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

Correct 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 \( 1 \, \text{V} \).