Show how would you join three resistors, each of resistance 9Ω so that the equivalent resistance of the combination is (i) 13Ω (ii) 6Ω?

The possible way of the combination are:

The resistor of 9Ω can combine parallel and one in series.

$\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\right)+{R}_{3}\phantom{\rule{0ex}{0ex}}\left(\frac{1}{9}+\frac{1}{9}\right)+9\phantom{\rule{0ex}{0ex}}\left(\frac{2}{9}\right)+9\phantom{\rule{0ex}{0ex}}4.5\Omega +9\Omega =13.5\Omega $

(ii) 2 resistors connected in series and one in parallel.

R_{1} + R_{2} =

= 9Ω + 9Ω = 18Ω

Now

18Ω and 9Ω are connected

$\frac{18x9}{18+9}=6\Omega $