211EEE1301-Mesh and nodal.pdf.pptx for ccccccccc

211EEE1301-BEEE-M.Karuppasamypandiyan
211EEE1301-
BASIC ELECTRICAL AND ELECTRONICS
ENGINEERING
All you need to be an inventor is a good imagination and a pile of junk.-EDISON
UNIT-I
DC Circuits
DC Circuit analysis-Mesh and Nodal analysis
M.KARUPPASAMYPANDIYAN
AP/EEE
Kalasalingam Academy of Research and Education
1

© Kalasalingam academy of research and education
MESH-BASIC CONCEPT
✔ In formulating mesh analysis we assign a mesh current to each mesh.
✔ Mesh Analysis is developed by applying KVL around mesh in the circuit.
✔ +ve sign : Both current pass through common resistance in the same direction.
✔ -ve sign: Both the current pass through the common resistance in opposite
direction.
✔ Loop (mesh) analysis results in a system of linear equation which must be solved
for unknown currents.
✔ Number of mesh equation needed to solve a circuit with b number of branches, n
number of node is b-n+1
Mesh = “Mesh is defined as a loop which is not contain any other loops with in it .”
211EEE1301-BEEE-M.Karuppasamypandiyan 2

Mesh Current Analysis-PROCEDURE
Step 1: Count the number of meshes.
Step 2: Define mesh current on each mesh. Let the values
be I1, I2, I3 and so on. Label each unknown current
in each mesh, going clockwise.
Step 3: Use Kirchhoff's voltage law (KVL) on each mesh,
generating mesh equations. In this case, use V=I*R.
When resistors are in both meshes, I=(I1-I2) or I=I1+I2
Step 4: Solve the equations for unknown currents (Apply Cramer’s
rule)

Methods of Solving Sets of Equations
✔ Linear Algebra
✔ Substitution
✔ Graphing
✔ Euclid’s Method
✔ Matrix algebra-Cramer’s rule

Mesh Analysis: Example 1
+
10
V
4 ohm
2 ohm
6 ohm
7 ohm
2
V
I1 I2
+ _
+
20V
_
_
Figure 1 : Circuit for Example
Write the mesh
equations and solve
for the currents I1,
and I2.
5

4 ohm
2 ohm
+
10
V
Write the mesh
equations and
solve for the
currents I1, and
I2.
6 ohm 7
ohm
2
V
I1 I2
+ _
+
20V
_
_
Figure 2: Circuit for Example
Mesh 1 4I1 + 6(I1 – I2) =10 - 2
Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20
Eq (1.1)
Eq (2.1)

The previous equations can be written in matrix form as:
⎣ ⎢22⎥
⎡10
⎦⎣2 ⎦ ⎣ ⎦
⎢
− 6 15 ⎥⎢
I
− 6
⎤⎡ I
1 ⎤ ⎡8
I1 = 2.2105
I2 = 2.3509

Write the mesh equations
and solve for the currents
I1, and I2

•Apply KVL (and Ohm’s law) to mesh 1:
15 + 10i2 = 5i1 + 10i1 + 10
▪Apply KVL (and Ohm’s law) to mesh 2:
10 + 10i1 = 10i2 + 6i2 + 4i2
Next we use algebra to simplify our equations.
For mesh 1:
15 + 10i2 = 5i1 + 10i1 + 10
becomes 15 1−10 2=5 (1)
𝑖 𝑖
For mesh 2:
10 + 10i1 = 10i2 + 6i2 + 4i2
becomes 10 1−20 2=−10 (2)
𝑖 𝑖
We now have our two equations in two variables.

The previous equations can be written in matrix form as:
⎣ ⎢-10⎥
⎡15
⎦⎣ 2 ⎦ ⎣
⎢
− 10 -20 ⎥⎢
I
− 10
⎤⎡ I
1 ⎤ = ⎡5
⎤
I1 = 1A
I2 = 1 A

Solve for the mesh currents in the circuit below.

⮚Write KVL, clockwise, for each mesh.
(1)
(2)
(3)
Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 +
10
Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 -
8
Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 +
8
12

Clearing Equations (1), (2) and (3) gives,
Standard Equation form In matrix form:
The Solution:
I1= 2.473 A ; I2= 0.594 A; I3= 1.709 A
20I1 – 4I2 – 10I3 = 30
-4I1 + 18I2 – 11I3 = -18
-10I1 – 11I2 + 30I3 = 20
13

Node Voltages Analysis
 It is generally used to find unknown node voltage.
 It involves the application of KCL equations, instead of KVL.
 One of the nodes is taken as reference or datum or ground node. It is
better to select the one that has maximum number of branches
connected.
 The reference node is assumed to be at ground or zero potential.

Node Voltages Analysis-Procedure
⮚Choose a reference node.
⮚Assign node voltages to the other nodes.
⮚Apply KCL to each node other than the reference node;
express currents in terms of node voltages. (Assume all the
currents are away from the nodes)
⮚Solve the resulting system of linear equations.

Node Voltages Analysis-Example 1
Solve the circuit given, using the node voltage method.

Solution : It has only two nodes. Node 2 has been taken as reference node. The
currents in various branches have been assumed. Writing the KCL equations,

Find V1, V2, and V3 using Nodal Analysis.
500Ω
500Ω
1kΩ
500Ω
500Ω
I1
1 2 3
V1 V2 V3
18

500Ω
V1
500Ω
V1 V2
19

KCL at Node 1
500Ω
500Ω
V1 V2
20

KCL at Node 2
500Ω
1kΩ
500Ω V2 V3
V1
21

KCL at Node 3
500Ω
500Ω
I2
V2 V3
22

Node Voltages Analysis-
Node 1:
Node 2:
Node 3:

Node Voltages Analysis-
The three equations can be combined into a single matrix/vector equation.
Solution:
V1 = 1.33V, V2=1.17V, V3=1.58V

211EEE1301-Mesh and nodal.pdf.pptx for ccccccccc

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211EEE1301-Mesh and nodal.pdf.pptx for ccccccccc