Question

A battery of 9 V is connected in series with resistors of $0.2\Omega ,0.3\Omega ,0.4\Omega ,0.5\Omega and12\Omega .$
How much current would flow through the $12\Omega$ resistor?

Answer 1

9 V is connected in series with resistors of 0.2Ω, 0.3Ω, 0.4Ω, 0.5Ω and 12Ω
so,
$R_{eq}=0.2+0.3+0.4+0.5+12=13.4$

so current $I$ is given as,

$I=\frac{V}{R}=\frac{9}{13.4}=0.67A$