connection diagram for this method is
shown in Figure for a star-connected load.
(i) Total power = sum of wattmeter readings =
P1 + P2 The power factor may be
determined from:
tanφ = √3 (P1 − P2/ P1 + P2 )

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 Comparison of star and delta connections

Loads connected in delta dissipate three times more power than when connected in star to
the same supply.

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(ii) For the same power, the phase currents must be the same for both delta and star
connections (since power=3Ip2Rp), hence the line current in the delta-connected system is
greater than the line current in the corresponding star-connected system. To achieve the

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same phase current in a star-connected system as in a delta-connected system, the line
voltage in the star system is √3 times the line voltage in the delta system. Thus for a given
power transfer, a delta system is associated with larger line currents (and thus larger
conductor cross-sectional area) and a star system is associated with a larger line voltage
(and thus greater insulation).
Advantages of three-phase systems

Advantages of three-phase systems over single-phase supplies include: (i) For a

given amount of power transmitted through a system, the three-phase system requires
conductors with a smaller crosssectional area. This means a saving of copper (or
aluminium) and thus the original installation costs are less. (ii) Two voltages are
available (iii) Three-phase motors are very robust, relatively cheap, generally
smaller, have self-starting properties, provide a steadier output and require little
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maintenance compared with single-phase motors.
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Q)
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Phasor Diagram
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