Solution 2.9.6.8
R
i(t)
e(t)
L
i
L
(0)
Figure 1: Paralell RL circuit
For the circuit shown in Figure 1 we knowfromKircho 's currentlawthat
The di erential equation governing the circuit is
i(t)=i
R
(t)+i
L
(t):
Now
i
R
(t)=
v(t)
R
+
Acos!t
R
:
Applying the Laplace transform wehave
I
R
(s)=
As
R(s
2
+ !
2
:
For the inductor we knowthat
v(t)=v
L
(t)=L
d
dt
i
L
(t):
Applying the Laplace trasform, wehave
V (s)=
As
s
2
+ !
2
= L[sI
L
(s); i
L
(0)] = LsI
L
(s) ; Li
L
(0):
Thus
I
L
(s) =
1
Ls
"
As +(s
2
+ !
2
)Li
L
(0)
s
2
+ !
2
#
=
(A=L)s +(s
2
+ !
2
)i
L
(0)
s(s
2
+ !
2
)
=
1
Then the total currentis
I(s) = I
R
(s)+I
L
(s)
=
As
R(s
2
+ !
2
)
+
(A=L)s+(s
2
+ !
2
)i
L
(0)
s(s
2
+ !
2
)
=
(A=R)s
2
+(s
2
+ !
2
)i
L
(0)+ (A=L)s
s(s
2
+ !
2
)
=
B
s
+
M
s ; j!
+
M
s + j!
:
All that remains is to nd B and M.Tothat end,
B =
"
s[(A=R)s
2
+(s
2
+ !
2
)i
L
(0)+ (A=L)s]
s(s
2
+ !
2
)
#
j
s=0
=
i
L
(0)!
2
!
2
= i
L
(0)
M =
"
(s ; j!)[(A=R)s
2
+(s
2
+ !
2
)i
L
(0)+ (A=L)s]
s(s ; j!)(s+ j!)
#
j
s=j!
=
;(A=R)!
2
+(;!
2
+ !
2
)i
L
(0)+ (A=L)j!]
j!(j2!)
=
;(A=R)!
2
+(A=L)j!
;2!
2
=
A
2R
;
A
j2!L
= (A=2)(R+ j=!L)
= (A=2)
p
R
2
+ !
2
6
;;
where
=tan
;1
(L=R):
Then
i(t)=[i
L
(0)+ A
p
R
2
+ !
2
cos(!t+ )]1(t):
2
