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Chapter 24: Problem 46

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

Short Answer

Expert verified
In parallel: 0.0002548 J, 18.2 碌C. In series: 0.0000532 J, 3.808 碌C.

Step by step solution

01

Calculate Capacitance for Parallel Configuration

For capacitors in parallel, simply add the capacitances together. The formula is \( C_{parallel} = C_1 + C_2 \). Therefore, \( C_{parallel} = 0.45\,\mu\text{F} + 0.20\,\mu\text{F} = 0.65\,\mu\text{F} \).
02

Calculate Energy Expended (Parallel)

Using the formula for energy in a capacitor, \( E = \frac{1}{2}CV^2 \), where \( C = 0.65\,\mu \mathrm{F} = 0.65 \times 10^{-6}\,\mathrm{F} \) and \( V = 28\,\mathrm{V} \), we find: \( E = \frac{1}{2}(0.65 \times 10^{-6})(28^2) = 2.548 \times 10^{-4}\,\mathrm{J} \).
03

Calculate Capacitance for Series Configuration

For capacitors in series, use the formula \( \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} \). Calculating this gives: \( \frac{1}{C_{series}} = \frac{1}{0.45\,\mu\mathrm{F}} + \frac{1}{0.20\,\mu\mathrm{F}} = \frac{1}{0.00045} + \frac{1}{0.00020} \). Therefore, \( C_{series} \approx 0.136\,\mu\mathrm{F} \).
04

Calculate Energy Expended (Series)

Using \( E = \frac{1}{2}CV^2 \) with \( C = 0.136 \times 10^{-6}\,\mathrm{F} \) and \( V = 28\,\mathrm{V} \), we calculate: \( E = \frac{1}{2}(0.136 \times 10^{-6})(28^2) = 5.32 \times 10^{-5}\,\mathrm{J} \).
05

Calculate Charge in Parallel

For parallel, the total charge \( Q = CV \). With \( C = 0.65 \mu\mathrm{F} \) and \( V = 28\,\mathrm{V} \), \( Q = 0.65 \times 10^{-6} \times 28 = 18.2 \times 10^{-6}\,\mathrm{C} \).
06

Calculate Charge in Series

For series, the total charge remains the same for each capacitor, hence calculate as \( Q = C_{series} \times V = 0.136 \times 10^{-6} \times 28 = 3.808 \times 10^{-6}\,\mathrm{C} \).

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Key Concepts

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

Parallel Circuits
When dealing with parallel circuits, it's essential to remember that the total capacitance (\( C_{parallel} \)) is simply the sum of all individual capacitances. This happens because each capacitor has the same voltage across its terminals, which leads to an increase in the total charge storage capacity of the circuit.
To calculate the overall capacitance in a parallel configuration, use the formula:
  • \( C_{parallel} = C_1 + C_2 + \ldots + C_n \)
For the given exercise, two capacitors with values of \(0.45\,\mu\text{F}\)and \(0.20\,\mu\text{F}\)were placed in parallel, resulting in a combined capacitance of \(0.65\,\mu\text{F}\).
This increase in capacitance allows the circuit to store more energy from the battery, using the energy formula for capacitors:
  • \( E = \frac{1}{2} C V^2 \)
Series Circuits
In a series circuit, things work a bit differently when it comes to capacitors. The total capacitance decreases, because the charge must flow through each capacitor in sequence. As a result, the reciprocal of the total capacitance (\( C_{series} \)) is the sum of the reciprocals of each individual capacitance.
Thus, the formula used is:
  • \( \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \)
For the two capacitors in the exercise:\( \frac{1}{C_{series}} = \frac{1}{0.45\,\mu\mathrm{F}} + \frac{1}{0.20\,\mu\mathrm{F}} \)resulting in a total series capacitance of approximately \(0.136\,\mu\mathrm{F}\).
This reduced capacitance means that the energy stored:
  • \( E = \frac{1}{2} C V^2 \)
will be smaller in a series circuit compared to a parallel arrangement.
Capacitance
The concept of capacitance is fundamental to understanding both series and parallel circuits. Capacitance (\( C \)) is defined as the ability of a system to store charge per unit voltage. It is measured in farads (\( F \)). The higher the capacitance, the more charge it can store at a given voltage.
When capacitors are charged by a voltage source, like a battery, they store electrical energy. The energy stored in a capacitor is given by the formula:
  • \( E = \frac{1}{2} C V^2 \)
Where \( E \) is the energy in joules, \( C \) is the capacitance in farads, and \( V \) is the voltage in volts.
This formula shows why the arrangement of capacitors (parallel or series) affects the energy stored in the circuit.
Higher capacitance in parallel allows more energy storage, while lower capacitance in series allows less.

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Most popular questions from this chapter

(II) \((a)\) Show that each plate of a parallel-plate capacitor exerts a force $$F=\frac{1}{2} \frac{Q^{2}}{\epsilon_{0} A}$$ on the other, by calculating dW/dx where \(d W\) is the work needed to increase the separation by \(d x\) . (b) Why does using \(F=Q E,\) with \(E\) being the electric field between the plates, give the wrong answer?

For commonly used CMOS (complementary metal oxide semiconductor) digital circuits, the charging of the component capacitors \(C\) to their working potential difference \(V\) accounts for the major contribution of its energy input requirements. Thus, if a given logical operation requires such circuitry to charge its capacitors \(N\) times, we can assume that the operation requires an energy of \(N\left(\frac{1}{2} C V^{2}\right) .\) In the past 20 years, the capacitance in digital circuits has been reduced by a factor of about 20 and the voltage to which these capacitors are charged has been reduced from \(5.0 \mathrm{~V}\) to \(1.5 \mathrm{~V} .\) Also, present-day alkaline batteries hold about five times the energy of older batteries. Two present-day AA alkaline cells, each of which measures \(1 \mathrm{~cm}\) diameter by \(4 \mathrm{~cm}\) long, can power the logic circuitry of a hand-held personal digital assistant (PDA) with its display turned off for about two months. If an attempt was made to construct a similar PDA (i.e., same digital capabilities so \(N\) remains constant) 20 years ago, how many (older) AA batteries would have been required to power its digital circuitry for two months? Would this PDA fit in a pocket or purse?

(II) Six physics students were each given an air filled capacitor. Although the areas were different, the spacing between the plates, \(d,\) was the same for all six capacitors, but was unknown. Each student made a measurement of the area \(A\) and capacitance \(C\) of their capacitor. Below is a Table for their data. Using the combined data and a graphing program or spreadsheet, determine the spacing \(d\) between the plates.

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as \(35,000,000 \mathrm{~V}\). The bottoms of thunderclouds are typically \(1500 \mathrm{~m}\) above the Earth, and may have an area of \(120 \mathrm{~km}^{2}\). Modeling the Earth-cloud system as a huge capacitor, calculate ( \(a\) ) the capacitance of the Earth-cloud system, (b) the charge stored in the "capacitor," and (c) the energy stored in the "capacitor."

A \(0.50-\mu \mathrm{F}\) and a \(0.80-\mu \mathrm{F}\) capacitor are connected in series to a 9.0-V battery. Calculate (a) the potential difference across each capacitor and \((b)\) the charge on each. (c) Repeat parts \((a)\) and \((b)\) assuming the two capacitors are in parallel.
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