Question

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 ______ $\text{V}$.

Answer 1

Analyze the Symmetry of the Cube:
By symmetry, the current through the branches $e-b$ and $g-d$ is zero, as these branches are equidistant from points $a$ and $c$.
Thus, we can ignore these branches in our analysis.

Determine the Equivalent Resistance of the Cube:
After ignoring the branches $e-b$ and $g-d$, the remaining network of resistances can be simplified. The equivalent resistance $R_{\text{eq}}$ between points $a$ and $c$ is:
$R_{\text{eq}} = \frac{3}{2} \, \Omega$

Calculate the Current Through the Battery:
The total current $I$ supplied by the battery with emf $6 \, \text{V}$ is:
$I = \frac{V}{R_{\text{eq}}} = \frac{6}{\frac{3}{2}} = 4 \, \text{A}$

Determine the Current Through Each Branch:
Due to the symmetry of the cube, the current divides equally among the paths. The current $i_2$ through each resistor in the branches involving $e$ and $f$ is:
$i_2 = \frac{4}{8} \times 2 = 1 \, \text{A}$

Calculate the Voltage Difference Between Points $e$ and $f$:
The voltage difference $\Delta V$ between points $e$ and $f$ across a single $2 \, \Omega$ resistor is:
$\Delta V = i_2 \times R = 1 \times 1 = 1 \, \text{V}$

Conclusion:
The voltage difference between $e$ and $f$ is $1 \, \text{V}$.