As shown in diagram the loop of resistance shown by arrow is repeated at times, let us assume the equivalent resistance is x, so by adding one more loop the resistance will remain as x. x = 2 + $\frac{x}{1+x}$ (1 + x)x = 2 (1 + x) + x x + $x^2$ = 2 + 3 x $x^2$ – 2x – 2 = 0 x = $\frac{2\pm\sqrt{4-4\times 1(-2)}}{2}$ = $(1+\sqrt{3})$ Ω = 2.732 Ω Now, current drawn I = $\frac{12}{3.23}$ = 3.71 A