
The charge $Q$ is given as:
\[ Q = ne = \text{area under } I-t \text{ graph} \]
Substituting the values: \[ n = \frac{5000 \times 10^{-3}}{1.6 \times 10^{-19}} = 3.125 \times 10^{19} \]


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 \( r \).