A battery of 9V is connected in series with resistors of 0.2`Omega,3Omega,0.4Omega` and `12Omega` respectively. How much current would flow through the 12`Omega` resistors ?

Explanation

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm.s law as
V=IR
I=V/R
Where.
R is the equivalent resistance of resistances `0.2Omega,0.3Omega,0.4Omega,0.05Omega` and `12Omega`. These are connected in series. Hence, the sum of the resistances will give the value of R
`R=0.2+0.3+0.4+0.5+12=13.4Omega`
Potential difference, V=9V
I=9/13.4=0.671A
Therefore, the current that would flow through the `12Omega` resistor is 0.671A