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?

Since all the resistors are connected in series, the current flowing through all resistors will be the same. 

Given, 

$$\begin{gathered} {\mathrm{R}}_1=0.2 \Omega ; \\ {\mathrm{R}}_2=0.3 \Omega ; \\ {\mathrm{R}}_3=0.4 \Omega ; \\ {\mathrm{R}}_4=0.5 \Omega \& \\ {\mathrm{R}}_5=12 \Omega \end{gathered}$$

And voltage, $\mathrm{V} =9 \mathrm{V}$

Equivalent resistance,

                                        $\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3+{\mathrm{R}}_4+{\mathrm{R}}_5$

Substituting the values,

                                        $\mathrm{R}=0.2+0.3+0.4+0.5+12 = 13.4 \mathrm{\Omega }$

By Ohm's law,

Current flowing through the circuit, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Substituting the values in the above equation,

                                         $\mathrm{I}=\frac{9}{13.4}=0.672 \mathrm{A}$

Therefore, the current that would flow through the $12 \mathrm{\Omega }$ resistor is $0.672 \mathrm{A}$.