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One of the fundamental principles of calculus is a process called一体化。这一原则对于理解是很重要的,因为它在电感的行为中表现出来。值得庆幸的是,还有更熟悉的物理系统,也表现出集成过程,使理解更容易。
如果我们将恒定的水流介绍进入圆柱形罐中,水位内部将以恒定的速度升高:
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In calculus terms, we would say that the tankintegrateswater flow into water height. That is, one quantity (flow) dictates the rate-of-change over time of another quantity (height).
像水箱,电气电感also exhibits the phenomenon of integration with respect to time. Which electrical quantity (voltage or current) dictates the rate-of-change over time of which other quantity (voltage or current) in an inductance? Or, to re-phrase the question, which quantity (voltage or current), when maintained at a constant value, results in which other quantity (current or voltage) steadily ramping either up or down over time?
在A.电感,电流是电压的时间积分。也就是说,电感器上的施加电压决定了通过电感器随时间通过电感的变化率。
Challenge question: can you think of a way we could exploit the similarity of inductive voltage/current integration tosimulate水箱填充的行为,或相同数学关系描述的任何其他物理过程?
The concept of integration doesn’t have to be overwhelmingly complex. Electrical phenomena such as电容电感可以作为学生可以探索和理解微积分原则的优秀背景。您选择投入对此问题的讨论的时间将取决于您学生的数学方式。
Suppose a mass is connected to a winch by means of a cable, and a person turns the winch drum to raise the mass off the ground:
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物理学家可能会像能源交换一样看这种情况:转动鼓的人是消耗的能量,这反过来就是存在stored在潜在形式的质量。
Suppose now that the person stops turning the drum and instead engages a brake mechanism on the drum so that it reverses rotation and slowly allows the mass to return to ground level. Once again, a physicist would view this scenario as an exchange of energy: the mass is now释放energy, while the brake mechanism is converting that released energy into heat:
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在上述每个场景中,绘制描绘两个力的方向的箭头:块在滚筒上施加的力,以及鼓在质量上施加的力。将这些力方向与每个场景中的运动方向进行比较,并解释这些方向如何与质量和鼓的方式交替充当能量source和能量加载。
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后续问题:虽然可能是显而易见的,但这个问题与电路组件之间的能量交换密切相关!解释这个类比。
学生通常会发现能量流的概念对电气组件相令人困惑。我试图通过使用机械类比来制作这个概念,其中力和运动充当电压和电流(或Visa Versa)的模拟量。
Draw the direction of current in this circuit, and also identify the polarity of the voltage across the battery and across the电阻器。然后,将电池的极性与电流的方向进行比较,电阻器的极性与电流方向通过它。
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您如何注意到这两种不同类型组件的电压极性和电流方向之间的关系?确定这两个组件之间的基本区别,导致他们表现不同。
Here I show the answer in two different forms: current shown aselectron flow(left) and current shown asconventional flow(right).
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无论您选择如何遵循所述电路的分析,理解应该是相同的:电阻器和电池两端的原因电压极性尽管通过两者的电流相同,但电流方向相同。电池充当了source, while the resistor acts as a加载。
这种区别在物理学的研究中非常重要,其中必须确定机械系统是否是做工作或者是否work is being done on it。A clear understanding of the relationship between voltage polarity and current direction for sources and loads is very important for students to have before they study reactive devices such as inductors and capacitors!
画出的模式磁场通过电流通过直线和通过线圈产生:
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使用任何一个解释你的答案R.ight-hand rule(常规流程)或左手规则(电子流量)。
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In your students’ research, they will encounter a “right-hand rule” as well as a “left-hand rule” for relating electric current with magnetic field directions. The distinction between the two rules depends on whether the text uses “conventional flow” notation or “electron flow” notation to denote the movement of electrical charge through theconductors。可悲的是,这是电力中的那些概念中的另一个,这是由于电流的两个“标准”概念的普及而不必要地混淆。
当电流通过线圈时,它产生磁场。如果该电流的大小随时间变化,则磁场的强度也会变化。
我们还知道随时间变化的磁场通量将沿线线圈的长度诱导电压。解释如何互补原则电磁学and electromagnetic induction manifest themselves simultaneously in the same wire coil to produceself-induction。
此外,解释Lenz的定律如何涉及线圈自诱导电压的极性。
通过线圈的变化电流产生a电压下降that opposes the direction of change.
