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`. How much current would flow through the `12 Omega` resistor ?

Explanation

There is no current division occurring in a series circuit. Current flow throught the component is the same, given by Ohm's law as
V=IR
I=V/R
Where,
R is the equivalent resistance of resistances `0.2 Omega, 0.3 Omega, 0.4 Omega, 0.5 Omega and 12 Omega`. These are connected in series. Hence, the sum of the resistances will give the volume of R. `R=0.2+0.3+0.4+0.5+12=13.4 Omega`
Potantial difference, V = 9 V
I=9/13.4=0.671 A
Therefore, the current that would flow through the `12Omega` resistor is 0.671 A.