Question

A wire of resistance 15$\cdot 0\Omega$ is bent to form a regularhexagon $ABCDEFA$ . Find the equivalent resistance ofthe loop between the points $\left(a\right)A$ and $B,$ (b) $A$ and $C$and $\left(c\right)A$ and $D.$

Answer 1

a) $R_{eff}=\frac{\frac{15\times 5}{6}\times \frac{15}{6}}{\frac{15\times 5}{6}+\frac{15}{6}}=\frac{\frac{15\times 5\times 15}{6\times 6}}{\frac{75+15}{6}}$

$=\frac{15\times 5\times 15}{6\times 90}=\frac{25}{12}=2.08\Omega$

b) Across AC,

$R_{eff}=\frac{\frac{15\times 4}{6}\times \frac{15\times 2}{6}}{\frac{15\times 4}{6}+\frac{15\times 2}{6}}=\frac{\frac{15\times 4\times 15\times 2}{6\times 6}}{\frac{60+30}{6}}$

$=\frac{15\times 4\times 15\times 2}{6\times 90}=\frac{10}{3}=3.33\Omega$

c) Across AD,

$R_{eff}=\frac{\frac{15\times 3}{6}\times \frac{15\times 3}{6}}{\frac{15\times 3}{6}+\frac{15\times 3}{6}}=$

$=\frac{15\times 3\times 15\times 3}{6\times 90}=\frac{15}{4}=3.75\Omega$