Case I: initially, the switch is at position 1, and by applying KVL, we get the following form:
60
10
120
it
di
dt
()
+=
Applying the Laplace transform, the equation is written as follows:
60
10
120
IssIs
s
()()
+=
or
()
()6010
120
+=
sIs
s
or
Is
ss
()
()
=
+
120
1060
=
+
=
+
=+
+
=-
+
120
106
12
6
6
22
6
ss
ss
A
s
B
s
ss
()
()
Taking the inverse Laplace transform, the equation is given as follows:
ite
t
()=-
-
22
6
= 2(1 -e
-6t
) A
Steady state currentLt=
=-
=
→∞
-∞
()
()
t
it
e21
2A
This current will act as the initial current for case II.
Case II: when the switch is at position 2, t= 0
+
. By applying KVL, we get the following form:
()
()60100++
⋅=
Rit
di
dt
Given
di
dt
=12
(60 ...
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