Problem 53 - Fundamentals of Physics

Question

A \(120 \mathrm{~V}\) potential difference is applied to a space heater that dissipates \(500 \mathrm{~W}\) during operation. (a) What is its resistance during operation? (b) At what rate do electrons flow through any cross section of the heater element?

Step-by-Step Solution

Answer

(a) Resistance is 28.8 Ω. (b) Electron flow rate is approximately 2.61 × 10¹⁹ electrons per second.

Step 1: Identify Given Values

We are given the voltage \(V = 120 \, \text{V}\) and the power \(P = 500 \, \text{W}\). These will be used to find the resistance and the rate of electron flow.

Step 2: Determine Resistance

Use the formula \(P = \frac{V^2}{R}\) to find the resistance \(R\). Rearrange the formula to \(R = \frac{V^2}{P}\). Substitute the given values to find \(R\): \(R = \frac{(120)^2}{500} = \frac{14400}{500} = 28.8 \, \Omega\).

Step 3: Find Current through the Heater

Use the power formula \(P = VI\) to solve for current \(I\). Rearrange to \(I = \frac{P}{V}\) and substitute the given values: \(I = \frac{500}{120} = 4.17 \, \text{A}\).

Step 4: Calculate Electron Flow Rate

The rate of electron flow, also known as the current, can be expressed in terms of the charge of the electrons: \(I = \frac{n \cdot e}{t}\). We have already found \(I = 4.17 \, \text{A}\), where \(e = 1.6 \times 10^{-19} \, \text{C}\). The number of electrons \(n\) crossing any section per second is \(n = \frac{I \cdot t}{e}\). For 1 second, \(n = \frac{4.17 imes 1}{1.6 \times 10^{-19}} = 2.61 \times 10^{19} \, \text{electrons per second}\).

Key Concepts

Electric Power

Electric power is the rate at which electrical energy is converted into another form of energy, such as heat in a space heater. It is expressed in watts (W). This calculation is crucial in understanding how much electricity an appliance uses over time.
To find electric power, we use the formula \(P = VI\), where \(P\) is power, \(V\) is voltage, and \(I\) is the current.
In the example given, a space heater operates at a voltage of \(120 \text{ V}\) and power of \(500 \text{ W}\). These values help in deriving other important electrical parameters, such as resistance and current.
Electric power lets us know how efficient and effective these devices are. If a heater uses more power than necessary, it could mean wasted energy and money. Understanding electric power helps make informed decisions about energy use and efficiency.

Resistance Calculation

Resistance is a measure of how much an object opposes the flow of electric current. It's denoted by \(R\) and measured in ohms (\(\Omega\)). Calculating resistance is crucial for designing electrical devices that operate safely and efficiently.
Ohm's Law gives us the relationship between voltage \(V\), current \(I\), and resistance \(R\): \(V = IR\). However, when power \(P\) and voltage are given, another formula is used: \(P = \frac{V^2}{R}\). By rearranging, the resistance can be found with \(R = \frac{V^2}{P}\).
In this exercise, a space heater with a voltage of \(120 \text{ V}\) and power of \(500 \text{ W}\) has a resistance of \(28.8 \Omega\). This calculation is practical for ensuring the heater operates efficiently without causing excessive wear or safety issues.

Electron Flow Rate

The flow rate of electrons through a conductor is known as the electron flow rate, directly related to the electrical current. This rate informs us about the number of electrons passing a cross section of a conductor per second, often quantified as current in amperes (\(\text{A}\)).
We use the formula \(I = \frac{n \cdot e}{t}\), where \(I\) is the current, \(n\) is the number of electrons, \(e\) is the electron charge \((1.6 \times 10^{-19} \text{ C})\), and \(t\) is time (in seconds).
Given a current of \(4.17 \text{ A}\) through the heater, the number of electrons flowing through per second is \(n = \frac{4.17 \times 1}{1.6 \times 10^{-19}} = 2.61 \times 10^{19}\) electrons per second.
Understanding the electron flow rate is vital for diagnosing potential problems in electrical circuits, ensuring components are not overloaded and maintaining safe operation.