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Chapter 19: Problem 37

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 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 is approximately 0.451 A.

Step by step solution

01

Understand the Given Information

We have a battery rated at a voltage of \(3.70 \mathrm{~V}\) and an energy capacity of \(3.15 \times 10^{4} \mathrm{~J}\). This energy allows the phone to operate for 5.25 hours. We need to determine the average current the phone draws.
02

Convert Time to Seconds

Since current is typically calculated in terms of seconds, convert the operation time from hours to seconds. Multiply 5.25 hours by 3600 seconds per hour:\[5.25 \text{ hours} \times 3600 \text{ seconds/hour} = 18900 \text{ seconds}\]
03

Use Formula for Energy

The energy stored in a battery can also be expressed as the product of voltage, current, and time:\[E = V \times I \times t\]Where \(E\) is the energy in joules, \(V\) is the voltage in volts, \(I\) is the current in amperes, and \(t\) is the time in seconds.
04

Rearrange the Formula for Current

Solve for the current \(I\) using the formula:\[I = \frac{E}{V \times t}\]
05

Substitute the Known Values

Substitute \(E = 3.15 \times 10^{4} \mathrm{~J}\), \(V = 3.70 \mathrm{~V}\), and \(t = 18900 \text{ seconds}\) into the equation:\[I = \frac{3.15 \times 10^{4}}{3.70 \times 18900}\]
06

Calculate the Final Answer

Perform the computation to find the current:\[I = \frac{3.15 \times 10^{4}}{3.70 \times 18900} \approx 0.451 \text{ A}\]Thus, the average current the phone draws is approximately \(0.451\) amperes.

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Key Concepts

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

Battery Voltage
Battery voltage is a crucial factor in determining how much energy a battery can provide at any given time. It is measured in volts (V) and represents the potential difference across the battery's terminals. For the cell phone battery described in the problem, the voltage is given as 3.70 V. This means that when the phone is operating, the battery is delivering electrical energy with a pressure of 3.70 volts.

Understanding battery voltage helps us know how "strong" the electrical force is. Higher voltage often means the battery can push more power into the device faster. However, it works in conjunction with how much total energy it can store, known as the energy capacity.
Energy Capacity
Energy capacity is described as the total amount of energy a battery can store and deliver. It is usually measured in joules (J) or watt-hours (Wh). In this problem, the energy capacity of the phone battery is specified as \(3.15 \times 10^4 \text{ J}\). This value tells us how much total energy the battery can deliver before it needs to be recharged.

A battery with a larger energy capacity can last longer or provide more power to a device. It plays an important role when determining how long a device can run on a single charge, contributing directly to the concept of operational time.
  • More energy capacity = Longer usage time
  • Allows for stronger power output
Operational Time
Operational time refers to the duration a device can function on its battery. In this scenario, the cell phone can operate for 5.25 hours before its battery depletes. To use this information in equations related to current, the time is usually converted into seconds. This conversion is done because the standard unit of time for electrical calculations is seconds.

To convert 5.25 hours to seconds, multiply by 3600 (the number of seconds in one hour):
  • 5.25 hours \(\times\) 3600 seconds/hour = 18900 seconds
Operational time tells us how long a battery can support a device without interruption, making it essential for calculating the average current drawn.
Ohm's Law
Ohm's Law is a fundamental principle used to understand the relationship between voltage, current, and resistance in electrical circuits. Although in this exercise we're more directly concerned with the equation for energy \(E = V \times I \times t\), Ohm鈥檚 Law itself states:
  • \(V = I \times R\)
  • Where \(V\) is voltage, \(I\) is current, and \(R\) is resistance.
For this problem, we're rearranging the energy equation to solve for the current \(I\), using:
  • \(I = \frac{E}{V \times t}\)
  • Where \(E\) is energy, \(V\) is voltage, and \(t\) is time.
By substituting known values for \(E, V,\) and \(t\), we calculate the average current the phone draws, which gives us an understanding of how much electrical flow powers the device over the operational period.

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Most popular questions from this chapter

This procedure is not recommended! You'll see why after you work the problem. You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver that has a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of \(1.0 \mathrm{k} \Omega .\) (a) If you accidentally touch the "hot" wire connected to the \(120 \mathrm{~V}\) line, how much current will pass through your body? Is this amount enough to be dangerous? (The maximum safe current is about \(5 \mathrm{~mA} .\) ) (b) How much electric power is delivered to your body?

An iron wire has a temperature coefficient of resistivity of \(+0.0050\left(\mathrm{C}^{\circ}\right)^{-1} .\) When it is placed into boiling water, it has a resistance of \(1000 \Omega\). What is its resistance when it is placed into an ice-water bath?

Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic-that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a \(5 \mathrm{~mm}\) length of thread between two electrical contacts. The researchers stretched the thread in \(1 \mathrm{~mm}\) increments to more than twice its original length, and then allowed it to return to its original length, again in \(1 \mathrm{~mm}\) increments. Some of the resistance measurements are given in the table: $$\begin{array}{l|cccccccc}\begin{array}{l}\text { Resistance of } \\\\\text { thread }\left(10^{9} \Omega\right)\end{array} & 9 & 19 & 41 & 63 & 102 & 76 & 50 & 24 \\ \hline \begin{array}{l}\text { Length of } \\\\\text { thread }(\mathrm{mm})\end{array} & 5 & 7 & 9 & 11 & 13 & 9 & 7 & 5\end{array}$$ What is the maximum current that flows in the thread during this experiment if the voltage source is a \(9 \mathrm{~V}\) battery? A. about \(1 A\) B. about \(0.1 A\) C. about \(1 \mu A\) D. about 1 nA

An idealized voltmeter is connected across the terminals of a \(15.0 \mathrm{~V}\) battery, and a \(75.0 \Omega\) appliance is also connected across its terminals. If the voltmeter reads \(11.3 \mathrm{~V}\) : (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

A toaster using a Nichrome heating element operates on \(120 \mathrm{~V}\). When it is switched on at \(20^{\circ} \mathrm{C}\), the heating element carries an initial current of 1.35 A. A few seconds later, the current reaches the steady value of 1.23 A. (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the temperature range from \(20^{\circ} \mathrm{C}\) to the final temperature of the element is \(4.5 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1} .\) (b) What is the power dissipated in the heating element (i) initially; (ii) when the current reaches a steady value?
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