/*! 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 22 (I) \((a)\) Six \(3.8-\mu\) F ca... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 22

(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?

Short Answer

Expert verified
(a) 22.8 碌F; (b) 0.6333 碌F.

Step by step solution

01

Understanding Capacitors in Parallel

When capacitors are connected in parallel, the total or equivalent capacitance is simply the sum of the individual capacitances. If we have six capacitors each with a capacitance of \(3.8\,\mu\text{F}\), then the equivalent capacitance \(C_{eq}\) is calculated as: \[ C_{eq} = C_1 + C_2 + \ldots + C_6 = 6 \times 3.8\,\mu\text{F} \]
02

Calculating Parallel Equivalent Capacitance

Using the formula for capacitors in parallel: \(C_{eq} = 6 \times 3.8\,\mu\text{F} = 22.8\,\mu\text{F}\). Therefore, the equivalent capacitance of the capacitors when connected in parallel is \(22.8\,\mu\text{F}\).
03

Understanding Capacitors in Series

When capacitors are connected in series, the reciprocal of the equivalent capacitance \(C_{eq}\) is the sum of the reciprocals of the individual capacitances. This is expressed as: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_6} = 6 \times \frac{1}{3.8\,\mu\text{F}} \]
04

Calculating Series Equivalent Capacitance

Calculate \(\frac{1}{C_{eq}} = 6 \times \frac{1}{3.8} = \frac{6}{3.8}\). Therefore, \(C_{eq} = \frac{3.8}{6} = 0.6333\,\mu\text{F}\). To find the equivalent capacitance for the series connection: \(C_{eq} = 0.6333\,\mu\text{F}\).

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.

Capacitors in Parallel
When capacitors are connected in parallel, it's like connecting the channels of multiple barrels together. Each capacitor adds its own capacity to hold charge, enhancing the total capacitance. That makes the math quite simple: Just add up the capacitance values of all capacitors involved.

For instance, if you have six capacitors in parallel, each with a capacitance of \(3.8\,\mu\text{F}\), you simply multiply that capacitance by six.
  • Each capacitor's value: \(3.8\,\mu\text{F}\)
  • Number of capacitors: 6
  • Total or equivalent capacitance (\(C_{eq}\)): \(6 \times 3.8 = 22.8\,\mu\text{F}\)
This principle makes it easy to scale up a system's total capacitance by simply attaching more capacitors in parallel. The equivalent capacitance increases because each capacitor is adding more potential for storing charge on its own.
Capacitors in Series
Connecting capacitors in series is quite different. It narrows down the pathway, much like connecting several straws end-to-end. This setup reduces the total capacitance because the total distance for the charges to move increases.

Here, the reciprocal of the total or equivalent capacitance (\(C_{eq}\)) is the sum of the reciprocals of each capacitor's capacitance. So, if all capacitors have the same value, like in our example where each has \(3.8\,\mu\text{F}\), the calculation becomes straightforward.
  • Find the reciprocal of one capacitor鈥檚 value: \(\frac{1}{3.8}\)
  • Multiply by the number of capacitors: \(6 \times \frac{1}{3.8}\)
  • The total is the reciprocal of \(C_{eq}\): \(\frac{1}{C_{eq}} = \frac{6}{3.8}\)
  • Finally, invert the total to find \(C_{eq}\): \(0.6333\,\mu\text{F}\)
In essence, connecting in series divides the load among the capacitors, resulting in a lower equivalent capacitance.
Capacitance Calculation
Whether capacitors are connected in series or parallel, calculating their equivalent capacitance is crucial. It helps to determine how much energy the system can store.

When calculating for capacitors in parallel, you add the capacitance values because each one contributes individually to the total potential. For series arrangements, however, you face the reduction in total capacitance, as the overall pathway becomes longer.

Each setup has distinct uses:
  • Parallel: Boosts total capacitance, beneficial for increasing a circuit's energy storage.
  • Series: Reduces total capacitance, useful for reducing the total energy storage and increasing voltage tolerance.
Understanding these calculations allows for effective design and optimization of electronic circuits, using the inherent properties of capacitors to achieve desired electrical outcomes.

