Problem 27 - A Complete Resource Book in Physics for JEE Main

Question

A wire with resistance \(12 \Omega\) is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is (A) \(12 \Omega\) (B) \(24 \Omega\) (C) \(6 \Omega\) (D) \(3 \Omega\)

Step-by-Step Solution

Answer

The effective resistance between the two points on any diameter of the circle is 3 Ω, as found by dividing the wire into two equal halves (each with a resistance of 6 Ω) and using the formula for resistances in parallel. The correct answer is (D) \(3 \Omega\).

Step 1: Divide the wire into two halves

Consider the wire bent in the form of a circle. Let's divide it into two halves. Each half will have half the resistance of the total wire, which means each half has a resistance of 6 Ω.

Step 2: Find the resistances in parallel

The two halves of the wire are now in parallel connection between the diameter endpoints. To find the effective resistance, we must use the formula for resistances in parallel. The formula for resistances in parallel, for two resistances R1 and R2, is given by: \[ R_\text{eff} = \frac{R1 \cdot R2}{R1 + R2} \]

Step 3: Apply the resistances in parallel formula

We will apply the given formula using R1 = R2 = 6 Ω: \[ R_\text{eff} = \frac{6 \cdot 6}{6 + 6} \]

Step 4: Calculate the effective resistance

Simplifying the expression, we get: \[ R_\text{eff} = \frac{36}{12} = 3 Ω\] So, the effective resistance between the two points on any diameter of the circle is 3 Ω. The correct answer is (D) \(3 \Omega\).