The potential difference across the terminals of the cell when the battery is short-circuited is zero.

The circuit diagram of a potentiometer for determining the $\mathrm{emf} \text{'}\mathrm{E}\text{'}$ of a cell of negligible internal resistance is as shown in figure:

Derive an expression to find out internal resistance of a cell.

The variation of terminal potential difference $\left(V\right)$ with current flowing through as cell is as shown

The $\mathrm{emf}$ and internal resistance of the cell are

Two cells of the same EMF $E$ but different internal resistances $r_1$ and $r_2$ are connected in series with an external resistance $R$ as shown in the figure. The terminal potential difference across, the second cell is found to be zero. The external resistance $R$ must then be:

A battery of e.m.f $12 \mathrm{V}$ and internal resistance $0.5 \mathrm{\Omega }$ is charged by a battery charger which supplies a $132 \mathrm{V}$ d.c. supply using a series resistance of $11.5 \mathrm{\Omega }$. What is the terminal voltage of the battery during charging?

A battery of $emf 12V$ and internal resistance $3\Omega$ is connected to a resistor. If the current in the circuit is $0.5A$,.Calculate terminal voltage of the battery when the circuit is closed.