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

A high-voltage supply can be constructed from a variable capacitor with interleaving plates which can be rotated as in Fig. \(24-36 .\) A version of this type of capacitor with more plates has a capacitance which can be varied from \(10 \mathrm{pF}\) to \(1 \mathrm{pF}\). (a) Initially, this capacitor is charged by a \(7500-\mathrm{V}\) power supply when the capacitance is \(8.0 \mathrm{pF}\). It is then disconnected from the power supply and the capacitance reduced to \(1.0 \mathrm{pF}\) by rotating the plates. What is the voltage across the capacitor now? (b) What is a major disadvantage of this as a high-voltage power supply?

Paper has a dielectric constant \(K=3.7\) and a dielectric strength of \(15 \times 10^{6} \mathrm{~V} / \mathrm{m}\). Suppose that a typical sheet of paper has a thickness of \(0.030 \mathrm{~mm}\). You make a "homemade" capacitor by placing a sheet of \(21 \times 14 \mathrm{~cm}\) paper between two aluminum foil sheets (Fig. \(24-41)\). The thickness of the aluminum foil is \(0.040 \mathrm{~mm}\). (a) What is the capacitance \(C_{0}\) of your device? (b) About how much charge could you store on your capacitor before it would break down? (c) Show in a sketch how you could overlay sheets of paper and aluminum for a parallel combination. If you made 100 such capacitors, and connected the edges of the sheets in parallel so that you have a single large capacitor of capacitance \(100 C_{0}\), how thick would your new large capacitor be? ( \(d\) ) What is the maximum voltage you can apply to this \(100 C_{0}\) capacitor without breakdown?

(II) A \(2.20-\mu \mathrm{F}\) capacitor is charged by a \(12.0 . \mathrm{V}\) battery. It is disconnected from the battery and then connected to an uncharged \(3.50-\mu \mathrm{F}\) capacitor (Fig. \(20 ) .\) Determine the total stored energy \((a)\) before the two capacitors are connected, and (b) after they are connected. (c) What is the change in energy?

(III) The quantity of liquid (such as cryogenic liquid nitrogen) available in its storage tank is often monitored by a capacitive level sensor. This sensor is a vertically aligned cylindrical capacitor with outer and inner conductor radii \(R_{\mathrm{a}}\) and \(R_{\mathrm{b}},\) whose length \(\ell\) spans the height of the tank. When a and \(R_{b},\) whose length \(\ell\) spans the height of the tank. When a nonconducting liquid fills the tank to a height \(h(5 \ell)\) from the tank's bottom, the dielectric in the lower and upper region between the cylindrical conductors is the liquid \(\left(K_{\text { liq }}\right)\) and its vapor \(\left(K_{\mathrm{v}}\right),\) respectively (Fig, \(33 ) .\) (a) Determine a formula for the fraction \(F\) of the tank filled by liquid in terms of the level-sensor capacitance \(C .[\) Hint: Consider the sensor as a combination of two capacitors. \(.\) (b) By connecting a capacitance-measuring instrument to the level sensor, \(F\) can be monitored. Assume the sensor dimensions are \(\ell=2.0 \mathrm{m}, \quad R_{\mathrm{n}}=5.0 \mathrm{mm}, \quad\) and \(\quad R_{\mathrm{b}}=4.5 \mathrm{mm} .\) For liquid nitrogen \(\left(K_{\mathrm{liq}}=1.4, \quad K_{\mathrm{V}}=1.0\right),\) what values of \(C\) (in pF) will correspond to the tank being completely full and completely empty?

(II) Compact "ultracapacitors" with capacitance values up to several thousand farads are now commercially available. One application for ultracapacitors is in providing power for electrical circuits when other sources (such as a battery) are turned off. To get an idea of how much charge can be stored in such a component, assume a \(1200-\mathrm{F}\) ultracapacitor is initially charged to 12.0 \(\mathrm{V}\) by a battery and is then disconnected from the battery. If charge is then drawn off the plates of this capacitor at a rate of \(1.0 \mathrm{mC} / \mathrm{s},\) say, to power the backup memory of some electrical gadget, how long (in days) will it take for the potential difference across this capacitor to drop to 6.0 \(\mathrm{V}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Astrophysics

Read Explanation

Classical Mechanics

Read Explanation

Mechanics and Materials

Read Explanation

Turning Points in Physics

Read Explanation

Space Physics

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.