Solution:

To find the internal resistance ($r$) of the battery, we use the following equations based on the given information:

1. Case 1: Current flows from negative to positive terminal

The potential difference is:

$$V_1 = E - I_1 r$$

Substitute

$V_1 = 10 \, \text{V}$

,

$I_1 = 3 \, \text{A}$

:

$$10 = E - 3r \tag{1}$$

2. Case 2: Current flows in reverse (from positive to negative terminal)

The potential difference is:

$$V_2 = E + I_2 r$$

Substitute

$V_2 = 15 \, \text{V}$

,

$I_2 = 2 \, \text{A}$

:

$$15 = E + 2r \tag{2}$$

3. Solve the two equations

From Equation (1):

$$E = 10 + 3r$$

Substitute

$E = 10 + 3r$

into Equation (2):

$$15 = (10 + 3r) + 2r$$

Simplify:

$$15 = 10 + 5r$$

$$5r = 5 \implies r = 1 \, \Omega$$

Final Answer:

The internal resistance of the battery is:

$$\boxed{1 \, \Omega}$$