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CIRCUIT ANALYSIS FOR COMPLETE IDIOTS
电路分析完全傻瓜书

Circuit Analysis for Complete IDIOTS
by David Smith
电路分析完全傻瓜书 由大卫·史密斯撰写

All Rights Reserved.
版权所有。

No part of this publication may be reproduced in any form or by any means, including scanning, photocopying, or otherwise without prior written permission of the copyright holder.
本出版物的任何部分不得以任何形式或任何手段复制,包括扫描、复印或其他方式,未经版权持有者事先书面许可。

Copyright © 2019
版权 © 2019
Other books in the Series:
系列中的其他书籍:

Arduino for Complete Idiots
Arduino 完全傻瓜指南

Digital Signal Processing for Complete Idiots
数字信号处理入门

Control Systems for Complete Idiots Basic Electronics for Complete Idiots
完全傻瓜的控制系统 完全傻瓜的基础电子学

Digital Electronics for Complete Idiots Electromagnetic Theory for Complete Idiots
完全傻瓜的数字电子 完全傻瓜的电磁理论

Table of Contents
目录

PREFACE
前言

  1. INTRODUCTION
    介绍
  2. VOLTAGE & CURRENT LAWS
    电压和电流定律
  3. BASIC ANALYSIS TECHNIQUES
    基本分析技术
  4. NETWORK THEOREMS
    网络定理
  5. CAPACITANCE
    电容
  6. INDUCTANCE
    电感
  7. AC FUNDAMENTALS
    交流电基础知识
  8. AC CIRCUITS
    交流电路
  9. ANALYSIS TECHNIQUES (FOR AC).
    分析技术(用于交流电)。
  10. NETWORK THEOREMS (FOR AC).
    网络定理(交流电)。
  11. LAPLACE TRANSFORM
    拉普拉斯变换
  12. TRANSIENT ANALYSIS
    瞬态分析
  13. 3-PHASE SYSTEMS
    三相系统
REFERENCES
参考文献

CONTACT
联系

PREFACE
前言

In today’s world, there’s an electronic gadget for everything and inside these gadgets are circuits, little components wired together to perform some meaningful function.
在当今世界,几乎每样东西都有电子设备,这些设备内部有电路,小组件相互连接以执行某些有意义的功能。

Have you wondered how a led display sign works or how a calculator works or toy cars work?
你有没有想过 LED 显示标志是如何工作的,计算器是如何工作的,玩具车是如何工作的?

How is it possible?? Answer, all because of electrical circuits.
这怎么可能??答案,全都因为电路。

These tiny components when arranged in certain manner can do wonders. Fascinating isn’t it?
这些微小的组件以某种方式排列时可以创造奇迹。很迷人,不是吗?

Our fascination with gadgets and reliance on machinery is only growing day by day and hence from an engineering perspective, it is absolutely crucial to be familiar with the analysis and designing of such Circuits, at least identify components.
我们对小工具的迷恋和对机器的依赖与日俱增,因此从工程的角度来看,熟悉此类电路的分析和设计,至少识别组件是绝对重要的。

Circuit analysis is one of basic subjects in engineering and particularly important for Electrical and Electronics students.
电路分析是工程学中的基本学科之一,对于电气和电子专业的学生尤其重要。

So circuit analysis is a good starting point for anyone wanting to get into the field.
因此,电路分析是任何想要进入该领域的人的一个良好起点。

It is a very easy subject to learn and understand, but messing up these ideas or misunderstanding them, will lead to a lot of headache in other subjects.
这是一个非常容易学习和理解的科目,但搞混这些概念或误解它们,会在其他科目中带来很多头痛。

In this book we provide a concise introduction into basic Circuit analysis.
在本书中,我们提供了对基本电路分析的简明介绍。

A basic knowledge of Calculus and some Physics are the only prerequisites required to follow the topics discussed in the book.
学习本书讨论的主题只需要基本的微积分知识和一些物理知识。

We’ve tried to explain the various fundamental concepts of Circuit theory in the simplest manner without an over reliance on math.
我们尝试以最简单的方式解释电路理论的各种基本概念,而不依赖过多的数学。

Also, we have tried to connect the various topics with real life situations wherever possible.
此外,我们尽可能将各种主题与现实生活中的情况联系起来。

This way even first timers can learn the basics of Circuit theory with minimum effort.
这样,即使是初学者也能以最小的努力学习电路理论的基础知识。

Hopefully the students will enjoy this different approach to Circuit Analysis.
希望学生们会喜欢这种不同的电路分析方法。

The various concepts of the
各种概念的

subject are arranged logically and explained in a simple reader-friendly language with illustrative figures.
主题以逻辑方式安排,并用简单易懂的语言和插图进行解释。

This book is not meant to be a replacement for those standard Circuit theory textbooks, rather this book should be viewed as an introductory text for beginners to come in grips with advanced level topics covered in those books.
这本书并不是为了替代那些标准的电路理论教科书,而是应该被视为一本入门书籍,帮助初学者掌握那些书中涵盖的高级主题。

This book will hopefully serve as inspiration to learn Circuit theory in greater depths.
这本书希望能激励人们更深入地学习电路理论。

Readers are welcome to give constructive suggestions for the improvement of the book and please do leave a review.
欢迎读者提出建设性的建议以改进本书,并请留下评论。

1. INTRODUCTION
1. 引言

1.1 ELECTRICAL CHARGE
1.1 电荷

Have you ever wondered what Electricity is and where it comes from?
你有没有想过电是什么以及它来自哪里?

