https://file.aiquickdraw.com/mathbot/file/17271763192805puq0txu.jpg

The problem involves a conductor with a resistance of 4 ohms, which is formed into a circular ring. Another conductor, made from the same material and having the same cross-sectional area, is connected across the diameter of this ring. The question asks for the resistance between the terminals of the conductor's diameter.

### Step-by-step solution:

1. **Initial Setup:**
   - The ring has a total resistance of 4Ω. The ring is a full circle, and the resistance will be uniformly distributed around the circumference.
   - When a conductor is connected across the diameter, the circle is divided into two equal halves.

2. **Resistance of Each Half:**
   - Since the ring has a total resistance of 4Ω, each half of the ring will have a resistance of 2Ω (as the ring is divided equally into two halves along the diameter).

3. **Effective Resistance of Two Halves in Parallel:**
   - The two halves of the ring are connected in parallel because they share the same two terminals (the endpoints of the diameter). For two resistances \( R_1 \) and \( R_2 \) in parallel, the equivalent resistance \( R_{eq} \) is given by:
   \[
   \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
   \]
   Since both resistances are equal (2Ω each), we can compute the parallel combination:
   \[
   \frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1
   \]
   Thus, \( R_{eq} = 1\Omega \).

4. **Addition of the Conductor:**
   - The new conductor is also placed across the diameter of the circle. Since it has the same cross-sectional area as the ring conductor and is made of the same material, its resistance depends on its length.
   - The length of this conductor is equal to the diameter of the circle, which is half the circumference of the original ring.
   - Let the total circumference of the ring be \( 2\pi r \), then the diameter is \( \pi r \), and the resistance of this new conductor is proportional to its length, hence it is \( \frac{4}{\pi} \Omega \).

5. **Final Resistance Calculation:**
   - Now, we have two resistances in parallel: the equivalent resistance of the ring's two halves (1Ω) and the resistance of the new conductor (\( \frac{4}{\pi} \Omega \)).
   - The total resistance \( R_{total} \) between the two terminals is given by the parallel combination of these two resistances:
   \[
   \frac{1}{R_{total}} = \frac{1}{1} + \frac{\pi}{4}
   \]
   Simplifying:
   \[
   \frac{1}{R_{total}} = 1 + \frac{\pi}{4}
   \]
   Therefore:
   \[
   R_{total} = \frac{1}{1 + \frac{\pi}{4}} = \frac{4}{4 + \pi}
   \]

Thus, the correct answer is option **(D) \( \frac{4}{4 + \pi} \)**.

---

Would you like further details on any specific step or concept?

Here are five related questions:
1. What happens if the resistance of the ring is not uniform?
2. How does the material of the conductor affect the resistance?
3. What if the additional conductor had a different cross-sectional area?
4. How does the length of a conductor influence its resistance?
5. How can we generalize this setup for an elliptical ring?

**Tip:** Parallel resistors always reduce the overall resistance—combining two equal resistors in parallel halves the resistance.

A conductor of resistance 4Ω is formed as a circular ring. Another conductor, made from the same material and with the same cross-sectional area, is connected across the diameter of the ring. What is the resistance between the terminals of the conductor's diameter?

Learn how to calculate the resistance between terminals of a circular conductor with a diameter connection. This problem involves applying the concept of parallel resistance, circle geometry, and proportional resistance calculations. Suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation