Page 61 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

v 0 = 0 = 60 2cos – 135q + k + k 2

1
C
and since cos – 135q = – 2 2 the above expression reduces to

e
k + k = 60 (2)
2
1
Differentiating (1) we get
dv – 2t – 3t
C
-------- = 60 2sin t + 45q + – 2k e – 3k e

2
1
dt
and
dv
-------- C = 60 2sin 45q – 2k 3k 2
–

1
dt
t = 0
dv
–
–
C
-------- = 60 2k 3k (3)
dt 1 2
t = 0
dv i i
C
C
Also, -------- = ---- = ---- L and at t = 0
dt C C

dv i 0
L
C
-------- = --------------- = 0 (4)
dt t = 0 C
Equating (3) and (4) we get
2k + 3k = 60 (5)
2
1
Simultaneous solution of (2) and (5) yields k = 120 and k = – 60 . Then, by substitution into (1)
2
1
2t
–
v t = 60 2cos t – 135q + 120e – 60e – 3t

C
5.
3 : t = 0 R 6 :
A S
B
3H i t
L
+ + `
2 : v t
C
12 V 112 F
e

R
We must first find the value of before we can establish initial conditions for i 0 L = 0 and
2 2
e
v 0 C = 0 . The condition for critical damping is D – Z = 0 where D P = G2C = 12R'C
e
P
0
1-49 Circuit Analysis II with MATLAB Applications
Orchard Publications