PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

188 Practical MATLAB® Applications for Engineers

ANALYTICAL Solution

The loop differential equation is given by

V

di t

dt

1 

^20 Vf^2 it ort^0

()

()

with

A and ii() 0 ( ) A

10

2

5

20

2



 

 10

Assuming a solution for i(t) of the form i(t) = A + Be−t/τ, where τ = L/R = 0.5 s, and by

using the initial and fi nal current conditions, the constants A and B can be evaluated.

By following this process the current is then i(t) = 10 − 5 e−^2 t A, and the corresponding

voltages are vR(t) = 20 − 10 e−^2 t V and vL(t) = 10e−^2 t V, for t ≥ 0.

MATLAB Solution

% Script file: RC _ IC

syms t;

% symbolic solution

it _ 1 = dsolve(‘Dy+2*y=20’,’y(0)=5’,’t’);

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

figure(1)

subplot(3,1,1);

ezplot (it _ 1,[0 2]);ylabel(‘i(t) in amps’);

title(‘plots of: [i(t), vr(t) and vL(t)] vs t’);

subplot(3,1,2);

vtr _ 1 = it _ 1*2;

ezplot(vtr _ 1,[0 2]);ylabel(‘vr(t)’);title(‘ ‘);

subplot (3,1,3);

vl_1 = 20-vtr_1;

ezplot (vl _ 1,[0 2]);ylabel(‘vL(t)’);title(‘ ‘);

xlabel (‘time in sec.’);

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

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

disp(‘The i(t) for t >0 is:’);

it _ 1

disp(‘The voltage across R=2 ohms is :’)

vtr _ 1

disp(‘The voltage across L=1 henry is :’)

vl _ 1

disp(‘********Verify solution******************’);

echo on

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

verify = diff(it _ 1,t)+2*it _ 1

echo off

% numerical solution

clear;

% solution using ode45

t0 = 0;R=2;

tf = 2;

ts = [t0 tf];

i0 = 5; % initil cond.

t =linspace(0,2,100);

[t,it] = ode45(‘f’,ts,i0);