PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Alternating Current Analysis 293

fprintf(‘The current i1(t)=%fcos(10t %f)amps\n’,I1mag,I1ang)

fprintf(‘The current i2(t)=%fcos(10t %f)amps\n’,I2mag,I2ang)

fprintf(‘The current i3(t)=%fcos(10t+ %f)amps\n’,I3mag,I3ang)

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

The script fi le loops is executed and the results are indicated as follows:

>> loops

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

System Matrices

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

The impedance matrix is given by:

Z =

30.0000 - 0.0500i -20.0000 0

-20.0000 20.0000 + 29.9750i 0 + 0.0250i

0 0 + 0.0250i 10.0000 - 0.0250i

The voltage matrix is given by:

V =

5.0000

0

6.3640 + 6.3640i

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

Loop Currents

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

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

phasor domain

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The magnitude of the Current I1 (in amps) is :0.195625

and its phase angle in degrees is :-21.209060

The magnitude of the current I2 (in amps) is :0.109148

and its phase angle in degrees:-77.628332

The magnitude of the current I3 (in amps)is :0.899767

and its phase angle in degrees is:45.152608

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

time domain

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The current i1(t) = 0.195625cos(10t -21.209060)amps

The current i2(t) = 0.109148cos(10t -77.628332)amps

The current i3(t) = 0.899767cos(10t+ 45.152608)amps

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

Example 3.15

Analyze the circuit diagram shown in Figure 3.73, and obtain by hand the system node

equations, as well as the matrix node equation (all elements are given as admittances).

Create the script fi le nodes that returns

1. The system matrices Y and I


2. The node voltages V 1 and V 2 in phasor form


3. The instantaneous node voltages v 1 (t) and v 2 (t)