Five resistors carrying $\frac{1}{5} \mathrm{\Omega }$ resistance each are connected in series. Find the net equivalent resistance.

If resistances in a circuit are connected in series, the total resistance is equal to the sum of the individual resistances. 

So, the total resistance (R) will be 
$\mathrm{R} =\mathrm{ }{\mathrm{R}}_1\mathrm{ }+\mathrm{ }{\mathrm{R}}_2\mathrm{ }+\mathrm{ }{\mathrm{R}}_3+\mathrm{ }{\mathrm{R}}_4+{\mathrm{R}}_5$   

Where ${\mathrm{R}}_1$_, ${\mathrm{R}}_2$_, and ${\mathrm{R}}_3$ , ${\mathrm{R}}_4 \mathrm{and} {\mathrm{R}}_5$ are individual resistance of the resistors. In this case 
Total resistance $\left(\mathrm{R}\right)$ $=\mathrm{ }\frac{1}{5}\mathrm{ }+\mathrm{ }\frac{1}{5}\mathrm{ }+\mathrm{ }\frac{1}{5}\mathrm{ }+\mathrm{ }\frac{1}{5}\mathrm{ }+\mathrm{ }\frac{1}{5}$
                                 $=\frac{5}{5}$
                                 $=1 \mathrm{\Omega }$
Five resistors carrying $\frac{1}{5} \mathrm{\Omega }$ resistance each are connected in series. The equivalent resistance will be$1\mathit{ \Omega }$.