Direct Current and Transient Analysis
179
ANALYTICAL Solution
Part a
The loop differential equation, as well as its current solution i(t), is given as follows:
L
di t
dt
Ri t
V
( )
( )
ϩ
ϭ 0
then
i t
V
R
V
R
e
t
R
t
( ) ϭ
Ն
ϭ
0
0
0
Ϫ
⍀
Ϫ ր
for
for
1, 2, and 5
where τ = L/R.
MATLAB Solution
% Script file: RL
% parts ( b and c )
itR _ 1 = dsolve(‘Dy+y =10’,’y(0) = 0’,’t’);
itR _ 2 = dsolve(‘Dy+2*y=10’,’y(0)=0’,’t’);
itR _ 5 = dsolve(‘Dy+5*y=10’,’y(0)=0’,’t’);
disp(‘***************************’);
disp(‘The solutions i(t) for R=1,2 and 5 (in Ohms) are:’);
disp(‘***************************’);
disp(‘***************************’);
itR _ 1
disp(‘***************************’);
itR _ 2
disp(‘***************************’);
itR _ 5
disp(‘***************************’);
disp(‘***************************’);
ezplot(itR _ 5,[0 3]);
hold on;
grid on; ezplot(itR _ 2,[0 3]);
hold on;
ezplot(itR _ 1,[0 3]);
hold on;
title(‘i(t) vs t for L=1 & R=1,2 and 5’)
axis([0 3 0 11]);
xlabel(‘time t’);ylabel(‘i(t)’)
FIGURE 2.81
Network of Example 2.23.
V 0 = 10 V
R
L = 1 H
Switch closes at t = 0
i (t )
CRC_47760_CH002.indd 179
CRC_47760_CH002.indd 179
7/23/2008 1:38:55 PM
7/23/2008 1:38:55 PM