Correct Option: B

This problem can be easily solve by using superposition theorem
Case I. Take current source, the circuit becomes

I1 = Is ×		(1 + 1 || 1)	=	Is . 3 / 2
		(2 + 1 + 1 || 1)		2 + 3 / 2

=	3 Is	amp.
	7

Case II. Take voltage source, the circuit becomes

I =	V
	Req

where R_eq = equivalent resistance as seen from the terminal AB

R eq = 3 || 1 + 1 =		3	+ 1 =	7
		4		4

V = 2V

I =		2	=	8	amp.
		7/4		7

I 2 = 1 ×		1	=	8	.	1	=	2
		1 + 3		7		4		7

Now, the net current I = I₁ + I₂

I =		3Is	+	2
		7		7

But given that

I = 2 =		3Is	+	2
		7		7

or 7 × 2 = 3 I_s + 2
or 14 – 2 = 3 I_s
or I_s = 4 amp.