1. ELECTRICAL
CIRCUITS
PRESENTED BY :-
1. NIKHIL ROHILLA
2. RAJESH KUMAR
3. RAHUL NEGI
4. RISHABH DEV

2. CONTENT…!!!
• Introduction
• Electrical Circuits
• Electrical Circuits in Linear Algebra
• Series and Parallel Circuits
• Nodal Voltage Analysis and Current Analysis
• Gaussian Elimination
• Truss Analysis
• Spring Mass System
• Idea of Linear Algebra In Physics
• Velocity Of Rocket
• Some Examples

3. INTRODUCTION
• This presentation is mainly about to let us all
know that how electrical circuits works on
applications of LINEAR ALGEBRA.
• All you need to be a inventor is a good
imagination and a pile of junk.
by:-
THOMAS EDISON

4. Electrical Circuits
• Electrical circuit is nothing but just a
combination of transistor, capacitor, diodes,
etc. including some logic gates.
• Each component has it’s 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 source
flow.

6. Linear Algebra in Electrical Circuits
• Linear Algebra most apparently uses by
electrical engineers.
• Wherever there is system of linear equation
arises the concept of linear algebra.
• Various electrical circuits solution like
Kirchhoff's law , Ohm’s law are conceptually
arise linear algebra.

7. Continued…
• To solve various linear equations we need to
introduce the concept of linear algebra.
• Using Gaussian Elimination not only
computer engineers but most of daily
computational work minimized .
• Now we don’t have to use extremely large
number of pages to calculate complex system
of linear equations.

8. Simple Series or Parallel Circuits
• For simple circuits, such as those used in math
textbooks to introduce systems of equations,
it is often sufficient to use series and parallel
relationships to simplify circuits.

9. • With this done, Ohm’s Law (V=IR) can be used to find
voltages or currents.
• Vp=Vs Vs=Rs*I I=I1=I2=I3=I4 V=IR
• 6V=6V 6V=3(ohm)*I 2A=2A=2A=2A=2A V1=20V
I=2A V2=30V
V3=30V
V4=20V
• Larger circuits though, are a problem, as this method is
no longer efficient. It becomes far too time consuming
to analyze and reduce circuits equations. Instead a new
method of determining voltages and currents is used
called Nodal Voltage Analysis and Loop Current Analysis

10. Nodal Voltage Analysis and Loop
Current Analysis
• Using Nodal or Loop Analysis, we end up with
systems of equations with unknown variables.
i1+25(i1-i2)+50(i1-i2)=0 76i1-25i2-50i2=0
25(i2-i1)+30i2+(i1-i2)=0 -25i2+56i2-i2=0
50(i2-i1)+(i2-i2)+55i2=0 -50i1-i2+106i2=0

11. • By simplifying and manipulating these equations
i1=(1/76)(25i2+50i3+10) ->
25((1/76)(25i2+50i3+10))+ 56 i2 - i3 = 0
• (-625/76) i2 – (1250/76) i3 – (250/76) + 56i2 - i3 = 0
• (3631/76) i2 – (663/38) i3 = (250/76) -> i2 =
(1326/3631) i3 + (250/76) -50(1/76)(25 [(1326/3631)
i3 + (250/76)] + 50i3+10) - (1326/3631)i3 + (250/76)
+ 106i3 = 0
• i3 = 0.117 , i2 = 0.111 , i1 = 0.245
• This method too, has its pitfalls, as circuits with many
loops or nodes will require many substitutions, not
to mention the large task of keeping track of all the
variables.

12. Gaussian Elimination
To fix all the assertion that we have performed
earlier we use Gaussian elimination.
In this method we need to keep all eqs. into matrix
form, for e.g.
Since the columns are of same variable it’s easy to
do row operation to solve for the unknowns.

