In the above circuit the current in each resistance is:

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

1. Potentials Along the Bottom Rail

The first battery ($$2\text{V}$$) is encountered with the positive terminal on the right. Potential at the bottom of the first resistor = $$0\text{V} + 2\text{V} = 2\text{V}$$.

The second battery further increases the potential. Potential at the bottom of the second resistor = $$2 + 2 = 4\text{V}$$.

The third battery further increases the potential. Potential at the bottom of the third resistor = $$4\text{V} + 2\text{V} = 6\text{V}$$.

2. Potentials Along the Top Rail

The first battery ($$2\text{V}$$) is encountered with the positive terminal on the right. Potential at the top of the first resistor = $$0\text{V} + 2\text{V} = 2\text{V}$$.

The second battery ($$2\text{V}$$) follows. Potential at the top of the second resistor = $$2\text{V} + 2\text{V} = 4\text{V}$$.

The third battery ($$2\text{V}$$) follows. Potential at the top of the third resistor = $$4\text{V} + 2\text{V} = 6\text{V}$$.

3. Current in each resistor:

Resistor 1: $$\Delta V = V_{top} - V_{bottom} = 2\text{V} - 2\text{V} = 0\text{V} \implies I_1 = 0\text{A}$$

Resistor 2: $$\Delta V = V_{top} - V_{bottom} = 4\text{V} - 4\text{V} = 0\text{V} \implies I_2 = 0\text{A}$$

Resistor 3: $$\Delta V = V_{top} - V_{bottom} = 6\text{V} - 6\text{V} = 0\text{V} \implies I_3 = 0\text{A}$$