Three equal-resistance resistors are first connected in parallel combination and then in series combination. What is the ratio of equivalent resistance in a series combination to the equivalent resistance in a parallel combination?

Given that,
${\mathrm{R}}_1={\mathrm{R}}_2={\mathrm{R}}_3=\mathrm{R}$
The equivalent resistance when 3 resistors are connected in series:${\mathrm{R}}_{\text{ser }}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3=\mathrm{R}+\mathrm{R}+\mathrm{R}=3\mathrm{R}$     

The equivalent resistance when 3 resistors are connected in parallel:
$\frac{1}{{\mathrm{R}}_{\text{para }}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}$                                
$\frac{1}{R_{\text{para }}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}$
$\therefore {\mathrm{R}}_{\text{para }}=\mathrm{R}/3$            

Ratio of equivalent resistances in a series combination to a parallel combination:

$\frac{{\mathrm{R}}_{\mathrm{ser}}}{{\displaystyle {\mathrm{R}}_{\mathrm{para}}}}= \frac{3\mathrm{R}}{{\displaystyle \frac{\mathrm{R}}{3}}}$

$\frac{{\mathrm{R}}_{\mathrm{ser}}}{{\displaystyle {\mathrm{R}}_{\mathrm{para}}}}= \frac{9}{{\displaystyle 1}}$

Therefore, the ratio of equivalent resistances in a series combination to a parallel combination will be 9:1.