Ch 25: Current, Resistance, and EMF

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 25: Current, Resistance, and EMF

Problem 22

Chapter 25, Problem 22

A hollow aluminum cylinder is $2.50$ m long and has an inner radius of $2.75$ cm and an outer radius of $4.60$ cm. Treat each surface (inner, outer, and the two end faces) as an equipotential surface. At room temperature, what will an ohmmeter read if it is connected between (a) the opposite faces and (b) the inner and outer surfaces?

Verified step by step guidance

First, understand that the problem involves calculating the resistance of a hollow aluminum cylinder. The resistance can be calculated using the formula:

R = \frac{ρ L}{A}

, where

ρ

is the resistivity of aluminum,

L

is the length of the cylinder, and

A

is the cross-sectional area through which current flows.

For part (a), calculate the resistance between the opposite faces. The cross-sectional area

A

is the area of the annular ring formed by the outer and inner radii. Use the formula for the area of an annular ring:

A = π (r^{2} - r^{1})

, where

r^{2}

is the outer radius and

r^{1}

is the inner radius.

Substitute the values for the radii and length into the resistance formula for part (a). Convert the radii from centimeters to meters before substituting:

R = \frac{ρ \times 2.50}{π ((0.046)^{2} - (0.0275)^{2})}

For part (b), calculate the resistance between the inner and outer surfaces. Here, the current flows radially through the cylindrical shell. The resistance formula for radial flow is:

R = \frac{ρ \times L}{π \times (r^{2} - r^{1})}

, where

L

is the length of the cylinder.

Substitute the values for the radii and length into the resistance formula for part (b). Again, ensure the radii are in meters:

R = \frac{ρ \times 2.50}{π \times (0.046 - 0.0275)}

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Resistance in Cylindrical Conductors

The resistance of a cylindrical conductor depends on its length, cross-sectional area, and the material's resistivity. For a hollow cylinder, the cross-sectional area is the difference between the outer and inner areas. The formula R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area, is used to calculate resistance.

Equipotential Surfaces

Equipotential surfaces are surfaces where the electric potential is constant. In this problem, the inner, outer, and end faces of the cylinder are equipotential, meaning no potential difference exists across these surfaces. This concept is crucial for understanding how current flows and how resistance is measured in different configurations.

Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance, expressed as V = IR. This principle is essential for determining the readings of an ohmmeter, which measures resistance by applying a known voltage and measuring the resulting current.

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Resistance and Ohm's Law

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