Self-induction不是一个难以把握的概念了吗possesses a good understanding of electromagnetism, electromagnetic induction, and Lenz’s Law. Some students may struggle understanding self-induction, because it is probably the first application they’ve seen where these three phenomena inter-relate simultaneously.
∫f(x) dx微积分警报! |
在简单的电阻电路中,可以通过使用电阻除施加的电压来计算电流:
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虽然对这次电路的分析可能似乎微不足道,但我想鼓励你从新鲜的角度来看这里发生的事情。在物理学研究中观察到的重要原则是平衡,当时自然地“寻求”的平衡状态。该简单电路所寻求的余量是电压的平等:电阻上的电压必须与源的电压输出相同的值:
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If the resistor is viewed as a source of voltage seeking equilibrium with the voltage source, then current必须converge at whatever value necessary to generate the necessary balancing voltage across the resistor, according to Ohm’s Law (V = IR). In other words,the resistor’s current achieves whatever magnitude it has to in order to generate a voltage drop equal to the voltage of the source。
这似乎是一种分析这种简单电路的奇怪方式,电阻“寻求”以产生等于源的电压降,以及当前“神奇地”假设它必须实现该电压平衡,但它是有帮助的在理解其他类型的电路元件中。
例如,在这里,我们具有通过开关连接到大线圈的直流电压源。假设电线线圈具有可忽略的阻力(0Ω):
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与电阻电路一样,线圈将“寻求”,一旦开关关闭,一旦开关就达到电压源的电压平衡。然而,我们知道在线圈中感应的电压与电流的电流不正常成比例 - 而是线圈的电压下降与R.ate of change of magnetic flux over timeas described by Faraday’s Law of electromagnetic induction:
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在哪里,
V.coil=瞬时感应电压,在伏特中
N= N.umber of turns in wire coil
\(\ frac {dφ} {dt} \)=磁通量的瞬时变化率,每秒韦尔斯
假设线圈电流和磁通量之间的线性关系(即,当我双打时,φ加倍),在开关关闭后,通过时间内容描述这种简单的电路的电流。
当开关关闭时,电流随时间以线性速率稳步增加:
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挑战的问题:真正的线线圈包含电工实习ical resistance (unless they’re made of superconducting wire, of course), and we know how voltage equilibrium occurs in resistive circuits: the current converges at a value necessary for the resistance to drop an equal amount of voltage as the source. Describe, then, what the current does in a circuit with aR.eal线圈,不是超导线圈。
尚未理解电感概念的学生可能倾向于表明该电路中的电流将是无限的欧姆的法律(I = E/R). One of the purposes of this question is to reveal such misunderstandings, so that they may be corrected.
该电路提供了微积分原理的优秀示例一体化,在跨电感器上施加稳定的电压导致稳定的电压越来越多当前的。Whether or not you should touch on this subject depends on the mathematical aptitude of your students.
电感is a very important property in many types of electric circuits. Define what “inductance” is, and what causes it.
“电感“is the capacity of a conductor to store energy in the form of a magnetic field, resulting from an applied current. You may also find a definition of “inductance” stated in terms of opposition to change in applied current over time.
电感is caused by the establishment of a magnetic field around a conductor.
询问学生表达了哪些测量电感。另外,如果他们认为任何给定导体的电感随着所施加的电流或存储的能量而变化,或者如果电感是无关的,则无关。
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如果电磁铁线圈中的匝数增加三倍,则假设没有其他变量的磁通量(φ)的幅度发生在磁通量(φ)的幅度发生(电流通过线圈,磁路的磁阻的磁阻,等。)?
如果电感器中的电线匝数增加三倍,则诱导电压的幅度发生在给定的磁通量率随时间变化\(\ frac {dφ} {dt} \)?
If the number of turns of wire in an inductor is tripled, what happens to the magnitude of its inductance, measured in Henrys? Explain your answer.
如果n三级,那么φ三元脉冲,所有其他因素都是相等的。
If \(\frac{dφ}{dt}\) triples, then e triples, all other factors being equal.
如果n三元,则L增加了一个因素九, all other factors being equal.
This question presents an interesting problem in qualitative mathematics. It is closely related to the “chain rule” in calculus, where one function y = f(x) is embedded within another function z = f(y), such that \(\frac{dz}{dy}\)\(\frac{dy}{dx}\) = \(\frac{dz}{dx}\). The purpose of this exercise is for students to gain a conceptual grasp of why inductance does not vary linearly with changes in N.
