This presentation discusses the application of linear algebra concepts in electrical circuits. It provides examples of how linear algebra is used to analyze both simple and complex circuit diagrams. For a mobile phone circuit diagram, the group sets up a system of linear equations using Kirchhoff's laws and solves it using Gaussian elimination to find the node voltages. Similarly, for a drone circuit, nodal analysis is performed to obtain a system of linear equations, which is then solved using Cramer's rule to determine the node voltages. The presentation notes that while linear algebra can be applied to analyze many circuit types, it has limitations for complicated non-linear circuits involving capacitors and inductors. The differential and integral equations for these components introduce non

1. East West University
PRESENTATION
GROUP 3
COURSE CODE: MAT205
COURSE TITLE: LINEAR
ALGEBRA AND COMPLEX
VARIABLES
SECTION: 3
1

2. Group 3 Members
NAME: MD. RIAZUL ISLAM
ID: 2016-1-80-028
NAME: UTHPOL KISOR MITHU
ID: 2016-2-55-009
NAME: PRANTO SAHA UTTSOB
ID: 2016-2-80-019
NAME: ABU MD.MAHIR HASSAN
ID: 2016-1-55-012
NAME: KM. RAKIBUR RAHMAN
ID: 2015-2-53-032
2

3. Introduction
This presentation is mainly
about to let us all know that
the application of linear
algebra in the field of
electrical circuits.
3

4. What is Linear Algebra?
Linear algebra is the branch of mathematics
concerning vector spaces and linear mappings
between such spaces.
4

5. Electrical circuits
 Electrical Circuit is nothing but just a combination resistor,
capacitor, inductor, diodes, etc. including some logic gates.
 Each component has its own specification.
 And through which we get to know what currents and
voltages are.
 An electrical circuit is a path in which electrons from a
voltage or current flows.
5

6. Basic Electrical Circuits
6

7. 7

8. Electrical
Circuit
Components
8

9. Linear Algebra in Linear Circuits
 Linear algebra most apparently uses by electrical engineers.
 When ever there is system of linear equation arises the
concept of linear algebra.
 Various electrical circuits solution like Kirchhoff's law, Ohms
law are conceptually arise linear algebra.
9

10. Internal Linear Circuit In Mobile
phone
10

11. 11

12. After Using Nodal analysis We get the following 4 equations
Again After simplifying the following equations we get,
380V1-150V2= 555
55V1-118V2+55V3= 0
15V2-380V3+15V4= 0
5V1+10V3-24V4= 0
12

13. 
















0
24
10
0
5
0
15
380
15
0
0
0
55
118
55
555
0
0
150
380
augment















0
8
.
4
2
0
1
0
15
380
15
0
0
0
55
118
55
4473
.
1
0
0
3947
.
0
1

















4473
.
1
8
.
4
2
3947
.
0
0
0
15
380
15
0
0
0
55
71
.
139
0
4473
.
1
0
0
3947
.
0
1
R1’=R1/380 &
R4’=R4/5
R2’=R2-55*R1 & R4’=R4-R1
13

14. 














66
.
3
16
.
12
36
.
4
0
0
0
1
93
.
24
0
0
0
0
3936
.
0
1
0
4473
.
1
0
0
3947
.
0
1















83
.
0
78
.
2
1
0
0
0
04
.
0
1
0
0
0
0
3936
.
0
1
0
4473
.
1
0
0
3947
.
0
1
R2’=R2/-139.71 ; R3’=R3/15
& R4’=R4/-0.3947
R3’=R3-R2 & R4’=R4-R2
R3’=R3/-24.93 &
R4’=R4/4.36















66
.
3
16
.
12
06
.
5
1
0
0
1
33
.
25
1
0
0
0
3936
.
0
1
0
4473
.
1
0
0
3947
.
0
1
14

15. 













83
.
0
74
.
2
0
0
0
0
04
.
0
1
0
0
0
0
3936
.
0
1
0
4473
.
1
0
0
3947
.
0
1
R4’= R4+R3
Now Backward Substituting the Equation from
Augment matrix we get .. .. . . .. . . .
V4= 0.3029 Volt
V3=0.04967 Volt
V2=0.8626 Volt
V1=1.8010 Volt
15

16. Now we have found the lower voltage as
we need for our external device(Flash light;
Camera; Speaker or any other else) of
mobile phone .For doing these we had
some linear equations and we already have
solved them using Gaussian Elimination
method. That is the applicable relation with
Linear Algebra in Circuit of mobile phone.
16

