Answer:
[tex]r = 28.6 ohm[/tex]
Explanation:
As we know that when battery is in open circuit then the potential difference of the cell is known as EMF
so EMF is given as
[tex]EMF = 4.00 V[/tex]
now when the battery is connected across a resistance of 200 ohm then there is current flowing through the battery
it is given as
[tex]i = \frac{V}{R + r}[/tex]
[tex]17.5 \times 10^{-3} = \frac{4}{200 + r}[/tex]
[tex]200 + r = 228.6 [/tex]
[tex]r = 28.6 ohm[/tex]
