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Chapter 21: Problem 4

In 1 second, a battery charger moves 0.60 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery. a. How much work does the charger do? b. What is the power output of the charger in watts?

Short Answer

Expert verified
a. The charger does 0.90 Joules of work. b. The power output of the charger is 0.90 Watts.

Step by step solution

01

Derive Work

Given the charge (\(Q\)) is 0.60 Coulombs (C) and the voltage (\(V\)) is 1.5 Volts (V), the work (\(W\)) done by the charger is obtained by the equation \(W = Q \times V\). Substitute the given values to derive the work done.
02

Compute for Work

Multiplying 0.60 C by 1.5 V gives a work done of 0.90 Joules (J).
03

Derive Power

Next, we calculate power. Power (\(P\)) can be derived using the formula \(P = \frac{W}{t}\), where \(W\) is the work and \(t\) is the time. Given the work done from step 2 and the given time of 1 second, substitute the values to derive power.
04

Compute for Power

Dividing 0.90 J by 1 s results to a power output of 0.90 Watts (W).

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Key Concepts

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

Electric Potential
Electric potential is like the energy each unit of charge gets in an electric field. Imagine electricity flowing like water through pipes. Much like how water flows from high to low elevation, electric charge moves from high to low potential.
In a battery, the potential difference is set by the battery's voltage. This potential difference pushes charge through a circuit. For a 1.5 V AA battery, this means each coulomb of charge gains 1.5 Joules as it moves from the negative to the positive terminal.
Electric potential is measured in volts (V), and it plays a crucial role in determining how much work an electrical source does.
Charge Movement
Charge movement is the flow of electric charge, which is primarily done by electrons in a conductor. These electrons carry energy from one part of the circuit to another.
In our exercise, 0.60 coulombs of charge move from the negative to the positive terminal of the battery in one second. This movement is facilitated by the potential difference, encouraging the charges to flow and do useful work, like causing a light bulb to shine.
Understanding charge movement is essential, as it helps in figuring out how quickly energy is being transferred in a circuit.
Coulombs
Coulombs are the units used to measure electric charge. It's the amount of charge transferred by a steady current of one ampere in one second. Simply put, coulombs tell us how much charge is moving in an electric circuit.
In our problem, 0.60 coulombs of charge are transferred, which helps us calculate the work done by the battery. Knowing the amount of charge and the potential difference allows us to determine how much energy is used or produced in a circuit.
Coulombs are critical in understanding electric circuits since they directly relate to the amount of work or power produced by the movement of charge.
Voltage Calculation
Voltage calculation is vital to understanding how much work an electrical circuit or device can do. The basic formula to remember is:
  • Work (W) = Charge (Q) x Voltage (V)
In our example, the work done by the charger is calculated by multiplying 0.60 coulombs of charge by the 1.5 volts of the battery. This yields 0.90 joules of work, meaning 0.90 joules of energy are used to move the charge from one terminal to another.
Voltage calculation indicates how efficiently energy is transferred and how much power an electrical device can produce. Understanding this concept helps in designing circuits and using devices at their peak efficiency.

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Most popular questions from this chapter

Moving a charge from point A, where the potential is \(300 \mathrm{V}\), to point \(\mathrm{B},\) where the potential is \(150 \mathrm{V},\) takes \(4.5 \times 10^{-4} \mathrm{J}\) of work. What is the value of the charge?

Two 2.0 -cm-diameter disks spaced \(2.0 \mathrm{mm}\) apart form a parallel- plate capacitor. The electric field between the disks is \(5.0 \times 10^{5} \mathrm{V} / \mathrm{m}\) a. What is the voltage across the capacitor? b. How much charge is on each disk? c. An electron is launched from the negative plate. It strikes the positive plate at a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What was the electron's speed as it left the negative plate?

An uncharged capacitor is connected to the terminals of a \(3.0 \mathrm{V}\) battery, and \(6.0 \mu \mathrm{C}\) flows to the positive plate. The \(3.0 \mathrm{V}\) battery is then disconnected and replaced with a \(5.0 \mathrm{V}\) battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate?

A flying hummingbird picks up charge as it moves through the air. This creates a potential near the bird. What is the "voltage of a hummingbird"? Assume that the bird acquires a charge of \(+200 \mathrm{pC}\), a typical value, and model the bird as a sphere of radius \(3 \mathrm{cm}\).

A proton with an initial speed of \(800,000 \mathrm{m} / \mathrm{s}\) is brought to rest by an electric field. a. Did the proton move into a region of higher potential or lower potential? b. What was the potential difference that stopped the proton? c. What was the initial kinetic energy of the proton, in electron Volts?
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