Of course, students can obtain the same (third) answer just by looking at the inductance formula (in terms of N, μ, A, and l), without all the conceptual work. It would be good, in fact, if a student happens to derive the same answer by inspection of this formula, just to add variety to the discussion. But the real purpose of this question, again, is a conceptual understanding of that formula.
可以通过以下等式计算线圈中固有的电感量:
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在哪里,
l =电感in Henrys
n =围绕核心缠绕的电线数量“转弯”
μ=芯材的渗透率(绝对,不相对)
A = Core area, in square meters
L.= Length of core, in meters
计算必须在直径的中空,非磁性(空气)芯2cm的绕线和10cm长时间缠绕的匝数,以产生22mH的电感。您可以使用自由空间的渗透率(μ0.)对于空气芯的μ值。
接下来,计算使用相同尺寸的固体铁芯产生相同的电感的所需匝数,假设铁具有相对渗透率(μR.)4000。
Finally, knowing that the formula for the area of a circle is πr2,重新写入电感方程,以接受电感半径而不是电感区域的值。换句话说,substitute该等式中的区域(A)的半径(R),使其仍然提供一种用于电感的准确图。
Approximately 2360 turns of wire for the air core, and approximately 37 turns of wire for the iron core.
新电感方程:
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这个问题首先是代数操作练习:鉴于其他变量的值来解决n。学生应该能够研究μ的值0.quite easily, being a well-defined physical constant.
Note that in this equation, the Greek letter “mu” (μ) is not a metric prefix, but rather an actual variable! This confuses many students, who are used to interpreting μ as the metric prefix “micro” \(\frac{1}{1,000,000}\).
Note also how the re-written equation puts pi (π) ahead of all the variables in the numerator of the fraction. This is not absolutely necessary, but it is conventional to write constants before variables. Do not be surprised if some students ask about this, as their answers probably looked like this:
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Suppose you wished to build a component with no other purpose than to provide inductance in an electric circuit (an电感器)。您如何设计这样的设备以执行此功能,以及如何最大限度地提高其电感?
我会让您确定如何构建电感,从您自己的研究中是如何构建的。
To increase inductance:
这些因素对于理解可变电感器的功能很重要。务必在与学生讨论中培养可变电感器的主题。
像所有领域一样的磁场有两个基本措施:字段force和领域flux。在A.电感器, which of these field quantities is directly related to current through the wire coil, and which is directly related to the amount of energy stored?
基于这种关系,当铁的一条靠近线圈时,磁场量改变,连接到线圈,连接到恒定电流的源极?
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场力是线圈电流的直接函数,并且场通量是存储能量的直接函数。
如果铁杆靠近连接到恒流源的线圈,则由线圈产生的磁场力将保持不变,而磁场通量将增加(以及它,存储在中的能量的量磁场)。
The concept of a场地是非常摘要的,但至少磁场是大多数人体验领域内的东西。这个问题对于帮助学生区分现场力和现场助焊剂,他们应该理解(通过线圈的恒定电流,与由磁场通量产生的吸引力)。
假设电感直接连接到可调电流源,并且稳定的该源的电流是稳定的增加over time. We know that an increasing current through an inductor will produce a magnetic field of increasing strength. Does this increase in magnetic field constitute anaccumulationof energy in the inductor, or a释放来自电感的能量?在这种情况下,电感器是否充当了加载或者作为A.sourceof electrical energy?
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Now, suppose the adjustable current source is steadily减少over time. We know this will result in a magnetic field of decreasing strength in the inductor. Does this decrease in magnetic field constitute anaccumulationof energy in the inductor, or a释放来自电感的能量?在这种情况下,电感器是否充当了加载或者作为A.sourceof electrical energy?
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For each of these scenarios, label the inductor’s voltage drop polarity.
随着应用电流增加,电感器的行为s as a load, accumulating additional energy from the current source. Acting as a load, the voltage dropped by the inductor will be in the same polarity as across a resistor.
