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 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

A proton traveling along the \(x\) -axis at a speed of \(1.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.0-\mathrm{cm}-\) wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.0 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2}\). How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

A capacitor with a vacuum between its plates is connected to a battery and then the gap is filled with Mylar. By what percentage is its energy-storing capacity increased?

The capacitance of a spherical capacitor consisting of two concentric conducting spheres with radii \(r_{1}\) and \(r_{2}\) \(\left(r_{2}>r_{1}\right)\) is given by \(C=4 \pi \epsilon_{0} r_{1} r_{2} /\left(r_{2}-r_{1}\right) .\) Suppose that the space between the spheres, from \(r,\) up to a radius \(R\) \(\left(r_{1}

Thermocoax is a type of coaxial cable used for high-frequency filtering in cryogenic quantum computing experiments. Its stainless steel shield has an inner diameter of \(0.35 \mathrm{~mm},\) and its Nichrome conductor has a diameter of \(0.17 \mathrm{~mm}\). Nichrome is used because its resistance doesn't change much in going from room temperature to near absolute zero. The insulating dielectric is magnesium oxide \((\mathrm{MgO}),\) which has a dielectric constant of \(9.7 .\) Calculate the capacitance per meter of Thermocoax.

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

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