The current $i$ in the circuit is determined by the formula:
\[ i = \frac{V_0}{R_\text{eff}} \] where $R_\text{eff}$ is the effective resistance of the circuit.
For the given configuration: \[ R_\text{eff} = R + 2R = 3R \]
A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).