Electrical circuits in concept of linear algebra

ELECTRICAL CIRCUITS

PRESENTED BY :-1. NIKHIL ROHILLA 2. RAJESH KUMAR3. RAHUL NEGI4. RISHABH DEV

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

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

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.

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.

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.

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.

• 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

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=025(i2-i1)+30i2+(i1-i2)=0 -25i2+56i2-i2=050(i2-i1)+(i2-i2)+55i2=0 -50i1-i2+106i2=0

• 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.

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.

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

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

• 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

• 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

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

• 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

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.

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

• 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.

• 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

• 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 -kNow if m1 = 2kg , m2 = 3kg , m3= 2.5kg,

k’s=10kg/sec2

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 0.1inv[K] = 0.1 0.15 0.15 0.1 0.15 0.25

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.

• 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.

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 + a3Find the values of a1, a2 and a3

time t, (seconds)

Velocity v, (meters/second)

5 106.8

8 177.2

12

279.2

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

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

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

THANK YOU…!!!