/*! 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

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The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce \(3.15 \times 10^4\) 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 0.450 A.

Step by step solution

01

Understand the Given Data

The problem provides us with the following information: the voltage of the battery is rated at 3.70 V, the total energy produced by the battery is \(3.15 \times 10^4\) J, and the total time of operation is 5.25 hours before recharging is needed.
02

Convert Hours into Seconds

Since energy and current calculations are typically done in seconds, we need to convert 5.25 hours into seconds. There are 3600 seconds in an hour, so the total time in seconds is \(5.25 \text{ h} \times 3600 \text{ s/h} = 18900 \text{ s}\).
03

Use the Formula for Power

Power (P) is defined as energy (E) divided by time (t), so we use the formula \(P = \frac{E}{t}\). Here, \(E = 3.15 \times 10^4\) J and \(t = 18900\) s. Therefore, \(P = \frac{3.15 \times 10^4}{18900}\) W.
04

Calculate the Power

Substitute the given values into the formula to get the power: \(P = \frac{3.15 \times 10^4}{18900} \approx 1.6667\) W.
05

Use the Formula for Current

We have the power and the voltage, and we can find the average current (I) using the formula \(P = IV\), where \(I\) is the current and \(V\) is the voltage (3.70 V). Rearranging for \(I\), we have \(I = \frac{P}{V}\).
06

Calculate the Average Current

Using the power calculated in Step 4, substitute \(P = 1.6667\) W and \(V = 3.70\) V into the formula to find the current: \(I = \frac{1.6667}{3.70} \approx 0.450\) A.

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Key Concepts

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

Average Current
The average current is the constant flow of electric charge over a certain period. In the context of this problem, it's the steady flow of electrons that keeps the cell phone powered during usage. To determine this quantity, we first need the power of the device, which depends on both the total energy the battery can deliver and the time it operates.

The average current can be calculated using the formula:
  • \( P = IV \)
  • \( I = \frac{P}{V} \)
Here, \( I \) is the current, \( P \) is the power, and \( V \) is the voltage. By rearranging and solving \( I = \frac{P}{V} \), we find that the average current is the amount of electric charge passing through the circuit per unit of time, measured in amperes (A). For this cellphone, the average current is found to be approximately 0.450 A.
Voltage
Voltage, often referred to as electric potential difference, is an essential element of electric circuits. It's the force that pushes electrical charges through a conducting loop. In other words, voltage causes electric current to flow in a circuit. Voltage is measured in volts (V).For the cell phone problem, the battery provides a voltage of 3.70 V. This means that each charge moving through the circuit gains 3.7 joules of energy for every coulomb of charge. Understanding voltage helps in identifying how much energy is provided by the battery to move electric charges to operate the device. Knowing the voltage is crucial for calculating power and, subsequently, the average current, using the relationship \( P = IV \).
Power Formula
Power in electrical circuits is a measure of how much energy is converted by the circuit each second. It's defined as the rate at which work is done or energy is transferred and is measured in watts (W).
The power formula is important and is expressed as:
  • \( P = \frac{E}{t} \)
where \( E \) is the total energy in joules, and \( t \) is the time in seconds. In our exercise, the battery's energy conversion rate tells us how efficiently the battery transfers its stored energy into the circuit to power the phone. With given values, power helps us understand how quickly the energy is being used. If it's constant at 1.6667 W, it means that much energy is being consumed every second.
Energy Conversion
Energy conversion in electrical circuits is the process of changing stored energy in the battery into usable electrical energy. This is essential for any device, like a cell phone, to perform its functions.
In the provided exercise, the cell phone's battery converts a large amount of stored energy, \(3.15 \times 10^4\) J, over 5.25 hours of operation. This conversion is a demonstration of how chemical energy inside the battery is turned into electrical energy to power the phone. Understanding energy conversion helps in analyzing how long devices can run before needing a recharge, based on the power usage derived from the battery's energy and time of operation, emphasizing the importance of efficient energy transfer.

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Most popular questions from this chapter

You apply a potential difference of 4.50 V between the ends of a wire that is 2.50 m in length and 0.654 mm in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?

A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at 25 V for a very short time, usually about 3.0 ms. (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

A typical small flashlight contains two batteries, each having an emf of 1.5 V, connected in series with a bulb having resistance \(17 \Omega\). (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for 5.0 h, what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

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