Chapter 3 - Methods of Analysissesd.pptx

Methods of Analysis Debretabor University 1
Methods of Analysis
By:-
Tenenet A. (MSc.)
Electromechanical Engineering Department
2018 E.C.

Outline
• Introduction
• Nodal analysis
• Nodal analysis with voltage source
• Mesh analysis
• Mesh analysis with current source
• Nodal and mesh analyses by inspection
• Nodal versus mesh analysis

 Introduction
• Fundamental laws of circuit theory includes (Ohm’s law and Kirchhoff’s laws),
• By applying these laws we develop two powerful techniques for circuit analysis:
 Nodal analysis: which is based on a systematic application of Kirchhoff’s current law (KCL),
 Mesh analysis: which is based on a systematic application of Kirchhoff’s voltage law (KVL).
 Based on those techniques obtaining a set of simultaneous equations that are solved to obtain the
required values of current or voltage.
 The methods of solving simultaneous equations involves;
o Elimination
o Cramer’s rule

• It provides a general procedure for analyzing circuits using node voltages as the circuit
variables.
• Choosing node voltages instead of element voltages as circuit variables is convenient and
reduces the number of equations one must solve simultaneously.
• In this section the circuits do not contain voltage sources.
• Nodal analysis is also known as the node-voltage method.
• In nodal analysis, we are interested in finding the node voltages.
• Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit
involves taking the following three steps.
Nodal Analysis

Cont …
• Steps to Determine Node Voltages:
1. Select a node as the reference node. Assign voltage v1, v2, …vn-1 to the remaining
n-1 nodes. The voltages are referenced with respect to the reference node.
2. Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express
the branch currents in terms of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknown node
voltages.
Note: The number of non-reference nodes is equal to the number of independent
equations that we will derive.

Debretabor University
Methods of Analysis
6
The reference node is commonly called the ground since it is assumed to have zero
potential.
Common symbols for indicating a reference node,
(a) common ground, (b) ground, (c) chassis.

Figure: Typical circuit for nodal analysis
Methods of Analysis 7

 Current flows from a higher potential to a lower potential in a resistor
R
v
v
i lower
higher 

2
3
3
3
2
3
2
1
2
2
2
2
1
2
1
1
1
1
1
1
or
0
)
(
or
or
0
v
G
i
R
v
i
v
v
G
i
R
v
v
i
v
G
i
R
v
i










At node 1, applying KCL gives At node 2,
Based on Ohm’s Law

Cont …
substitute
3
2
2
2
1
2
2
2
1
1
1
2
1
R
v
R
v
v
I
R
v
v
R
v
I
I








2
3
2
1
2
2
2
1
2
1
1
2
1
)
(
)
(
v
G
v
v
G
I
v
v
G
v
G
I
I














 


















2
2
1
2
1
3
2
2
2
2
1
I
I
I
v
v
G
G
G
G
G
G

Example 1
• Calculus the node voltage in the circuit shown in Fig. (a)

Solution:
 At node 1: KCL
2
0
4
5 1
2
1
3
2
1







v
v
v
i
i
i
6
0
4
5 2
1
2
5
1
4
2








v
v
v
i
i
i
i
 At Node 2: KCL
 The assignment of labels (V1, V2, V3, etc.) to the non-reference nodes is entirely arbitrary.
 These labels are just variables used to set up the system of equations.

Cont …
• Method 2: Using the elimination technique
• Substituting v2 = 20 V
•

Cont .
Method 2: Use Cramer’s rule
• In matrix form:

Practice Problem: Obtain the node voltages in the circuit in Fig.
Answer: v1 = -2 V, v2 = -14 V.

Example 2: Determine the voltage at the nodes in Fig. (a)

Solution:
• At node 1: KCL
• At node 2: KCL
• At node 3: KCL

Method 2:To use Cramer’s rule in matrix form.

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛=
[
5 −2 − 3
4 5 0
3 − 6 − 4 ][
𝑣1
𝑣2
𝑣3
]=
[
6 0
0
0 ]

Nodal Analysis with Voltage Sources
• Case 1: The voltage source is connected between a non-reference node & the reference node:
• The non-reference node voltage is equal to the magnitude of voltage source and the number
of unknown non-reference nodes is reduced by one. For example; V1= 10V
• Case 2: The voltage source is connected between two non-reference nodes: a generalized
node (super-node) is formed.
5
6
0
8
0
4
2
3
2
3
2
3
1
2
1
3
2
4
1














v
v
v
v
v
v
v
v
i
i
i
i

• A super-node is formed by enclosing a (dependent or independent) voltage source
connected between two non-reference nodes and any elements connected in parallel with it.
• The required two equations obtained by the KCL of the super-node and the relationship of
node voltages due to the voltage source.
Super-node
2
0
4
2
7
2
0
2
1
7
2
2
1
2
1











v
v
v
v
i
i

Example: Find the node voltages in the circuit below.

Apply KVL for super nodes

Mesh Analysis
• Mesh analysis: based on the KVL
• A Mesh is a loop which does not contain any other loops within it.
• It is applicable to only planar circuit.

