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Loop (Mesh) Analysis Part 2-Basic Electronics-Lecture Slides, Slides of Fundamentals of Electronics

Bengal Engineering & Science UniversityFundamentals of Electronics

This lecture was delivered by Kantimoy Baag at Bengal Engineering and Science University for Fundamentals of Electronics course. Its main points are: Loop, Mesh, Analysis, Current, Source, Network, Direction, Clockwise, Counter-clockwise, KVL

Typology: Slides

2011/2012

Uploaded on 07/19/2012

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Lecture๎˜ƒThirteen๎˜ƒ
Loop๎˜ƒ(Mesh)๎˜ƒAnalysis๎˜ƒ
Independent๎˜ƒCurrent๎˜ƒSources๎˜ƒ
Example๎˜ƒ
Find๎˜ƒV0๎˜ƒusing๎˜ƒmesh๎˜ƒanalysis๎˜ƒ
After๎˜ƒidentifying๎˜ƒloop๎˜ƒcurrents,๎˜ƒfrom๎˜ƒdiagram๎˜ƒ
1
2
03
4mA
2mA
V6 3
I
I
kI
=
=โˆ’
=โˆ’
๎˜ƒ
For๎˜ƒloop๎˜ƒ3,๎˜ƒthe๎˜ƒKVL๎˜ƒequation๎˜ƒis๎˜ƒ
32 31 3
4( ) 2( ) 6 3kI I kI I kI
โˆ’
+โˆ’+=๎˜ƒ
Putting๎˜ƒvalue๎˜ƒof๎˜ƒI1๎˜ƒand๎˜ƒI2,๎˜ƒwe๎˜ƒget๎˜ƒ
30.25mAI
=
๎˜ƒ
Therefore,๎˜ƒV0๎˜ƒ=๎˜ƒโ€1.5V๎˜ƒ
Previously๎˜ƒwe๎˜ƒsaw๎˜ƒthat๎˜ƒnodal๎˜ƒanalysis๎˜ƒis๎˜ƒeasier๎˜ƒfor๎˜ƒcircuits๎˜ƒwith๎˜ƒ
voltage๎˜ƒsources๎˜ƒonly.๎˜ƒNow๎˜ƒwe๎˜ƒhave๎˜ƒlearnt๎˜ƒthat๎˜ƒloop/๎˜ƒmesh๎˜ƒ
analysis๎˜ƒsimplifies๎˜ƒthe๎˜ƒprocedure๎˜ƒfor๎˜ƒfinding๎˜ƒloop๎˜ƒcurrents๎˜ƒif๎˜ƒonly๎˜ƒ
current๎˜ƒsources๎˜ƒare๎˜ƒpresents๎˜ƒin๎˜ƒa๎˜ƒcircuit.๎˜ƒ๎˜ƒ
Example:๎˜ƒ
Find๎˜ƒI0๎˜ƒin๎˜ƒthe๎˜ƒcircuit๎˜ƒshown:๎˜ƒ
From๎˜ƒfigure,๎˜ƒit๎˜ƒis๎˜ƒobvious๎˜ƒthat๎˜ƒI1๎˜ƒ=๎˜ƒ2๎˜ƒmA๎˜ƒand๎˜ƒI2โ€I3๎˜ƒ
=๎˜ƒ4๎˜ƒmA๎˜ƒ
Writing๎˜ƒthe๎˜ƒKVL๎˜ƒequations๎˜ƒfor๎˜ƒloops๎˜ƒ2๎˜ƒand๎˜ƒ3๎˜ƒwe๎˜ƒ
get๎˜ƒ
๎˜ƒ
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Download Loop (Mesh) Analysis Part 2-Basic Electronics-Lecture Slides and more Slides Fundamentals of Electronics in PDF only on Docsity!

Lecture Thirteen

Loop (Mesh) Analysis

Independent Current Sources

Example

Find V 0 using mesh analysis

After identifying loop currents, from diagram

1 2 0 3

4 mA 2mA V 6 3

I

I

kI

For loop 3, the KVL equation is

4 ( k I 3 (^) โˆ’ I 2 (^) ) + 2 ( k I 3 (^) โˆ’ I 1 (^) ) + 6 kI 3 = 3

Putting value of I 1 and I 2 , we get

I 3 (^) =0.25 mA

Therefore, V 0 = โ€1.5V

Previously we saw that nodal analysis is easier for circuits with voltage sources only. Now we have learnt that loop/ mesh analysis simplifies the procedure for finding loop currents if only current sources are presents in a circuit.

Example:

Find I 0 in the circuit shown: From figure, it is obvious that I 1 = 2 mA and I 2 โ€I (^3) = 4 mA Writing the KVL equations for loops 2 and 3 we get

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2 2 1 4 mA 3 3 1 2 2 1 3 3 1 4 mA

loop kI k I I V kI k I I kI k I I loop kI k I I V

โŽฌโ‡’^ +^ โˆ’^ =^ โˆ’^ โˆ’^ โˆ’

Simplifying and putting the value of I 1 , we get 2 I 2 (^) + I 3 =6 mA

Solving simultaneously we get (^2 32 ) 2 3

4mA (^10 ) mA; mA 2 6 mA 3 3

I I

I I

I I

โŽฌโ‡’^ =^ = โˆ’

From figure 0 1 2 4 mA 3

I = I โˆ’ I = โˆ’

Dependent Voltage/ Current Sources

Example:

Calculate the loop currents

From figure it is obvious that

3 1 4 2

x 2 (^ ) x

V k I I I I I

Writing KVL equations for the loops identified

1 2 2 3 1 2 3 3 1 3 4 2 3 4 4 3 4 2 2 3 4

4mA 3 :2 ( ) 1 ( ) 1 3 2 8 mA 4 :1 ( ) 1 ( ) 12 2 12mA

x x

x

loop I mA loop I V V mA k I I I I I loop k I I k I I kI I I I loop k I I k I I I I I

Solving using matrix algebra

1 2 2 3 3 4 4

1 3 2 8 mA 1 3 2 8 mA 10 mA 1 1 2 12 1 1 2 12 18

I I

I I

I I

โŽก โˆ’^ โŽค โŽก โŽค โŽก โŽค โŽก โŽค โŽก โˆ’ โŽค^ โˆ’ โŽก โŽค โŽก โŽค

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