Page 63 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

0
With the initial condition v 0 = 7.2 V relation (1) becomes 7.2 = e k + 0 or k = 7.2 V

C
1
1
and (1) simplifies to
v t = e – 2t 7.2 + k t (2)
C
2
Differentiating (2) we get
dv – 2t – 2t
C
-------- = k e 2e – 7.2 + k t
2
2
dt
and
dv
C
=
-------- k – 27.2 + 0 = k – 14.4 (3)

2
2
dt
t = 0
dv i
C
Also, -------- = ---- and at t = 0
C
dt C
dv i 0 0
-------- C = ------------ = ---- = 0 (4)
C
dt C C
t = 0
because at t = 0 the capacitor is an open circuit.
Equating (3) and (4) we get k – 14.4 = 0 or k = 14.4 and by substitution into (2)
2
2
v t = e – 2t 7.2 + 14.4t = 7.2e – 2t 2t + 1

C
We find i t from i t + i t + i t = 0 or i t = – i t – i t where i t = Cdv e dt

C
C
C
L
C
R
R
L
L
and i t = v t R 1 + 2 = v t 3 . Then,
e
e
C
R
1 7.2 –
i t = – ------ – 12 14.4e – 2t 2t + 1 + 14.4e – 2t -------e – 3 2t 2t + 1 = – 2.4e – 2t t + 1

L
6.

At t = 0 the circuit is as shown below where i 0 L = 12 2 = 6A , v 0 C = 12 V , and thus
e
the initial conditions have been established.
2 : 4 :
+
A B
12 V ` 2H +
C
v 0
e
i 0 L 14 F
1-51 Circuit Analysis II with MATLAB Applications
Orchard Publications