Question

The ends of six wires, each of resistance R (= 10 \(\Omega\)) are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

Hint:

In symmetric circuits, identify series and parallel combinations systematically to simplify the network step by step.

Answer 1

The given circuit is a symmetric triangle, where each side has a resistance of \( R \). We need to find the effective resistance between points A and B.
Step 1: Resistors in Series
Each set of two resistors in series will give: \[ R_{\text{eq1}} = R + R = 2R \]
Step 2: Resistors in Parallel
The three \( 2R \) resistors are now in parallel. Using the formula for resistors in parallel: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{3}{2R} \] Thus, the effective resistance is: \[ R_{\text{eq}} = \frac{2R}{3} \]
Step 3: Substituting the value of \(R\)
Given that \( R = 10 \, \Omega \), we have: \[ R_{\text{eq}} = \frac{2 \times 10}{3} = 6.67 \, \Omega \]