The correct answer is option C, 1.13 A.
Given,
The resistance of the resistor, R=5 Ω
The internal resistance of the battery, r=3 Ω
The emf of the battery, E=9 V
The current through the battery is given by Ohm's law. According to whch,
[tex]I=\frac{E}{R+r}[/tex]Where I is the current through the resistor R.
On substituting the know values,
[tex]\begin{gathered} I=\frac{9}{5+3} \\ =1.13\text{ A} \end{gathered}[/tex]Therefore the current in the 5 Ω resistor is 1.13 A