Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

You have an electric device containing a \(10.0-\mu \mathrm{F}\) capacitor, but an application requires an \(18.0-\mu \mathrm{F}\) capacitor. What modification can you make to your device to increase its capacitance to \(18.0-\mu \mathrm{F} ?\)

Short Answer

Expert verified
Answer: The additional capacitor should have a capacitance of 8.0 μF.

Step by step solution

01

STEP 1: Determine the approach

As mentioned in the analysis, we need to connect an additional capacitor in parallel with the original capacitor to increase the capacitance. For parallel connection, the total capacitance is the sum of the individual capacitances: \(C_T = C_1 + C_2\).
02

STEP 2: Setup the equation

We need to find the capacitance of the additional capacitor \(C_2\) that must be connected in parallel with the original capacitor with capacitance \(C_1 = 10.0\,\mu\mathrm{F}\) to get the total capacitance \(C_T = 18.0\,\mu\mathrm{F}\). Using the formula for parallel capacitors, we get: $$C_T = C_1 + C_2.$$
03

STEP 3: Solve for the unknown

Substituting the known values of \(C_T\) and \(C_1\), we have: $$18.0\,\mu\mathrm{F} = 10.0\,\mu\mathrm{F} + C_2.$$ Solve for \(C_2\): $$C_2 = 18.0\,\mu\mathrm{F} - 10.0\,\mu\mathrm{F} = 8.0\,\mu\mathrm{F}.$$
04

STEP 4: State the modification

To increase the capacitance of the device to \(18.0\,\mu\mathrm{F}\), an additional capacitor with a capacitance of \(8.0\,\mu\mathrm{F}\) should be connected in parallel to the original capacitor.

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 Vaia!

Key Concepts

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

Electric Capacitance
Electric capacitance is a fundamental property of capacitors that determines their ability to store electrical energy. It is measured in farads (F), where one farad is the capacitance of a capacitor that can store one coulomb of charge with an electric potential difference of one volt across its plates.

In layman's terms, you can think of capacitance as a container for electrical charge; the larger the capacitance, the more charge it can hold for a given voltage. The capacitance of a capacitor is usually determined by its physical characteristics, such as the area of the plates, the distance between them, and the material of the dielectric (the insulating layer) between the plates. Understanding this concept is crucial when working with electronic circuits and in designing systems that depend on energy storage and timing applications.
Parallel Connection of Capacitors
When capacitors are connected in parallel, their total capacitance is the sum of their individual capacitances. This connection is akin to adding more storage bins side by side — the overall storage capacity increases because you now have more containers to store charges.

The formula for the total capacitance in a parallel circuit is expressed as:
\[C_{\text{T}} = C_1 + C_2 + C_3 + \ldots + C_n\]
where \(C_{\text{T}}\) is the total capacitance, and \(C_1\), \(C_2\), \(C_3\), ..., \(C_n\) are the capacitances of individual capacitors. One major consequence of this is that the voltage across each capacitor in parallel is the same and is equal to the voltage of the power source. In practical applications, connecting capacitors in parallel is an effective method to achieve the desired capacitance without having to manufacture a custom capacitor.
Solving for Unknown in Electric Circuits
To solve for an unknown component in an electric circuit, such as a missing capacitor value, we use the established relationships between different elements in the circuit—Ohm’s law for resistors, Kirchhoff’s rules for circuits, and the parallel formula for capacitors, as illustrated in the given exercise.

In our case, we addressed the unknown by setting up an equation based on the formula for capacitors in parallel. By rearranging the equation to isolate the unknown variable, we could solve for it arithmetically. This process of problem-solving is a fundamental aspect of electrical engineering and physics, providing a systematic approach to diagnose and design circuitry for various applications. It's crucial for students not only to memorize these equations but also to understand the principles behind them to apply them effectively to real-world scenarios.

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 distance between the plates of a parallel plate capacitor is reduced by half and the area of the plates is doubled. What happens to the capacitance? a) It remains unchanged. b) It doubles. c) It quadruples. d) It is reduced by half.

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

A parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) 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.

Sign-up for free