PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 139

MATLAB Solution

% Script file: charge

I = 1;

T = 2*60;

Q = I*T;

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

disp(‘The charge of 1 amp passing through a point during 2 minutes is :’);

disp(Q);

disp(‘Coulombs’)

format long

Ne=Q*(1/(1.602*10e-19));

disp(‘The number of electrons is:’); disp(Ne)

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

>> charge

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

The charge of 1 amp passing through a point during 2 minutes is :

120 Coulombs

The number of electrons is:

7.490636704119849e+ 019

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

Example 2.2

Create the MATLAB script fi le series that returns the equivalent resistance (across termi-

nals AB) shown in Figure 2.38, where R 1 = 10 Ω, R 2 = 10 Ω, R 3 = 20 Ω, and R 4 = 30 Ω.

MATLAB Solution

% Script file: series

Resist = [10 10 20 30];

Requi = sum(Resist);

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

disp(‘The equivalent resistance is=’); disp(Requi);

disp(‘ Ohms’)

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

>> series

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

The equivalent resistance is =

70 Ohms

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

FIGURE 2.38

Series connections of R 1 , R 2 , R 3 , and R 4 of Example 2.2.

R 1 = 10 Ω R 2 = 10 Ω R 3 = 20 Ω R 4 = 30 Ω

A

B