Biomedical Engineering Reference
In-Depth Information
Solution
For
t
0, the circuit is redrawn for analysis in the following figure.
C
L
400
Ω
100
Ω
+
+
i
C
i
L
+
−
15 V
v
C
5
m
F
v
L
10 mH
500
Ω
−
−
Summing the currents leaving node C gives
v
C
15
v
C
þ
v
C
v
L
10
6
þ
5
100
¼
0
400
which simplifies to
v
C
þ
2500
v
C
2000
v
L
¼
7500
Summing the currents leaving node L gives
10
3
Z
t
0
v
L
v
C
100
1
v
L
d
l þ
i
L
0ðÞþ
v
L
þ
500
¼
0
10
which, after multiplying by 500 and differentiating, simplifies to
10
3
6
v
L
þ
50
v
L
5
v
C
¼
0
Using the
D
operator method, the two differential equations are written as
Dv
C
þ
2500
v
C
2000
v
L
¼
7500 or
ð
D
þ
2500
Þ
v
C
2000
v
L
¼
7500
v
L
10
3
10
3
6
Dv
L
þ
50
v
L
5
Dv
C
¼
0or 6
D
þ
50
5
Dv
C
¼
0
We then solve for
v
L
from the first equation,
v
C
10
3
v
L
¼
0
:
5
D
þ
1
:
25
3
:
75
and then substitute
v
L
into the second equation, giving
v
L
0
v
C
10
3
10
3
10
3
6
D
þ
50
5
Dv
C
¼
6
D
þ
50
:
5
D
þ
1
:
25
3
:
750
5
Dv
C
¼
0
Reducing this expression yields
2
10
3
10
6
10
6
D
v
C
þ
10
:
417
Dv
C
þ
20
:
83
v
C
¼
62
:
5
Continued
