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Chapter 27: Problem 9

(a) In electron-volts, how much work does an ideal battery with a \(12.0 \mathrm{~V}\) emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If \(3.40 \times 10^{18}\) electrons pass through each second, what is the power of the battery in watts?

Short Answer

Expert verified
(a) 12.0 eV per electron. (b) 6.54 watts.

Step by step solution

01

Understanding Work Done by Battery

To find the work done by a battery on an electron, we need to use the relationship between voltage, charge, and work. The work done, measured in electron-volts (eV), is simply the voltage across which the charge is moved, since it is applied to a single electron, its charge is 1e (electron charge). Thus, the work done on one electron when it moves through a potential difference \( V \) is \( W = e \times V \). Here, \( e \) is the charge of an electron \( (1.602 \times 10^{-19} \text{ C}) \).
02

Calculating Work Done in Electron-Volts

The problem asks for the work in electron-volts (eV), which is the energy gained or lost by an electron moving through a potential difference of 1 volt. Thus, for a 12.0 V battery, the work done on an electron is \( 12.0 \text{ eV} \), since \( 1 ext{ V} = 1 ext{ eV} \).
03

Understanding the Current and Power Relationship

The power of the battery is related to both the current flowing and the voltage. Power \( P \) is given by the product of voltage \( V \) and current \( I \): \( P = V \times I \). We already have the voltage \( V = 12.0 \text{ V} \). Next, find the current using the relation \( I = \frac{q}{t} \), where \( q \) is the charge flowing in a given time \( t \).
04

Calculating Current

Given that \( 3.40 \times 10^{18} \) electrons pass per second, we can find the current. Each electron carries a charge \( e = 1.602 \times 10^{-19} \text{ C} \), so the total charge passing per second is \( q = 3.40 \times 10^{18} \times 1.602 \times 10^{-19} \text{ C} = 0.54468 \text{ C/s} = 0.54468 \text{ A} \) since \( 1 \text{ C/s} = 1 \text{ A} \).
05

Calculating Power in Watts

Substitute the values into the power equation: \( P = 12.0 \text{ V} \times 0.54468 \text{ A} = 6.53616 \text{ W} \). The power of the battery, given this flow of electrons, is approximately \( 6.54 \text{ W} \) when rounded to three significant figures.

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Key Concepts

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

Work and Energy
Work and energy are closely related concepts in physics. At its core, "work" is the process of energy transfer to an object via force and motion across a distance.
Specifically, when we talk about work done by a battery, we refer to the energy transferred to the charge as it moves through the battery's electric potential. This process can be simplified by the formula; Work is equal to voltage (V) multiplied by charge (q):
  • Equation for work: \( W = V \times q \)
  • Where \( V \) is the potential difference in volts, and \( q \) is the charge in coulombs.
To get this in electron-volts, remember that one electron with a charge moves through 1 volt gains 1 electron-volt of energy. Thus, a battery with a 12 V potential provides 12 electron-volts of energy to each electron passing through it.
In practical terms, this means that if an electron moves from one terminal of a battery to the other, it acquires energy equal to 12 eV for this specific case.
Electric Current
An electric current is essentially a flow of electric charge, usually in the form of electrons moving through a conductor like a wire. The rate at which this charge flows is defined as the electric current.
  • Measured in amperes (A), current is determined by the amount of charge passing through a point in the circuit per unit time.
  • The equation for current \( I \) is given by \( I = \frac{q}{t} \), where \( q \) is the charge and \( t \) is the time interval.
In the exercise provided, \( 3.40 \times 10^{18} \) electrons flow through the circuit per second. Knowing each electron has a charge of approximately \( 1.602 \times 10^{-19} \text{ C} \), you can calculate the current to be about \( 0.54468 \text{ A} \).
This current tells us how much charge is moving through the circuit, which is crucial for understanding how much power the battery produces.
Electric Charge
Electric charge is one of the fundamental properties of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electrical charges: positive and negative.
In practical applications:
  • Electrons, which are negatively charged, have a charge of approximately \( -1.602 \times 10^{-19} \text{ C} \).
  • Protons, which are positively charged, have an equal and opposite charge.
In our original problem, knowing the charge of an electron helps us determine how much charge flows over time in a circuit, contributing to calculations of current and power.
When electrons move from one terminal to another in a battery, they carry their charge through the electrical circuit. This is what makes electronics, devices, and light bulbs work, for example, by bridging the positive and negative charges to produce energy that runs the device.

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Most popular questions from this chapter

When the lights of a car are switched on, an ammeter in series with them reads \(10.0 \mathrm{~A}\) and a voltmeter connected across them reads \(12.0 \mathrm{~V}\) (Fig. 27-60). When the electric starting motor is turned on, the ammeter reading drops to \(8.00 \mathrm{~A}\) and the lights \(\operatorname{dim}\) somewhat. If the internal resistance of the battery is \(0.0500 \Omega\) and that of the ammeter is negligible, what are (a) the emf of the battery and (b) the current through the starting motor when the lights are on?

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.50 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time \(t,(\mathrm{c})\) the potential difference \(V_{C}\) across the capacitor, (d) the potential difference \(V_{R}\) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

A capacitor with initial charge \(g_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first one-third of its charge and (b) two-thirds of its charge?

Suppose that, while you are sitting in a chair, charge separation between your clothing and the chair puts you at a potential of \(200 \mathrm{~V}\), with the capacitance between you and the chair at \(150 \mathrm{pF}\). When you stand up, the increased separation between your body and the chair decreases the capacitance to \(10 \mathrm{pF}\). (a) What then is the potential of your body? That potential is reduced over time, as the charge on you drains through your body and shoes (you are a capacitor discharging through a resistance). Assume that the resistance along that route is \(300 \mathrm{G} \Omega\). If you touch an electrical component while your potential is greater than \(100 \mathrm{~V}\), you could ruin the component. (b) How long must you wait until your potential reaches the safe level of \(100 \mathrm{~V}\) ? If you wear a conducting wrist strap that is connected to ground, your potential does not increase as much when you stand up; you also discharge more rapidly because the resistance through the grounding connection is much less than through your body and shoes. (c) Suppose that when you stand up, your potential is \(1400 \mathrm{~V}\) and the chair-to-you capacitance is \(10 \mathrm{pF}\). What resistance in that wrist-strap grounding connection will allow you to discharge to \(100 \mathrm{~V}\) in \(0.30 \mathrm{~s}\), which is less time than vou would need to reach for, say, your computer?

A \(120 \mathrm{~V}\) power line is protected by a 15 A fuse. What is the maximum number of \(500 \mathrm{~W}\) lamps that can be simultaneously operated in parallel on this line without "blowing" the fuse because of an excess of current?
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