Determine the current in the three resistors depicted in the figure.

Let the potential of the bottom-left node be $0\text{ V}$.

Since the top-left node is connected to the bottom-left node by an ideal connecting wire, its potential is also $0\text{ V}$.

Moving along the top branch from left to right, the potential decreases by $2\text{ V}$ across each battery because the positive terminal (longer line) is on the left.
Potential at the top of the first resistor $= 0 - 2 = -2\text{ V}$.
Potential at the top of the second resistor $= -2 - 2 = -4\text{ V}$.
Potential at the top of the third resistor $= -4 - 2 = -6\text{ V}$.

Similarly, moving along the bottom branch from left to right, the potential also decreases by $2\text{ V}$ across each battery.
Potential at the bottom of the first resistor $= 0 - 2 = -2\text{ V}$.
Potential at the bottom of the second resistor $= -2 - 2 = -4\text{ V}$.
Potential at the bottom of the third resistor $= -4 - 2 = -6\text{ V}$.

The potential difference across each of the three resistors is:
$\Delta V_1 = -2\text{ V} - (-2\text{ V}) = 0\text{ V}$
$\Delta V_2 = -4\text{ V} - (-4\text{ V}) = 0\text{ V}$
$\Delta V_3 = -6\text{ V} - (-6\text{ V}) = 0\text{ V}$

Since the potential difference across each resistor is zero, the current flowing through all three resistors is $0\text{ A}$.

Answer: $0\text{ A}$