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Chapter 24: Problem 46

How much energy must a 28-V battery expend to charge a \(0.45-\mu \mathrm{F}\) and a \(0.20-\mu \mathrm{F}\) capacitor fully when they are placed \((a)\) in parallel, \((b)\) in series? \((c)\) How much charge flowed from the battery in each case?

Short Answer

Expert verified
In parallel, energy = 0.254 J; in series, energy 鈮� 0.032 J. Charge in parallel = 18.2 碌C; charge in series 鈮� 9 碌C.

Step by step solution

01

Understand the Problem

We have a 28-V battery and two capacitors: a 0.45-碌F capacitor and a 0.20-碌F capacitor. We need to know the energy expended by the battery to fully charge the capacitors in two different configurations: first in parallel and then in series.
02

Calculating Capacitance - Parallel Configuration

When capacitors are connected in parallel, their equivalent capacitance is the sum of their individual capacitances: \( C_{eq} = C_1 + C_2 = 0.45 \, \mu F + 0.20 \, \mu F = 0.65 \, \mu F \).
03

Calculating Energy - Parallel Configuration

The energy stored in capacitors is calculated using the formula \( E = \frac{1}{2} C V^2 \). For the equivalent parallel capacitance, \( E = \frac{1}{2} \times 0.65 \, \mu F \times (28 \, V)^2 \). Calculate the result to find the energy.
04

Calculating Capacitance - Series Configuration

For capacitors in series, the equivalent capacitance is given by \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \), so \( \frac{1}{C_{eq}} = \frac{1}{0.45 \, \mu F} + \frac{1}{0.20 \, \mu F} \). Find \( C_{eq} \) by calculating the reciprocal sum and then take the reciprocal.
05

Calculating Energy - Series Configuration

Use the same energy formula \( E = \frac{1}{2} C V^2 \) with \( C_{eq} \) from the series calculation: \( E_{series} = \frac{1}{2} C_{series} (28 \, V)^2 \). Calculate to find the energy.
06

Calculating Charge Flow - Parallel Configuration

For parallel capacitors, charge \( Q = C_{eq} \times V \). Using \( C_{eq} = 0.65 \, \mu F \), calculate \( Q_{parallel} = 0.65 \, \mu F \times 28 \, V \).
07

Calculating Charge Flow - Series Configuration

For series capacitors, charge remains the same across each capacitor as \( Q = C_{eq} \times V \). Use the \( C_{eq} \) from series to find \( Q_{series} = C_{series} \times 28 \, V \).

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Key Concepts

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

Energy Calculation in Capacitor Circuits
When dealing with capacitors, calculating the energy stored is pivotal. The formula used is \( E = \frac{1}{2} C V^2 \), representing the energy \( E \) stored in a capacitor with capacitance \( C \) and voltage \( V \).
For capacitors in parallel, you first sum their capacitances to find the equivalent capacitance \( C_{eq} \).
Then, substitute this equivalent capacitance into the energy formula. In our exercise, the parallel configuration gives \( C_{eq} = 0.65 \, \mu F \). Therefore, the energy expended when a 28-V battery charges them is:
  • \( E_{parallel} = \frac{1}{2} \times 0.65 \, \mu F \times (28 \, V)^2 \)
Calculating this gives the total energy the battery needs to supply. By following the same process with the equivalent capacitance for the series configuration, you can determine the energy stored in that setup as well.
Charge Calculation in Capacitor Configurations
Understanding charge flow is essential in analyzing capacitor circuits. Charge \( Q \) in a capacitor is calculated using \( Q = C \times V \), where \( C \) is capacitance and \( V \) is voltage.
In a parallel configuration, the equivalent capacitance \( C_{eq} = 0.65 \, \mu F \), and the charge is calculated as:
  • \( Q_{parallel} = C_{parallel} \times 28 \, V \)
Using the series equivalent capacitance, you can similarly calculate the charge flow in series. Despite differences in capacitance, the same formula applies to both parallel and series configurations.
This ensures that the total charge the battery provides is clear in both arrangements.
Parallel and Series Configuration of Capacitors
Capacitors can be arranged in either parallel or series, each affecting the total capacitance differently.
In a parallel configuration, the capacitors' capacitances simply add up:
  • \( C_{eq} = C_1 + C_2 \)
This increases the total capacitance, impacting energy and charge calculations. For the series configuration, calculate the reciprocal of the sum of reciprocals:
  • \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \)
This results in a smaller equivalent capacitance than any individual capacitor in the series.
Both arrangements significantly impact how energy is stored and how much charge flows, demonstrating how configuration modifies circuit properties.

