Page 37 - Circuit Analysis II with MATLAB Applications

Response of Parallel GLC Circuits with DC Excitation

dv
–
------ = – 6.4kcos M 4.8ksin M
dt t = 0

By substitution of (1.70), the above expression simplifies to
dv
–
------ = – 32 4.8ksin M (1.72)
dt
t = 0
dv dv i
C
Also, i = C------ or ------ = ---- and
C
dt
dt
C
+ + +
dv = i 0 I i 0 – R i 0 –
L
C
S
------
-------------------------------------------
--------------- =
dt C C
t = 0 +
or
–
e
dv = I – v 0 R – i 0 10 5 50 – 2 7.9 u 640 = 5056 (1.73)
e
S
L
C
------------------------------------------------- =
------
------------------------------- =
dt C 1640
e
t = 0
Equating (1.72) with (1.73) we get
– 32 – 4.8ksin M = 5056
or
ksin M = – 1060 (1.74)
The phase angle can be found by dividing (1.74) by (1.70). Then,
M
ksin M – 1060
--------------- = tan M = --------------- = – 212
kcos M 5
or
M = – 1 – tan 212 = – 1.566 rads = – 89.73 deg

The value of the constant is found from (1.70) as
k
kcos – 1.566 = 5

or
5
k = ------------------------------ = 1042
cos – 1.566
and by substitution into (1.69), the total solution is
vt = 1042e – 6.4t cos 4.8t – 89.73q (1.75)
The plot is shown in Figure 1.16 where we have used the following MATLAB code:

1-25 Circuit Analysis II with MATLAB Applications
Orchard Publications