A battery of emf $E$ and internal resistance $r$ is connected to a rheostat. When a current of 2 A is drawn from the battery, the potential difference across the rheostat is $\mathbf{5 ~ V}$. The potential difference becomes 4 V when a current of 4 A is drawn from the battery. Calculate the value of $E$ and $r$.

Briefly explain why and how a galvanometer is converted into an ammeter.

The potential difference applied across a given conductor is doubled. How will this affect (i) the mobility of electrons and (ii) the current density in the conductor? Justify your answers.