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

Answer 2

Step 1: Given information.
Twelve wires, each having resistance \( 2 \, \Omega \), are connected to form a cube.
A battery of \( 6 \, \text{V} \) is connected across the opposite vertices \( a \) and \( c \).
We are required to find the potential difference between points \( e \) and \( f \).

Step 2: Concept – Symmetry of the cube.
In a cube made of identical resistors, when a potential difference is applied between opposite vertices, all points that are symmetrically equivalent have the same potential.
Thus, the cube can be divided into three equipotential levels:
- Level 1: Vertex \( a \) (connected to +6 V terminal).
- Level 2: Vertices that are one edge away from \( a \): \( b, d, e \).
- Level 3: Vertices that are one edge away from \( c \): \( f, g, h \).
- Level 4: Vertex \( c \) (connected to 0 V terminal).

By symmetry, potentials of all points in the same level are equal.

Step 3: Equivalent resistance between opposite corners of a cube.
The equivalent resistance between opposite corners of a cube (each edge having resistance \( R \)) is given by: \[ R_{\text{eq}} = \frac{5}{6}R \] Substitute \( R = 2 \, \Omega \):
\[ R_{\text{eq}} = \frac{5}{6} \times 2 = \frac{5}{3} \, \Omega \]
Step 4: Total current through the cube.
The total current from the 6 V battery is: \[ I = \frac{V}{R_{\text{eq}}} = \frac{6}{5/3} = \frac{18}{5} = 3.6 \, \text{A} \]
Step 5: Potential distribution using symmetry.
Due to symmetry, the potential difference between successive levels (a → e → f → c) divides proportionally according to the symmetry of current distribution.
Hence, potential drops in equal steps from \( a \) to \( c \).
The potential at \( a \) is 6 V and at \( c \) is 0 V, so the intermediate potentials (by symmetry) are approximately: \[ V_a = 6 \, \text{V}, \quad V_e = 4 \, \text{V}, \quad V_f = 3 \, \text{V}, \quad V_c = 0 \, \text{V}. \]
Step 6: Voltage difference between e and f.
\[ V_e - V_f = 4 - 3 = 1 \, \text{V} \]
Final Answer:
\[ \boxed{1 \, \text{V}} \]