Page 59 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

`
100 : 20 H

+
+ v t
C
100 V 1 120 F
e

The general form of the differential equation that describes this circuit is same as in Exercise 1,
that is,
2
d v R dv 1 100
C
C
---------- + --- -------- + -------v = --------- t ! 0
C
dt 2 L dt LC LC
2
d v dv C
---------- + 5 -------- + 6v = 600
C
C
dt 2 dt
2
From the characteristic equation s + 5s + 6 = 0 we find that s = – 2 and s = – 3 and the total
1
2
response for the capacitor voltage is
s t s t
v t = v + v Cn = 100 + k e 1 + k e 2 = 100 + k e – 2t + k e – 3t (1)
Cf
C
1
2
2
1
Using the initial condition V = 80 V we get
0
0 0
v 0 C = V = 80 V = 100 + k e + k e
1
2
0
or
k + k = – 20 (2)
1
2
Differentiation of (1) and evaluation at t = 0 yields
dv
-------- C = – 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 0.2
L
C
-------- = --------------- = ---------------- = 24 (4)
dt C 1120
e
t = 0
Equating (3) and (4) we get
– 2k – 3k = 24 (5)
2
1
and simultaneous solution of (2) and (5) yields k = – 36 and k = 16 .
1
2
1-47 Circuit Analysis II with MATLAB Applications
Orchard Publications