Given data
The resistance of the first resistor is R1 = 4 ohm
The resistance of the second resistor is R2 = 5 ohm
The potential difference of the battery is V = 9 V
The resistors are connected in series. The expression for the equivalent resistance is given as:
[tex]\begin{gathered} R=R_1+R_2_{} \\ R=4\text{ }\Omega+5\text{ }\Omega \\ R=9\Omega \end{gathered}[/tex]
The expression for the current in the 4-ohm resistor is given as:
[tex]\begin{gathered} I=\frac{V}{R} \\ I=\frac{9\text{ V}}{9\text{ }\Omega} \\ I=1\text{ A} \end{gathered}[/tex]
Thus, the magnitude of the current flows through the 4-ohm resistor is 1 A.
