An infinite ladder network is built from identical sections. Each section has a 1 Ω resistor in series along the top rail, followed by a 2 Ω resistor…

Let $R$ be the equivalent resistance of the infinite ladder. Removing the first section leaves an identical ladder, so $R=1+\dfrac{2R}{2+R}$. This simplifies to $R^2-R-2=0$, whose positive root is $R=2\ \Omega$. The battery current is $I=12/2=6\ \mathrm{A}$. The first series $1\ \Omega$ resistor drops $6\ \mathrm{V}$, leaving $6\ \mathrm{V}$ across the first shunt $2\ \Omega$ resistor and across the rest of the ladder (also $2\ \Omega$ equivalent in parallel). The current through the first shunt is $6/2=3\ \mathrm{A}$, and the other $3\ \mathrm{A}$ continues into the remaining ladder, consistent with $6/2$ for the sub-ladder. A numerical truncation at 80 sections reproduces $R=2\ \Omega$.