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当施加的电流减小时,电感器充当源,将累积的能量释放到电路的其余部分,好像它是电流源本身的卓越电流。用作源,电感器下降的电压将与电池电量相同,电荷供电。
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将电感器两端的电压极性与随时间随时间的变化,是许多学生的复杂概念。由于它涉及随着时间的推移而变化的速度,它是引入微积分概念的绝佳机会([D / DT])。
对学生对暴露于增加或降低电流的电感的概念理解至关重要是电能之间的区别source与A加载。Students need to think “battery” and “resistor,” respectively when determining the relationship between direction of current and voltage drop. The complicated aspect of inductors (and capacitors!) is that they may switch character in an instant, from being a source of energy to being a load, and visa-versa. The relationship is not fixed as it is for resistors, which are always energy负荷。
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欧姆的法律tells us that the amount of voltage dropped by a fixed resistance may be calculated as such:
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然而,固定电感的电压和电流之间的关系是完全不同的。电感的“欧姆法”公式如此:
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在使用电流(i)和电压(e)的小写变量中使用什么意义?另外,表达式[di / dt]是什么意思?注意:如果您认为d's是变量,并且应该在这个分数中取消,再次思考:这不是普通的商品!D字母代表了一个称为a的微积分概念迪fferential, and a quotient of two d terms is called aderivative。
Lower-case variables represent瞬间值,而不是平均值。表达式\(\ frac {di} {dt} \)表示随着时间的推移电流变化的瞬时速度。
随访问题:操纵此等式以解决其他两个变量(\(\ frac {di} {dt} \)= ...; l = ...)。
我发现电容和电感的主题是优异的背景,其中将微积分基本原则引入学生。花时间讨论这个问题和问题,就像它会根据你的学生的数学能力而有所不同。
即使您的学生尚未准备好探索微积分,讨论如何为电感的电流与电压之间的关系涉及的仍然是一个好主意time。This is a radical departure from the time-independent nature of resistors, and of Ohm’s Law!
通过代替正确的电变量(电压,电流,电阻,电感)来完成此声明:
Inductors oppose changes in当前的, reacting to such changes by producing aV.oltage。
强调你的学生,电感是一个基本上R.eactive财产,相反随着时间的推移变化。电感器反应不是稳定的电流,只改变电流。
多年前,我决定通过将电磁铁制作电线的电磁铁来试验电磁。我将钢螺栓放在卷轴的中心,以便具有高渗透率的核心,并通过电线通过电池通过电线进行磁场。没有任何“跳线”导线,我将线轴的线端保持在每只手中的9伏电池端子接触。
电磁铁工作得很好,我能够用它产生的磁场移动一些钢纸夹。但是,当我通过释放其中一个电线从电池端子释放到电池端子时,我接受了一个小电击!这里显示的是我的示意图,在电路中:
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At the time, I didn’t understand how inductance worked. I only understood how to make magnetism with electricity, but I didn’t realize a coil of wire could generate (high voltage!) electricity from its own magnetic field. I did know, however, that the 9 volts output by the battery was much too weak to shock me (yes, I touched the battery terminals directly to verify this fact), so某物在电路中必须产生大于9伏的电压。
如果你去过那里来解释刚刚发生在我身上,你会说什么?
有几种不同的方法来解释电磁铁线圈如何产生比它从(电池)通电的电压更大的电压。One way is to explain the origin of the high voltage using Faraday’s Law of electromagnetic induction (e = N\(\frac{dφ}{dt}\), or e = L\(\frac{di}{dt}\)). Another way is to explain how it is the nature of an inductor to oppose any改变in current over time. I’ll leave it to you to figure out the exact words to say!
一种方法来帮助了解电感如何产生这种大电压是将其视为一个临时电流源, which will output as much voltage as necessary in an effort to maintain constant current. Just as ideal current sources are dangerous to open-circuit, current-carrying inductors are likewise capable of generating tremendous transient voltages.
虽然我的实验没有真正的安全危险,但可能已经提供了不同的情况。与您的学生讨论有必要创造实际安全危险所需的内容。
焊接到印刷电路板的部件通常具有“流浪”电感,也称为寄生电感。遵守该电阻,焊接到电路板:
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寄生电感来自哪里?关于电阻的电阻是什么,安装在电路板上,产生(非常)少量电感?如何最大限度地减少这种电感,以防情况下对电路的操作有害?
电感自然存在于任何导体上。导体越长,电感越多,所有其他因素都是平等的。
In high-frequencyAC circuits(例如计算机电路,其中电压脉冲在每秒数百万个循环中振荡,电路板上的甚至短的线或迹线都可能借助于其杂散电感而存在大量的麻烦。通过适当的电路板组件可以减少一些寄生电感,通过电路板上的组件布局重新设计。
According to an article inIEEE频谱杂志(“把被动放在他们的地方“, July 2003, Volume 40, Number 7, page 29), the transient currents created by fast-switching logic circuits can be as high as 500 amps/ns, which is a [di/dt] rate of 500billion每秒安培!!在这些水平下,即使是沿着组件引线和电路板迹线的寄生电感的几种皮高,也将导致显着的电压下降。
Many precision resistors utilize a线绕construction, where the resistance is determined by the type and length of wire wrapped around a spool. This form of construction allows for high precision of resistance, with low temperature sensitivity if certain metal alloys are used for the wire.
