Question

Find the equivalent resistance between two opposite corners of a cube made of wires, each having resistance \(R\).

• A. \( \dfrac{R}{6} \)
• B. \( \dfrac{5R}{6} \)
• C. \( \dfrac{R}{3} \)
• D. \( \dfrac{2R}{3} \)

Hint:

In symmetric resistor networks like cubes or tetrahedrons, nodes at equal potential can be grouped together. This reduces the circuit into simple series–parallel combinations.

Answer 1

The Correct Option is B

Concept: A cube has 12 identical resistors \(R\). When resistance is measured between opposite corners, symmetry can be used to simplify the circuit. All three vertices adjacent to the first corner are at the same potential, and similarly the three vertices adjacent to the opposite corner are at another common potential. Thus the circuit reduces to a simpler combination of series and parallel resistances.
Step 1: {Group symmetric points.} Let the opposite corners be \(A\) and \(B\). Due to symmetry: - The three vertices connected to \(A\) have equal potential. - The three vertices connected to \(B\) also have equal potential. Thus the cube effectively reduces to three resistors \(R\) from \(A\) to the first group.
Step 2: {Find equivalent resistance from \(A\) to the first symmetric group.} Three resistors \(R\) are in parallel: \[ R_1 = \frac{R}{3} \]
Step 3: {Resistance between the two middle groups.} There are six resistors connecting these groups, all in parallel: \[ R_2 = \frac{R}{6} \]
Step 4: {Resistance from the second group to \(B\).} Again three resistors in parallel: \[ R_3 = \frac{R}{3} \]
Step 5: {Add series resistances.} \[ R_{eq} = R_1 + R_2 + R_3 \] \[ = \frac{R}{3} + \frac{R}{6} + \frac{R}{3} \] \[ = \frac{2R}{6} + \frac{R}{6} + \frac{2R}{6} \] \[ = \frac{5R}{6} \]