Determine all the values of equivalent capacitance you can create using any combination of three identical capacitors with capacitance \(C\).

Short Answer

Expert verified
Answer: The possible equivalent capacitance values are: 1. All capacitors in series: \(C_{eq1} = \frac{C}{3}\) 2. Two capacitors in parallel, and the third one in series: \(C_{eq2} = \frac{2C}{3}\) 3. All capacitors in parallel: \(C_{eq3} = 3C\)

Step by step solution

01

Case 1: All capacitors in series

When all three capacitors are connected in series, the equivalent capacitance C_eq1 can be found using the formula: \(\frac{1}{C_{eq1}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C}\) Which simplifies to: \(\frac{1}{C_{eq1}} = \frac{3}{C}\) And then: \(C_{eq1} = \frac{C}{3}\)
02

Case 2: Two capacitors in parallel, and the third one in series

When two capacitors are connected in parallel and the third one is connected in series with this pair, the equivalent capacitance C_eq2 can be found using the following steps: 1. Find the equivalent capacitance of the two capacitors in parallel (C_12). The formula for the equivalent capacitance of capacitors in parallel is: \(C_{12} = C + C = 2C\) 2. Now, find the equivalent capacitance of C_12 and the third capacitor C connected in series. Using the formula for the equivalent capacitance of capacitors in series: \(\frac{1}{C_{eq2}} = \frac{1}{C_{12}} + \frac{1}{C}\) Substitute the value of C_12: \(\frac{1}{C_{eq2}} = \frac{1}{2C} + \frac{1}{C}\) Simplifying, we get: \(\frac{1}{C_{eq2}} = \frac{3}{2C}\) And then: \(C_{eq2} = \frac{2C}{3}\)
03

Case 3: All capacitors in parallel

When all three capacitors are connected in parallel, the equivalent capacitance C_eq3 can be found using the formula: \(C_{eq3} = C + C + C = 3C\)
04

Summary

There are three possible equivalents capacitances that can be created using any combination of three identical capacitors with capacitance C: 1. All capacitors in series: \(C_{eq1} = \frac{C}{3}\) 2. Two capacitors in parallel, and the third one in series: \(C_{eq2} = \frac{2C}{3}\) 3. All capacitors in parallel: \(C_{eq3} = 3C\) These are all the possible values of equivalent capacitance that can be created using three identical capacitors with capacitance C.

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Most popular questions from this chapter

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?

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.

A parallel plate capacitor with square plates of edge length \(L\) separated by a distance \(d\) is given a charge \(Q\), then disconnected from its power source. A close-fitting square slab of dielectric, with dielectric constant \(\kappa\), is then inserted into the previously empty space between the plates. Calculate the force with which the slab is pulled into the capacitor during the insertion process.

A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?

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?

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