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Ch 28: Fundamentals of Circuits
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 28: Fundamentals of CircuitsProblem 80
Chapter 28, Problem 80

You’ve made the finals of the Science Olympics! As one of your tasks, you’re given 1.0 g of aluminum and asked to make a wire, using all the aluminum, that will dissipate 7.5 W when connected to a 1.5 V battery. What length and diameter will you choose for your wire?

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Step 1: Calculate the resistance of the wire using the formula for power dissipation: \( P = \frac{V^2}{R} \). Rearrange the formula to solve for \( R \): \( R = \frac{V^2}{P} \). Substitute \( V = 1.5 \, \text{V} \) and \( P = 7.5 \, \text{W} \) into the equation.
Step 2: Use the formula for resistance in terms of resistivity: \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity of aluminum, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. Rearrange the formula to express \( L \) in terms of \( R \) and \( A \): \( L = \frac{R \cdot A}{\rho} \).
Step 3: Determine the cross-sectional area \( A \) of the wire using the formula for the area of a circle: \( A = \pi \frac{d^2}{4} \), where \( d \) is the diameter of the wire. Substitute this expression for \( A \) into the formula for \( L \).
Step 4: Use the density of aluminum (\( \rho_{\text{density}} = 2.7 \, \text{g/cm}^3 \)) to calculate the volume of the aluminum wire: \( V = \frac{m}{\rho_{\text{density}}} \), where \( m = 1.0 \, \text{g} \). Relate the volume to the length and cross-sectional area: \( V = L \cdot A \). Substitute \( A \) and \( L \) into this equation to express \( d \) in terms of \( R \), \( \rho \), and \( \rho_{\text{density}} \).
Step 5: Solve the system of equations to find the values of \( L \) and \( d \). Use the known values for the resistivity of aluminum (\( \rho = 2.82 \times 10^{-8} \, \Omega \cdot \text{m} \)) and the density of aluminum to complete the calculations.

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Key Concepts

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

Ohm's Law

Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This relationship is expressed as V = IR. Understanding this law is crucial for determining how the wire will behave when connected to a voltage source.
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Resistivity

Resistivity is a material property that quantifies how strongly a given material opposes the flow of electric current. It is denoted by the symbol ρ (rho) and is dependent on the material's nature and temperature. For aluminum, the resistivity is relatively low, which means it is a good conductor. This concept is essential for calculating the resistance of the wire based on its dimensions.
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Power Dissipation

Power dissipation in an electrical circuit refers to the rate at which electrical energy is converted into heat energy, typically measured in watts (W). It can be calculated using the formula P = IV, where P is power, I is current, and V is voltage. In this scenario, understanding how to manipulate the wire's dimensions to achieve the desired power dissipation is key to solving the problem.
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Related Practice
Textbook Question

A 15 μF capacitor charged to 12 V is discharged through a resistor. The energy stored in the capacitor decreases by 50% in 0.25 s. What is the value of the resistance?

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Textbook Question

The capacitor in FIGURE P28.73 begins to charge after the switch closes at t = 0 s. Find an expression for the current I at time t. Graph I from t = 0 to t = 5τ.

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Textbook Question

The flash on a compact camera stores energy in a 120 μF capacitor that is charged to 220 V. When the flash is fired, the capacitor is quickly discharged through a lightbulb with 5.0 Ω of resistance. At what rate is the lightbulb dissipating energy 250 μs after the flash is fired?

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