Direct Current and Transient Analysis
195
with the boundary conditions given by
L
di t
dt
v
t
C
( )
( )
ϭ ϭ
ϭ
0
0
10V
or
di t
dt
i
t
( )
( )
ϭ ϭ
ϭ
0
40
0
0
V and
A
Note that L
di t
dt
C
i t
t
( )
( )
ϩ
ϭ
Ն
1
0
0
∫
for
satisfying KVL
MATLAB Solution
>> current _ it = dsolve(‘D2y+144*y=0’,’y(0)=0,Dy(0)=40’,’t’)
current _ it =
10/3*sin(12*t)
>> ezplot(it,[0 1])
% the plot is shown in Figure 2.96
>> xlabel(‘time t in sec’)
>> ylabel(‘i(t) in amps’)
>> title(‘i(t) vs t’)
FIGURE 2.95
Network of Example 2.28.
FIGURE 2.96
Plot of i(t) of Example 2.28.
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−4
−3
−2
−1
0
1
2
3
4
time t in sec
i(t) versus t
i(t) in amps
V 0 = 10 V
Switch moves down at t = 0
+
+
−
−
C 1 = 1/36 F
R = 2 Ω
i (t )
L = 1/4 H
CRC_47760_CH002.indd 195
CRC_47760_CH002.indd 195
7/23/2008 1:38:59 PM
7/23/2008 1:38:59 PM