Determine the currents passing through the resistances in the...

Assuming the diodes to be ideal, they act as short circuits when forward-biased and open circuits when reverse-biased.

In circuit (a), both diodes are forward-biased as their p-sides are connected towards the positive terminal of the battery. They act as short circuits. The current is $I_a = \dfrac{V}{R} = \dfrac{2}{2} = 1\text{ A}$.

In circuit (b), the first diode is forward-biased but the second diode is reverse-biased. The reverse-biased diode acts as an open circuit, breaking the series circuit. Thus, the current is $I_b = 0$.

In circuit (c), both diodes are in parallel and both are forward-biased. They act as short circuits. The equivalent resistance of the diode combination is zero. The current is $I_c = \dfrac{V}{R} = \dfrac{2}{2} = 1\text{ A}$.

In circuit (d), the top diode is forward-biased (acts as a short circuit) and the bottom diode is reverse-biased (acts as an open circuit). The current flows entirely through the top branch. The equivalent resistance of the parallel combination is zero. The current is $I_d = \dfrac{V}{R} = \dfrac{2}{2} = 1\text{ A}$.

The currents are $1\text{ A}$, $0$, $1\text{ A}$, and $1\text{ A}$ respectively.