To answer these questions, we have to start with the atom.
要回答这些问题,我们必须从原子开始。

Although we are more interested in the properties of electricity than the phenomenon itself, it wouldn’t hurt us to quickly discuss the basics.
虽然我们对电的性质比现象本身更感兴趣,但快速讨论一下基础知识也无妨。

Everything in the universe is made of atoms and every atom consists of 3 types of particles, neutrons, protons and electrons.
宇宙中的一切都是由原子构成的,每个原子由三种粒子组成:中子、质子和电子。

Neutrons and protons are packed together in the nucleus and make up the center of an atom, whereas the electrons move around the nucleus in a constant motion.
中子和质子聚集在原子核中,构成原子的中心,而电子则以恒定的运动围绕原子核移动。

For this discussion, we are only concerned about protons and electrons or more specifically, a property these two particles possess called the Electric Charge.
在本次讨论中,我们只关注质子和电子,或者更具体地说,这两个粒子所具有的一个属性,称为电荷。

Although it is very unlikely you’ll ever come across a proper definition for charge, the best we can come up with is, that charge is a form of electrical energy.
虽然你很可能永远不会遇到电荷的正式定义,但我们能想到的最好解释是,电荷是一种电能形式。

Protons have a positive charge and Electrons have a negative charge.
质子带有正电荷,电子带有负电荷。

In a normal atom, the number of protons is equal to the number of electron and thus the atom as a whole is electrically neutral.
在一个正常的原子中,质子的数量等于电子的数量,因此整个原子是电中性的。

Neutral objects aren’t of much interest to us, we are more interested in charged bodies.
中性物体对我们并不太感兴趣,我们更关注带电物体。

Electric Charge is denoted by the letter Q.
电荷用字母 Q 表示。

The SI unit of electric charge is Coulomb © and it is the charge possessed by 6.24 × 10 18 6.24 × 10 18 6.24 xx10^(18)6.24 \times 10^{18} electrons.
电荷的国际单位是库仑(C),它是 6.24 × 10 18 6.24 × 10 18 6.24 xx10^(18)6.24 \times 10^{18} 个电子所拥有的电荷。

1.2 CURRENT
1.2 当前

Previously we mentioned that free electrons are responsible for the flow of Electric Current.
之前我们提到,自由电子负责电流的流动。

The concept behind this phenomenon is very simple, whenever a charged particle moves, it produces an Electric Current.
这个现象背后的概念非常简单,任何带电粒子移动时,它都会产生电流。

Obviously the protons can’t move, because they are inside the nucleus.
显然,质子不能移动,因为它们在原子核内。

And the electrons close to the nucleus are held tightly by the force of attraction, so they can’t move either.
而靠近原子核的电子被吸引力紧紧束缚,因此它们也无法移动。

So the only way an Electric current is produced is through movement of outer electrons, called the free electrons (it’s a little different in electronics though).
因此,电流产生的唯一方式是通过外层电子的运动,称为自由电子(不过在电子学中稍有不同)。
To understand this better, consider the inside section of a Conductor as shown below.
要更好地理解这一点,请考虑下面所示的导体内部部分。

Conductors have tons of free electrons and they keep moving in random direction (due to thermal energy), and each of these small movements contribute to an Electric current.
导体中有大量自由电子,它们在随机方向上不断移动(由于热能),这些微小的运动共同产生电流。

You might be thinking, if an electric current is produced this easily in a conductor, why do we need batteries and generators and power plants and stuff.
你可能在想,如果在导体中如此容易产生电流,为什么我们还需要电池、发电机和电厂等东西。

Can’t we just hook up a small piece of copper wire to a bulb and be done with it.
我们不能只把一小段铜线连接到灯泡上就完成了吗?

Unfortunately, that won’t work.
不幸的是,那行不通。

That’s because the currents produced by each free electron are in random direction (in accordance with the direction of their motion) and when we consider the conductor as a whole, these currents cancel each other out and net current is zero.
这是因为每个自由电子产生的电流方向是随机的(与它们的运动方向一致),当我们将导体视为一个整体时,这些电流相互抵消,净电流为零。

The way out of this problem is to make all the free electrons drift in one direction and thus the net Electric Current adds up to a non-zero value.
解决这个问题的方法是使所有自由电子朝一个方向漂移,从而净电流累加到一个非零值。

To do this all we need is a little effort, a force of sorts, called the EMF or the Electromotive Force.
要做到这一点,我们只需要一点努力,一种被称为电动势(EMF)的力量。

We will discuss more about the EMF in the next section.
我们将在下一节中讨论更多关于 EMF 的内容。
So Electric Current can be defined as the flow of charge (electrons) when subjected to an EMF.
电流可以定义为在电动势作用下电荷(电子)的流动。

Or the more accurate definition would be, Current is the rate of flow of charge.
或者更准确的定义是,电流是电荷流动的速率。

Mathematically, Current I is equal to,
在数学上,电流 I 等于,

The unit of current is Ampere, named after French mathematician and physicist André-Marie Ampère.
电流的单位是安培,以法国数学家和物理学家安德烈-玛丽·安培的名字命名。

One ampere of current represents one coulomb of electrical charge moving past a specific point in one second.
一安培的电流表示在一秒钟内有一个库仑的电荷经过一个特定点。

1.3 EMF

EMF stands for Electromotive force.
EMF 代表电动势。

The name may give you the impression that electromotive force is a type of force.
这个名称可能会给你留下电动势是一种力的印象。

Actually, it is not.
实际上,情况并非如此。

As mentioned in the previous section, EMF or the Electromagnetic force is an energy that can cause current to flow in an electrical circuit or device.
如前一节所述,电磁场或电磁力是一种能在电路或设备中引起电流流动的能量。

This means that a current can flow in a circuit or a device, only if an EMF is provided.
这意味着电流只能在电路或设备中流动,如果提供了电动势。

Sources of EMF can be batteries, solar cells, generators etc. EMF is denoted by the symbol E and is measured in unit Volt (V).
电磁场的来源可以是电池、太阳能电池、发电机等。电磁场用符号 E 表示,单位为伏特(V)。