Unfortunately, though, wrapping wire around a spool forms a coil, which will naturally possess a significant amount of inductance. This is generally undesirable, as we would like to have resistors possessing只有抵抗, with no “parasitic” properties.
然而,存在线圈可以缠绕的特殊方法,以便具有几乎没有电感。这种方法被称为bifilarwinding, and it is common in wire-wound resistor construction. Describe how bifilar winding works, and why it eliminates parasitic inductance.
我不会直接描述如何制作一个双向绕组,但我会给你一个暗示。比较直线直线的电感,与一半折叠成一个:
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Now, how could a non-inductivecoilof wire be made using the same principle?
This technique is very useful in reducing or eliminating parasitic inductance. Typically, parasitic inductance is not a problem unless very high rates of current change are involved, such as in high-frequency AC circuits (radio, high-speed digital logic, etc.). In such applications, knowing how to control stray inductance is very important to proper circuit operation.
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包括计算机内部工作的数字逻辑电路基本上只不过是由名为的半导体元件制成的开关阵列。晶体管。作为开关,这些电路具有两个状态:开启和关闭,分别表示1和0的二进制状态。
The faster these switch circuits are able to change state, the faster the computer can perform arithmetic and do all the other tasks computers do. To this end, computer engineers keep pushing the limits of transistor circuit design to achieve faster and faster switching rates.
这种速度的种族导致计算机电源电路的问题,因为目前的“浪涌”(技术称为瞬态) created in the conductors carrying power from the supply to the logic circuits. The faster these logic circuits change state, the greater the [di/dt] rates-of-change exist in the conductors carrying current to power them. Significant voltage drops can occur along the length of these conductors due to their parasitic inductance:
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假设当从“关闭”状态切换到Ön“状态时,逻辑门电路在从”关闭“状态切换时,每个纳秒(175A / ns)产生175个放大器的瞬态电流。如果电源导体的总电感是10微米(9.5 pH),电源电压为5伏DC,则在其中一个“浪涌”期间逻辑门的电源端子的电压保持多大电压?
在当前瞬时= 3.338 V期间剩余的逻辑门端子处的电压
学生可能会惊叹于每纳秒的175安培的[DI / DT]速率,这相当于175billion每秒安培。不仅是这个数字现实,它也是一些估计也很低(见IEEE频谱杂志,2003年7月,第40卷,7号,在文章中“把被动放在他们的地方“)。Some of your students may be very skeptical of this figure, not willing to believe that ä computer power supply is capable of outputting 175 billion amps?!”
This last statement represents a very common error students commit, and it is based on a fundamental misunderstanding of [di/dt]. “175 billion amps per second” is not the same thing as “175 billion amps”. The latter is an absolute measure, while the former is a随着时间的推移变化率。It is the difference between saying “1500 miles per hour” and “1500 miles”. Just because a bullet travels at 1500 miles per hour does not mean it will travel 1500 miles! And just because a power supply is incapable of outputting 175 billion amps does not mean it cannot output a current that改变s每秒以1750亿安培的速度!
电感有一个紧密的机械类比:惯性。Explain what mechanical “inertia” is, and how the quantities of velocity and force applied to an object with mass are respectively analogous to current and voltage applied to an inductance.
当物体受到恒定的不平衡力时,其速度以线性速率变化:
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在哪里,
f =应用于对象的净力
m = Mass of object
V.= Velocity of object
t = Time
In a similar manner, a pure inductance experiencing a constant voltage will exhibit a constant rate of current change over time:
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向您的学生解释惯性和电感之间的相似性如何如此接近,电感器可用于电模拟机械惯性。
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电感器以磁场的形式存储能量。我们可以通过将电感电压和电感器电流(P = IV)的产品集成随时间的时间来计算存储在电感中的能量,因为我们知道电源是完成工作(W)的速率,并且完成的工作量对于从零电流拍摄的电感到一些非零电流量构成能量存储(U):
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找到一种方法来将电感(L)和电流(I)替换为Integrand,因此您可以集成以找到描述存储在电感器中的能量量的等式,以用于任何给定的电感和电流值。
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The integration required to obtain the answer is commonly found in calculus-based physics textbooks, and is an easy (power rule) integration.
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