Question

A battery of $9V$ is connected in series with resistors of $0.2$ ohms, $0.3$ ohms, $0.4$ ohms, $0.5$ ohms and $12$ ohms respectively. How much current would flow through the $12\Omega$ resistor?

Answer 1

Step 1. Given data:

Battery voltage = $9V$

Resistors in series of = $R_1$=$0.2$ ohms,$R_2$= $0.3$ ohms,$R_3$= $0.4$ ohms,$R_4$= $0.5$ ohms, and $R_5$=$12$ ohms

We have to find the current through $12\Omega$ resistor.

Step 2. Formula used:

We know, that the net resistance in a series circuit is given by:

$R_{net}=R_1+R_2+R_3+....$

And according to ohm's law, $V=IR$, where $V=$voltage, $I=$current, $R=$Net resistance of series combination,

Step 3. Calculations:

Let R be the equivalent resistance in series

$$\begin{gathered} R=R_1+R_2+R_3+R_4+R_5 \\ R=0.2+0.3+0.4+0.5+12=13.4\Omega \end{gathered}$$

When resistors are connected in series, the current is the same in all the resistors.

Now, current through $12\Omega$ resistor, $I=\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$R_{}$}\right.=\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$13.4$}\right.=0.671A$

Hence, the current in $12\Omega$ resistor is $0.671A$.