/*! 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 17 Two capacitors with capacitances... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 17

Two capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in series. Show that, no matter what the values of \(C_{1}\) and \(C_{2}\) are, the equivalent capacitance is always less than the smaller of the two capacitances.

Short Answer

Expert verified
Question: Prove that the equivalent capacitance of two capacitors connected in series is always less than the smaller of the two capacitances. Answer: By using the formula for equivalent capacitance in series, $C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}}$, we can show that if we assume $C_1 \leq C_2$, then $C_{eq} < C_{1}$. This proves that the equivalent capacitance is always less than the smaller capacitance.

Step by step solution

01

Write the formula for equivalent capacitance in series

The formula for the equivalent capacitance \(C_{eq}\) when two capacitors are connected in series is given by: $$ \frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} $$
02

Rewrite the expression for equivalent capacitance

To make the comparison with \(C_1\) and \(C_2\) easier, let's rewrite the expression for \(C_{eq}\): $$ C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} $$
03

Compare the equivalent capacitance to the smaller capacitance

Now we will compare \(C_{eq}\) to \(C_1\) and \(C_2\) separately. Without loss of generality, let's suppose that \(C_1 \leq C_2\). We will demonstrate that \(C_{eq} < C_1\): $$ C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} < \frac{C_{1}C_{2}}{C_{1}} = C_{2} $$ Since \(C_1 \leq C_2\), we have \(C_{eq} < C_{1}\). Hence, the equivalent capacitance is always less than the smaller capacitance.

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!

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

Two circular metal plates of radius \(0.61 \mathrm{~m}\) and thickness \(7.1 \mathrm{~mm}\) are used in a parallel plate capacitor. A gap of 2.1 mm is left between the plates, and half of the space (a semicircle) between the plates is filled with a dielectric for which \(\kappa=11.1\) and the other half is filled with air. What is the capacitance of this capacitor?

Design a parallel plate capacitor with a capacitance of \(47.0 \mathrm{pF}\) and a capacity of \(7.50 \mathrm{nC}\). You have available conducting plates, which can be cut to any size, and Plexiglas sheets, which can be cut to any size and machined to any thickness. Plexiglas has a dielectric constant of 3.40 and a dielectric strength of \(4.00 \cdot 10^{7} \mathrm{~V} / \mathrm{m}\). You must make your capacitor as compact as possible. Specify all relevant dimensions. Ignore any fringe field at the edges of the capacitor plates.

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 \(18.0-\mu \mathrm{F} ?\)

Does it take more work to separate the plates of a charged parallel plate capacitor while it remains connected to the charging battery or after it has been disconnected from the charging battery?

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Astrophysics

Read Explanation

Energy Physics

Read Explanation

Modern Physics

Read Explanation

Particle Model of Matter

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.