4. A wire has resistance 12 ohms. It is bent and its two ends are connected to form a
circle. The effective resistance between the two points on any diameter of the circle is
a. 3 ohms
b. 6 ohms
c. 12 ohms
d. 24 ohms
The effective resistance between two points on any diameter of a circle formed by a 12-ohm wire is 3 ohms. This classic physics problem tests understanding of resistance distribution in a uniform wire bent into a circular loop. Current flows through two equal paths, forming a parallel combination.
Problem Breakdown
A uniform wire with total resistance R = 12 Ω is bent into a circle, joining its ends seamlessly. Points A and B lie on a diameter, dividing the circle into two semicircles. Each semicircle spans half the wire length, so the resistance per semicircle is R/2 = 6 Ω.
From A to B, current splits equally through both semicircles. These act as two 6 Ω resistors in parallel. The equivalent resistance Req is given by
1/R_eq = 1/6 + 1/6 = 1/3, so Req = 3 Ω.
Option Analysis
- a. 3 ohms: Correct. Parallel combination of two 6-ohm paths yields 3 ohms, as verified by calculation.
- b. 6 ohms: Incorrect. Represents single semicircle resistance, ignoring the parallel path.
- c. 12 ohms: Incorrect. Equals total wire resistance, applicable only when used end-to-end before bending.
- d. 24 ohms: Incorrect. Has no physical basis here and may reflect confusion with series doubling, but in this setup the paths are parallel.
Key Insight
Resistance scales linearly with length in a uniform wire, so the two semicircles share the current equally. The parallel combination formula halves the effective resistance compared to a single path, a standard result in circular conductor problems often appearing.
