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Chapter 20: Problem 26

An MP3 player operates with a voltage of 3.7 V, and is using 0.095 W of power. Find the current being supplied by the player鈥檚 battery.

Short Answer

Expert verified
The current is approximately 0.02568 A.

Step by step solution

01

Understanding the Formula

We need to find the current supplied by the battery. The relationship between power, voltage, and current is given by the formula: \[ P = V imes I \]where \( P \) is power (in watts), \( V \) is voltage (in volts), and \( I \) is current (in amperes).To find the current, rearrange this formula to solve for \( I \).
02

Rearranging the Formula

To solve for current \( I \), rearrange the formula:\[ I = \frac{P}{V} \]This shows that current is the power divided by the voltage.
03

Substituting the Values

Substitute the given values into the rearranged formula:\[ I = \frac{0.095}{3.7} \]This calculation will give you the current.
04

Calculating the Current

Perform the division:\[ I = 0.02568 ext{ A} \]Thus, the current supplied by the battery is approximately 0.02568 A (25.68 mA).

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Key Concepts

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

Ohm's Law
To fully understand how current, voltage, and resistance interact in an electrical circuit, we need to explore Ohm's Law. This fundamental principle states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The law is represented by the equation: \[ V = I \times R \]where:
  • \( V \) is the voltage in volts,
  • \( I \) is the current in amperes, and
  • \( R \) is the resistance in ohms.
This indicates that if you increase the voltage, the current will also increase, assuming the resistance remains constant. Conversely, increasing the resistance will cause the current to decrease if the voltage stays the same. Ohm's Law is crucial for designing and analyzing circuits.
power formula
The power formula is an essential tool to calculate electricity usage in various devices. It connects power, voltage, and current through the relationship \[ P = V \times I \] where:
  • \( P \) is the power in watts,
  • \( V \) is the voltage in volts, and
  • \( I \) is the current in amperes.
This formula shows how power, which is the rate at which energy is used or transferred, depends on the current and voltage in a system. By rearranging this formula, you can find any one of these quantities if the other two are known. For example, to find the current (\( I \)), you divide the power by the voltage: \[ I = \frac{P}{V} \] This calculation is especially helpful in determining how much current a device draws based on its power needs and operating voltage.
electrical circuits
An electrical circuit is a path through which electricity flows. It's made up of various elements that work together to transmit power effectively. Understanding circuits is key to analyzing any electrical system. Here's how it works:
  • Components: Circuits include components like resistors, capacitors, inductors, batteries, and switches.
  • Path: Electricity travels in a closed-loop, which may contain multiple paths if it's a parallel circuit.
  • Types: There are two main types of circuits 鈥� series and parallel. In series circuits, components are connected in a line, so the same current flows through each. In parallel circuits, the voltage across each component is the same, but the current can vary.
  • Behavior: Current flows from a power source (like a battery) through conductors (wires), and is regulated by resistive components. This flow is what powers devices.
Understanding these basics helps you predict circuit behavior and solve complex problems.
MP3 player battery
Batteries are vital for portable devices like MP3 players. They store chemical energy and convert it into electrical energy to power the device. Here's what you should know about the MP3 player battery from the exercise:
  • Voltage: The MP3 player battery operates at 3.7 V, which means it provides this amount of potential difference for the current to flow.
  • Power Consumption: The power used by the MP3 player is 0.095 W. This tells us how much energy the device uses per second.
  • Current Supply: By using the power formula, the current drawn from the battery can be found. This is crucial because the battery must supply enough current to meet the player's needs without depleting too quickly.
  • Efficiency: Understanding power and current helps in assessing how efficiently the battery supports the device.
A well-matched battery ensures the MP3 player has optimal performance and battery life.

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Most popular questions from this chapter

A piece of Nichrome wire has a radius of \(6.5 \times 10^{-4} \mathrm{m} .\) It is used in a laboratory to make a heater that uses \(4.00 \times 10^{2} \mathrm{W}\) of power when connected to a voltage source of 120 \(\mathrm{V}\) . Ignoring the effect of temperature on resistance, estimate the necessary length of wire.

A battery delivering a current of 55.0 \(\mathrm{A}\) to a circuit has a terminal voltage of 23.4 \(\mathrm{V}\) . The electric power being dissipated by the internal resistance of the battery is 34.0 \(\mathrm{W}\) . Find the emf of the battery.

A \(14-\Omega\) coffee maker and a \(16-\Omega\) frying pan are connected in series across a \(120-\mathrm{V}\) source of voltage. \(A 23-\Omega\) bread maker is also connected across the \(120-\mathrm{V}\) source and is in parallel with the series combination. Find the total current supplied by the source of voltage.

An \(86-\Omega\) resistor and a \(67-\Omega\) resistor are connected in series across a battery. The voltage across the \(86-\Omega\) resistor is 27 V. What is the voltage across the \(67-\Omega\) resistor?

A cylindrical aluminum pipe of length 1.50 \(\mathrm{m}\) has an inner radius of \(2.00 \times 10^{-3} \mathrm{m}\) and an outer radius of \(3.00 \times 10^{-3} \mathrm{m}\) . The interior of the pipe is completely filled with copper. What is the resistance of this unit? (Hint: Imagine that the pipe is connected between the terminals of a battery and decide whether the aluminum and copper parts of the pipe are in series or in parallel.)
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