17. What is Drone?
17

18. This is a Drone 18

19. Internal Linear Circuit In
Drones
19

20. CIRCUIT DIAGRAM 20

21. FINDING NODE EQUATIONS
NODE EQUATIONS:
For Node V1:
(V1-8)/2 + V1/3 + V1/4 + (V1-V2)/2 = 0
For Node V2:
(V2-V1)/2 + V2/3 + (V2-V3)/2 = 0
For Node V3:
(V3-V2)/2 + V3/3 + (V3-V4)/2 = 0
For Node V4:
(V4-V3)/2 + V4/3 +V4/5 + (V4-8)/2=0
21

22. So, the equation becomes,
19V1-6V2 = 48
-3V1+8V2-3V3 = 0
-3V2 + 8V3-3V4 = 0
-15V3+46V4 = 120
22

23. Finding Node Voltages Using Cramer’s Rule
The equation becomes, 19V1-6V2 = 48
-3V1+8V2-3V3 = 0
-3V2 + 8V3-3V4 = 0
-15V3+46V4 = 120
 Let,
A =


















46
15
0
0
3
8
3
0
0
3
8
3
0
0
6
19
A1=

















46
15
0
120
3
8
3
0
0
3
8
0
0
0
6
48
A2 =
















46
15
120
0
3
8
0
0
0
3
0
3
0
0
48
19
A3=
















46
120
0
0
3
0
3
0
0
0
8
3
0
48
6
19
A4=

















120
15
0
0
0
8
3
0
0
3
8
3
48
0
6
19
23

24. NOW,
V1=
|𝑨𝟏|
|𝑨|
=
𝟏𝟏𝟎𝟔𝟒𝟎
𝟑𝟓𝟒𝟏𝟔
=3.1240 V
V2=
|𝑨𝟐|
|𝑨|
=
𝟔𝟕𝟎𝟑𝟐
𝟑𝟓𝟒𝟏𝟔
=1.8927 V
V3=
|𝑨𝟑|
|𝑨|
=
𝟔𝟖𝟏𝟏𝟐
𝟑𝟓𝟒𝟏𝟔
=1.8949 V
V4=
|𝑨𝟒|
|𝑨|
=
𝟏𝟏𝟒𝟔𝟎𝟎
𝟑𝟓𝟒𝟏𝟔
=3.2358 V
24

25. CRAMER’S RULE
 An explicit formula for the solution of a system of linear equations
with as many equations as unknowns, valid whenever the system
has a unique solution.
 Expresses the solution in terms of the determinants of the square
matrix and of matrices obtained from it by replacing one column
by the vector of right-hand-sides of the equations.
25

26. APPLICATION
 If the network is complex, the number of equation
(unknowns) increase. In such case, the solution of
simultaneous equations can be obtained by Cramer's Rule
for determinants.
 We can apply this application to solve Kirchhoff's voltage
law and determine the value of current flowing through
different branches.
26

27. EXAMPLE
 Apply Kirchhoff's current law and voltage law to the circuit and
Indicate the various branch currents (I1 & I2).
27

28. Applying KVL in loop 1 : A-B-E-F-A.,
- 15 i1 + 20 i2 + 50 = 0 ...........(1)
And for loop 2 : B-C-D-E-B.,
-30 (i1 - i2 ) - 100 + 20 i2 = 0 ............(2)
Rewriting all the equations, taking constants on one side.
15 i1 + 20 i2 = 50 …......(3)
- 30 i1 + 50 i2 = 100 ........(4)
28

29. I1=D1/D= 500/1350 = 0.37 A
I2= D2/D = 3000/1350 = 2.22 A
∴ I1 = 0.37 A and I2 = 2.22 A
29

30. BENEFIT TO USE LINEAR ALGEBRA IN LINEAR CIRCUIT
 In the linear circuit there are lots of node voltages ,
which we need to know .If we express them by linear
equation then the finding result become much easier.
 It can also be used to find the result of mesh current
by using linear equation.
30

31. Complicated non-linear circuit
31

32. Limitations
The equations become for each passive component i(c)= 𝒊(𝒕) 𝒅𝒕 +i(0)
v(C)=C
𝒅𝒗
𝒅𝒕
these are the differentiate and integral equation so they are not linear as we see …
If we also try to convert them in phasor domain then they become
v(L)=i(L)jѠL
v(C)=i(L)
𝟏
𝒋Ѡ𝑪
here we also see that there are some complex entry.
Because j= −𝟏
That’s why these complicated capacitor and inductive Circuit cannot be solved by
using of the method of Linear Algebra.So this is the simple Limitation of relation
between Linear algebra and non-Linear Circuit.
32

33. Thank you all……….
Are there any question?
33