Question

Three resistors with resistances of 9 Ω, 18 Ω, and 30 Ω are in a series circuit with a 12 volt battery. What is the current flow through the resistors

Answer 1

Given:

The resistances in the series are,

$\begin{gathered} R_1=9\text{ }\Omega \\ R_2=18\text{ }\Omega \\ R_3=30\text{ }\Omega \end{gathered}$

The potential of the battery is,

$V=12\text{ V}$

To find:

The current flow through the resistors

Explanation:

The equivalent resistance is,

$\begin{gathered} R=R_1+R_2+R_3 \\ =9+18+30 \\ =57\text{ }\Omega \end{gathered}$

The current in all the resistors will be the same as the resistances are in a series combination.

The current is,

$\begin{gathered} i=\frac{V}{R} \\ =\frac{12}{57} \\ =0.21\text{ A} \end{gathered}$

Hence, the current is 0.21 A.