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Chapter 26: Problem 2

How many electrons flow through a light bulb each second if the current through the light bulb is \(0.75 \mathrm{~A}\) ? From \(I=q / t\), the charge flowing through the bulb in \(1.0 \mathrm{~s}\) is $$ q=I t=(0.75 \mathrm{~A})(1.0 \mathrm{~s})=0.75 \mathrm{C} $$ But the magnitude of the charge on each electron is \(e=1.6 \times 10^{-19}\) C. Therefore, $$ \text { Number }=\frac{\text { Charge }}{\text { Charge } / \text { electron }}=\frac{0.75 \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=4.7 \times 10^{18} $$

Short Answer

Expert verified
Approximately \(4.7 \times 10^{18}\) electrons flow through the bulb each second.

Step by step solution

01

Identify the Given Information

We know that the current through the light bulb is given as \(I = 0.75 \text{ A}\). This is the amount of charge flowing per second.
02

Determine Charge Flow Over Time

Using the formula \(q = It\), where \(q\) is the charge, \(I\) is the current, and \(t\) is the time (\(1.0 \text{ s}\)), we calculate the charge: \( q = (0.75 \text{ A})(1.0 \text{ s}) = 0.75 \text{ C} \).
03

Use the Charge of an Electron

The charge for one electron is given as \(e = 1.6 \times 10^{-19} \text{ C}\). We need this to find out how many electrons make up the \(0.75 \text{ C}\).
04

Calculate the Number of Electrons

To find how many electrons correspond to the total charge, divide the total charge by the charge of one electron: \[ \text{Number of Electrons} = \frac{0.75 \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} = 4.7 \times 10^{18} \] electrons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Flow
When we think about how electricity works in simple terms, it can help to imagine it as a flow of tiny particles. Electrons, which are negatively charged subatomic particles, are responsible for carrying electrical energy through wires and other conductors.
  • These electrons move through the conducting material, creating a flow known as an "electric current".
  • In a metal wire, which is a common conductor, electrons move from one atom to the next.
  • This movement happens because of the electric field applied by a power source, like a battery.
The flow of electrons is crucial because it's how we harness electrical energy to power devices, lights, and other everyday items. Understanding electron flow gives us the basis for more complex topics in electricity and electronics.
Charge Calculation
The concept of charge calculation is essential in understanding how much electrical energy is moving through a circuit at any given moment. To find out how much charge is transferred, we use the formula:\[ q = I imes t \]Where:
  • q is the charge in coulombs (C).
  • I is the current in amperes (A).
  • t is the time in seconds (s).
By calculating the charge, you can understand how many electrons are flowing through a conductor. The charge flow gives us concrete data through which we can compute the number of electrons involved in the current. This fundamental calculation is the building block for more advanced electrical studies and applications.
Electric Current
Electric current is a fascinating phenomenon that describes the flow of electric charge in a circuit. It is measured in amperes (A), often referred to as "amps." An electric current is essentially how many charges pass through a given point in a circuit per second.
Here are some easy ways to visualize and understand electric current:
  • Think of it as water flowing through a pipe, where the pipe is the conductor and water represents the electric charge.
  • The flow is driven by a difference in electric potential, much like water flows downhill due to gravity.
  • This flow powers our electric devices, lights, and appliances every day.
In our exercise, knowing that the current is 0.75 A, we determine how many electrons are moving through a given point each second. Having a clear grasp of what electric current represents can shed light on the broader workings of electrical circuits and systems.

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Most popular questions from this chapter

A direct-current generator has an emf of \(120 \mathrm{~V}\); that is, its terminal voltage is \(120 \mathrm{~V}\) when no current is flowing from it. At an output of \(20 \mathrm{~A}\), the terminal potential is \(115 \mathrm{~V}\). (a) What is the internal resistance \(r\) of the generator? \((b)\) What will be the terminal voltage at an output of \(40 \mathrm{~A}\) ? The situation is much like that shown in Fig. \(26-3 .\) Now, however, \(\varepsilon=120 \mathrm{~V}\) and \(I\) is no longer \(25 \mathrm{~A}\). (a) In this case, \(I=20 \mathrm{~A}\) and the p.d. from \(A\) to \(B\) is \(115 \mathrm{~V}\). Therefore, $$ 115 \mathrm{~V}=+120 \mathrm{~V}-(20 \mathrm{~A}) r $$ from which \(r=0.25 \Omega\). (b) Now \(I=40\) A. So Terminal p.d. \(=\varepsilon-I r=120 \mathrm{~V}-(40 \mathrm{~A})(0.25 \Omega)=110 \mathrm{~V}\)

The resistivity of aluminum is \(2.8 \times 10^{-8} \Omega \cdot \mathrm{m}\). How long a piece of aluminum wire \(1.0 \mathrm{~mm}\) in diameter is needed to give a resistance of \(4.0 \Omega\) ?

An electric utility company runs two \(100 \mathrm{~m}\) copper wires from the street mains up to a customer's premises. If the wire resistance is \(0.10 \Omega\) per \(1000 \mathrm{~m}\), calculate the line voltage drop for an estimated load current of \(120 \mathrm{~A}\).

An electron gun in a TV set shoots out a beam of electrons. The beam current is \(1.0 \times 10^{-5} \mathrm{~A}\). How many electrons strike the TV screen each second? How much charge strikes the screen in a minute?

When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is \(1.0\) megohm \(\left(10^{6} \Omega\right)\). How much current passes through the insulation of the motor if the test voltage is \(1000 \mathrm{~V}\) ?
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