Chapter 3 Methods of Analysis: 2) Mesh Analysiswaleedeid.tripod.com/Lecture5_cir_analysis.pdf · Chapter 3 | Methods of Analysis: 2) Mesh Analysis ... Faculty of Electronic Engineering,

Mesh Analysis Procedures Mesh Analysis with Current Sources Conclusions

Chapter 3 — Methods of Analysis:2) Mesh Analysis

Dr. Waleed Al-Hanafywaleed [email protected]

Faculty of Electronic Engineering, Menoufia Univ., Egypt

MSA Summer Course:Electric Circuit Analysis I (ESE 233) — Lecture no. 5

July 27, 2011

Dr. Waleed Al-Hanafy MSA Summer Course: Electric Circuit Analysis I (ESE 233) — Lecture no. 5

Chapter 3 — Methods of Analysis: 2) Mesh Analysis


Overview

1 Mesh Analysis Procedures

2 Mesh Analysis with Current Sources

3 Conclusions

Reference:[1] Alexander Sadiku, Fundamentals of Electric Circuits, 4th ed.McGraw-Hill, 2009.




Introduction

Mesh analysis provides another general procedure for analysing circuits,using mesh currents as the circuit variables

Using mesh currents instead of element currents as circuit variables isconvenient and reduces the number of equations that must be solvedsimultaneously

Recall that a loop is a closed path with no node passed more than once.A mesh is a loop that does not contain any other loop within it

Nodal analysis applies KCL to find unknown voltages in a given circuit,while mesh analysis applies KVL to find unknown currents

Mesh analysis is not quite as general as nodal analysis because it is onlyapplicable to a circuit that is planar

A planar circuit is one that can be drawn in a plane with no branchescrossing one another; otherwise it is nonplanar

A circuit may have crossing branches and still be planar if it can beredrawn such that it has no crossing branches




Example




What is a Mesh?

A mesh is a loop which does not contain any other loops within it

Paths abefa and bcdeb are meshes, but path abcdefa is not a mesh




Mesh Method Steps

Steps to Determine the Mesh Currents:

1 Assign mesh currents i1, i2,· · · , in to the n meshes

2 Apply KVL to each of the n meshes. Use Ohm’s law toexpress the voltages in terms of the mesh currents

3 Solve the resulting n simultaneous equations to get the meshcurrents.




Example-1

Find the branch currents I1, I2, and I3using mesh analysis

For mesh 1, applying KVL

−15 + 5i1 + 10(i1 − i2) + 10 = 0

or 3i1 − 2i2 = 1

For mesh 2,

6i2 + 4i2 + 10(i2 − i1)− 10 = 0

or i1 = 2i2 − 1

Solving for i1 and i2 results in i1 = i2 = 1A.

Thus I1 = i1 = 1A, I2 = i2 = 1A, and

I3 = i1 − i2 = 0




Exercise-1

Calculate the mesh currents i1 andi2 in the circuit shownAnswer: i1 = 2

3 A, i2 = 0




Example-2

Use mesh analysis to find the current i0 in the circuit shown

Formesh 1, −24+ 10(i1 − i2) + 12(i1 − i3) = 0

=⇒ 11i1 − 5i2 − 6i3 = 12

Formesh 2, 24i2 + 4(i2 − i3) + 10(i2 − i1) = 0

=⇒ −5i1 + 19i2 − 2i3 = 0

Formesh 3, 4i0 + 12(i3 − i1) + 4(i3 − i2) = 0

Since i0 = i1 − i2, then =⇒ −i1 − i2 + 2i3 = 0

Finally: i1 = 2.25A, i2 = .75A, and i3 = 1.5A, thus

i0 = i1 − i2 = 1.5A




Applying mesh analysis to circuits containing current sources(dependent or independent) may appear complicated. But it isactually much easier than what we encountered in the previoussection, because the presence of the current sources reduces thenumber of equations.Case 1: When a current source exists only in one mesh: Considerthe circuit below, we set i2 = −5A and write a mesh equation forthe other mesh in the usual way, that is,−10 + 4i1 + 6(i1 + i2) = 0⇒ i1 = −2A




Case 2: When a current source exists between two meshes:Consider the circuit below we create a supermesh by excluding thecurrent source and any elements connected in series with it, asshown in Fig. (b). Thus, A supermesh results when two mesheshave a (dependent or independent) current source in common.Thus −20 + 6i1 + 10i2 + 4i2 = 0⇒ 6i1 + 14i2 = 20. Since byapplying KCL at node 0, i2 = i1 + 6, then i1 = −3.2A andi2 = 2.8A.




Example-3

For the circuit shown, find i1 to i4 using meshanalysisNote that meshes 1 and 2 form a supermeshsince they have an independent current sourcein common. Also, meshes 2 and 3 form anothersupermesh because they have a dependent cur-rent source in common. The two supermeshesintersect and form a larger supermesh as shown.

For the larger supermesh, we have 2i1 + 4i3 +

8(i3 − i4) + 6i2 = 0. We also have at node P,

i2 = i1 + 5 and at node Q i2 = i3 + 3i0. But

i0 = −i4. At mesh 4, 2i4 + 8(i4 − i3) + 10 = 0.

Solving results in i1 = −7.5 A, i2 = −2.5 A,

i3 = 3.93 A, and i2 = 2.143 A.




Homework

Use mesh analysis to determine i1,i2, and i3 in the circuit shownAnswer: i1 = 3.474 A, i2 =.4737 A, and i3 = 1.1052 A




Conclusion

Concluding remarks

Mesh analysis method is studied as a key tool to analyse anycircuit

Basic mesh analysis steps is introduced highlighted by someexamples

The case of supermesh is also given with examples

.



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Chapter 3 Methods of Analysis: 2) Mesh Analysiswaleedeid.tripod.com/Lecture5_cir_analysis.pdf · Chapter 3 | Methods of Analysis: 2) Mesh Analysis ... Faculty of Electronic Engineering,