December 2015
Intermediate to advanced
737 pages
23h 39m
English
Content preview from Network Analysis and Synthesis, 1st Edition by Pearson
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Transient Response of Circuits Using Differential Equations 253
Now, differentiating equation (6.19), we can write the equation as in the following:
di
dt
Ce Ce
tt
=− −
−−
05 45
1
05
2
45
..
..
Now applying the condition, the following can be obtained:
At
we get
t
di
dt
=
=
0
24
By substituting the values in the equation, we get the form as follows:
24
05 45
12
=− −
.,.
CC
(6.21)
By solving equation (6.20) and (6.21), we can get the value as in the following:
CC
12
0+=
(6.22)
−− =
05 45 24
12
..CC (6.23)
From equation (6.22),
CC
21
=−
; by substituting this value in equation (6.23), we get the value
as follows:
−+ =
05 45
24
11
..
CC
or
424
1
C
=
or
C
1
6
=
From equation (6.20), the v
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