Log out Go to App
Go to App

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Calculate the capacitance of the Earth. Treat the Earth as an isolated spherical conductor of radius \(6371 \mathrm{~km}\).

The Earth can be thought of as a spherical capacitor. If the net charge on the Earth is \(-7.8 \cdot 10^{5} \mathrm{C},\) find \((\) a) the capacitance of the Earth and (b) the electric potential energy stored on the Earth's surface

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?

An isolated solid spherical conductor of radius \(5.00 \mathrm{~cm}\) is surrounded by dry air. It is given a charge and acquires potential \(V\), with the potential at infinity assumed to be zero. a) Calculate the maximum magnitude \(V\) can have. b) Explain clearly and concisely why there is a maximum.

Considering the dielectric strength of air, what is the maximum amount of charge that can be stored on the plates of a capacitor that are a distance of \(15 \mathrm{~mm}\) apart and have an area of \(25 \mathrm{~cm}^{2}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Force

Read Explanation

Modern Physics

Read Explanation

Waves Physics

Read Explanation

Electrostatics

Read Explanation

Astrophysics

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.

Sign-up for free