Given:
The resistances in the series are,
[tex]\begin{gathered} R_1=9\text{ }\Omega \\ R_2=18\text{ }\Omega \\ R_3=30\text{ }\Omega \end{gathered}[/tex]The potential of the battery is,
[tex]V=12\text{ V}[/tex]To find:
The current flow through the resistors
Explanation:
The equivalent resistance is,
[tex]\begin{gathered} R=R_1+R_2+R_3 \\ =9+18+30 \\ =57\text{ }\Omega \end{gathered}[/tex]The current in all the resistors will be the same as the resistances are in a series combination.
The current is,
[tex]\begin{gathered} i=\frac{V}{R} \\ =\frac{12}{57} \\ =0.21\text{ A} \end{gathered}[/tex]Hence, the current is 0.21 A.
