The resistance of the battery is 4.50 Ω.
Given data:
The emf of battery is e=15 V.
The resistance of external load is R= 18 Ω.
The voltage of battery is V=12 V.
The current passed by the battery can be calculated as,
[tex]\begin{gathered} I=\frac{V}{R} \\ I=\frac{12\text{ V}}{18\text{ }\Omega} \\ I=0.667\text{ A} \end{gathered}[/tex]The internal resistance of the battery is given by,
[tex]r=\frac{e-V}{I}[/tex]Substitute the values in above equation,
[tex]\begin{gathered} r=\frac{15\text{ V-12 V}}{0.667} \\ r=4.50\text{ }\Omega \end{gathered}[/tex]Thus, the resistance of the battery is 4.50 Ω.
