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A battery-powered global positioning system (GPS) receiver operating on 9.0 \(\mathrm{V}\) Uraws a current of 0.13 \(\mathrm{A}\) . How much electrical energy does it consume during 1.5 \(\mathrm{h} ?\)

Short Answer

Expert verified
The GPS receiver consumes 6318 joules of energy in 1.5 hours.

Step by step solution

01

Understand the Problem

We need to find out how much electrical energy is consumed by a GPS receiver that operates with a voltage of 9.0 \(\mathrm{V}\) and draws a current of 0.13 \(\mathrm{A}\) over a period of 1.5 hours.
02

Convert Time to Seconds

First, convert the time of operation from hours to seconds because energy calculations are typically done in seconds. \[1.5 \; \text{hours} = 1.5 \times 60 \times 60 = 5400 \; \text{seconds}\]
03

Calculate Power Consumption

To find the power consumed by the device, use the formula \(P = V \times I\),where \(V\) is voltage and \(I\) is current.\[P = 9.0 \; \text{V} \times 0.13 \; \text{A} = 1.17 \; \text{W} \; (\text{watts})\]
04

Calculate Electrical Energy Consumption

Electrical energy consumed can be found by multiplying the power by the time the device was operating. Using the formula \(E = P \times t\), where \(E\) is energy, \(P\) is power, and \(t\) is time:\[E = 1.17 \; \text{W} \times 5400 \; \text{s} = 6318 \; \text{J} \; (\text{joules})\]
05

Conclusion

The GPS receiver consumes 6318 joules of electrical energy when operating for 1.5 hours at the given voltage and current.

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Key Concepts

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

GPS Receiver
A GPS receiver is a device used to receive and decode signals sent by satellites in orbit around Earth to determine its precise location. These devices have become integral parts of many gadgets, from smartphones to dedicated navigation systems in cars. The main function of a GPS receiver is to persistently monitor satellite signals to calculate the geographical position of the device.

A critical aspect of GPS receivers is their dependency on battery power. Since they need to be used in various environments, often without access to power sources, efficient energy consumption is crucial. They tap into their internal batteries to remain functional.
  • They work primarily by obtaining positioning data via satellite communication.
  • Due to their constant operation, especially in high-demand applications, understanding their power consumption is important to maintain sustained use.
By optimizing a GPS receiver's power use, battery life can be extended, making them more dependable for longer periods.
Power Consumption Formula
The power consumption formula is a straightforward way to calculate how much energy a device uses. The fundamental equation is:

\[ P = V \times I \]
Where:
  • \( P \) is the power measured in watts (W).
  • \( V \) is the voltage measured in volts (V).
  • \( I \) is the current measured in amperes (A).
By using this formula, you can easily determine the power that any electrical device, including a GPS receiver, consumes. This calculation is useful to understand how long a battery will last under specific conditions.

Knowing the power consumption helps in designing batteries or power systems that align with the energy demands of the GPS receiver. Engineers often aim to reduce \( P \) without sacrificing performance, to optimize device efficiency.
Energy in Joules
Energy is the capacity to do work, and in the context of electrical devices, it refers to the capability to enable electronic circuits to function. Joules (J) are the SI unit of energy, providing a standard measure to quantify electrical energy usage.

When we talk about the energy consumed by a GPS receiver, we usually mean how much work it can do or how long it can operate on a given power source. The formula to find energy consumption is:

\[ E = P \times t \]

Here:
  • \( E \) is energy in joules (J).
  • \( P \) is power in watts (W).
  • \( t \) is time in seconds (s).
Using this formula helps us calculate how much energy a device like a GPS receiver consumes over time. For instance, if a GPS receiver uses 1.17 watts of power for 5400 seconds (1.5 hours), it consumes 6318 joules of energy.

This understanding allows for the effective management of battery resources and ensures users plan their GPS use efficiently, without unexpected interruptions.

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Most popular questions from this chapter

A copper wire has a square cross section 2.3 \(\mathrm{mm}\) on a side. The wire is 4.0 \(\mathrm{m}\) long and carries a current of 3.6 \(\mathrm{A}\) . The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3}\) . Find the magnitudes of (a) the current density in the wire and \((b)\) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

A \(1.50-m\) cylindrical rod of diameter 0.500 \(\mathrm{cm}\) is connected to a power supply that maintains a constant potential difference of 15.0 \(\mathrm{V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) the ammeter reads 18.5 \(\mathrm{A}\) , while at \(92.0^{\circ} \mathrm{C}\) it reads 17.2 \(\mathrm{A}\) . You can ignore any thermal expansion of the rod. Find (a) the resistivity and (b) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

The potential difference across the terminals of a battery is 8.4 \(\mathrm{V}\) when there is a current of 1.50 \(\mathrm{A}\) in the battery from the negative to the positive terminal. When the current is 3.50 \(\mathrm{A}\) in the reverse direction, the potential difference becomes 9.4 \(\mathrm{V}\) . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

A \(3.00-\mathrm{m}\) length of copper wire at \(20^{\circ} \mathrm{C}\) has a \(1.20-\mathrm{m}\) -long section with diameter 1.60 \(\mathrm{mm}\) and a \(1.80-\mathrm{m}\) -long section with diameter 0.80 \(\mathrm{mm}\) . There is a current of 2.5 \(\mathrm{mA}\) in the \(1.60-\mathrm{mm}-\) diameter section. (a) What is the current in the 0.80 -mm-diameter section? (b) What is the magnitude of \(\vec{E}\) in the 1.60 -mm- diameter section? (c) What is the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the 0.80 \(\mathrm{cmm}\) - diameter section? (d) What is the potential difference between the ends of the 3.00 -m length of wire?

The potential difference between points in a wire 75.0 \(\mathrm{cm}\) apart is 0.938 \(\mathrm{V}\) when the current density is \(4.40 \times 10^{7} \mathrm{A} / \mathrm{m}^{2}\) . What are (a) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the wire and \((\mathrm{b})\) the resistivity of the material of which the wire is made?

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