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Question

In the circuit shown, the switch S is open for a long time and is closed at t = 0. The current i(t) for t0+ is

A
i(t)=0.50.125 e1000tA
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B
i(t)=1.50.125 e1000tA
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C
i(t)=0.50.5 e1000tA
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D
i(t)=0.375 e1000tA
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Solution

The correct option is A i(t)=0.50.125 e1000tA
Method 1:
At t=0, the circuit is shown below:

Current will divide equally in both 10 Ω resistors as shown in diagram.
At t0+, the circuit in s-domain is shown below:

[L=15 mH,i(0+)=i(0)=0.75 A]
The above circuit can be rearranged as given below:

Applying nodal analysis,
V(s)1.5s10+V(s)10+V(s)+15×103×0.7510+15×103 s=0
V(s)10+V(s)10+V(s)10+15×103 s
=1510s15×103×0.7510+15×103 s
V(s)[10+15×103 s+55(10+15×103 s)]
=15[10+15×103 s10 s×103×0.7510 s(10+15×103 s)]
(1+103 s)V(s)=(10+7.5×103 s)2 s[s+10001000]
V(s)=10000+7.5 s2000 s
V(s)=10000+7.5 s2 s(s+1000)
=100002×1000s+{10000+7.5(1000)}2(1000)(s+1000)
=5s1.25s+1000
Taking inverse Laplace transform,
v(t)=51.25 e1000tV
i(t)=V(t)10
=0.50.125 e1000tA
Method 2:
i(0+)=0.752=0.375 A
i()=0.5 A
i(t)=i(){i()i(0+)}eRt/L
where, R = equivalent resistance seen across L with current source opened

R=10+(10||10)=15 Ω
i(t)=0.5{0.50.375}e15t/15×103
=0.50.125 e1000tA

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