13. Continued…
• This method is known as Gaussian Elimination. Now, for large
circuits, this will still be a long process to row reduce to echelon
form.
• With the help of a computer and the right software, ridiculously
large circuits consisting of hundreds of thousands of components
can be analyzed in a relatively short span of time.
• Today’s computers can perform billions of operations within a
second, and with the developments in parallel processing, analyses
of larger and larger electrical systems in a short time frame are very
feasible

14. Example - Loop Current Analysis
Using Gaussian Elimination
• Loop Equations:-
• 1i1 + 25(i1-i2) + 50(i1-i3) = 10
• 25(i2- i1) + 30(i2- i4) + 1(i2- i3) = 0
• 50(i3- i1) + 1(i3- i2) + 55(i3-i4) = 0
• 55(i4- i3) + 30(i4- i2) + 25(i4-i5) + 50(i4- i6) = 0
• 25(i5- i4) + 30 i5+ 1(i5-i6) = 0
• 50(i6- i4) + 1(i6- i5) + 55 i6 = 0

15. • Collect terms(simplified form):
1. 76i1 - 25i2 – 50i3 + 0 i4 + 0i5 + 0 i6 = 10
2. -25i1 + 56 i2 – 1 i3 – 30 i4 + 0i5 + 0 i6 = 0
3. -50i1 –1 i2 + 106 i3 – 55 i4 + 0 i5 + 0 i6 = 0
4. 0i1 –30 i2 – 55 i3 +160 i4 – 25i5 – 50 i6 = 0
5. 0i1 + 0 i2 + 0 i3 –25 i4 + 56i5 – 1i6 = 0
6. 0i1 + 0i2 + 0i3 – 50 i4 – 1i5 + 106 i6 = 0
Now write them as augmented matrix:
76 –25 –50 0 0 0 10
-25 56 –1 –30 0 0 0
-50 –1 106 –55 0 0 0
0 –30 –55 160 –25 –50 0
0 0 0 –25 56 –1 0
0 0 0 –50 –1 106 0

16. • Row reduce :-
1 0 0 0 0 0 .478
0 1 0 0 0 0 .348
0 0 1 0 0 0 .353
0 0 0 1 0 0 .239
0 0 0 0 1 0 .109
0 0 0 0 1 0 .114
i1 = .478 A, i2 = .348 A, i3 = .353 A, i4 = .239A,
i5 = .109 A, i6 = .114 A

17. Truss Analysis
• Linear Algebra is quite
generally used in Structural
Engineering. The analysis of
a structure in equilibrium
involves writing down many
equations in many
unknowns. Often these
equations are linear, even
when material deformation
(i.e. bending) is considered.
This is exactly the sort of
situation for which linear
algebra is the best
technique. Consider, for the
following two dimensional
truss : ring

18. • The beams are joined together by smooth pins
and supports are fastened.
• Any external force act on a joint, the truss is
stable if and only if the vertical and horizontal
components of the forces at each joint sum to
zero.
Horizontal –A+C(sinθ)+Fe cosΦ
Vertical B+C(cosθ)-Fe sinΦ

19. Spring Mass System
• Spring-mass Systems play an important role in
mechanical and other engineering systems. Such a
system is shown in the figure . It composed of three
masses, suspended vertically by a series of spring.
• The left portion of the diagram indicates the state of the
system before release (i.e., the condition in which spring
are neither stretched nor compressed). However, after
the masses are released, they are pulled downward by
the force of gravity.
• The resulting displacement of each spring is measured
with respect to along local coordinates referenced to its
initial position, as shown on right side of the diagram.

20. For each mass, Newton's second Law of motion
(i.e., F=ma) can be applied in conjunction with force
balances to develop the mathematical model of the
system:
m*d2x/dt2 = FD - FU

21. • In order to analyze we’ll apply Hook’s Law
Therefore net force on Mass m1
m1*d2x/dt2=m1.g+2k(x2-x1)-kx1
Thus, we have derived a second order ordinary differential
equation to describe the displacement of the first mass
with respect to time. However, it can be noticed that
solution cannot be obtained because the model includes a
second dependent variable x2. Consequently, free body
diagrams must be developed for the masses m2 and m3.

