a battery of 9V is connected in series with resistors of 0.2Ω,0.3Ω,0.4Ω,0.5Ωand 12Ω resistors. How much current would flow through th 12Ω resistors?

current flowing through 12Ω = 0.67 A

The resistances are given by 0.2Ω,0.3Ω,0.4Ω,0.5Ωand 12Ω

in series, we add the resistances, so the total resistance is given by

R=0.2+0.3+0.4+0.5+12=13.4 Ω

using ohm's law V = i R

9 = i (13.4)

i = 9/13.4

i=0.67 A