ENGINEERING MATHEMATICS -I SYSTEM OF LINEAR ALGEBRAIC
Resulting equations are
(8.38)
EQUATIONS,
I|-b+ = 0 Ih= -1
2, + 313 = 18 21, + 12 = 18 :. I, =3
- 181, = -72
I =4
h= -l amp, h=3 amps and h = 4
Direction of h is opposite to that shown in amps.
the figure.
Ex. 3Determine the currents in the
network given in figure below. (Dec. 2019, Aug. 2022)
40 Q 20 V
I
30 2
40 2 40 V
Fig. 8.5
Sol. :Applying KCL and KVL we get the
or
equations
Ih -h + h=0
401, + 301, = 20 41 + 31, =2 ie
401, + 301, = 40 31, + 41a = 4
Consider, augmented matrix
[1 -1 1:0 1 -1 1 :07 [1 -1 1 : 0]
4 3 0:2 R2- 4R1 ~ 0 7 -4: 2 7R3 - 3R ~07 -4 : 2
03 4 : 4 Lo 3 4 : 4J Lo 0 40 : 22
Resulting equations are Ij - , + , = 0
84 6
12 =14010
1
712 - 4l; = 2
20
22 11
4013 = 22 I3 40 20
:. Ih=2o Ih=
6 11
10 and I3 = 20
Ex. 4 :Determine the currents in the network given in the Fig. 8.6 below.
19 Sol. : Applying KCL at the nodes P, Q, R
..(1)
19
I; +Is = L4 ...(2)
...(3)
R
5V
Applying KVL to three loops
29 h+ 21; = 2 ..4)
21, + 21, = 4 ...5)
4 \
Is = 5 ... (6)
Fig. 8.6
I; = 5 is known. We require four equations to obtain four unknowns I, l, b, L. We omit equation (3) and proceed to solve the
ystem of equations (1), (2), (4) and (5) we write down the system in matrix form
1 -1 1 0
-5
1 0 0 -1
1 20 0 2
011 0
-5
AX = B where, X= B=
2
L4. 2
ENGINEERINGMATHEMATICS -I
(8.39) SYSTEM OF LINEAR ALGEBRAIC EQUATIONS, ...
(onsider augmented matrix
C1 -1 1 0 : 0 1 -1 1 0: 0 1 -1 1 0 : 0
1 0 0 -1 : -5 R2 - R1
01 -1 -1 : -5 0 1 -1 -1: -5
[A:B] = R3 - R1
1 2 0 0: 2 00 2 3 : 17 Ra - R3 ~
R4 R1 00 2 3 : 17
Lo 1 1 0: 2 00 2 1: 7 J Lo 0 0 -2:-10
posulting system is lh -l2 + l3 = 0, Iz - I3 - L4 = 5, 213 + 3L4 = 17, -2L4 = -10
L4 = 5, 3 = 1, I, = 1, lh = 0, Is = 5
It satisfies equation (3) also.
Ex. 5: For a traffic floW problenm, ensuring that at each junction total inflow = total out flow of traffic, find the values of unknowns,
corresponding to flow pattern shown below.
60
A B
’70
80
+-100
80 90
Fig. 8.7
Sol.: At the junction A, X6 + 60 = X1 X1- X = 60 ... (1)
At the junction B, X1 - 70= %2 or X1 - X2= 70 ... (2)
At the junction C, X2 + 100= X3 or X3-X2= 100 ... (3)
At the junction D, X3 -90= X4 or X4 - X3=- 90 . .(4)
At the junction E, X4 + 80=Xs Xs - X4= 80 ... (5)
At the junction F, Xs- 80 = X6 or X6 - Xs-80 ... (6)
Matrix form of these equation is
1 0 0 0 0-1] X1 60
1 -10 0 0 0 X2 70
0-1 1 0 0 0 X3 100
00 1 -10 0 X4 90
0 0 0 -11 0 80
Lo 0 0 01 -1J 80
1 0 0 00 -1 : 60 1 00 0 0-1: 60
1 -1 0 0 0 0 : 70 -1 1 00-70
-1 1 0 00:10o-R2 -1 1 00 100
[A|B] = 0 0 1 -1 0 0 : 90 - Ra -1 1 00-90
-1 1 0 : 80 -1 1 0 80
0 1 -1 : 80 0 -1 1:-80
10 0 0 0 - 1 : 60
R2 + R1
0 1 0 0 0 -1 - 10
Ra + R2 90
00 1 0 0 0
R4 + R3 0
00 0 1 0
Ra + Rs 00 0 0 1 0 80
R6 + Rs 000 0 1
Resulting equation are X1 - X6 = 60 X5 = 80
X2 - X6 = -10 X6 = 0
X3 = 90 X1 = 60
X4 = 0 X2 = - 10
- e
ENGINEERING MATHEMATICS (8.40)
SYSTEM OF LINEAR
ALGEBRAIC EQUATIONS,
Ex. 6: Solve the following traffic problem.
400
200
300 D
A
400
500 B 600
500
+300
Fig. 8.8
Sol. : Traffic inflowat each junction = Traffic out flow
At junction A, 200 + 300 = X1 + X4
At junction B, X3 + X4 = 500 + 300
At junction C, 500 + 600 = x2 + X3
At junction D, X1 + X2 = 400 + 400
The system of equation is
X1 + Ox2 + Ox3 + XA = 500
X1 + X2 + Ox3 + Ox4 = 800
Ox1 + X2 + X3 + Ox4 = 1100
Ox1 + 0x1 + X3 + X4 = 800
C10 0 1: 500 10 0 1 : 500
R2 - Ri ~ 010-1: 300
1100: 800
[A:B] = Rs- R2 ~ 001 1 :800
011 0: 1100 R4 -R
Lo0 1 1: 800 Looo 0:0
We get X1 + X4 = 500 Let X4 = t X3 = 800-t
X2 - X = 300 X2 = 300 +t X1 = 500-t
X3 + X4 = 800
where t can take any value.
If t= 100, X1 = 400, X2 = 400, X3 = 700, X4 = 100
EXERCISES 8.11
1. Find the currents I, Iz and l in the circuit shown in the figure.
=6 B L R=4
A 4
R, =2 V,)10
Fig. 8.9
8
Ans. I, ==
AGINEERING MATHEMATICS
(8.41) SYSTEM OF LINEAR ALGEBRAIC EQUATIONS,
R, =3
R, = 2 I Rz =3
1
Ra =1
R, = 4
H
8 V Fig. 8.11
Fig. 8.10 Ans. Ih = 0, Iz =1, 3 =1
Ans. I1=1, l, = 2, I3 = 1
