Question

A \( 4 \, \Omega \) resistance wire is bent into a shape of a circle. What will be the effective resistance between the ends of its diameter?

Hint:

When a resistance wire is bent into a circle and the current is applied between the ends of the diameter, the resistance is halved for each half of the circle, and the total resistance is found using the formula for parallel resistances.

Answer 1

Effective Resistance of a Circular Wire:
Let's consider a resistance wire of total resistance \( R = 4 \, \Omega \) that is bent into the shape of a circle. The key point is that when current flows between the ends of the diameter, it flows through two equal halves of the wire. 1. Divide the wire into two equal halves: The wire is bent into a circle, and the current flows between the two ends of the diameter. This divides the circular wire into two equal halves, each with resistance \( R_1 \) and \( R_2 \). Since the wire is uniform, each half of the wire will have half the total resistance: \[ R_1 = R_2 = \frac{4 \, \Omega}{2} = 2 \, \Omega. \] 2. Calculate the total resistance: The two halves of the wire are effectively in parallel because the current splits between them when it reaches the diameter. The formula for the total (effective) resistance \( R_{\text{eff}} \) of two resistors in parallel is: \[ \frac{1}{R_{\text{eff}}} = \frac{1}{R_1} + \frac{1}{R_2}. \] Substituting \( R_1 = 2 \, \Omega \) and \( R_2 = 2 \, \Omega \): \[ \frac{1}{R_{\text{eff}}} = \frac{1}{2} + \frac{1}{2} = 1. \] Therefore, the effective resistance between the ends of the diameter is: \[ R_{\text{eff}} = 1 \, \Omega. \] Thus, the effective resistance between the ends of the diameter of the circular wire is \( 1 \, \Omega \).