Question

A regular hexagon is formed by six wires each of resistance \( r \,\Omega \) and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

• A. \( \dfrac{4}{5} r \)
• B. \( \dfrac{3}{4} r \)
• C. \( \dfrac{3}{5} r \)
• D. \( \dfrac{5}{8} r \)

Hint:

In symmetric resistor networks, always look for identical current paths—symmetry greatly simplifies calculations.

Answer 1

The Correct Option is A

To solve this problem, we need to understand the configuration of resistors in the hexagon. The hexagon is formed by six sides, each with a resistance of \(r \, \Omega\), and all vertices of the hexagon are connected to the center with the same resistance, \(r \, \Omega\).

1. We label the corners of the hexagon as \(A, B, C, D, E,\) and \(F\). Let the center of the hexagon be point \(O\).
2. We're asked for the equivalent resistance between two opposite corners, say \(A\) and \(D\), when current enters at \(A\) and leaves at \(D\).
3. The hexagon and its connections create a symmetric network. The symmetry helps in simplifying the problem by making intelligent guesses about equal potential points.
4. Since the arrangement is symmetric, the potential at \(B\) and \(F\) must be the same. Similarly, \(C\) and \(E\) will share the same potential.
5. The resistors between opposite pairs can be combined using series-parallel rules:
   • Resistors \(AB\), \(BO\), and \(OF\) are in series.
   • This series combination is parallel to \(BF\).
   • By symmetry and applying similar logic to other parts of the hexagon, we evaluate the net resistance between \(A\) and \(D\).
6. Applying Kirchhoff's laws or the star-delta transformation method can efficiently solve such circuits.

After solving through methods such as delta-star transformations, it turns out that the resistance between any two opposite corners, considering incoming and outgoing currents, results in an equivalent resistance of:

\(\dfrac{4}{5} r\).

Thus, the correct option is \(\dfrac{4}{5} r\).