Understanding Resistance in a Circular Coil

The question asks for the resistance between the ends of any diameter of a circular coil that has a total resistance of 20 Ω for its single turn.

When we consider the resistance between two points on a circular coil that are diametrically opposite (i.e., at the ends of a diameter), the coil is effectively split into two equal halves.

Let the total resistance of the single-turn coil be \(R_{total}\). We are given \(R_{total} = 20 \, \Omega\).

Since the coil is uniform, the resistance of any part of the coil is proportional to its length. When considering the resistance across a diameter, the coil is divided into two semicircular sections. Each semicircular section is half the total length of the coil.

Therefore, the resistance of each semicircular section will be half of the total resistance:

\(R_{half} = \frac{R_{total}}{2}\)

\(R_{half} = \frac{20 \, \Omega}{2} = 10 \, \Omega\)

Now, these two semicircular sections are connected between the same two points (the ends of the diameter). This means they are connected in parallel.

To find the equivalent resistance of two resistors connected in parallel, we use the formula:

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}\)

In this case, \(R_1 = R_{half} = 10 \, \Omega\) and \(R_2 = R_{half} = 10 \, \Omega\).

Substituting these values:

\(R_{eq} = \frac{10 \, \Omega \times 10 \, \Omega}{10 \, \Omega + 10 \, \Omega}\)

\(R_{eq} = \frac{100 \, \Omega^2}{20 \, \Omega}\)

\(R_{eq} = 5 \, \Omega\)

So, the equivalent resistance between the ends of any diameter of the coil is 5 Ω.

Step-by-Step Calculation

1. Identify the total resistance of the coil: \(R_{total} = 20 \, \Omega\).
2. Recognize that considering resistance across a diameter divides the coil into two equal semicircular parts.
3. Calculate the resistance of each semicircular part: \(R_{half} = R_{total} / 2 = 20 \, \Omega / 2 = 10 \, \Omega\).
4. Understand that these two parts are connected in parallel between the diameter's ends.
5. Calculate the equivalent resistance of two 10 Ω resistors in parallel using the formula \(R_{eq} = \frac{R_1 R_2}{R_1 + R_2}\).
6. Compute the final equivalent resistance: \(R_{eq} = \frac{10 \times 10}{10 + 10} = \frac{100}{20} = 5 \, \Omega\).

Result Summary

Revision Table: Key Concepts

Additional Information: Resistance and Geometry

The resistance of a conductor depends on its material, length, and cross-sectional area. The formula is \(R = \rho \frac{L}{A}\), where:

• \(\rho\) is the resistivity of the material.
• \(L\) is the length of the conductor.
• \(A\) is the cross-sectional area.

In this circular coil problem, the material (\(\rho\)) and cross-sectional area (\(A\)) are uniform throughout the coil. Therefore, the resistance is directly proportional to the length (\(L\)). When the coil is split into two equal halves by the diameter, the length of each half is \(L_{total}/2\), and consequently, the resistance of each half is \(R_{total}/2\).

Connecting these two halves across the diameter forms a parallel circuit because the current entering at one end of the diameter splits and flows through both semicircular paths simultaneously, recombining at the other end of the diameter. This parallel arrangement of the two 10 Ω resistances results in the lower equivalent resistance of 5 Ω.