Page 62 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

1
1
2
2
2
and Z = 1LC . Then, D = § © --------------------------- · 2 = Z = ---------------------- where R' = R + 2 : . Therefore,
e
P
0
0
112 ¹
1 12
e
2R' u
3 u
e
6
12
§
§ --------------------- · 2 = 4 , or ------------- · 2 = 4 , or R + 2 2 = 36 4 = 9 , or R + 2 = 3 and thus R = . 1
e
© 2R + 2 ¹ © R + ¹ 2

At t = 0 the circuit is as shown below.
6 :
+ v
3 : 1 : 6 :
i 0 L
+ +
v 0 C
12 V
From the circuit above
6
v 0 = v 6 : = --------------------- u 12 = 7.2 V
C
3 ++
6
1
and
v 7.2
6 :
i 0 L = --------- = ------- = 1.2 A
6
6
At t = 0 + the circuit is as shown below.
1 : 6 :

i t i t
L
R
i t 3H
C
+ `
2 : v t
C
112 F
e
Since the circuit is critically damped, the solution has the form

– D t
P
v t = e k + k t
2
1
C
1
·
where D P = § © --------------------------------------- = 2 and thus
21 +
112¹
e
2 u

v t = e – 2t k + k t (1)
C
1
2
1-50 Circuit Analysis II with MATLAB Applications
Orchard Publications