Twelve wires each having resistance $2\ \Omega$ are joined to form a cube. A battery of $6\ \text{V}$ emf is joined across points $a$ and $c$. The voltage difference between $e$ and $f$ is ______ V.
Numerical value type. Enter your answer.
Answer: 1
Take $V_a = 6\ \text{V}$ and $V_c = 0$. The cube is symmetric about the plane that bisects $ac$ perpendicularly; this swaps $a \leftrightarrow c$ and $h \leftrightarrow f$ while $b, d, e, g$ lie on that plane. So
$$V_b = V_d = V_e = V_g = 3\ \text{V},\qquad V_h + V_f = 6\ \text{V}$$
Node $h$ is joined to $a$, $e$ and $g$ by equal resistors, so (Kirchhoff's current law)
$$V_h = \frac{6 + 3 + 3}{3} = 4\ \text{V} \Rightarrow V_f = 2\ \text{V}$$
$$V_e - V_f = 3 - 2 = 1\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics