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Nodal Analysis Part 3-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: Nodal, Analysis, Dependent, Voltage, Current, Source, Simultaneous, Equations

Typology: Slides

2011/2012

Uploaded on 07/19/2012

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Lecture๎˜ƒEleven๎˜ƒ
Nodal๎˜ƒAnalysis๎˜ƒ
Dependent๎˜ƒ(Voltage/๎˜ƒCurrent)๎˜ƒSources๎˜ƒ
Example๎˜ƒ
Find๎˜ƒI0๎˜ƒin๎˜ƒthe๎˜ƒfigure๎˜ƒshown:๎˜ƒ
From๎˜ƒfigure,๎˜ƒit๎˜ƒis๎˜ƒobvious๎˜ƒthat๎˜ƒ๎˜ƒ
12 2
0and
2k 1k
x
VV V
II
โˆ’
==
๎˜ƒ
Writing๎˜ƒnodal๎˜ƒequation๎˜ƒat๎˜ƒnode๎˜ƒ1๎˜ƒand๎˜ƒ2๎˜ƒ
112
012
1: 2k 2
2: 4m 3
x
x
node V I V V
node I I V V
=โ‡’=
+=โ‡’โˆ’
๎˜ƒ๎˜ƒ
Solving๎˜ƒV1๎˜ƒ=๎˜ƒ16๎˜ƒV๎˜ƒand๎˜ƒV2๎˜ƒ=๎˜ƒ8๎˜ƒV๎˜ƒ
Therefore๎˜ƒ12
02k
VV
Iโˆ’
==๎˜ƒ4๎˜ƒmA๎˜ƒ
Example๎˜ƒ
Find๎˜ƒthe๎˜ƒcurrent๎˜ƒI0๎˜ƒin๎˜ƒthe๎˜ƒfigure๎˜ƒbelow:๎˜ƒ
From๎˜ƒfigure๎˜ƒit๎˜ƒis๎˜ƒobvious๎˜ƒthat๎˜ƒ
12
12
2
3
23
6V
x
x
VV V VV
VV
V
โˆ’= โŽซโ‡’=
โŽฌ
=โŽญ
=
๎˜ƒ
Since๎˜ƒthe๎˜ƒnode๎˜ƒvoltage๎˜ƒV3๎˜ƒis๎˜ƒknown,๎˜ƒtherefore๎˜ƒ
writing๎˜ƒnodal๎˜ƒequations๎˜ƒfor๎˜ƒnode๎˜ƒ1๎˜ƒand๎˜ƒ3๎˜ƒonly๎˜ƒ
yields๎˜ƒ
12k
6k
6k
2Vx
6V
12k
V1V2V3
๎˜ƒ
docsity.com
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Download Nodal Analysis Part 3-Basic Electronics-Lecture Slides and more Slides Fundamentals of Electronics in PDF only on Docsity!

Lecture Eleven

Nodal Analysis

Dependent (Voltage/ Current) Sources

Example

Find I 0 in the figure shown:

From figure, it is obvious that

0 1 2 and^2 2k^ x 1k

I = V^^ โˆ’ V^ I = V

Writing nodal equation at node 1 and 2

1 1 2 0 1 2

1: 2k 2 2: 4m 3

x x

node V I V V node I I V V

Solving V 1 = 16 V and V 2 = 8 V

Therefore 0 1 2 2k

I =^ V^ โˆ’ V = 4 mA

Example

Find the current I 0 in the figure below:

From figure it is obvious that

1 2 1 2 2 3

6V

x x

V V V

V V

V V

V

โŽฌโ‡’^ =

Since the node voltage V3 is known, therefore writing nodal equations for node 1 and 3 only yields

12k

6k

6k

2V x

6V

12k

V 1 V^2 V 3

(^11 ) 2 1 1 2 2 2 3 2

2 :^6 12 6

x

x

V

V

node V V^ V I k k V^ V^ V^ V node V^ I V^ V k^ k^ k^ k k k

+ โˆ’^ = โŽซ

โŽฌโ‡’^ +^ +^ =

+ = โˆ’^ โŽช

Simplifying this

V 1 (^) + V 2 = 6

Putting value of V2 we get

1 1

4 6 9 V

V = โ‡’ V =

Now 0 1 3 mA 12 8

I V

k

Example:

From Figure it is obvious that

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

x y y

V V

V V V V

V V V V

V V V V

V V

Since the nodal equation is known for node 1, so now for calculating from nodes 2, 3 and 4, we write following nodal equations

2 1 2 3

2 2 3 (^2 3) 0. (^2 3) 0. 0.2^4 4

y y y

x V x V V

node V^ V V^ V V V V node V^ V V I V V (^) V I node I V^ V^ V

โŽฌโ‡’^ +^ +

+ + โˆ’^ = โŽช

= I 0.2 Vy 14 4 2.5^1 +^ V^ + V^ โˆ’ V

Now the set of simultaneous equations is