Current Electricity · Example 3.23

Wire Bent into a Circle — Resistance Across a Diameter

A wire of resistance 6R is bent in the form of a circle. What is the effective resistance between the two ends of a diameter?

The diameter splits the ring into two equal semicircular arcs of 3R each.

Given

Asked

Effective resistance R_eff measured between the two end points of a diameter (the points A–A).

Formula

The two semicircular arcs connect the same two points, so they are in parallel:

\[ \frac{1}{R_{\text{eff}}} = \frac{1}{R_1} + \frac{1}{R_2} \quad\Longrightarrow\quad R_{\text{eff}} = \frac{R_1\,R_2}{R_1 + R_2} \]

Solution

STEP 1 — Split the ring

A diameter divides the circle into two equal halves, each carrying half the wire: \[ R_1 = R_2 = \frac{6R}{2} = 3R \]

STEP 2 — They are in parallel

Both arcs start at one terminal A and end at the opposite terminal A → identical end points → parallel.

STEP 3 — Combine

\[ R_{\text{eff}} = \frac{(3R)(3R)}{3R + 3R} = \frac{9R^2}{6R} = \frac{3R}{2} \]

Effective Resistance R_eff = 3R⁄2 = 1.5 R

Short Cut

For a uniform wire of total resistance R_total bent into a circle, the resistance across any diameter is always:

\[ R_{\text{diameter}} = \frac{R_{\text{total}}}{4} \]

Here: 6R ÷ 4 = 1.5R ✓ — no need to redo the parallel maths.

Why ÷4? Each half = R/2; two equal resistors in parallel halve again → (R/2) ÷ 2 = R/4.

General tap (not diameter): if the two points split the ring into arcs R₁ and R₂ (with R₁+R₂=R_total), use R_eff = R₁R₂ / R_total.

Six-step recall: Given → Asked → Formula → Solution → Short Cut