Current
当前

1.4 POTENTIAL DIFFERENCE
1.4 潜在差异

Both EMF and Potential Difference are closely related and are often used interchangeably in many places, but they aren’t the same quantities.
电动势和电位差密切相关,许多地方常常可以互换使用,但它们并不是相同的量。

When a current flows through a material, the electrons are accelerated due to the applied EMF.
当电流通过材料时,电子由于施加的电动势而加速。

But these electrons don’t gain much velocity, because they keep colliding with ions in the material and due to this, the kinetic energy of the electrons is converted to heat.
但是这些电子并没有获得太大的速度,因为它们不断与材料中的离子碰撞,因此电子的动能转化为热能。

What this means is that, the electrons at one of the material has more energy than the electrons at the other end, which leads to a potential difference.
这意味着,材料一端的电子比另一端的电子具有更高的能量,从而导致了电位差。

This obviously is a rough explanation, the actual physics behind phenomenon is more complex and beyond the scope of this book.
这显然是一个粗略的解释,现象背后的实际物理学更复杂,超出了本书的范围。

It is important to note that, Potential difference is always measured between 2 points and never at a single point.
重要的是要注意,电位差总是测量两个点之间的,而不是在单个点上。
To sum up, the EMF is the driving force that keeps electrons in motion and Potential difference is the difference in energy of the electrons as a current is passed through a material.
总之,电动势是使电子运动的驱动力,而电位差是电流通过材料时电子能量的差异。

Both EMF and Potential difference have the common unit Volt (V).
电动势和电位差的共同单位是伏特(V)。

The term Voltage can be used in place of Potential difference or EMF.
电压一词可以用来代替电位差或电动势。

1.5 OHM'S LAW
1.5 欧姆定律

From the previous sections itself, it must be pretty clear that the Voltage and the Current are two closely related quantities.
从前面的部分来看,电压和电流是两个密切相关的量。

They have a cause effect relation as given by this general equation:
它们之间存在因果关系,如下所示的通用方程:

Effect = Cause Opposition =  Cause   Opposition  =(" Cause ")/(" Opposition ")=\frac{\text { Cause }}{\text { Opposition }}
效果 = Cause Opposition =  Cause   Opposition  =(" Cause ")/(" Opposition ")=\frac{\text { Cause }}{\text { Opposition }}

Where the Voltage is the cause and the Current is the effect.
电压是原因,电流是结果。

Now the question is, what could possibly be the opposition to current?
现在的问题是,可能对电流的反对是什么?

This is where we introduce a quantity called Resistance.
这是我们引入一个叫做电阻的量的地方。

The concept of Resistance is analogous to friction in mechanics.
阻力的概念类似于力学中的摩擦。

Every material has a tendency to oppose current, but some more than the others.
每种材料都有抵抗电流的倾向,但有些材料的抵抗力更强。

Materials with large no.
材料数量大。

of free electrons like metals have low resistance or a low tendency to oppose current.
自由电子像金属一样具有低电阻或低反对电流的倾向。

Such materials are called Conductors. Whereas materials with small no.
这种材料被称为导体。而具有少量的材料。

of free electrons like plastic have high resistance. Such materials are called Insulators.
像塑料这样的自由电子具有高电阻。这种材料被称为绝缘体。

And some materials fall in between, they offer some resistance, but not very high either.
有些材料介于两者之间,它们提供了一定的阻力,但也不是很高。

They are called Semi-conductors.
它们被称为半导体。
Now let’s substitute the terms we introduced so far into our general equation from earlier.
现在让我们将到目前为止介绍的术语代入我们之前的通用方程中。

Current = Voltage Resistance =  Voltage   Resistance  =(" Voltage ")/(" Resistance ")=\frac{\text { Voltage }}{\text { Resistance }}
当前 = Voltage Resistance =  Voltage   Resistance  =(" Voltage ")/(" Resistance ")=\frac{\text { Voltage }}{\text { Resistance }}

I = V R I = V R I=(V)/(R)I=\frac{V}{R}
  • Ohm’s Law
    欧姆定律

The result is this beautiful equation called the Ohm’s Law, after the German physicist and mathematician Georg_Simon Ohm (weird name right??).
结果是这个美丽的方程,称为欧姆定律,以德国物理学家和数学家乔治·西蒙·欧姆的名字命名(名字很奇怪,对吧??)。

It’s one of the most fundamental things there is in electrical engineering. Get used to it,
这是电气工程中最基本的事情之一。习惯它吧,

because it will remain with you as long as you do anything electrical related.
因为只要你做任何与电相关的事情,它就会伴随你。
The Ohm’s law essentially implies that, the current flowing through a material/circuit is directly proportional to the Voltage applied across it, provided that the resistance of the material remain fixed.
欧姆定律基本上意味着,流经材料/电路的电流与施加在其上的电压成正比,前提是材料的电阻保持不变。

So if we were to apply twice the voltage across a bulb, twice the amount of current would flow through it or if we apply one third the voltage, then one third the current would flow.
所以如果我们在灯泡两端施加两倍的电压,流过它的电流将是两倍;如果我们施加三分之一的电压,那么流过的电流将是三分之一。

Graphically the Ohm’s law would look like,
欧姆定律在图形上看起来是这样的,

The Unit of Resistance is Ohm and is denoted by the Greek letter Ω Ω Omega\Omega.
电阻的单位是欧姆,用希腊字母 Ω Ω Omega\Omega 表示。

1.6 CONDUCTANCE
1.6 导电性

While we are at it, let’s define one more new quantity called Conductance.
在此期间,让我们定义一个新的量,称为电导。

Conductance is the inverse of Resistance.
导电性是电阻的倒数。

It’s a measure of how well a material allows current to flow
它是衡量材料允许电流流动的能力

through it. The Unit of Conductance is Siemens and is denoted by Ω 1 Ω 1 Omega^(-1)\Omega^{-1}.
通过它。电导的单位是西门子,用 Ω 1 Ω 1 Omega^(-1)\Omega^{-1} 表示。

1.7 RESISTOR
1.7 电阻器

Have you seen one of these tiny components in an electronic circuit before??
你见过电子电路中的这些微小组件吗?
Those are resistors. A Resistor is a device that provide resistance in an electrical circuit. WHAT??
那些是电阻器。电阻器是一个在电路中提供电阻的设备。什么??

