For the circuit shown below, the value of I1 = ?

The given circuit:

Applying KCL at node A, we get

V_A - 30	+	V_A - 36 - 6 l₁	= 2 . . . . . . .(i)
5		6

l₁ =	V_A - 30
	5

V_A = 5 l₁ + 30 . . . . . . . . . . (ii)
From equation (i) and (ii), we get

V_A	1	+	1	- l₁ = 14
	5		6

(5I₁ + 30)	6 + 5	- l₁ = 14
	30

or
55 I₁ + 330 – 30 I₁ = 420
or
25 I₁ = 420 – 330 = 90
or
I_{1_{= 90/25 = 18/5 = 3·6
Hence alternative (C) is the correct choice.}}