Page 65 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

dv i 0 6
L
-------- C = ------------ = ---------- = 24 (4)
dt C 14
e
t = 0
From (3) and (4)
k – – 2k = 24 (5)
1
2
and from (2) and (5) k = 48 and k = – 36 . By substitution into (1) we get
1
2
t –
v t = 48e – 36e – 2t
C
Then,
2
di d i
C
L
v AB = v t – v t = L------- – v t = LC ---------- – v t
C
L
C
C
dt
2
dt
t –
= 0.5 ------- 48e – § d 2 t – 36 e – 2t · – 48e – 36 e – 2t
© dt 2 ¹
t –
= 0.5 48e – t – 144 e – 2t – 48e – 36 e – 2t
t –
t –
= – 24e – 108e – 2t = – 24 e + 4.5e – 2t

1-53 Circuit Analysis II with MATLAB Applications
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