/*! 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 13 A \(2.0 \mu \mathrm{F}\) capacit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 13

A \(2.0 \mu \mathrm{F}\) capacitor and a \(4.0 \mu \mathrm{F}\) capacitor are connected in parallel across a \(300 \mathrm{~V}\) potential difference. (a) What is the total energy stored by them? (b) They are next connected in series across that potential difference. What is the ratio of the total energy stored by them in the parallel arrangement to that in the series arrangement?

Short Answer

Expert verified
(a) 0.27 J; (b) Ratio is 4.5.

Step by step solution

01

Calculating Total Capacitance in Parallel

When capacitors are connected in parallel, the total capacitance \(C_P\) is the sum of their individual capacitances. Thus, \(C_P = C_1 + C_2 = 2.0 \mu \mathrm{F} + 4.0 \mu \mathrm{F} = 6.0 \mu \mathrm{F}\).
02

Calculating Energy Stored in Parallel

The energy \(E\) stored in a capacitor is given by the formula \(E = \frac{1}{2} C V^2\). Substituting the values for parallel configuration, we have \(E_P = \frac{1}{2} \times 6.0 \times 10^{-6} \mathrm{~F} \times (300)^2 \mathrm{~V^2} = 0.27 \mathrm{~J}\).
03

Calculating Total Capacitance in Series

For capacitors in series, the reciprocal of the total capacitance \(C_S\) is the sum of the reciprocals of their individual capacitances: \(\frac{1}{C_S} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2.0 \mu \mathrm{F}} + \frac{1}{4.0 \mu \mathrm{F}} = \frac{3}{4} \mu \mathrm{F}^{-1}\). Therefore, \(C_S = \frac{4}{3} \mu \mathrm{F} = 1.33 \mu \mathrm{F}\).
04

Calculating Energy Stored in Series

Using the total capacitance in series, the energy stored is \(E_S = \frac{1}{2} C_S V^2 = \frac{1}{2} \times 1.33 \times 10^{-6} \mathrm{~F} \times (300)^2 \mathrm{~V^2} = 0.06 \mathrm{~J}\).
05

Calculating the Ratio of Energies

To find the ratio of energy stored in the parallel arrangement to the series arrangement, divide the energy in parallel by the energy in series: \(\text{Ratio} = \frac{E_P}{E_S} = \frac{0.27}{0.06} = 4.5\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Energy Stored in Capacitors
Capacitors are fundamental components in electrical circuits that have the ability to store energy. The energy (E) stored in a capacitor can be quantified using the formula:\[E = \frac{1}{2} C V^2\]where (C) is the capacitance and (V) is the voltage applied across the capacitor.

The beauty of capacitors lies in their capability to hold energy, which can be utilized in various electronic applications like timing circuits or smoothing out signals. This stored energy is particularly useful in circuits needing a temporary energy boost.

In our exercise, we calculate energy storage for both parallel and series configurations. We see significantly more energy stored in parallel than in series. This difference is due to the way capacitors interact with potential difference and capacitance in each setup.
Parallel and Series Configurations
The arrangement of capacitors greatly affects both total capacitance and energy storage. When connecting capacitors, they can either be arranged in parallel or in series, each configuration having a distinct effect:
  • **Parallel Configuration**: Here, the total capacitance ( C_P ) is the sum of individual capacitances. This simple addition results in a larger total capacitance, providing greater energy storage capacity. For example, in the given exercise, a capacitance of 6.0 碌F was achieved using a parallel arrangement.
  • **Series Configuration**: The total capacitance ( C_S ) in a series setup is governed by the reciprocal of the sum of individual reciprocals. This results in a smaller capacitance compared to individual values. This can be seen when a series arrangement resulted in a capacitance of 1.33 碌F.
Each setup has its applications, and the choice between series and parallel often depends on desired circuit characteristics and performance.
Capacitance Calculations
Calculating capacitance in different configurations requires understanding the relationships among capacitors:
  • In a **parallel configuration**, the total capacitance is simply the sum of the capacitances: \( C_P = C_1 + C_2 + ... + C_n \). This formula highlights why more capacitance results in higher energy storage.
  • In a **series configuration**, the calculation is slightly more complex, involving the reciprocals: \( \frac{1}{C_S} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \). After performing the calculation, remember to invert the final result to find the total capacitance \( C_S \).
These calculations form the bedrock of understanding how effective a capacitor configuration will be in storing energy. It鈥檚 crucial to apply proper formulas for accurate results, as seen in energy differences between parallel and series arrangements in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The space between two concentric conducting spherical shells of radii \(b=1.70 \mathrm{~cm}\) and \(a=1.20 \mathrm{~cm}\) is filled with a substance of dielectric constant \(\kappa=6.91\). A potential difference \(V=73.0 \mathrm{~V}\) is applied across the inner and outer shells. Determine (a) the capacitance of the device, (b) the free charge \(q\) on the inner shell, and (c) the charge \(q^{\prime}\) induced along the surface of the inner shell.

A parallel-plate capacitor has a capacitance of \(100 \mathrm{pF}\), a plate area of \(80 \mathrm{~cm}^{2}\), and a mica dielectric \((\kappa=5.4)\) completely filling the space between the plates. At \(85 \mathrm{~V}\) potential difference, calculate (a) the electric field magnitude \(E\) in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica.

Two parallel plates of area \(100 \mathrm{~cm}^{2}\) are given charges of equal magnitudes \(8.4 \times 10^{-7} \mathrm{C}\) but opposite signs. The electric field within the dielectric material filling the space between the plates is \(1.4 \times 10^{6} \mathrm{~V} / \mathrm{m}\). (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

A parallel-plate capacitor has square plates with edge length \(8.20 \mathrm{~cm}\) and \(1.30 \mathrm{~mm}\) separation. (a) Calculate the capacitance. (b) Find the charge for a potential difference of \(120 \mathrm{~V}\).

A parallel-plate capacitor has plates of area \(0.080 \mathrm{~m}^{2}\) and a separation of \(1.2 \mathrm{~cm}\). A battery charges the plates to a potential difference of \(120 \mathrm{~V}\) and is then disconnected. A dielectric slab of thickness \(4.0 \mathrm{~mm}\) and dielectric constant \(4.8\) is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge \(q\) (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation

Work Energy and Power

Read Explanation

Thermodynamics

Read Explanation

Geometrical and Physical Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.