Page 53 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

1.
i t
L
`
10 : 0.2 H +
+ v t
C
100u t V i 8mF
0

di
Ri + L----- + v = 100 t ! 0
dt C
dv
C
and since i = i = C-------- , the above becomes
C
dt
2
dv d v
C
C
RC -------- + LC ---------- + v = 100
dt dt 2 C
2
d v R dv C 1 100
C
---------- + --- -------- + -------v = ---------
C
dt 2 L dt LC LC
2
d v 10 dv 1 100
C
C
---------- + ------- -------- + --------------------------------- v = ---------------------------------
C
dt 2 0.2 dt 0.2 u 8 u 10 – 3 0.2 u 8 u 10 – 3
2
d v dv C
C
---------- + 50 -------- + 625 v = 62500
C
dt 2 dt
From the characteristic equation
2
s + 50s + 625 = 0
we get s = s = – 25 (critical damping) and D = R2L = 25
e
1
S
2
The total solution is
– D t – 25 t
S
v t = v + v Cn = 100 + e k + k t = 100 + e k + k t (1)

Cf
2
2
1
1
C
0

With the first initial condition v 0 C = 0 the above expression becomes 0 = 100 + e k + 0 or
1
k = – 100 and by substitution into (1) we get
1
v t = 100 + e – 25 t k t – 100 (2)
2
C

To evaluate k 2 we make use of the second initial condition i 0 L = 0 and since i = i C , and
L
dt
i = i = Cdv C e , we differentiate (2) using the following MATLAB code:
C
1-41 Circuit Analysis II with MATLAB Applications
Orchard Publications