But isn’t resistance a bad thing?
但抵抗不是一件坏事吗?

Yes, resistance does oppose current and it does cause energy loss.
是的,电阻确实会对电流产生阻碍,并导致能量损失。

But when used the right way it isn’t always a bad thing.
但当以正确的方式使用时,这并不总是一件坏事。

Do you know that resistance is the reason we have bulbs and heaters?
你知道电阻是我们有灯泡和加热器的原因吗?

Resistors are electrical components that help control the flow of current in a circuit.
电阻器是帮助控制电路中电流流动的电气元件。

A high resistance means there is less current available for a given voltage.
高电阻意味着在给定电压下可用的电流较少。

It is widely used in heating applications, for biasing, voltage dividers and tons of other applications.
它广泛用于加热应用、偏置、电压分压器以及许多其他应用。

The symbol for resistor is:
电阻器的符号是:

1.8 POWER
1.8 功率

Electrical power is defined as the rate at which electrical energy is transferred from an energy source to a circuit.
电力被定义为电能从能源源转移到电路的速率。

When current is passed through a resistor, energy is dissipated as heat.
当电流通过电阻时,能量以热的形式散失。

It is easy to calculate Electrical power, it is simply the product of the current (I) flowing through a component and the voltage ( V ) across the component.
计算电功率很简单,它只是流过一个元件的电流(I)与该元件两端电压(V)的乘积。
P = VI P = VI P=VI\mathrm{P}=\mathrm{VI}
Applying the Ohm’s law, 2 other forms of equation can be obtained,
应用欧姆定律,可以得到另外两种形式的方程
P = V 2 R P = I 2 R P = V 2 R P = I 2 R {:[P=(V^(2))/(R)],[P=I^(2)R]:}\begin{aligned} & P=\frac{V^{2}}{R} \\ & P=I^{2} R \end{aligned}
Unit of electrical power is Watts.
电功率的单位是瓦特。

2. VOLTAGE & CURRENT LAWS
2. 电压和电流定律

2.1 SERIES CIRCUIT
2.1 串联电路

A series circuit is a circuit in which any number of components are connected one after the other, such that there is a single path for the flow of current.
串联电路是指多个元件一个接一个连接在一起的电路,从而形成一个电流流动的单一路径。

For example, in the circuit shown in the figure below, the Resistors R 1 R 1 R_(1)R_{1} and R 2 R 2 R_(2)R_{2} are in series, because they are connected at a common point b. Similarly, Resistor R 2 R 2 R_(2)\mathrm{R}_{2} and the Voltage source are also in series, with the common point c.
例如,在下图所示的电路中,电阻器 R 1 R 1 R_(1)R_{1} R 2 R 2 R_(2)R_{2} 是串联的,因为它们连接在一个公共点 b。类似地,电阻器 R 2 R 2 R_(2)\mathrm{R}_{2} 和电压源也是串联的,公共点为 c。


C C CC
If there were any other components (that carry current) connected at any of these nodes ( a , b a , b a,ba, b or c c cc ), then this circuit wouldn’t be a series circuit anymore. For instance, if there
如果在这些节点( a , b a , b a,ba, b c c cc )的任何一个连接了其他任何带电流的组件,那么这个电路就不再是串联电路了。例如,如果有

had been a third resistor R 3 R 3 R_(3)\mathrm{R}_{3} connected between nodes a and b , as shown in the figure below, this is no longer a series circuit.
在下图中,节点 a 和 b 之间连接了一个第三个电阻 R 3 R 3 R_(3)\mathrm{R}_{3} ,这不再是一个串联电路。

Clearly there are 2 paths for the current to flow, through R 1 & R 3 R 1 & R 3 R_(1)&R_(3)R_{1} \& R_{3}.
显然,电流有两条路径可以流动,通过 R 1 & R 3 R 1 & R 3 R_(1)&R_(3)R_{1} \& R_{3}


C C CC

2.2 KIRCHHOFF'S VOLTAGE LAW (KVL)
2.2 基尔霍夫电压定律 (KVL)

Kirchhoff’s Law’s…Wait!! "Laws " you say?? You mean there’s more than one law??
基尔霍夫定律……等等!!“定律”你说??你是说不止一条定律??

Yes, there are 2 Kirchhoff’s Law’s: Kirchhoff’s Voltage law & the Kirchhoff’s Current Law.
是的,有两个基尔霍夫定律:基尔霍夫电压定律和基尔霍夫电流定律。

Kirchhoff’s laws are the most fundamental laws, next to the Ohm’s law, in Electrical engineering.
基尔霍夫定律是电气工程中最基本的定律,仅次于欧姆定律。

But fortunately, just like the Ohm’s law, these are 2 really simple laws. Even
但幸运的是,就像欧姆定律一样,这两个法则真的很简单。即使

simpler than the Ohm’s Law I would say, because there is no formula, just a simple statement.
比欧姆定律简单,我会说,因为没有公式,只有一个简单的陈述。

The entire basis of Circuit analysis are these 2 laws and the Ohm’s law.
电路分析的整个基础是这两个定律和欧姆定律。

They are basically spin offs to the energy and charge conservation laws.
它们基本上是能量和电荷守恒定律的衍生物。

