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Chapter 19: Problem 52

Refer to Multiple-Concept Example 3 to review the concepts that are needed here. A cordless electric shaver uses energy at a rate of \(4.0 \mathrm{~W}\) from a rechargeable \(1.5-\mathrm{V}\) battery. Each of the charged particles that the battery delivers to the shaver carries a charge that has a magnitude of \(1.6 \times 10^{-19} \mathrm{C}\). A fully charged battery allows the shaver to be used for its maximum operation time, during which \(3.0 \times 10^{22}\) of the charged particles pass between the terminals of the battery as the shaver operates. What is the shaver's maximum operation time?

Short Answer

Expert verified
The shaver's maximum operation time is 30 minutes.

Step by step solution

01

Understand the Concepts

The power consumption of the shaver is given as \( P = 4.0 \text{ W} \) and it operates using a \( 1.5\text{ V} \) battery. The total number of charged particles passing through is \( 3.0 \times 10^{22} \), each with a charge of \( 1.6 \times 10^{-19} \text{ C} \). We need to find the maximum operation time.
02

Calculate Total Charge Delivered

Find the total charge \( Q \) delivered by the battery by multiplying the number of charged particles by the charge of each particle:\[Q = (3.0 \times 10^{22}) \times (1.6 \times 10^{-19} \text{ C}) = 4.8 \times 10^{3} \text{ C}\]
03

Use Power Formula to Find Time

Use the formula for power \( P = \frac{V \times Q}{t} \) to solve for time \( t \). Rearranging the formula gives:\[t = \frac{V \times Q}{P}\]Substitute \( V = 1.5 \text{ V} \), \( Q = 4.8 \times 10^3 \text{ C} \), and \( P = 4.0 \text{ W} \) to find \( t \):\[t = \frac{1.5 \times 4.8 \times 10^3}{4.0} = 1800 \text{ s}\]
04

Convert Time to Minutes

Convert the operation time from seconds to minutes:\[t = \frac{1800 \text{ s}}{60} = 30 \text{ minutes}\]

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Key Concepts

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

Power Consumption in Electric Circuits
Power consumption is a fundamental concept in electric circuits, referring to the rate at which energy is used over time. In our exercise, the electric razor consumes power at a rate of 4.0 watts (W). But what does this really mean?
Simply put, power consumption indicates how much energy is being used by the device each second. For our shaver, it uses or 'consumes' 4.0 joules of energy per second. This helps us understand how much energy we must plan for when the device is in operation.
  • Power (\(P\) defined in watts) = Energy consumed (in joules) per second
  • In our scenario: 1 watt = 1 joule/second
Understanding the power consumption helps determine how long the device can function before it exhausts the available energy in the battery. Let's dive deeper into the role of charge and current in this process.
Understanding Charge and Current
Charge and current are closely intertwined concepts in electrical circuits. "Charge" refers to the electricity that is carried by particles like electrons. In our exercise, each particle has a charge amount of \(1.6 \times 10^{-19} \text{ C}\).
"Current," on the other hand, is the flow of electric charge and is measured in amperes (A). It shows how much charge flows through a circuit per second. If we know the number of particles moving through the circuit and the charge, we can determine the total charge using:
  • Total charge \( (Q) = \, \) Number of particles \( \times \, \) Charge per particle
In this case: \(Q = (3.0 \times 10^{22}) \times (1.6 \times 10^{-19} \, \text{C}) = 4.8 \times 10^3 \, \text{C}\) This tells us how much total charge is moved during the battery's operation.
Understanding charge and current is crucial for calculating how the battery's energy is utilized over time.
How Battery Operation Time is Determined
Battery operation time refers to the duration for which a battery can power a device before it needs recharging. Calculating this involves understanding the complete transfer of energy from the battery to the device. In our problem, the operation time can be determined using the power formula.
Power \( P\) is given by the equation:\[ P = \frac{V \times Q}{t} \]where:- \(V\) is the voltage (1.5 V in our problem)- \(Q\) is the total charge- \(t\) is the time in seconds the device operates
By rearranging the formula to solve for operation time (\( t\)), we have:\[ t = \frac{V \times Q}{P} \]Substituting in the values, \( t = \frac{1.5 \times 4.8 \times 10^3}{4.0} = 1800 \, \text{s} \) or 30 minutes.
This calculation highlights how efficiently energy is used and helps in planning the duration for which a device will function based on a full charge.

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Most popular questions from this chapter

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{~mm}\). When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{~V} / \mathrm{m}\). What is the magnitude of the potential difference \(\Delta V\) between the conductors?

An axon is the relatively long tail-like part of a neuron, or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{~m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that an axon can be treated like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{~m}^{2},\) what is its capacitance?

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{~V} ?\)

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{~kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{~kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{~m},\) the speed of the \(3.00 \times 10^{-3}-\mathrm{kg}\) particle is \(125 \mathrm{~m} / \mathrm{s} .\) Find the initial separation between the particles.

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed \(v_{\mathrm{e}}\), while the proton acquires a speed \(v_{\mathrm{p}}\). (a) As each particle accelerates from rest, it gains kinetic energy. Does it gain or lose electric potential energy? (b) Does the electron gain more, less, or the same amount of kinetic energy as the proton does? (c) Is \(v_{\mathrm{e}}\) greater than, less than, or equal to \(v_{\mathrm{p}}\) ? Justify your answers. Find the ratio \(v_{\mathrm{e}} / v_{\mathrm{p}}\). Verify that your answer is consistent with your answers to the Concept Questions.
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