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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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TOC.tableOfContents
- 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
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Young & Freedman Calc 15th EditionCh 25: Current, Resistance, and EMFProblema 19
Young & Freedman Calc 15th EditionCh 25: Current, Resistance, and EMFProblema 19Capitolo 25, Problema 19
In household wiring, copper wire mm in diameter is often used. Find the resistance of a -m length of this wire.
Guida verificata passo dopo passo1
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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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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Percorso guidato
05:25
Resistivity & Resistors in Circuits
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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05:25Resistivity & Resistors in Circuits
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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