Page 25 - Circuit Analysis II with MATLAB Applications

Response of Series RLC Circuits with AC Excitation

The total solution is

it = i t + i t = k e – 200t k e – 300t + 20cos + 10000t 87.15 – o (1.33)
n
2
1
f
The constants k 1 and k 2 are evaluated from the initial conditions. From (1.33) and the first initial
condition i 0 = 5A we get
L
0
i0 = k e + k e 0 20cos – + 87.15 o = 5
2
1
i0 = k + k + 20 u 0.05 = 5
2
1
k + k = 4 (1.34)
1
2
We need another equation in order to compute the values of k 1 and k 2 . This equation will make use
dv
of the second initial condition, that is, v 0 = 2.5 V . Since i t = it = C-------- C , we differentiate
C
C
dt
(1.33), we evaluate it at t = 0 , and we equate it with this initial condition. Then,
di – 200t – 300t 5 o
-----= – 200k e – 300k e – 2 u 10 sin 10000t 87.15 – (1.35)
dt 1 2
and at t = , 0

di 6 o 5
–
----- = – 200k 300k 2 u 10 sin – 87.15 = – 200k 300k + 2 u 10 (1.36)
–
–
dt t = 0 1 2 1 2
Also, at t = 0 +
+ di +
Ri 0 + L----- v 0 + = 200cos 0 = 200
dt c
t = 0 +
di
and solving for ----- we get
dt
t = 0 +
di = 200 0.5 u – 2.5 195000 (1.37)
–
5
-----
------------------------------------------ =
dt + 10 – 3
t = 0
Next, equating (1.36) with (1.37) we get
– 200k 300k = – 5000
–
2
1
or
k + 1.5k = 25 (1.38)
1
2

1-13 Circuit Analysis II with MATLAB Applications
Orchard Publications