Textbook Question
The current in a wire varies with time according to the relationship . What constant current would transport the same charge in the same time interval?
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Ch 25: Current, Resistance, and EMF
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 25, Problem 19
In household wiring, copper wire mm in diameter is often used. Find the resistance of a -m length of this wire.
Verified step by step guidance1
First, identify the formula for resistance in a wire, which is given by \( R = \frac{\rho L}{A} \), where \( R \) is the resistance, \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.
Determine the resistivity \( \rho \) of copper. This is a known value and can be found in tables of material properties. For copper, \( \rho \approx 1.68 \times 10^{-8} \text{ ohm meters} \).
Calculate the cross-sectional area \( A \) of the wire. Since the wire is cylindrical, use the formula for the area of a circle: \( A = \pi r^2 \), where \( r \) is the radius of the wire. Convert the diameter to radius by dividing by 2: \( r = \frac{2.05 \text{ mm}}{2} = 1.025 \text{ mm} \). Convert this to meters: \( r = 1.025 \times 10^{-3} \text{ m} \).
Substitute the radius into the area formula: \( A = \pi (1.025 \times 10^{-3})^2 \text{ m}^2 \). Calculate this value to find the cross-sectional area.
Finally, substitute the values for \( \rho \), \( L \), and \( A \) into the resistance formula: \( R = \frac{1.68 \times 10^{-8} \times 24.0}{A} \). Calculate \( R \) using the computed area from the previous step.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Resistance
Resistance is a measure of the opposition to the flow of electric current through a conductor. It is determined by the material's properties, length, and cross-sectional area. The formula for resistance is R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area.
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Resistivity
Resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is denoted by ρ and is measured in ohm-meters (Ω·m). Copper has a low resistivity, making it an excellent conductor for household wiring.
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Cross-sectional Area
The cross-sectional area of a wire affects its resistance. For a cylindrical wire, the area A can be calculated using the formula A = π(d/2)^2, where d is the diameter. A larger cross-sectional area results in lower resistance, allowing more current to flow through the wire.
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Related Practice
Textbook Question
A copper transmission cable km long and cm in diameter carries a current of A. How much electrical energy is dissipated as thermal energy every hour?
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Textbook Question
A ductile metal wire has resistance . What will be the resistance of this wire in terms of if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)
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Textbook Question
An idealized ammeter is connected to a battery as shown in Fig. E. Find the reading of the ammeter.
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Textbook Question
The current in a wire varies with time according to the relationship . How many coulombs of charge pass a cross section of the wire in the time interval between and ?
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Textbook Question
What is the resistance of a carbon rod at °C if its resistance is at °C?
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