Question

An electromotive force of a cell is .8 5 V and is connected to a load of 4.25Omega , the voltage drops to 2.0 V. Find the current and internal resistance of the cell.

Answer 1

$$I=\frac{\cal {E}}{R+r}$$

$$V=IR$$

Hence, the current

$$I=\frac{V}{R}=\frac{2.0}{4.25}=0.47\;\rm A$$

and, the internal resistance

$$r=\frac{\cal E}{I}-R=8.5/2.0-4.25=2.25\:\Omega$$