Basic Electric Circuits Session 9A

Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 2
Session 9A: Focus
 Mesh Analysis
◦ Introduction
◦ Planar Circuit
◦ Definition of Mesh
◦ Mesh Currents
 Mesh Analysis
◦ Example Problems
 Home Work Problems

Mesh Analysis - Introduction

Mesh Analysis … Introduction
 As we have seen, nodal analysis is a straightforward
analysis technique when only current sources are
present
◦ Though voltage sources are easily accommodated with the
supernode concept.
 Still, nodal analysis is based on KCL.
 If you wonder, isn’t there a similar approach based on
KVL?
 There is one and it’s known as mesh analysis
 Although only strictly applicable to what we will
shortly define as a planar circuit
 It is simpler to apply than nodal analysis, in many
cases

Planar Circuit

Planar Circuit
 If it is possible to draw the diagram of a circuit on a
plane surface in such a way that no branch passes
over or under any other branch, then that circuit is
said to be a planar circuit

Planar or non-planar Circuit?
 Is the circuit shown below a planar circuit?
Connections Crossing over
Each other
It is a Non-planar
Circuit

Planar or Non-Planar Circuit?
 Is the circuit shown below a planar or non-planar
circuit?
No connections are
crossing over Each other
- Planar Circuit

Definition of Mesh

Definitions: with Examples
 Mesh is a loop that does not contain any other loops
within it
Not a loop, since node A
is passed through twice
No paths or loops
A
Sample planar circuit
It is a loop, but not a mesh,
It encloses other loops
This is both a loop and
a mesh
This is both a loop and
a mesh
Is there a path or loop?

Planar Circuits with Meshes
 The mesh is a property of a planar circuit and is undefined for
a non-planar circuit
 Once a circuit has been drawn neatly in planar form, it often has
the appearance of a multi-paned window
 The boundary of each pane in the window may be considered
to be a mesh
 If a network is planar, mesh analysis can be used to accomplish
the analysis
 This technique involves the concept of a mesh current

Mesh Currents

Example: Mesh Currents
 Is the above circuit planar?
 How many meshes are there?
 Which are mesh currents?
 Mesh currents are shown below.
Yes, a Planar Circuit
Two meshes
i1 and i2

Mesh Currents
 A mesh current is indicated by a curved arrow that
almost closes on itself and is drawn inside the
appropriate mesh
 Although the directions are arbitrary, we will always
choose clockwise mesh currents
◦ To have consistency in the analysis and to minimize errors
while writing the mesh equations
 We no longer have a current or current arrow shown
directly on each branch in the circuit
 The current through any branch must be determined by
considering the mesh currents flowing in every mesh in
which that branch appears

Branches and Mesh Currents
 What is the maximum number of meshes that any
branch could appear on?
 Which branch (resistor) is appearing on two meshes?
 What is the current flowing left to right through 6 Ω ?
 What is the current flowing right to left through 4 Ω ?
 What is the current flowing down through 3 Ω ?
 What is the current flowing down through 10 V?
Two meshes
3Ω
(i1 - i2)
i1
-i2
i2

Example 1: Mesh Analysis
 Find i1 and i2 :
 Applying KVL to left mesh:
 Applying KVL to right mesh:
 Solving these two equations:
6i1 + 3i1 - 3i2 = 42
9i1 - 3i2 = 42
3i2 - 3i1 + 4i2 = 10
- 3i1 + 7i2 = 10
- 9i1 + 21 i2 = 30
18 i2 = 72
i1 = 6A
i2 = 4A
i1 = 6A, i2 = 4A
-42 + 6i1 + 3(i1 - i2) = 0
3(i2 - i1) + 4i2-10 = 0

Note: This is Practice problem 4.7 on page 96, Figure 4.20 in the Ref 1 book (by Hayt)
- 10 + 4(i1 -i2) + 3 = 0
4(i2 - i1) + 5i2 + 9i2 + 10(i2 - i3) = 0
4i1 - 4i2 = 7
KVL to mesh with i1 :
i2 = 460 mAi1 = 2.22 mA
-4i1+28i2-10i3 = 0
-3 + 10(i3 - i2) + 1i3 + 7i3= 0
- 10i2+18i3= 3
-2i1+14i2-5i3 = 0
i3

 Find i1 when A = 2i2 :
- 2+ 2i1+5(i1 -i2) – A = 0
-6+ 4i2 +5(i2 - i1) +3i2 = 0
7i1 - 5i2 -2i2 = 2
i1 = 1.35 A
-5i1+12i2 = 6
=2i2
7i1 - 7i2 = 2

Note: This is problem 37 on page 116, Figure 4.62 in the Ref 1 book (by Hayt)
 Find i :
- 2+ 3i1 + 4(i1 -i2) + 1(i1 -i3) = 0
4(i2 - i1) + 1i2 = 0
8i1 - 4i2 - i3 = 2
i = -261 mA
-4i1 + 5i2= 0
i = i3 – i2
How many meshes are there? 3
What is i in terms of mesh currents?
1(i3- i1) + 4i3 = 0
-i1 + 5i3= 0
Solving these three equations:
i1 = 434.78 mA i2 = 347.82 mA i3 = 86.95 m A
i = i3 – i2 i = -261 mA

Home Work Problems

S9A_Problem 1: Mesh Analysis
- 6 + 14i1 + 10(i1 - i2) = 0
24i1 - 10i2 = 6
10(i2 - i1) + 10i2 + 5 = 0
i1 = 184.2 mA, i2 = -157.9 mA
14i1 + 10i1 - 10i2 = 6
KVL to mesh with i1 : KVL to mesh with i2 :
-10i1 + 20i2 = -5
-2i1 + 4i2 = -1
38i2 = -6
i2 = -157.9 mA
i1 = 184.2 mA
-24i1 + 48i2 = -12

S9A_Problem 2: Mesh Analysis
 Find power supplied by the 2V source:
- 5 + 4i1 + 2(i1 - i2) – 2 = 0
6i1 - 2i2 = 7
2 + 2(i2 - i1) + 5i2 + 1 = 0
i1 = 1.132A, i2 = -0.1053A
4i1 + 2i1 - 2i2 = 7
KVL to mesh with i1 : KVL to mesh with i2 :
-2i1 + 7i2 = -3
-6i1 + 21i2 = -9
19i2 = -2
i2 = -0.1053 A
i1 = 1.132 A
Note: This is Example problem 4.7 on page 94, Figure 4.17 in the Ref 1 book (by Hayt)

Session 9A: Summary
 Mesh Analysis
◦ Introduction
◦ Planar Circuit
◦ Definition of Mesh
◦ Mesh Currents
 Mesh Analysis
◦ Example Problems
 Home Work Problems

References
Ref 1 Ref 2

Basic Electric Circuits Session 9A

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