/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 The battery for a certain cell p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The battery for a certain cell phone is rated at \(3.70 \mathrm{~V}\). According to the manufacturer it can produce \(3.15 \times 10^{4} \mathrm{~J}\) of electrical energy, enough for \(5.25 \mathrm{~h}\) of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Short Answer

Expert verified
The average current that the cell phone draws when turned on can be calculated by these steps.

Step by step solution

01

Using the formula for electrical power

To solve this problem, first calculate the power using the formula for electrical power, which is \( P = \frac{E}{t} \), where \(P\) is the power produced by the cell phone battery, \(E\) is the electrical energy (given as \(3.15 \times 10^{4}\) J), and \(t\) is the operational time (provided as \(5.25\) h). However, the upon calculation, the value of time needs to be converted from hours to seconds because the standard unit of energy (Joules) used in this formula is based on seconds. Conversion is done by multiplying \(5.25\) hr by \(3600 \, s/hr\).
02

Power Calculation

Substitute the given energy into the equation for power with \( t = 5.25 \times 3600\), which takes the time to seconds. The calculation would be \( P = \frac{3.15 \times 10^{4}}{5.25 \times 3600}\). Calculating this gives the power in Watts.
03

Calculating the Current

Next, use the power calculated in step 2 and the given voltage to find the current. The formula for power related to current and voltage is \( P = IV \), where \( I \) is the current we need to find. Re-arranging this equation gives \( I = \frac{P}{V}\). Substituting the power from step 2 and \(V = 3.70 \, V\), we get the value for the current which is in Amps.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electrical Energy
Electrical energy is a form of energy resulting from the flow of electric charge. This energy is utilized in many day-to-day devices, from tiny microchips to colossal industrial machines. In the context of a cell phone battery, we can think of electrical energy as the energy stored within the battery that enables the phone to operate without being connected to a wall socket.

When we say the cell phone battery can produce \(3.15 \times 10^{4} \mathrm{~J}\) of electrical energy, we're describing the total amount of work the battery can do before it needs to be recharged. This is akin to the fuel in a car's tank – once it's depleted, you need to 'refill' or in the case of the battery, recharge it. To find the average current drawn from such a battery, we need to understand how this stored energy is converted into power over the operation time, which then helps us calculate the current.
Power Consumption
Power consumption in electrical terms refers to the rate at which electrical energy is used by an appliance or device. It is commonly measured in watts (W), where one watt equals one joule per second. Power consumption can tell us how efficient a device like a cell phone is with its battery's energy, thereby giving us an insight into its performance and operational cost.

To calculate the power consumption from the battery of the cell phone, we divide the total electrical energy by the time of operation, as seen in the formula \( P = \frac{E}{t} \). After converting the time the cell phone operates from hours to seconds, we can find the power in watts. This calculation reveals how much energy the cell phone uses per second and sets the stage for determining the average current draw.
Ohm's Law
Ohm's Law is a fundamental principle in the field of electrical engineering and physics describing the relationship between voltage, current, and resistance in an electrical circuit. The law is typically stated as \( V = IR \), where \( V \) is the voltage across the circuit, \( I \) is the current flowing through it, and \( R \) is the resistance of the circuit.

In the case of our cell phone battery problem, we don't have an explicit resistance value but what we do have are voltage and power, from which we can derive the current. By reworking the Power equation \(P = IV\) to \(I = \frac{P}{V}\), we apply the spirit of Ohm's Law to indirectly solve for the current. Understanding this relationship is crucial for troubleshooting and designing electrical circuits, as it helps to ensure that the electrical devices function properly within their designed parameters.
SI Unit Conversion
In scientific calculations and technical disciplines, working with the International System of Units (SI) ensures consistency and clarity. This system includes units like meters for length, kilograms for mass, and seconds for time. In the context of electricity, key units include volts for voltage, watts for power, and amperes (amps) for current.

During calculations, it's important to convert all units into the SI system to maintain uniformity. For example, in the solution provided, to work with the power formula correctly, the time needs to be converted from hours to seconds, because power's SI unit (watt) is defined as joules per second. Multiplying the time by \(3600\, s/hr\) standardizes the units, allowing us to accurately calculate power and, subsequently, the current. This kind of unit conversion is an essential skill for anyone working within the sciences or engineering fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L\). The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L\). (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in the rod as a function of \(x\). (c) Find the electric potential \(V(x)\) in the rod as a function of \(x\). (d) Graph the functions \(\rho(x), E(x),\) and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\)

The magnetic flux through a coil is given by \(\Phi_{B}=\alpha t-\beta t^{3}\) where \(\alpha\) and \(\beta\) are constants. (a) What are the units of \(\alpha\) and \(\beta ?\) (b) If the induced emf is zero at \(t=0.500 \mathrm{~s},\) how is \(\alpha\) related to \(\beta ?\) (c) If the emf at \(t=0\) is \(-1.60 \mathrm{~V},\) what is the \(\mathrm{emf}\) at \(t=0.250 \mathrm{~s} ?\)

A circular loop of wire with a radius of \(12.0 \mathrm{~cm}\) and oriented in the horizontal \(x y\) -plane is located in a region of uniform magnetic field. A field of \(1.5 \mathrm{~T}\) is directed along the positive \(z\) -direction, which is upward. (a) If the loop is removed from the field region in a time interval of \(2.0 \mathrm{~ms}\), find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

A single loop of wire with an area of \(0.0900 \mathrm{~m}^{2}\) is in a uniform magnetic field that has an initial value of \(3.80 \mathrm{~T},\) is perpendicular to the plane of the loop, and is decreasing at a constant rate of \(0.190 \mathrm{~T} / \mathrm{s}\)(a) What emf is induced in this loop? (b) If the loop has a resistance of \(0.600 \Omega,\) find the current induced in the loop.

A battery has emf \(24.0 \mathrm{~V}\) and internal resistance \(3.00 \Omega .\) A resistor of resistance \(R\) is connected to the battery. What are the two values of \(R\) for which \(21.0 \mathrm{~W}\) of electrical power is consumed in the resistor?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.