We’ll get to the Kirchhoff’s Current Law in later section.
我们将在后面的章节中讨论基尔霍夫电流定律。

For now, we’ll focus on the Kirchhoff’s Voltage Law or the KVL.
目前,我们将重点关注基尔霍夫电压定律或 KVL。

Kirchhoff’s voltage law (KVL) states that “the algebraic sum of the potential rises and drops around a closed loop (or path) is zero”.
基尔霍夫电压定律(KVL)指出:“一个闭合回路(或路径)中电势升高和降低的代数和为零。”

Symbolically,
象征性地,

Closed Path
闭合路径
In layman’s terms Kirchhoff’s voltage law essentially means: “Voltage supplied = Voltage used up, around a closed loop”.
用通俗的话来说,基尔霍夫电压定律基本上意味着:“供电电压 = 使用电压,围绕一个闭合回路。”

Forming a KVL equation is really easy, start at a certain point of the circuit and note down all the potential changes (either rises or drops) in one particular direction, till the starting point is reached once again.
形成一个 KVL 方程真的很简单,从电路的某个点开始,记录在一个特定方向上的所有电位变化(无论是上升还是下降),直到再次到达起始点。

Then equate the resulting expression to zero. That’s it.
然后将结果表达式等于零。就这样。

For the above Circuit, KVL equation is E V 1 V 2 = 0 E V 1 V 2 = 0 E-V_(1)-V_(2)=0\mathbf{E}-\mathbf{V}_{\mathbf{1}}-\mathbf{V}_{\mathbf{2}} \mathbf{= 0} or E E E\mathbf{E} (Voltage supplied) = V 1 + V 2 = V 1 + V 2 =V_(1)+V_(2)=\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}} (Voltage Used up).
对于上述电路,KVL 方程是 E V 1 V 2 = 0 E V 1 V 2 = 0 E-V_(1)-V_(2)=0\mathbf{E}-\mathbf{V}_{\mathbf{1}}-\mathbf{V}_{\mathbf{2}} \mathbf{= 0} E E E\mathbf{E} (供电电压) = V 1 + V 2 = V 1 + V 2 =V_(1)+V_(2)=\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}} (消耗电压)。

Do note that KVL is applicable to all loops or closed paths, however complex the circuit maybe.
请注意,基尔霍夫电压定律适用于所有回路或闭合路径,无论电路多么复杂。

2.3 RESISTORS IN SERIES
2.3 串联电阻

When dealing with a circuit containing large no of components, it’s a smart thing to simplify the circuit.
在处理包含大量元件的电路时,简化电路是一个明智的选择。

This applies to resistors as well.
这同样适用于电阻。

A combination of resistors, be it series or parallel or otherwise can be replaced by a single resistance, called the equivalent or the effective resistance of the circuit.
一组电阻,无论是串联、并联还是其他方式,都可以用一个单一的电阻来替代,称为电路的等效电阻或有效电阻。

For a series combination of resistors, the equivalent resistance is found by simply adding the individual resistance values.
对于电阻的串联组合,等效电阻通过简单地将各个电阻值相加来找到。

Mathematically,
数学上,
R eq = R 1 + R 2 + . . + R N = i = 1 N R i R eq = R 1 + R 2 + . . + R N = i = 1 N R i {:[R_(eq)=R_(1)+R_(2)+dots..+R_(N)],[=sum_(i=1)^(N)R_(i)]:}\begin{aligned} \mathrm{R}_{\mathrm{eq}} & =\mathrm{R}_{1}+\mathrm{R}_{2}+\ldots . .+\mathrm{R}_{\mathrm{N}} \\ & =\sum_{i=1}^{N} R_{i} \end{aligned}
The proof for this is pretty straight forward.
这个证明相当简单。

Consider our example (first one) from section 2.1. Let V 1 & V 2 V 1 & V 2 V_(1)&V_(2)\mathrm{V}_{1} \& \mathrm{~V}_{2} are the voltages across the resistors R 1 R 1 R_(1)R_{1} and R 2 R 2 R_(2)R_{2} respectively. Using KVL , we know V = V 1 + V 2 V = V 1 + V 2 V=V_(1)+V_(2)\mathbf{V}=\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}}. Therefore,
考虑我们在 2.1 节中的例子(第一个)。设 V 1 & V 2 V 1 & V 2 V_(1)&V_(2)\mathrm{V}_{1} \& \mathrm{~V}_{2} 是电阻 R 1 R 1 R_(1)R_{1} R 2 R 2 R_(2)R_{2} 上的电压。根据基尔霍夫电压定律,我们知道 V = V 1 + V 2 V = V 1 + V 2 V=V_(1)+V_(2)\mathbf{V}=\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}} 。因此,
R eq = V I = V 1 + V 2 I = V 1 I + V 2 I R eq = R 1 + R 2 R eq = V I = V 1 + V 2 I = V 1 I + V 2 I R eq = R 1 + R 2 {:[R_(eq)=(V)/(I)=(V_(1)+V_(2))/(I)=(V_(1))/(I)+(V_(2))/(I)],[R_(eq)=R_(1)+R_(2)]:}\begin{gathered} \mathrm{R}_{\mathrm{eq}}=\frac{V}{I}=\frac{V_{1}+V_{2}}{I}=\frac{V_{1}}{I}+\frac{V_{2}}{I} \\ \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2} \end{gathered}

2.4 VOLTAGE DIVIDER RULE
2.4 电压分压规则

In the last section, we saw that in a series connection, the resistors share a common current, but have different voltage drops across them.
在最后一节中,我们看到在串联连接中,电阻器共享一个共同的电流,但它们之间的电压降却不同。

Now we will try to find out the exact magnitude of the voltage drops.
现在我们将尝试找出电压降的确切大小。

