Question

Twelve wires, each having equal resistance of 2 $\Omega$, are joined to form a cube. What is the equivalent resistance between the ends of a face diagonal?

• A. $\frac{3}{2}\Omega$
• B. $\frac{2}{3}\Omega$
• C. $\frac{5}{2}\Omega$
• D. $\frac{2}{5}\Omega$

Answer 1

The correct option is A

$\frac{3}{2}\Omega$

Im going to take a current I flowing in through A due to an externally connected cell V. Let the current going into AC be i. That must be the same current coming through EB into node B because those two arms are symmetric. I’m just calling a junction a node. Now, Currents through arms AD and AH must be the same. Let it be $\frac{(I-i)}{2}$. Now currents must divide at junction C symmetrically as $\frac{i}{2}$ and $\frac{i}{2}$. There must be no currents through the arms HF and DG as the currents are the same through AD and DB; and hence CF and FE.

Thus if there is a cell connected between A and B with voltage V,
Taking loop ABEA,
$\frac{2(I-i)}{2}+\frac{2(I-i)}{2}-V=0V=2(I-i)\to \left(1\right)NowtakeloopACFEBA,2i+\frac{2i}{2}+\frac{2i}{2}+2i-V=0V=6ii=\frac{V}{6}Pluggingthisbackinto\left(1\right),V=2I-\frac{V}{3}\frac{4V}{3}=2I\frac{V}{I}=\frac{3}{2}\Omega$
The resistance between points A and B is $\frac{3}{2}\Omega$.