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

(a) Length is 5.85 m for 75 V; (b) Length is 10.40 m for 100 V.

Step 1: Understand the Problem

We are given the voltage across a Nichrome wire, its cross-sectional area, and the resistivity of Nichrome. We need to determine the length of the wire to dissipate a specific power. We need to do this for two scenarios: one for 75 V and another for 100 V.

Step 2: Recall the Power and Resistance Formula

The formula for electric power dissipation in terms of voltage and resistance is given by:\[ P = \frac{V^2}{R} \]where \(P\) is the power, \(V\) is the voltage, and \(R\) is the resistance of the wire.

Step 3: Recall the Resistance Formulation

We calculate the resistance \(R\) of the wire using the formula:\[ R = \rho \frac{L}{A} \]where \(\rho\) is the resistivity of the wire, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.

Step 4: Solve for Length Using Power Dissipation for 75 V

First, we rearrange the power formula to find the resistance:\[ R = \frac{V^2}{P} = \frac{75^2}{5000} \]Calculate the resistance: \[ R \approx 1.125 \Omega \]Next, use the resistance formula to find length \(L\):\[ L = \frac{R \cdot A}{\rho} = \frac{1.125 \cdot 2.60 \times 10^{-6}}{5.00 \times 10^{-7}} \]Calculate the length: \[ L \approx 5.85 \text{ m} \]

Step 5: Solve for Length Using Power Dissipation for 100 V

Apply the same steps with 100 V. First find the resistance:\[ R = \frac{100^2}{5000} = 2 \Omega \]Then use this resistance to find the wire length:\[ L = \frac{2 \cdot 2.60 \times 10^{-6}}{5.00 \times 10^{-7}} \]Calculate the length: \[ L \approx 10.40 \text{ m} \]

Key Concepts

Resistance Calculation

When it comes to electric circuits, understanding how to calculate resistance is fundamental. Resistance, denoted as \( R \), is a measure of how difficult it is for an electric current to flow through a material. The formula for resistance, given that you have the resistivity \( \rho \), the length of the wire \( L \), and the cross-sectional area \( A \), is expressed as:

• \( R = \rho \frac{L}{A} \)

Here's how each component relates to the formula:

• \( \rho \) (resistivity) indicates how strongly a material opposes the flow of electric current. A higher resistivity means more resistance.
• \( L \) (length) matters because the longer a wire is, the more material the electricity has to pass through, increasing resistance.
• \( A \) (cross-sectional area) affects resistance inversely; a larger area allows current to pass through more easily, reducing resistance.

In our exercise, we calculated the resistance of a Nichrome wire, which required us to rearrange the power formula to solve for \( R \) first, then find \( L \). Calculating resistance was crucial because it allowed us to know exactly how long a wire needs to be to achieve the desired power dissipation.

Voltage and Power Relationship

Voltage and power in an electric circuit are closely intertwined. The power dissipated by a resistive element is the product of the potential difference across the element and the current flowing through it. However, when resistance \( R \) is known, a more convenient formula is:

• \( P = \frac{V^2}{R} \)

Where:

• \( P \) is the power (in watts, W)
• \( V \) is the voltage (in volts, V)
• \( R \) is the resistance (in ohms, \( \Omega \))

This relationship shows how power depends not just on voltage but also on resistance:

• If voltage increases with constant resistance, the power increases significantly because it is proportional to the square of the voltage.
• Conversely, if you know the power and voltage, you can determine the resistance using \( R = \frac{V^2}{P} \).

In the given problem, understanding this relationship was key to finding resistance under different voltages (75 V and 100 V), ensuring that the power dissipated by the wire remained at the specified 5000 W.

Resistivity

Resistivity is a property intrinsic to materials that quantifies how strongly they resist current flow. The formula for resistance includes resistivity, and it plays a vital role in determining how suitable a material is for certain applications:

• \( \rho \) is the symbol for resistivity, and its unit is \( \Omega \cdot m \).
• Materials with low resistivity are good conductors (like copper), while those with high resistivity are good insulators (like rubber).

Nichrome wire, used in our exercise, has a resistivity of \( 5.00 \times 10^{-7} \Omega \cdot m \). This value influences how much length of Nichrome wire is needed to achieve specific resistance and power outcomes. High resistivity means that even short lengths of wire provide substantial resistance. Therefore, understanding resistivity and its effects helps in selecting the right materials for specific electrical tasks.
This understanding is pivotal not only for academic purposes but also in practical situations where the efficiency of devices depends on these calculations.