In the given circuit, the terminal potential difference of the cell is :

First, find the equivalent resistance of the external circuit. The two resistors of $$4 \Omega$$ are connected in parallel. Let this equivalent resistance be $$R_{eq}$$.

$$ \frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} $$

$$ \frac{1}{R_{eq}} = \frac{2}{4} = \frac{1}{2} $$

$$ R_{eq} = 2 \Omega $$

Next, find the total resistance of the entire circuit. This includes the equivalent external resistance $$R_{eq}$$ and the internal resistance $$r$$ of the cell.

$$ R_{total} = R_{eq} + r $$

$$ R_{total} = 2 + 1 $$

$$ R_{total} = 3 \Omega $$

Now, calculate the total current $$I$$ flowing through the main circuit using the electromotive force (EMF) $$\varepsilon$$ of the cell.

$$ I = \frac{\varepsilon}{R_{total}} $$

$$ I = \frac{3}{3} $$

$$ I = 1 \text{ A} $$

The terminal potential difference $$V$$ is the voltage across the external circuit. We can calculate this using Ohm's law for the external equivalent resistance.

$$ V = I \times R_{eq} $$

$$ V = 1 \times 2 $$

$$ V = 2 \text{ V} $$

Alternatively, the terminal potential difference can also be calculated by subtracting the voltage drop across the internal resistance from the cell's EMF.

$$ V = \varepsilon - I \times r $$

$$ V = 3 - (1 \times 1) $$

$$ V = 3 - 1 $$

$$ V = 2 \text{ V} $$