Question

Emf of a cell is 1.5V and its internal resistance is 1ohm.For what current drawn from the cell will the terminal potential difference be half of its emf?

Answer 1

Terminal potential of the battery is given by:
V = E - Ir
where E is the emf of the battery
I is the current
r is the internal resistance of the battery.

$$\begin{gathered} V=\frac{E}{2} \\ r=1ohm \\ \frac{E}{2}=E-I\times 1 \\ I=E-E/2 \\ E=1.5v \\ I=1.5/2=0.75A \end{gathered}$$