/*! 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 25 A group of identical capacitors ... [FREE SOLUTION] | 91Ó°ÊÓ

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A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group?

Short Answer

Expert verified
There are 10 capacitors in the group.

Step by step solution

01

Understand the Series Capacitance

When identical capacitors are connected in series, the total capacitance (\(C_{series}\) is calculated using the formula \(\frac{1}{C_{series}} = \frac{1}{C} + \frac{1}{C} + ... + \frac{1}{C}\), where \(C\) is the capacitance of a single capacitor and there are \(n\) capacitors. Simplify this to \(\frac{1}{C_{series}} = \frac{n}{C}\), hence, \(C_{series} = \frac{C}{n}\).
02

Understand the Parallel Capacitance

When identical capacitors are connected in parallel, the total capacitance \(C_{parallel}\) is the sum of each individual capacitance. For \(n\) capacitors, the formula is \(C_{parallel} = C + C + ... + C = nC\).
03

Set Up the Ratio of Parallel and Series Capacitance

According to the problem, \(C_{parallel}\) is 100 times \(C_{series}\), so we set up the equation: \(C_{parallel} = 100 \times C_{series}\). Replacing the formulas derived in steps 1 and 2: \(nC = 100 \times \frac{C}{n}\).
04

Solve for the Number of Capacitors

Cancelling out the capacitance \(C\) from both sides of the equation since it's the same for each capacitor and non-zero, gives us \(n^2 = 100\). Taking the square root of both sides yields \(n = \sqrt{100} = 10\). Therefore, there are 10 capacitors in the group.

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Key Concepts

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

Series Capacitance Calculation
Understanding how to calculate the total capacitance when capacitors are connected in series is crucial for anyone dealing with electronics. Imagine lining up capacitors one after another so that the charge must pass through each capacitor sequentially. In the series configuration, the inverse of the total capacitance is the sum of the inverses of each individual capacitor's capacitance. Formally, for identical capacitors, the relationship is expressed as
\[\begin{equation}\frac{1}{C_{series}} = \frac{1}{C} + \frac{1}{C} + \ ... + \frac{1}{C}\end{equation}\]
where \(C\) represents the capacitance of a single capacitor and \(n\) is the number of capacitors. Simplification yields
\[\begin{equation}C_{series} = \frac{C}{n}\end{equation}\]
With each additional capacitor added in series, the total capacitance decreases, which can be a non-intuitive concept for beginners.
Parallel Capacitance Calculation
When capacitors are connected in parallel, the situation is quite different compared to a series connection. Here, the total capacitance is straightforward: it is simply the sum of all capacitances. For parallel configurations, imagine the capacitors sharing the same two connection points, offering multiple pathways for the charge. The calculation for identical capacitors in parallel is given as:
\[\begin{equation}C_{parallel} = C + C + ... + C = nC\end{equation}\]
The physical interpretation of this is that connecting capacitors in parallel effectively increases the plate's area, thereby increasing the total capacitance. This is in direct contrast to series connection, where the total capacitance decreases as more capacitors are added.
Capacitors in Circuits
Incorporating capacitors into circuits is fundamental in controlling and managing energy storage and release within the system. Capacitors can act as temporary storage devices, holding onto charge until it is needed elsewhere in the circuit. Their behavior drastically changes depending on whether they are in series or parallel, impacting their collective capacitance and the circuit's overall response. It is important to recognize that in actual circuits, capacitors also have attributes such as equivalent series resistance (ESR), which can affect their performance, especially at high frequencies or in power applications.
Electrical Capacitance
Electrical capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is an intrinsic property of the capacitor, determined by the physical characteristics such as the area of the plates, the distance between them, and the dielectric material used. In practical applications, capacitance is a crucial aspect as it dictates how much energy can be stored and how quickly it can be released back into the circuit. Understanding the principles underpinning capacitance allows students and engineers alike to design and troubleshoot circuits effectively.

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Most popular questions from this chapter

An isolated capacitor of unknown capacitance has been charged to a potential difference of 100 \(\mathrm{V}\) . When the charged capacitor is then connected in parallel to an uncharged \(10.0-\mu \mathrm{F}\) capacitor, the potential difference across the combination is 30.0 \(\mathrm{V}\) . Calculate the unknown capacitance.

A uniform electric field \(E=3000 \mathrm{V} / \mathrm{m}\) exists within a certain region. What volume of space contains an energy equal to \(1.00 \times 10^{-7} \mathrm{J}\) ? Express your answer in cubic meters and in liters.

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of \(1.80 \mathrm{mm} .\) A \(20.0-\mathrm{V}\) potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

A 10.0 - \(\mu\) F capacitor has plates with vacuum between them. Each plate carries a charge of magnitude 1000\(\mu \mathrm{C}\) . A particle with charge \(-3.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-16} \mathrm{kg}\) is fired from the positive plate toward the negative plate with an initial speed of \(2.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Does it reach the negative plate? If so, find its impact speed. If not, what fraction of the way across the capacitor does it travel?

As a person moves about in a dry environment, electric charge accumulates on his body. Once it is at high voltage, either positive or negative, the body can discharge via sometimes noticeable sparks and shocks. Consider a human body well separated from ground, with the typical capacitance 150 \(\mathrm{pF}\) . (a) What charge on the body will produce a potential of 10.0 \(\mathrm{kV} ?\) (b) Sensitive electronic devices can be destroyed by electrostatic discharge from a person. A particular device can be destroyed by a discharge releasing an energy of 250\(\mu \mathrm{J}\) . To what voltage on the body does this correspond?

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