PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

294 Practical MATLAB® Applications for Engineers

ANALYTICAL Solution

The two node equations are

for node V 1 : (2 − j)V 1 + jV 2 = 5 ∠0°

for node V 2 : jV 1 + (4 − j4)V 2 = − 6 ∠−90°

The matrix node equation is given by

2

44

50

690

1

2









jj

jj

V

V

























∠

∠











* 

The aforementioned matrix equation can be expressed as [Y] * [V] = [I], where

Y

jj

jj





2

44













and

I





50

690

∠

∠













where the phasor current sources are converted to complex numbers as

5 ∠ 0 ° = 5 and − 6 ∠− 90 ° = −(−j6) = j6

MATLAB Solution

% Script file: nodes

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

disp(‘ System Matrices ’)

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

disp(‘The admittance matrix Y is given by:’)

Y = [2-j j;j 4-4*j]

disp(‘The current matrix I is given by:’)

I = [5;j*6]

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

disp(‘ Node voltages ’)

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

2

4

V 1 V 2

5

6

j 3

−j 4

j −j 3

−j

0 °A

− 90 °A

FIGURE 3.73
Network of Example 3.15.