AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The figure below represents the given circuit.
The current in the circuit is,
I(t)=I(∞)−[I(∞)−I(0)]e−t∕τ‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
The time constant, τ=‌

L

R

=‌

5

10

=0.5s
At t→∞, inductor behaves as short circuit hence circuit will be
I(∞)=‌

15

10

=1.5A‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
At t→0, inductor behaves as open circuit, I(0)=0
I(t)=1.5−[1.5−0]e‌

−t

(1∕2t)

At t=1 sec
I(1)=1.5(1−e−2t)‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(III)
From equation (II) and (III),
‌

I(∞)

I(0)

=‌

1.5

1.5(1−e−2)

‌

I(∞)

I(0)

=‌

e2

e2−1