For that we use the Voltage Divider Rule.
为此,我们使用电压分压规则。

From Ohm’s law,
根据欧姆定律,
I = E R 1 + R 2 + R 3 = E R T I = E R 1 + R 2 + R 3 = E R T I=(E)/(R_(1)+R_(2)+R_(3))=(E)/(R_(T))I=\frac{E}{R_{1}+R_{2}+R_{3}}=\frac{E}{R_{T}}
Then the Voltage drops across the resistors are:
然后电阻上的电压降为:
V 1 = I R 1 , V 2 = I R 2 , V 3 = I R 3 V 1 = E R 1 R T , V 2 = E R 2 R T , V 3 = E R 3 R T V 1 = I R 1 , V 2 = I R 2 , V 3 = I R 3 V 1 = E R 1 R T , V 2 = E R 2 R T , V 3 = E R 3 R T {:[V_(1)=IR_(1)","V_(2)=IR_(2)","V_(3)=IR_(3)],[V_(1)=(ER_(1))/(R_(T))","V_(2)=(ER_(2))/(R_(T))","V_(3)=(ER_(3))/(R_(T))]:}\begin{aligned} & \mathrm{V}_{1}=I \mathrm{R}_{1}, \mathrm{~V}_{2}=I \mathrm{R}_{2}, \mathrm{~V}_{3}=I \mathrm{R}_{3} \\ & \mathrm{~V}_{1}=\frac{E R_{1}}{R_{T}}, \mathrm{~V}_{2}=\frac{E R_{2}}{R_{T}}, \mathrm{~V}_{3}=\frac{E R_{3}}{R_{T}} \end{aligned}
To sum up, the Voltage drop across a Resistor in series connection is given,
总之,串联电阻上的电压降是给定的,
V R = ( Voltage across combination ) × ( Resistance R ) Total Resistance V R = (  Voltage across combination  ) × (  Resistance  R )  Total Resistance  V_(R)=((" Voltage across combination ")xx(" Resistance "R))/(" Total Resistance ")\mathrm{V}_{\mathrm{R}}=\frac{(\text { Voltage across combination }) \times(\text { Resistance } \mathrm{R})}{\text { Total Resistance }}

2.5 PARALLEL CIRCUIT
2.5 并联电路

A parallel circuit is a circuit in which any number of components are connected across 2 common terminals, such that they share a common voltage.
并联电路是指任何数量的元件连接在两个公共端子之间的电路,使它们共享一个共同的电压。

For example, in the circuit shown in the figure below, the Resistors R 1 R 1 R_(1)R_{1} and R 2 R 2 R_(2)R_{2} are in parallel, because they are connected between the same terminals a and b .
例如,在下图所示的电路中,电阻器 R 1 R 1 R_(1)R_{1} R 2 R 2 R_(2)R_{2} 是并联的,因为它们连接在相同的端子 a 和 b 之间。

The current will be divided amongst the resistors, according as their resistance values.
电流将根据电阻值在电阻器之间分配。

2.6 KIRCHHOFF'S CURRENT LAW (KCL)
2.6 基尔霍夫电流定律 (KCL)

According to the Kirchhoff’s Current Law, the algebraic sum of the currents entering and leaving a node or a junction of a circuit is zero.
根据基尔霍夫电流定律,进入和离开电路节点或连接点的电流代数和为零。

It’s easily evident that this law is derived from the Law of conservation of charge.
显然,这条法律源于电荷守恒定律。

The idea is really simple, once a current is generated in a circuit, it is distributed throughout the circuit.
这个想法非常简单,一旦在电路中产生电流,它就会在整个电路中分布。

It cannot just accumulate in a wire or vanish into thin air.
它不能仅仅积累在一根电线上或消失在空气中。
Symbolically,
象征性地,
I entering = I leaving I entering  = I leaving  sumI_("entering ")=sumI_("leaving ")\sum I_{\text {entering }}=\sum I_{\text {leaving }}
Consider the example shown below and let’s formulate the KCL equation for node a.
考虑下面的示例,让我们为节点 a 公式化 KCL 方程。

At node a, there are 3 currents, one entering and 2 leaving. Hence the KCL equation is, I = I 1 I = I 1 I=I_(1)\mathrm{I}=\mathrm{I}_{1} + I 2 + I 2 +I_(2)+\mathrm{I}_{2}.
在节点 a,有 3 个电流,一个进入,两个离开。因此,KCL 方程是, I = I 1 I = I 1 I=I_(1)\mathrm{I}=\mathrm{I}_{1} + I 2 + I 2 +I_(2)+\mathrm{I}_{2}

2.7 RESISTORS IN PARALLEL
2.7 并联电阻

For a parallel combination of resistors, the reciprocal of the equivalent resistance is the sum of the reciprocals of the
对于电阻的并联组合,等效电阻的倒数是各个电阻倒数的总和

