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 would flow through the 12 Ω resistor?

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 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. 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 Ω

Potential difference, *V*= 9 *V*

*I*= 9/13.4 = 0.671 A

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.