Every resistor depicted in the figure possesses a resistance of 10...

Let us analyze both circuits one by one.

For Circuit (a):
Resistor $a$ is connected directly in parallel with the left battery. Since the battery has an emf of $10\text{ V}$, the potential difference across resistor $a$ is $10\text{ V}$.
The current through resistor $a$ is:
$I_a = \dfrac{V}{R} = \dfrac{10\text{ V}}{10\text{ }\Omega} = 1\text{ A}$

Resistor $b$ is connected across the series combination of the two batteries. Observing the battery symbols, the left battery has its positive terminal on the left and negative on the right, while the right battery has its negative terminal on the left and positive on the right. They are connected such that their negative terminals face each other (opposing polarities).
The net potential difference across the combination of the two batteries is:
$V_{net} = 10\text{ V} - 10\text{ V} = 0\text{ V}$
Since resistor $b$ is connected across these two points, the potential difference across it is $0\text{ V}$.
The current through resistor $b$ is:
$I_b = \dfrac{0\text{ V}}{10\text{ }\Omega} = 0\text{ A}$

For Circuit (b):
Let the continuous bottom wire be the reference node at $0\text{ V}$.
The left battery connects the top-left node to the reference node with its positive terminal upwards. Thus, the potential at the top-left node is $10\text{ V}$.
Resistor $a$ is connected between the top-left node and the reference node ($0\text{ V}$).
The current through resistor $a$ is:
$I_a = \dfrac{10\text{ V} - 0\text{ V}}{10\text{ }\Omega} = 1\text{ A}$

Similarly, the right battery connects the top-right node to the reference node with its positive terminal upwards. Thus, the potential at the top-right node is also $10\text{ V}$.
Resistor $b$ is connected between the top-left node ($10\text{ V}$) and the top-right node ($10\text{ V}$).
The potential difference across resistor $b$ is:
$\Delta V_b = 10\text{ V} - 10\text{ V} = 0\text{ V}$
The current through resistor $b$ is:
$I_b = \dfrac{0\text{ V}}{10\text{ }\Omega} = 0\text{ A}$

Therefore, in both circuits, the current in $a$ is $1\text{ A}$ and the current in $b$ is $0\text{ A}$.

Answer: $1 \text{ A}$ in $a$ and $0 \text{ A}$ in $b$ in both circuits