individual resistances. Mathematically,
个体电阻。数学上,
1 R eq = 1 R 1 + 1 R 2 + . + 1 R N = i = 1 N 1 R i 1 R eq = 1 R 1 + 1 R 2 + . + 1 R N = i = 1 N 1 R i {:[(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))+dots.+(1)/(R_(N))],[=sum_(i=1)^(N)(1)/(R_(i))]:}\begin{aligned} \frac{1}{\mathrm{R}_{\mathrm{eq}}} & =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\ldots .+\frac{1}{\mathrm{R}_{N}} \\ & =\sum_{i=1}^{N} \frac{1}{\mathrm{R}_{i}} \end{aligned}
Consider our example from section 2.1. Let I 1 & I 2 I 1 & I 2 I_(1)&I_(2)\mathrm{I}_{1} \& \mathrm{I}_{2} be the currents flowing through the resistors R1 and R2 respectively. Using KCL, we know I = I 1 + I 2 I = I 1 + I 2 I=I_(1)+I_(2)\mathbf{I}=\mathbf{I}_{\mathbf{1}}+\mathbf{I}_{\mathbf{2}}.
考虑我们在第 2.1 节中的例子。设 I 1 & I 2 I 1 & I 2 I_(1)&I_(2)\mathrm{I}_{1} \& \mathrm{I}_{2} 为流过电阻 R1 和 R2 的电流。根据基尔霍夫电流定律,我们知道 I = I 1 + I 2 I = I 1 + I 2 I=I_(1)+I_(2)\mathbf{I}=\mathbf{I}_{\mathbf{1}}+\mathbf{I}_{\mathbf{2}}
1 R eq = I V = I 1 + I 2 V = I 1 V + I 2 V 1 R eq = 1 R 1 + 1 R 2 1 R eq = I V = I 1 + I 2 V = I 1 V + I 2 V 1 R eq = 1 R 1 + 1 R 2 {:[(1)/(R_(eq))=(I)/(V)=(I_(1)+I_(2))/(V)=(I_(1))/(V)+(I_(2))/(V)],[(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))]:}\begin{gathered} \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{I}{V}=\frac{I_{1}+I_{2}}{V}=\frac{I_{1}}{V}+\frac{I_{2}}{V} \\ \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \end{gathered}

2.8 CURRENT DIVIDER RULE
2.8 电流分配规则

The Current Divider Rule is used to determine the magnitude of current entering each branch of a parallel connection.
电流分配规则用于确定进入并联连接每个支路的电流大小。

From Ohm’s law,
根据欧姆定律,
I = E R T I = E R T I=(E)/(R_(T))I=\frac{E}{R_{T}}
Then the Currents flowing through resistors are:
然后通过电阻的电流是:
I 1 = E R 1 , I 2 = E R 2 , I 1 = E R 3 I 1 = IR T R 1 , I 2 = IR T R 2 , I 1 = I IR T R 3 I 1 = E R 1 , I 2 = E R 2 , I 1 = E R 3 I 1 = IR T R 1 , I 2 = IR T R 2 , I 1 = I IR T R 3 {:[I_(1)=(E)/(R_(1))","I_(2)=(E)/(R_(2))","I_(1)=(E)/(R_(3))],[:.I_(1)=(IR_(T))/(R_(1))","I_(2)=(IR_(T))/(R_(2))","I_(1)=(IIR_(T))/(R_(3))]:}\begin{aligned} \mathrm{I}_{1} & =\frac{\mathrm{E}}{\mathrm{R}_{1}}, \mathrm{I}_{2}=\frac{\mathrm{E}}{\mathrm{R}_{2}}, \mathrm{I}_{1}=\frac{\mathrm{E}}{\mathrm{R}_{3}} \\ \therefore \mathrm{I}_{1} & =\frac{\mathrm{IR}_{\mathrm{T}}}{\mathrm{R}_{1}}, \mathrm{I}_{2}=\frac{\mathrm{IR}_{\mathrm{T}}}{\mathrm{R}_{2}}, \mathrm{I}_{1}=\frac{I \mathrm{IR}_{\mathrm{T}}}{\mathrm{R}_{3}} \end{aligned}
To sum up, the Current flowing through a Resistor in parallel connection is given by,
总之,流过并联连接中的电阻器的电流由以下公式给出,
I R = ( Total Current ) x ( Total Resistance ) Resistance R I R = (  Total Current  ) x (  Total Resistance  )  Resistance  R I_(R)=((" Total Current ")x(" Total Resistance "))/(" Resistance "R)\mathrm{I}_{\mathrm{R}}=\frac{(\text { Total Current }) \mathrm{x}(\text { Total Resistance })}{\text { Resistance } \mathrm{R}}

2.9 OPEN & SHORT CIRCUIT
2.9 开路与短路

Short Circuit is a condition where two points in a circuit are directly connected to each other through a path of zero resistance.
短路是指电路中的两个点通过零电阻的路径直接连接在一起的状态。

The voltage across the 2 points will be always zero in case of a short circuit.
在短路的情况下,两个点之间的电压将始终为零。
Short Circuit
短路

Open Circuit
开路

Open Circuit is exactly the opposite condition as short circuit.
开路正好与短路相反。

In case of an open circuit, there is no connection between two points in a circuit and hence no current flows between the 2 points.
在开路的情况下,电路中的两个点之间没有连接,因此在这两个点之间没有电流流动。

3. BASIC ANALYSIS TECHNIQUES
3. 基本分析技术

3.1 ENERGY SOURCES
3.1 能源来源

There are basically 2 types of energy sources: Voltage source & Current source.
基本上有两种类型的能源来源:电压源和电流源。

Again they can be classified as ideal & practical sources.
它们可以再次被分类为理想和实用的来源。

First we’ll discuss ideal sources then consider practical sources.
首先我们将讨论理想的来源,然后考虑实际的来源。

3.1.1 Voltage Source
3.1.1 电压源

An ideal Voltage source is an Energy source which gives constant Voltage across its terminals irrespective of the current drawn by the load connected to its terminals.
理想电压源是一个能量源,它在其端子之间提供恒定电压,而不管连接到其端子的负载所抽取的电流。

At any instant of time, the voltage across the terminals remain the same.
在任何时刻,端子之间的电压保持不变。

Thus the V-I Characteristics of an ideal voltage source is a straight line as shown.
因此,理想电压源的电压-电流特性是一条直线,如图所示。


But it is not possible to make such Voltage sources in practice.
但在实践中不可能制造这样的电压源。

Practically, all Voltage sources have a small internal resistance.
实际上,所有电压源都有一个小的内阻。

