Page 64 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

For t ! 0 the circuit is as shown below.

i t
L
R 1 2 : R 2 4 :

A B
L ` 2H 14 F + v t
e
C C

For this circuit
di
L
R + R 2 i + v + L------- = 0
C
1
L
dt
and with i = i = Cdv e dt the above relation can be written as

C
L
C
2
dv d v
C
C
R + R 2 C-------- + LC---------- + v = 0
C
1
dt
2
dt
2
d v R + R 2 dv C 1
1
---------- + ---------------------- -------- + -------v = 0
C
C
dt 2 L dt LC
2
d v dv
C
C
---------- + 3 -------- + 2v = 0
C
dt 2 dt
The characteristic equation of the last expression above yields s = – 1 and s = – 2 and thus
2
1
t –
v t = k e + k e – 2t (1)
1
C
2

With the initial condition v 0 = 12 V and (1) we get
C
k + k = 12 (2)
1
2
Differentiating (1) we get
dv t – – 2t
-------- = k – e – 2k e
C
dt 1 2
and
dv
C
-------- = k – – 2k (3)
dt 1 2
t = 0
dv i i
C
C
Also, -------- = ---- = ---- L and at t = 0
dt C C
1-52 Circuit Analysis II with MATLAB Applications
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