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Most popular questions from this chapter

Let us try to estimate the maximum "static electricity" charge that might result during each walking step across an insulating floor. Assume the sole of a person's shoe has area \(A \approx 150 \mathrm{~cm}^{2},\) and when the foot is lifted from the ground during each step, the sole acquires an excess charge \(Q\) from rubbing contact with the floor. (a) Model the sole as a plane conducting surface with \(Q\) uniformly distributed across it as the foot is lifted from the ground. If the dielectric strength of the air between the sole and floor as the foot is lifted is \(E_{\mathrm{S}}=3 \times 10^{6} \mathrm{~N} / \mathrm{C},\) determine \(Q_{\max },\) the maximum possible excess charge that can be transferred to the sole during each step. (b) Modeling a person as an isolated conducting sphere of radius \(r \approx 1 \mathrm{~m},\) estimate a person's capacitance. \((c)\) After lifting the foot from the floor, assume the excess charge \(Q\) quickly redistributes itself over the entire surface area of the person. Estimate the maximum potential difference that the person can develop with respect to the floor.

To make a \(0.40-\mu F\) capacitor, what area must the plates have if they are to be separated by a 28 -mm air gap?

A multilayer film capacitor has a maximum voltage rating of \(100 \mathrm{~V}\) and a capacitance of \(1.0 \mu \mathrm{F}\). It is made from alternating sheets of metal foil connected together, separated by films of polyester dielectric. The sheets are \(12.0 \mathrm{~mm}\) by \(14.0 \mathrm{~mm}\) and the total thickness of the capacitor is \(6.0 \mathrm{~mm}\) (not counting the thickness of the insulator on the outside). The metal foil is actually a very thin layer of metal deposited directly on the dielectric, so most of the thickness of the capacitor is due to the dielectric. The dielectric strength of the polyester is about \(30 \times 10^{6} \mathrm{~V} / \mathrm{m}\). Estimate the dielectric constant of the polyester material in the capacitor.

(II) \(\mathrm{A} 2.70-\mu \mathrm{F}\) capacitor is charged to 475 \(\mathrm{V}\) and a \(4.00-\mu \mathrm{F}\) capacitor is charged to 525 \(\mathrm{V}\) . (a) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

Capacitors can be used as "electric charge counters." Consider an initially uncharged capacitor of capacitance \(C\) with its bottom plate grounded and its top plate connected to a source of electrons. \((a)\) If \(N\) electrons flow onto the capacitor's top plate, show that the resulting potential difference \(V\) across the capacitor is directly proportional to \(N .\) (b) Assume the voltage- measuring device can accurately resolve voltage changes of about \(1 \mathrm{mV}\). What value of \(C\) would be necessary to detect each new \(\begin{array}{lllll}\text { collected electron? } & \text { (c) Using modern semiconductor }\end{array}\) technology, a micron-size capacitor can be constructed with parallel conducting plates separated by an insulating oxide of dielectric constant \(K=3\) and thickness \(d=100 \mathrm{nm}\). To resolve the arrival of an individual electron on the plate of such a capacitor, determine the required value of \(\ell(\) in \(\mu \mathrm{m}\) ) assuming square plates of side length \(\ell\)
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