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

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 0.2 Ohm, 0.3 Ohm, 0.4Ohm, 0.5 Ohm and 12 Ohm. 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 = 13.4 Ohm
Potential difference, V= 9 V
I= 9/13.4 = 0.671 A
Therefore, the current that would flow through the 12 Ohm resistor is 0.671 A.

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