An infinite ladder is constructed with 1\ and 2\ resistors. It...

Let the equivalent resistance of the infinite ladder across the input terminals be $R_{eq}$.

Since the ladder is infinite, removing the first repeating unit leaves the equivalent resistance unchanged. Thus, the circuit can be modeled as a $1\ \Omega$ resistor in series with a parallel combination of a $2\ \Omega$ resistor and $R_{eq}$.

The equivalent resistance is given by:

$R_{eq} = 1 + \dfrac{2 R_{eq}}{2 + R_{eq}}$

$R_{eq}(2 + R_{eq}) = 2 + R_{eq} + 2 R_{eq}$

$R_{eq}^2 + 2 R_{eq} = 2 + 3 R_{eq}$

$R_{eq}^2 - R_{eq} - 2 = 0$

$(R_{eq} - 2)(R_{eq} + 1) = 0$

Since resistance cannot be negative, $R_{eq} = 2\ \Omega$.

The total current drawn from the $6\text{ V}$ battery is:

$I = \dfrac{V}{R_{eq}} = \dfrac{6}{2} = 3\text{ A}$

This total current flows through the first $1\ \Omega$ resistor. At the first junction, the current splits into two parallel branches: the first $2\ \Omega$ resistor and the rest of the infinite ladder.

The resistance of the rest of the infinite ladder is also $R_{eq} = 2\ \Omega$. Since the two parallel branches have equal resistance ($2\ \Omega$ each), the current divides equally between them.

The current passing through the $2\ \Omega$ resistor closest to the battery is:

$I_{2\Omega} = \dfrac{I}{2} = \dfrac{3}{2} = 1.5\text{ A}$