Question

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

Answer 1

For resistors in series,

$R_{eq}=R_1+R_2+R_3+R_4+R_5$

$=0.2+0.3+0.4+0.5+12$

$=13.4\Omega$

By ohm's Law:

$V=I{R}_{eq}$

$9=13.4I$

$I=0.67A$

When resistors are connected in series, current is same in all the resistors. Hence, current in $12\Omega$ resistor $=0.67A$.