Problem 47 - Fundamentals of Physics

Question

A heating element is made by maintaining a potential difference of \(75.0 \mathrm{~V}\) across the length of a Nichrome wire that has a \(2.60 \times 10^{-6} \mathrm{~m}^{2}\) cross section. Nichrome has a resistivity of \(5.00 \times 10^{-7} \Omega \cdot \mathrm{m} .\) (a) If the element dissipates \(5000 \mathrm{~W}\), what is its length? (b) If \(100 \mathrm{~V}\) is used to obtain the same dissipation rate, what should the length be?

Step-by-Step Solution

Answer

For 75 V, the wire length is approximately 5.85 mm; for 100 V, it is approximately 10.4 mm.

Step 1: Understanding the Problem

We need to find the length of the Nichrome wire when a certain power dissipation occurs, first with a voltage of 75.0 V and then with 100 V, both giving a power dissipation of 5000 W. We can use resistance formulas and P = VI (dissipated power) for this.

Step 2: Using Power and Ohm's Law

Power (P) is given by P = VI and also V^2/R. For a wire, resistance (R) can be given by R = \( \frac{\rho L}{A} \), where \( \rho \) is resistivity, L is length, and A is cross-sectional area. Rearrange to find L: \( L = \frac{R A}{\rho} \).

Step 3: Calculating Resistance for 75 V

Since P = 5000 W and V = 75 V, we have \( 5000 = 75 \cdot I \). Find current, \( I = \frac{5000}{75} = 66.67 \) A. \( P = \frac{V^2}{R} = \frac{75^2}{R} = 5000 \). Solve for R: \( R = \frac{75^2}{5000} = 1.125 \Omega \).

Step 4: Finding Length for 75 V

With R known, use \( R = \frac{\rho L}{A} \) to find L: \( 1.125 = \frac{5.00 \times 10^{-7} \times L}{2.60 \times 10^{-6}} \). Solving gives \( L \approx 5.85 \times 10^{-3} \) m or 5.85 mm.

Step 5: Repeating For 100 V

For V = 100 V and P = 5000 W, find current \( I = \frac{5000}{100} = 50 \) A and R: \( R = \frac{100^2}{5000} = 2 \Omega \).

Step 6: Finding Length for 100 V

Use R = \( 2 = \frac{5.00 \times 10^{-7} \times L}{2.60 \times 10^{-6}} \) to find L. Solve \( 2 = \frac{5.00 \times 10^{-7} \times L}{2.60 \times 10^{-6}} \) getting \( L \approx 1.04 \times 10^{-2} \) m or 10.4 mm.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in the field of electricity and electronics. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. The formula representing Ohm's Law is expressed as:

• \( V = IR \)

Here, \( V \) is the voltage in volts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms. This relation is pivotal for solving problems involving electrical circuits.
For example, if you know the voltage and resistance, you can rearrange the formula to calculate the current:

• \( I = \frac{V}{R} \)

Similarly, if you know the power dissipation and resistance, you can find voltage using the power formula:

• \( P = VI \)
• \( V = \frac{P}{I} \)

Understanding how to manipulate these formulas allows you to solve for unknowns in a circuit, which is crucial in practical scenarios like designing electrical systems.

Nichrome Wire

Nichrome wire is widely used in applications where heat generation is needed, such as in toasters or heating elements. It is an alloy of nickel and chromium, which gives it specific properties:

• High resistivity: This makes it better for generating heat when electrical current passes through.
• Corrosion resistance: It withstands damage from oxidation, maintaining its performance over time.
• Stability: It remains stable even at the high temperatures it generates due to its high melting point.

For the exercise, the Nichrome wire has a given resistivity which is essential in calculating resistance in conjunction with the wire's length and cross-section area. The formula for resistance for a piece of wire is:

• \( R = \frac{\rho L}{A} \)

where \( \rho \) represents the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Its consistent and predictable resistivity makes Nichrome an ideal choice for heating applications.

Power Dissipation

Power dissipation in an electrical circuit is the process where electric energy is converted into heat energy. This concept is pivotal in understanding how circuits work and ensuring that they function within safe limits.
Power can be calculated using different variations, depending on the known quantities, with the most common being:

• \( P = VI \surmise\)
• \( P = I^2R \surmise\)
• \( P = \frac{V^2}{R} \surmise\)

In the exercise, power dissipation is specified to be 5000 watts. By using these formulas, you can determine other unknowns in the circuit such as the resistance or the length of the wire when subjected to different voltages. Understanding power dissipation helps in designing circuits that meet specified power requirements without overheating.

Resistivity

Resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current.
It is denoted by the symbol \( \rho \) and is expressed in ohm-meters \( \Omega \cdot m \).

• Lower resistivity indicates a material that easily allows electric current to flow.
• Higher resistivity means the material resists the flow of electricity to a greater extent.

Resistivity is influenced by factors such as temperature and the material's inherent properties. In the context of the exercise, known resistivity of the Nichrome wire is used to calculate its electrical resistance.

• The relationship between resistance, resistivity, length, and cross-sectional area is given by:\( R = \frac{\rho L}{A} \)

This expression shows how a material's resistivity directly affects the resistance of a wire. By solving for \( L \), the length of the wire, we can see how changing the applied voltage while maintaining constant power dissipation affects the wire's length.