/*! 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 54 To provide the pulse of energy n... [FREE SOLUTION] | 91Ó°ÊÓ

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To provide the pulse of energy needed for an intense bass, some car stereo systems add capacitors. One system uses a 2.0 F capacitor charged to 24 V, double the normal 12 V provided by the car's battery. How much energy does the capacitor store at \(12 \mathrm{V}\) ? At \(24 \mathrm{V} ?\)

Short Answer

Expert verified
The energy stored in the capacitor at 12V is 144J, while at 24V it is 576J.

Step by step solution

01

Calculate the energy at 12V

Insert the given values of \(C = 2.0F\) and \(V = 12V\) into the formula to find the energy. This gives \(E = 0.5*2.0*(12)^2 = 144J\).
02

Calculate the energy at 24V

Now do the same for the given value of \(V = 24V\). This gives \(E = 0.5*2.0*(24)^2 = 576J\).
03

Interpret the result

This shows that the energy stored in the capacitor at 24V is four times the energy stored at 12V, due to the square relationship between voltage and energy.

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Key Concepts

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

Energy Storage in Capacitors
Think of a capacitor as a temporary battery. It stores energy and releases it quickly when needed. This is especially useful in car audio systems to deliver sudden bursts of power for high bass notes. The amount of energy a capacitor can store depends on its capacitance and voltage. The formula for the energy (E) stored in a capacitor is given by: \(E = 0.5 \cdot C \cdot V^2\), where \(C\) is the capacitance in Farads, and \(V\) is the voltage in Volts.

- **High capacitance**: A capacitor with a larger \(C\) can store more energy.- **High voltage**: Increasing \(V\) also increases the stored energy exponentially, as it involves squaring the voltage.Capacitors in car audio systems can quickly release the stored energy to maintain sound quality and avoid distortion during powerful audio moments. They essentially work to enhance the system's performance, allowing for smoother bass output.
Voltage and Capacitor Energy Relationship
The relationship between voltage and the energy stored in a capacitor is quadratic. This means that doubling the voltage results in four times the energy stored. Let's break down why this happens.- **Formula Insight**: From the formula \(E = 0.5 \cdot C \cdot V^2\), you can see that the voltage is squared. This squared relationship means small changes in voltage heavily impact energy storage.- **Example Calculation**: In the exercise, when the voltage is increased from 12V to 24V, the calculated energy increases from 144J to 576J. Notice that while the voltage only doubled, the energy quadrupled.This quadratic nature is crucial for car audio systems, allowing small increases in voltage to significantly boost the energy available to power the audio output, delivering clearer and more powerful sound.
Physics of Car Audio Systems
Car audio systems are interesting applications of physics in everyday life. The role of capacitors in these systems is to complement the car's electrical system by providing the necessary power during peak audio demands. - **Power Demand**: Bass notes require sudden bursts of power. Capacitors can instantaneously supply this power without draining the car battery. - **Avoiding Distortion**: By supplying additional power, capacitors prevent the sound quality from dropping during high-power audio peaks. This keeps the music smooth and enjoyable. Overall, capacitors bridge the gap between the car’s power supply and the audio system's needs. This ensures there is always enough energy available for dynamic soundscapes, enhancing the listener’s experience. Understanding this relationship helps in designing more efficient and powerful car audio systems.

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Most popular questions from this chapter

In 1 second, a battery charger moves 0.60 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery. a. How much work does the charger do? b. What is the power output of the charger in watts?

Moving a charge from point A, where the potential is \(300 \mathrm{V}\), to point \(\mathrm{B},\) where the potential is \(150 \mathrm{V},\) takes \(4.5 \times 10^{-4} \mathrm{J}\) of work. What is the value of the charge?

Guiana dolphins are one of the few mammals able to detect electric fields. In a test of sensitivity, a dolphin was exposed to the variable electric field from a pair of charged electrodes. The magnitude of the electric field near the sensory organs was measured by detecting the potential difference between two measurement electrodes located \(1.0 \mathrm{cm}\) apart along the field lines. The dolphin could reliably detect a field that produced a potential difference of \(0.50 \mathrm{mV}\) between these two electrodes. What is the corresponding electric field strength?

A capacitor consists of two 6.0 -cm-diameter circular plates separated by \(1.0 \mathrm{mm} .\) The plates are charged to \(150 \mathrm{V},\) then the battery is removed. a. How much energy is stored in the capacitor? b. How much work must be done to pull the plates apart to where the distance between them is \(2.0 \mathrm{mm} ?\)

A 1.2 nF parallel-plate capacitor has an air gap between its plates. Its capacitance increases by \(3.0 \mathrm{nF}\) when the gap is filled by a dielectric. What is the dielectric constant of that dielectric?

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