PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

156 Practical MATLAB® Applications for Engineers

The matrix equation is then given by

15 5

535

10

0

1

2









































I

I

where the 2 × 2 matrix is the resistance matrix R and [10 0]T represents the column vec-

tor V, as indicated by

[]RI V*[]
[]

then

[I] = [R] \ [V]

MATLAB Solution

>> % part (i) matrix solution

>> R = [15 -5;-5 35];

>> V = [10;0];

>> I = R\V;

>> Result = [I(1) I(2) I(1)-I(2)]; % results are printed

>> disp(‘**************************************’);

>> disp(‘**********R E S U L T S**************’);

>> disp(‘ Matrix current solutions’);

>> disp(‘**************************************’);

>> disp (‘The currents I1, I2, I3 are given by:’); disp(Result)

>> disp (‘ amp. amp. amp.’);

>> disp(‘**************************************’)

************************************************

***************R E S U L T S******************

Matrix current solutions

***********************************************

The currents I1, I2, I3 are given by:

0.7000 0.1000 0.6000

amp. amp. amp.

************************************************

>> % Part (ii); symbolic solution

>> sym I1 I2 I3;

>> disp(‘**************************************’)

FIGURE 2.53
Network of Example 2.14.

A

10 V^5 Ω

10 Ω

−

+

I

I 2

10 Ω

20 Ω

I 1

Loop # 1 Loop # 2