Mesh Analysis
 Mesh is a loop that does not contain any other loop within it.
 Using mesh currents as the circuit variables.
 Using mesh currents instead of element currents as circuit variables is convenient and
reduces the number of equations that must be solved simultaneously.
 Mesh analysis applies KVL to find unknown currents.
 It is only applicable to a circuit that is planar.
 It is also known as loop analysis or the mesh-current method

Mesh Analysis
 Non Planar circuit: branches crossing one another. Fig (a)
 Planar circuit: no branches crossing one another. Fig (b)

• Steps to determine Mesh Currents:
1. Identify the total number of meshes
2. Assign mesh currents i1, i2, .., in to the n meshes.
3. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in
terms of the mesh currents.
4. Solve the KVL equation to get the mesh currents.
A circuit with two meshes

Cont’d…
• Apply KVL to each mesh. For mesh 1,
• For mesh 2,
• Solve for the mesh currents.
• Use i for a mesh current and I for a branch current. It’s evident from Fig. above
that
1
2
3
1
3
1
2
1
3
1
1
1
)
(
0
)
(
V
i
R
i
R
R
i
i
R
i
R
V








2
2
3
2
1
3
1
2
3
2
2
2
)
(
0
)
(
V
i
R
R
i
R
i
i
R
V
i
R

































2
1
2
1
3
2
3
3
3
1
V
V
i
i
R
R
R
R
R
R
2
1
3
2
2
1
1 ,
, i
i
I
i
I
i
I 




Example 2: Find the branch current I1, I2, and I3 using mesh analysis. And
the power dispated at 4Ω resistor
For mesh 1,
1
2
3
0
10
)
(
10
5
15
2
1
2
1
1








i
i
i
i
i
For mesh 2,
1
2
0
10
)
(
10
4
6
2
1
1
2
2
2







i
i
i
i
i
i
 We can find i1 and i2 by substitution method or Cramer’s rule. Then,
2
1
3
2
2
1
1 ,
, i
i
I
i
I
i
I 




Method 1: Elimination 6i2 − 3 − 2i2 = 1, i2 = 1 A i1 = 2i2 − 1 = 2 − 1 = 1 A.
Thus, I1 = i1 = 1 A, I2 = i2 = 1 A, I3 = i1 − i2 = 0
Method 2 : Cramer’s rule

Example 3: Use mesh analysis to find the current I0 in the circuit of Fig. below
 Apply KVL to each mesh. For mesh
1,
12
6
5
11
0
)
(
12
)
(
10
24
3
2
1
3
1
2
1









i
i
i
i
i
i
i
 Apply KVL to each mesh. For mesh
2,
0
2
19
5
0
)
(
10
)
(
4
24
3
2
1
1
2
3
2
2









i
i
i
i
i
i
i
i

• For mesh 3,
• In matrix from Eqs. (1) to (3) become
•
we can calculate i1, i2 and i3 by Cramer’s rule, and find I0.
0
2
0
)
(
4
)
(
12
)
(
4
,
A,
node
At
0
)
(
4
)
(
12
4
3
2
1
2
3
1
3
2
1
2
1
0
2
3
1
3
0

















i
i
i
i
i
i
i
i
i
i
I
I
i
i
i
i
I































0
0
12
2
1
1
2
19
5
6
5
11
3
2
1
i
i
i

Mesh Analysis with Current Sources
 Case 1
● Current source exist only in one mesh,
Write a mesh equation for the other mesh in the usual way, that is,
● One mesh variable is reduced

Super-mesh
Case 2: Current source exists between two meshes, a super-mesh is obtained.
• A super-mesh when current source is presented between two meshes, we remove the branches
results when two meshes have a (dependent, independent) current source in common.
• Properties of a supermesh:
1. The current source in the supermesh is not completely ignored; it provides the
constraint equation necessary to solve for the mesh currents.
2. A supermesh has no current of its own.
3. A supermesh requires the application of both KVL and KCL.

Example: determine the power dispated at 4Ω resistor in the fig below.
applying KVL to the supermesh in
Fig (b) gives
6i1 + 14i2 = 20 ……(1)
We apply KCL to a node in the
branch where the two meshes
intersect in fig (a) gives. i2 = i1 + 6
we get i1 = −3.2 A, i2 = 2.8 A P=
=4=31.36w

• Example: For the circuit in Fig. below,, find i1 to i4 using mesh analysis.
Similarly, a super-mesh formed from three meshes needs three equations: one is from the
super-mesh and the other two equations are obtained from the two current sources.

 Similarly, a super-mesh formed from three meshes needs three equations: one is from the super-mesh and the
other two equations are obtained from the two current sources
0
10
2
)
(
8
5
0
6
)
(
8
4
2
4
4
3
4
3
2
2
1
2
4
3
3
1















i
i
i
i
i
i
i
i
i
i
i
i
i
Applying KVL in mesh 4,
2i4 + 8(i4 − i3) + 10 = 0 or
5i4 − 4i3 = −5
Hint:
Determinant, Δ=14
Δ1 = -105
Δ2 = -35
Δ3 = 55
Δ4 = 30

Nodal Versus Mesh Analysis
• Both nodal and mesh analyses provide a systematic way of analyzing a complex
network.
• The choice of the better method dictated by two factors.
• First factor: nature of the particular network. The key is to select the method
that results in the smaller number of equations.
• Second factor: information required.

Summary
1. Nodal analysis: the application of KCL at the non-reference nodes
• A circuit has fewer node equations
2. A super-node: two non-reference nodes
3. Mesh analysis: the application of KVL
• A circuit has fewer mesh equations
4. A super-mesh: two meshes

Chapter 3 - Methods of Analysissesd.pptx

More Related Content

Similar to Chapter 3 - Methods of Analysissesd.pptx

Recently uploaded

Chapter 3 - Methods of Analysissesd.pptx