2.The figure below shows cross sections through three long conductors of the same length and material, with circular cross-sections of edge as shown. Conductor B(outer radius 2R) will fit snugly within conductor A(outer radius 3R, and conductor C (radius R) will fit snugly within conductor B. Rank the three conductors according to their end-to-end resistance. A B с OO

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
icon
Related questions
Question
### Conductors and Resistance

#### Problem Statement
The figure above shows cross sections of three long conductors, each having the same length and material. The conductors are depicted with circular cross-sections, detailed as follows:

- **Conductor A**: Outer radius of 3R.
- **Conductor B**: Outer radius of 2R, fits snugly within conductor A.
- **Conductor C**: Radius R, fits snugly within conductor B.

#### Task
Rank the three conductors according to their end-to-end resistance.

#### Analysis
The resistance \( R \) of a conductor is given by the formula:

\[ R = \frac{\rho L}{A} \]

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

Since all conductors have the same length and material:
- The resistivity (\( \rho \)) and length (\( L \)) are constant for all three.

Therefore, resistance is inversely proportional to the cross-sectional area (\( A \)).

#### Calculating Cross-Sectional Areas
- **Conductor A**: \( A_A = \pi (3R)^2 = 9\pi R^2 \)
- **Conductor B**: \( A_B = \pi (2R)^2 = 4\pi R^2 \)
- **Conductor C**: \( A_C = \pi R^2 \)

#### Ranking Resistance
The resistance is inversely proportional to the cross-sectional area, so:

- **Conductor C** (smallest area) has the highest resistance.
- **Conductor B** has a higher resistance than **Conductor A**, but lower than **Conductor C**.
- **Conductor A** (largest area) has the lowest resistance.

#### Conclusion
The order of the conductors from highest to lowest resistance is:
1. Conductor C
2. Conductor B
3. Conductor A

This sequence provides an understanding of how cross-sectional area impacts resistance in conductors of identical material and length.
Transcribed Image Text:### Conductors and Resistance #### Problem Statement The figure above shows cross sections of three long conductors, each having the same length and material. The conductors are depicted with circular cross-sections, detailed as follows: - **Conductor A**: Outer radius of 3R. - **Conductor B**: Outer radius of 2R, fits snugly within conductor A. - **Conductor C**: Radius R, fits snugly within conductor B. #### Task Rank the three conductors according to their end-to-end resistance. #### Analysis The resistance \( R \) of a conductor is given by the formula: \[ R = \frac{\rho L}{A} \] Where: - \( \rho \) is the resistivity of the material. - \( L \) is the length of the conductor. - \( A \) is the cross-sectional area. Since all conductors have the same length and material: - The resistivity (\( \rho \)) and length (\( L \)) are constant for all three. Therefore, resistance is inversely proportional to the cross-sectional area (\( A \)). #### Calculating Cross-Sectional Areas - **Conductor A**: \( A_A = \pi (3R)^2 = 9\pi R^2 \) - **Conductor B**: \( A_B = \pi (2R)^2 = 4\pi R^2 \) - **Conductor C**: \( A_C = \pi R^2 \) #### Ranking Resistance The resistance is inversely proportional to the cross-sectional area, so: - **Conductor C** (smallest area) has the highest resistance. - **Conductor B** has a higher resistance than **Conductor A**, but lower than **Conductor C**. - **Conductor A** (largest area) has the lowest resistance. #### Conclusion The order of the conductors from highest to lowest resistance is: 1. Conductor C 2. Conductor B 3. Conductor A This sequence provides an understanding of how cross-sectional area impacts resistance in conductors of identical material and length.
Expert Solution
Step 1

Determine the cross-sectional area of conductor A.

AA=π(RA2RB2)AA=π[(3R)2(2R)2]AA=5πR2

steps

Step by step

Solved in 4 steps

Blurred answer
Knowledge Booster
DC circuits
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON