A battery of emf 12 V and internal resistance 1 Ω is connected to an infinite ladder network. Each section consists of a series resistor of 1 Ω…

Let $R_x$ be the resistance of the whole ladder; removing one section leaves an identical ladder, so $R_x=1+\dfrac{2R_x}{2+R_x}$, which gives $R_x^2-R_x-2=0$ and $R_x=2\ \Omega$. Total resistance $=2+1=3\ \Omega$, so $I=12/3=4\ \mathrm{A}$. In the first section the series resistor carries $4$ A, dropping $4$ V. The node voltage after it is $V_1=12-4\times1-4\times1=4\ \mathrm{V}$ (internal drop $4$ V, series drop $4$ V). The first shunt $2\ \Omega$ and the rest of the ladder ($2\ \Omega$) are in parallel across $V_1$, so each carries $2$ A. Each following section repeats this: the current entering is halved at every shunt. The shunt currents are $2,\ 1,\ 0.5$ A, so the third shunt carries $0.5$ A and dissipates $P=I^2R=0.25\times2=0.50\ \mathrm{W}$.