Two wires of the same metal have same length, but their cross-sections are in the ratio $$3:1.$$  They are joined in series. The resistance of thicker wire is $$10\,\Omega .$$  The total resistance of the combination will be

Solution :
Length of each wire $$ = \ell ;$$  Area of thick wire $$\left( {{A_1}} \right) = 3A;$$   Area of thin wire $$\left( {{A_2}} \right) = A$$   and resistance of thick wire $$\left( {{R_1}} \right) = 10\,\Omega .$$
Resistance $$\left( R \right) = \rho \frac{\ell }{A} \propto \frac{1}{A}\left( {{\text{if}}\,\ell \,{\text{is constant}}} \right)$$
$$\eqalign{ & \therefore \frac{{{R_1}}}{{{R_2}}} = \frac{{{A_2}}}{{{A_1}}} = \frac{A}{{3A}} = \frac{1}{3} \cr & {\text{or,}}\,{R_2} = 3{R_1} = 3 \times 10 = 30\,\Omega \cr} $$
The equivalent resistance of these two resistors in series
$$ = {R_1} + {R_2} = 30 + 10 = 40\,\Omega .$$