Correct Option: A

When the switch in the position B, as shown below.

Writing voltage equation.

Ri + L	di	= E2 ....(A)
	dt

Taking Laplace transform

RI(s) + 2[sI(s) – I(0)] =	E2
	s

or (R + 2) I(s) =	E2	– 16 (since, I(0) = – 8A)
	s

or I(s) =	E2	–	16
	s(R + 2s)		(R + 2s)

or I(s) =	E2		1	–	2		–	16
	R		s		R + 2s			(R + 2s)

(By using partial fraction)
Taking inverse Laplace transform
i(t) = E₂ R 1 – e^{[–(R/2) t]} – 8e^{–(R/2) t}....(B)

Given: i(t) =	E2	= 4 ....(i)
	R

Again	di(0)	= 3 =	E2	.	R	+ 4R ....(ii)
	dt		R		2

Solving equations (i) and (ii), we get
R = 0.5 ohm