A semi-infinite ladder network is made of identical sections. Each section has a 1 Ω resistor in series along the top rail followed by a 2 Ω resistor…

Let the input resistance of the infinite ladder be $R_x$. Removing one section leaves an identical ladder, so $R_x=1+\dfrac{2R_x}{2+R_x}$, giving $R_x^2-R_x-2=0$ and $R_x=2\ \Omega$. The battery current is $I=\dfrac{12}{1+2}=4\ \mathrm{A}$ and the terminal voltage is $12-4\times1=8\ \mathrm{V}$. The first series resistor drops $4\times1=4\ \mathrm{V}$, so the voltage at the first node (across the first $2\ \Omega$ shunt resistor) is $8-4=4\ \mathrm{V}$. Check: the shunt $2\ \Omega$ is in parallel with the remaining ladder of $2\ \Omega$, giving $1\ \Omega$, and $4\ \mathrm{A}\times1\ \Omega=4\ \mathrm{V}$. The power in the first shunt resistor is $P=\dfrac{V^2}{R}=\dfrac{4^2}{2}=8\ \mathrm{W}$.