/*! 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 52 Solve each equation for the indi... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each equation for the indicated variable. See Example 7 . \(W=\frac{C E^{2}}{2}\) for \(C\) (Electronics: energy stored in a capacitor)

Short Answer

Expert verified
The solution for \( C \) is \( C = \frac{2W}{E^{2}} \).

Step by step solution

01

Understanding the Equation

The given equation is for the energy stored in a capacitor: \[ W = \frac{C E^{2}}{2} \] We are asked to solve this equation for \( C \), the capacitance.
02

Isolate the Fraction involving C

Multiply both sides of the equation by 2 to eliminate the denominator: \[ 2W = C E^{2} \] This step simplifies the equation to a form where \( C \) is being multiplied directly by other terms.
03

Solve for C

To solve for \( C \), divide both sides of the equation by \( E^{2} \): \[ C = \frac{2W}{E^{2}} \] Now, the capacitance \( C \) is expressed in terms of energy \( W \) and voltage squared \( E^2 \).

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Key Concepts

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

Energy in a Capacitor
In the realm of electronics, capacitors are essential components that store electrical energy. The energy stored in a capacitor can be represented by the equation \( W = \frac{C E^{2}}{2} \). Here, \( W \) stands for energy, which is measured in joules, \( C \) is the capacitance in farads, and \( E \) is the voltage across the capacitor in volts. This equation highlights how the energy stored is directly proportional to both the capacitance and the square of the voltage.

To grasp the importance of energy storage in capacitors, consider that they help maintain power supply during fluctuations. Capacitors are used in various applications, like filtering noise from power lines and in timer circuits. Understanding their energy equation is key to designing circuits that take advantage of their storage and discharge properties.

Key points to note include:
  • Energy is proportional to both capacitance and voltage squared.
  • Higher voltage or capacitance leads to greater energy storage.
Capacitance
Capacitance, symbolized by \( C \), is a measure of a capacitor's ability to store an electric charge. In the context of the equation \( W = \frac{C E^{2}}{2} \), capacitance plays a critical role in determining energy storage efficiency.

Capacitance is expressed in farads, a unit that reflects the capacity of a capacitor to hold charge at one volt. Factors affecting capacitance include the physical characteristics of the capacitor like the surface area of the plates, the distance between them, and the dielectric material used.

Key characteristics of capacitance include:
  • The larger the surface area of the plates, the higher the capacitance.
  • The shorter the distance between plates, the higher the capacitance.
  • Dielectric materials with high permittivity increase capacitance.
Understanding capacitance helps in choosing the right capacitor for specific circuit needs, ensuring that it can adequately store and release energy as required.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and solving equations to isolate and solve for a specific variable. In the context of the energy stored in a capacitor, manipulating the equation \( W = \frac{C E^{2}}{2} \) is essential to solve for the capacitance \( C \).

To solve for \( C \), we perform the following steps:
  • Multiply both sides by 2 to eliminate the denominator, resulting in \( 2W = C E^{2} \).
  • Divide both sides by \( E^{2} \) to isolate \( C \), giving us \( C = \frac{2W}{E^{2}} \).
These steps illustrate the power of algebraic manipulation in turning complex equations into a form that is straightforward to interpret and understand.

Having mastery over algebra allows students to approach physics and electronics problems with confidence, breaking down intricate formulas and concepts into manageable pieces.

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Most popular questions from this chapter

Perform each indicated operation. The dose of medicine prescribed for a child depends on the child's age \(A\) in years and the adult dose \(D\) for the medication. Two expressions that give a child's dose are Young's Rule, \(\frac{D A}{A+12},\) and Cowling's Rule, \(\frac{D(A+1)}{24} .\) Find an expression for the difference in the doses given by these expressions.

Simplify. $$ \frac{3(a+1)^{-1}+4 a^{-2}}{\left(a^{3}+a^{2}\right)^{-1}} $$

Simplify each complex fraction. See Examples 1 and 2. $$ \frac{\frac{4}{5-x}+\frac{5}{x-5}}{\frac{2}{x}+\frac{3}{x-5}} $$

One of the great algebraists of ancient times a man named Diophantus. Litle is known of his life other than the lived and worked in Alexandria. Some historians believe he lived during the first century of the Christian era, about the time of Nero. The only che to his personal life is the following epigram found in a collection called the Palatine A nthology. God granted him youth for a sixth of his life and added a twelfth part to this. He clothed his cheeks in down. He lit him the light of wedlock after a seventh part and five years after his marriage. He granted him a son. Alas, lateborn wretched child. After attaining the measure of half his father's life, cruel fate overtook him, thus leaving Diophantus during the last four years of his life only such consolation as the science of numbers. How old was Diophantus at his death?" We are looking for Diophantus' age when he died, so let x represent that age. If we sum the parts of his life, we should get the total age. Parts of his life \(\left\\{\begin{array}{l}{\frac{1}{6} x+\frac{1}{12} x \text { is the time of his youth. }} \\ {\frac{1}{7} x \text { is the time between his youth and when }} \\ {\text { he married. }} \\ {5 \text { years is the time between his marriage }} \\ {\text { and the birth of his son. }} \\\ {\frac{1}{2} x \text { is the time Diophantus had with his son. }} \\ {4 \text { years is the time between his son's death }} \\ {\text { and his own. }}\end{array}\right.\) The sum of these parts should equal Diophantus' age when he died. $$ \frac{1}{6} \cdot x+\frac{1}{12} \cdot x+\frac{1}{7} \cdot x+5+\frac{1}{2} \cdot x+4=x $$ Solve the epigram.

\- Solve the following. See Examples I through 7. (Note: Some exercises can be modeled by equations without rational expressions.) It takes 9 hours for pump \(A\) to fill a tank alone. Pump \(B\) takes 15 hours to fill the same tank alone. If pumps \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are used, the tank fills in 5 hours. How long does it take pump \(\mathrm{C}\) to fill the tank alone?

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