/*! 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 91 Suppose two capacitors \(\left(C... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose two capacitors \(\left(C_{1}\right.\) and \(C_{2}\) ) are connected in series. Show that the sum of the energies stored in these capacitors is equal to the energy stored in the equivalent capacitor. [Hint: The energy stored in a capacitor can be expressed as \(\left.q^{2} /(2 C) .\right]\)

Short Answer

Expert verified
The sum of the energies stored in the capacitors is equal to the energy stored in the equivalent capacitor.

Step by step solution

01

Identify the Equivalent Capacitance

When capacitors are connected in series, their equivalent capacitance \(C_{eq}\) can be calculated using the formula: \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\] This formula allows us to find the equivalent single capacitor value for the series connection of \(C_1\) and \(C_2\).
02

Express the Energy Stored in Each Capacitor

The energy stored in a capacitor \(C\) with charge \(q\) is given by the formula:\[E = \frac{q^2}{2C}\]For capacitors in series, each capacitor has the same charge \(q\). Thus, the energy stored in \(C_1\) is \(E_1 = \frac{q^2}{2C_1}\) and in \(C_2\) is \(E_2 = \frac{q^2}{2C_2}\).
03

Calculate the Total Energy Stored in Series Capacitors

The total energy stored in the capacitors connected in series, \(E_{total}\), is the sum of energies stored in \(C_1\) and \(C_2\):\[E_{total} = E_1 + E_2 = \frac{q^2}{2C_1} + \frac{q^2}{2C_2}\]
04

Express the Energy Stored in the Equivalent Capacitance

The energy stored in the equivalent capacitor \(C_{eq}\) with the same charge \(q\) is:\[E_{eq} = \frac{q^2}{2C_{eq}}\]Substitute the equivalent capacitance relation from Step 1.
05

Show Equality of Energies

By substituting \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) from Step 1 into the expression for \(E_{eq}\), we get:\[E_{eq} = \frac{q^2}{2C_{eq}} = \frac{q^2}{2} \left( \frac{1}{C_1} + \frac{1}{C_2} \right)\]Thus, comparing \(E_{total}\) from Step 3 and \(E_{eq}\), they are equal, proving the given statement.

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Key Concepts

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

Equivalent Capacitance
When capacitors are connected in series, the overall capacitance differs from when they are connected in parallel. To find the equivalent capacitance, the formula is \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\]This equation is crucial for understanding how series capacitor systems behave. It shows that the total or equivalent capacitance \(C_{eq}\) is always less than the smallest individual capacitor in the series.
For example, if you have two capacitors in series with capacitances of 3 µF and 6 µF, the equivalent capacitance can be calculated as follows: - First, invert the individual capacitances: \(\frac{1}{3}\) and \(\frac{1}{6}\).- Add them: \(\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\).- The equivalent capacitance is then \( C_{eq} = 2 \) µF.
Understanding equivalent capacitance helps in designing circuits where specific voltage and energy characteristics are desired. This concept is particularly important in electronic devices where multiple capacitors are used to achieve a desired performance level.
Energy Stored in Capacitors
The energy stored in a capacitor is a measure of the capacitor's ability to hold charge. The formula for calculating this energy is :\[E = \frac{q^2}{2C}\]where \(E\) is the energy, \(q\) is the charge on the capacitor, and \(C\) is the capacitance.
For capacitors in series, each capacitor carries the same charge, unlike parallel connections where voltage is constant. Thus, the energy stored in each series capacitor can be expressed individually. This is done as:- For the first capacitor \(C_1\): \(E_1 = \frac{q^2}{2C_1}\)- For the second capacitor \(C_2\): \(E_2 = \frac{q^2}{2C_2}\)
Therefore, to find the total energy stored in capacitors arranged in series, we simply add the energies of each capacitor. This concept emphasizes that even in a series setup, the overall energy is the sum of individual energies.
Capacitor Energy Formula
The capacitor energy formula is foundational in understanding how energy is distributed across capacitors. For a capacitor, energy can be expressed using various formulas, one being:\[E = \frac{q^2}{2C}\]This tells us that the energy stored is directly dependent on the square of the charge \(q\) and inversely on the capacitance \(C\).
Additionally, for series capacitors, while charge \(q\) remains constant across all capacitors, the overall energy stored can be expressed in terms of the equivalent capacitance, \(C_{eq}\). The formula becomes:\[E_{eq} = \frac{q^2}{2C_{eq}}\]Using the required relational formula for equivalent capacitors in series \(\left(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\right)\), we confirm that the energy stored in the equivalent capacitance is the sum of the energies in each capacitor.
This energy balance not only verifies the mathematical calculations but also illustrates the energy conservation principle. Understanding this relationship is essential in circuit analysis, as it ensures accurate energy management and efficiency in electronic systems.

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Most popular questions from this chapter

A lightning bolt delivers a charge of \(35 \mathrm{C}\) to the ground in a time of \(1.0 \times 10^{-3} \mathrm{~s}\). What is the current?

A \(3.00-\mu F\) and a \(5.00-\mu\) F capacitor are connected in series across a \(30.0-V\) battery. A \(7.00-\mu \mathrm{F}\) capacitor is then connected in parallel across the \(3.00-\mu \mathrm{F}\) capacitor. Determine the voltage across the \(7.00-\mu \mathrm{F}\) capacitor.

Concept Questions Each resistor in the three circuits in the drawing has the same resistance \(R\), and the batteries have the same voltage \(V\). (a) How is the total power delivered by the battery related to the equivalent resistance connected between the battery terminals and to the battery voltage? (b) Rank the equivalent resistances of the circuits in descending order (largest first). (c) Rank the three values of the total power delivered by the batteries in descending order (largest first). Problem The values for \(R\) and \(V\) in the drawing are \(9.0 \Omega\) and \(6.0 \mathrm{~V}\), respectively. Determine the total power delivered by the battery in each of the three circuits. Be sure that your answer is consistent with your answer to Concept Question (c).

In measuring a voltage, a voltmeter uses some current from the circuit. Consequently, the voltage measured is only an approximation to the voltage present when the voltmeter is not connected. Consider a circuit consisting of two \(1550-\Omega\) resistors connected in series across a \(60.0-\mathrm{V}\) battery. (a) Find the voltage across one of the resistors. (b) A voltmeter has a full-scale voltage of \(60.0 \mathrm{~V}\) and uses a galvanometer with a full-scale deflection of \(5.00 \mathrm{~mA}\). Determine the voltage that this voltmeter registers when it is connected across the resistor used in part (a).

Three resistors, \(25,45,\) and \(75 \Omega,\) are connected in series, and a \(0.51-\mathrm{A}\) current passes through them. What is (a) the equivalent resistance and (b) the potential difference across the three resistors?

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