22. • The net force acting on masses m2 and m3 can
be expressed as
m2*d2x/dt2 = m2.g+k(x3-x2)-2k(x2-x1)
m3*d2x/dt2 = m3.g-k(x3-x2)
Finally we write all eqs. as
3kx1-2kx2 = m1.g
-2kx1+3kx2 = m2.g
-kx2+kx3 = m3.g

23. • In Matrix form :
[K].[X]=[Z]
[X]=inv[K].[W]
where [X] and [W] are column vectors of
unknown X and weight mg
3k -2k 0
[K]= -2k 3k 0
0 -k -k
Now if m1 = 2kg , m2 = 3kg , m3= 2.5kg,
k’s=10kg/sec2

24. 30 -20 0 19.6 Each element of
[X] = -20 30 -10 29.4 this matrix inv(K)
0 -10 10 24.5 tell us the
displacement of
x1 7.350 mass i due to a unit
[X]= x2 = 10.045 force imposed on
x3 12.495 mass j.
0.1 0.1 0.1
inv[K] = 0.1 0.15 0.15
0.1 0.15 0.25

25. Idea of Linear Algebra in Physics
• Thinking about a particle traveling through space,
we imagine that its speed and direction of travel
can be represented by a vector v in 3-dimensional
Euclidean space R3. Its path in time t might be
given by a continuously varying line — perhaps
with self-intersections — at each point of which
we have the velocity vector v(t).
• A static structure such as a bridge has loads which
must be calculated at various points. These are
also vectors, giving the direction and magnitude
of the force at those isolated points.

26. • In the theory of electromagnetism, Maxwell’s
equations deal with vector fields in 3-dimensional
space which can change with time. Thus at each
point of space and time, two vectors are
specified, giving the electrical and the magnetic
fields at that point.
• Given two different frames of reference in the
theory of relativity, the transformation of the
distances and times from one to the other is
given by a linear mapping of vector spaces.
• In quantum mechanics, a given experiment is
characterized by an abstract space of complex
functions. Each function is thought of as being
itself a kind of vector. So we have a vector space
of functions, and the methods of linear algebra
are used to analyze the experiment.

27. Velocity of Rocket
• The upward velocity of a rocket, measured at 3
different times, is shown in the following table
• The velocity over the time interval 5-12 is
approximated by a quadratic expression as
v(t) = a1*t2 + a2*t + a3
Find the values of a1, a2 and a3
time t,
(seconds)
Velocity v,
(meters/second)
5 106.8
8 177.2
12 279.2

28. Solution
• Substituting the values from the table into the quadratic eqs.
For v(t) gives :
106.8=25a1+5a2+a3 25 5 1 a1 106.8
177.2=64a1+8a2+a3 or 64 8 1 . a2 = 177.2
279.2=144a1+12a2+a3 144 12 1 a3 279.2
• By applying Gaussian elimination which is an aspect of linear
algebra we get the results as
a1=0.2905 a2=19.6905 a3=1.0857 to 4 d.p.
• We can also use the above relation to calculate approx pos of
the rocket for any time within time interval
5 <= t <= 12

29. Example - Nodal Voltage Analysis
Using Gaussian Elimination
• Node Equations:
(V1/30) + (V1-100)/5 + (V1-V3)/10 = 0
(V3 - V1)/10 + V3/10 + (V3-100)/20 = 0
• Collect terms(simplified form):
[(1/30) + (1/5) + (1/10)] V1 – (1/10) V3 = 20
(-1/10) V1 + [(1/10) + (1/10) + (1/20)] V3 = 5
=>
(1/3) V1 – (1/10) V3 = 20
(-1/10) V1 + (1/4) V3 = 5

30. Continued…
• Write as Augmented Matrix:
1/3 -1/10 20
-1/10 1/4 5
• Row Reduce to Echelon Form:
1 0 75
0 1 50
V1 = 75 V, V3 = 50 V

31. THANK YOU…!!!