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

For resistors in series, Req = R1 + R2 + R3 + R4 + R5 = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
By Ohm's Law: V = IReq
9 = 13.4I
I = 0.67 A
When resistors are connected in series, the current is the same in all the resistors. Hence, current in 12 Ω resistor = 0.67 A.