/*! 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 (II) When two capacitors are con... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 5.0 times greater than when they are connected in series and then connected to the same battery. What is the ratio of the two capacitances? (Before the battery is connected in each case, the capacitors are fully discharged.)

Short Answer

Expert verified
The ratio of the capacitances is approximately 2.618.

Step by step solution

01

Understand the Problem

We need to find the ratio of two capacitances, say \( C_1 \) and \( C_2 \), given that the total stored energy when they are connected in parallel is 5 times greater than when they are connected in series.
02

Formula for Stored Energy

The stored energy in a capacitor configuration is given by \( U = \frac{1}{2}CV^2 \), where \( C \) is the capacitance and \( V \) is the voltage. This formula will be used to compare the energies for parallel and series configurations.
03

Total Capacitance in Parallel

For two capacitors in parallel, the total capacitance is \( C_p = C_1 + C_2 \). The energy stored when connected to a battery with voltage \( V \) is \( U_p = \frac{1}{2} (C_1 + C_2) V^2 \).
04

Total Capacitance in Series

For two capacitors in series, the total capacitance is \( C_s = \frac{C_1 C_2}{C_1 + C_2} \). The energy stored is \( U_s = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V^2 \).
05

Energy Ratio Equation

According to the problem, \( U_p = 5 U_s \). Plugging the expressions for \( U_p \) and \( U_s \), we get: \( \frac{1}{2} (C_1 + C_2) V^2 = 5 \left( \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V^2 \right) \).
06

Simplify the Equation

Cancel \( \frac{1}{2}V^2 \) from both sides, resulting in \( (C_1 + C_2) = 5 \frac{C_1 C_2}{C_1 + C_2} \). This can be rearranged to \( (C_1 + C_2)^2 = 5C_1 C_2 \).
07

Solve for Ratio \( \frac{C_1}{C_2} \)

Let \( x = \frac{C_1}{C_2} \). Rewrite the equation in terms of \( x \): \( (xC_2 + C_2)^2 = 5xC_2^2 \) simplifies to \( (x+1)^2 = 5x \). Solve for \( x \), leading to \( x = \frac{3+\sqrt{5}}{2} \).
08

Final Result

The ratio of \( C_1 \) to \( C_2 \) is \( \frac{3+\sqrt{5}}{2} \), approximately 2.618.

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 91Ó°ÊÓ!

Key Concepts

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

Capacitance
Capacitance is a fundamental property of capacitors, defining their ability to store charge. When we talk about capacitance, we're referring to the quantity of charge a capacitor can store per volt across its plates. It is measured in farads (F), and the formula to calculate capacitance is given by:
  • \( C = \frac{Q}{V} \)
where \( C \) is the capacitance, \( Q \) is the charge, and \( V \) is the voltage across the capacitor.
Capacitors can store charge because of the electric field between their plates. The amount of charge a capacitor can hold depends on factors like the surface area of the plates, the distance between them, and the material of the dielectric. Larger plate areas and shorter distances between plates yield higher capacitance, allowing for more effective storage of electric charge.
Understanding capacitance is key to analyzing circuits, as it helps determine how energy is stored and released in electronic devices.
Parallel and Series Circuits
Understanding how capacitors behave in parallel and series circuits is crucial for analyzing electrical circuits. When capacitors are connected in parallel, the total capacitance is simply the sum of the individual capacitances. This is because the voltage across each capacitor remains the same, allowing you to add up their capacitances directly:
  • \( C_p = C_1 + C_2 \)
In parallel configurations, capacitors can store more charge at the same voltage because the effective plate area increases, bolstering overall storage capacity.
Meanwhile, when capacitors are connected in series, the situation is quite different. The total capacitance is found using the reciprocal formula:
  • \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \)
For just two capacitors, this simplifies to:
  • \( C_s = \frac{C_1 C_2}{C_1 + C_2} \)
In series, capacitors share the same charge. However, as a series combination results in a lower total capacitance than any individual capacitor, the energy stored is reduced in comparison to parallel arrangements.
Stored Energy in Capacitors
Capacitors store energy in the electric field created between their plates. The energy stored in a capacitor can be quantified with the formula:
  • \( U = \frac{1}{2} C V^2 \)
where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the voltage applied. This formula reveals that the energy stored is proportional to both the capacitance and the square of the voltage.
In parallel circuits, capacitors offer greater total capacitance, which significantly increases the stored energy. As in the given exercise where in a parallel arrangement, the stored energy is five times greater than in a series configuration, it highlights how differently these connections impact energy storage.
It's important to remember that energy storage in capacitors is instantaneous and can be released quickly, which makes capacitors useful in various electronic applications, from smoothing power supply outputs to flash photography.

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

(I) The potential difference between two short sections of parallel wire in air is 24.0 \(\mathrm{V}\) . They carry equal and opposite charge of magnitude 75 \(\mathrm{pC}\) . What is the capacitance of the two wires?

(I) \((a)\) Six \(3.8-\mu\) F capacitors are connected in parallel. What is the equivalent capacitance? (b) What is their equivalent capacitance if connected in series?

For commonly used CMOS (complementary metal oxide semiconductor) digital circuits, the charging of the component capacitors \(C\) to their working potential difference \(V\) accounts for the major contribution of its energy input requirements. Thus, if a given logical operation requires such circuitry to charge its capacitors \(N\) times, we can assume that the operation requires an energy of \(N\left(\frac{1}{2} C V^{2}\right) .\) In the past 20 years, the capacitance in digital circuits has been reduced by a factor of about 20 and the voltage to which these capacitors are charged has been reduced from \(5.0 \mathrm{~V}\) to \(1.5 \mathrm{~V} .\) Also, present-day alkaline batteries hold about five times the energy of older batteries. Two present-day AA alkaline cells, each of which measures \(1 \mathrm{~cm}\) diameter by \(4 \mathrm{~cm}\) long, can power the logic circuitry of a hand-held personal digital assistant (PDA) with its display turned off for about two months. If an attempt was made to construct a similar PDA (i.e., same digital capabilities so \(N\) remains constant) 20 years ago, how many (older) AA batteries would have been required to power its digital circuitry for two months? Would this PDA fit in a pocket or purse?

Capacitors can be used as "electric charge counters." Consider an initially uncharged capacitor of capacitance \(C\) with its bottom plate grounded and its top plate connected to a source of electrons. \((a)\) If \(N\) electrons flow onto the capacitor's top plate, show that the resulting potential difference \(V\) across the capacitor is directly proportional to \(N .\) (b) Assume the voltage- measuring device can accurately resolve voltage changes of about \(1 \mathrm{mV}\). What value of \(C\) would be necessary to detect each new \(\begin{array}{lllll}\text { collected electron? } & \text { (c) Using modern semiconductor }\end{array}\) technology, a micron-size capacitor can be constructed with parallel conducting plates separated by an insulating oxide of dielectric constant \(K=3\) and thickness \(d=100 \mathrm{nm}\). To resolve the arrival of an individual electron on the plate of such a capacitor, determine the required value of \(\ell(\) in \(\mu \mathrm{m}\) ) assuming square plates of side length \(\ell\)

(II) Six physics students were each given an air filled capacitor. Although the areas were different, the spacing between the plates, \(d,\) was the same for all six capacitors, but was unknown. Each student made a measurement of the area \(A\) and capacitance \(C\) of their capacitor. Below is a Table for their data. Using the combined data and a graphing program or spreadsheet, determine the spacing \(d\) between the plates.

See all solutions

Recommended explanations on Physics Textbooks

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.