For analysis purposes, we assume that this internal resistance is in series with the voltage source and is represented by R se R se R_(se)\mathrm{R}_{\mathrm{se}}. Because of R s e R s e R_(se)R_{s e}, the voltage across the terminals decreases slightly with the increase in the current.
为了分析的目的,我们假设这个内部电阻与电压源串联,并用 R se R se R_(se)\mathrm{R}_{\mathrm{se}} 表示。由于 R s e R s e R_(se)R_{s e} ,端子之间的电压随着电流的增加略微下降。


V-I Characteristics
V-I 特性
V L = V s I L R s e V L = V s I L R s e V_(L)=V_(s)-I_(L)R_(se)V_{L}=V_{s}-I_{L} R_{s e}
Usually, Voltage sources are manufactured keeping the internal resistance to the minimum, such that it acts more or less like an ideal voltage source (till a max load current limit).
通常,电压源的制造是将内部电阻保持在最低限度,使其在最大负载电流限制内表现得或多或少像一个理想电压源。

Batteries are an example of Voltage source.
电池是电压源的一个例子。

3.1.2 Current Source
3.1.2 电流源

No prizes for guessing what a current source is, an ideal current source is a power source that gives constant current, irrespective of the voltage appearing across its terminals
没有奖品可以猜测什么是电流源,理想电流源是一种提供恒定电流的电源,无论其端子之间出现的电压如何


V-I Characteristics
V-I 特性

But a practical Current source hardly ever functions this way.
但实际的电流源几乎从不以这种方式工作。

In a practical Current source, the current decreases slightly as the Voltage across the load terminals increase.
在实际的电流源中,随着负载端子上的电压增加,电流略微减少。

This behavior can be analyzed by considering a high internal resistance, represented by R sh R sh R_(sh)\mathrm{R}_{\mathrm{sh}} in parallel with the source.
这种行为可以通过考虑一个高内阻来分析,表示为 R sh R sh R_(sh)\mathrm{R}_{\mathrm{sh}} 与源并联。


V-I Characteristics
V-I 特性

I L = I S V L R s h I L = I S V L R s h I_(L)=I_(S)-(V_(L))/(R_(sh))I_{L}=I_{S}-\frac{V_{L}}{R_{s h}}

3.2 COMBINATION OF SOURCES
3.2 来源的组合

In many circuits, it is necessary to use multiple energy sources.
在许多电路中,有必要使用多个能源。

Analyzing such circuits directly is a bit of a mess.
直接分析这样的电路有点麻烦。

So what we usually do is to reduce the multiple sources to a single equivalent source, making the analysis a lot easier.
所以我们通常做的是将多个来源减少为一个等效来源,这样分析就容易多了。

Like the resistors and other circuit components, power sources too can have series or parallel combinations.
像电阻器和其他电路元件一样,电源也可以有串联或并联组合。

3.2.1 Combination of Voltage sources
3.2.1 电压源的组合

If two Voltage sources are in series i.e. they are connected back to back, the effective voltage is simply their algebraic sum.
如果两个电压源串联,即它们背对背连接,则有效电压就是它们的代数和。

It is important to consider their polarities while doing so.
在这样做时考虑它们的极性是很重要的。

If their polarities are the same, then the effective voltage is their sum and if their polarities are opposing, then the effective voltage is the difference of the 2 voltages.
如果它们的极性相同,则有效电压是它们的总和;如果它们的极性相反,则有效电压是两个电压的差。

Same
相同

Polarity
极性

Unlike a series connection, any two Voltage sources can’t be combined in parallel.
与串联连接不同,任何两个电压源不能并联。

Practically, only Voltage sources of the same magnitude are combined in parallel.
实际上,只有相同大小的电压源才能并联。

If 2 unequal Voltage sources are connected in parallel, there will be a circulating current between them.
如果两个不等的电压源并联连接,它们之间会有一个循环电流。

Essentially what happens is that, the smaller voltage source is acting as a load for the larger voltage source.
本质上发生的情况是,较小的电压源作为较大电压源的负载。

The magnitude of the current will depend on the value of the internal resistances of the 2 sources.
电流的大小将取决于两个电源的内部电阻值。

Since the internal resistance is usually very small, a very large current flows, leading to overheating and possibly irreparable damage.
由于内部电阻通常非常小,因此会产生非常大的电流,导致过热并可能造成无法修复的损坏。

Don’t even think about connecting 2 ideal voltage sources in parallel, results could be
不要想将两个理想电压源并联,结果可能是

catastrophic.
灾难性的。

And If you somehow manage to connect two voltage sources in parallel without damaging anything, the voltage across the combination will be somewhere between the 2 values depending on the internal resistances.
如果你以某种方式成功将两个电压源并联而不损坏任何东西,那么组合的电压将介于这两个值之间,具体取决于内部电阻。

If 2 equal voltage sources are connected in parallel, the single equivalent source will have the same voltage as the 2 sources.
如果两个相同电压的电源并联连接,则单个等效电源的电压将与这两个电源相同。

The only reason to do this would be if the load requires a higher current than the source can supply by itself.
这样做的唯一原因是负载需要的电流超过了电源自身能够提供的电流。

Other than that, no good can come from connecting 2 voltage sources in parallel.
除此之外,将两个电压源并联连接不会带来任何好处。

3.2.1 Combination of Current
3.2.1 组合电流

sources
来源

Connecting two Current sources in series is a bit like connecting two Voltage sources in parallel.
将两个电流源串联有点像将两个电压源并联。

It’s simply not a good idea.
这根本不是个好主意。

There are very few cases where such connection is required in practice, but that’s a rarity.
在实践中,这种连接所需的情况非常少,但这很少见。

In any case only 2 current sources of same magnitude are connected in series.
在任何情况下,只有两个相同大小的电流源串联连接。

The magnitude of single equivalent source will supply the same current as the individual sources.
单一等效源的大小将提供与各个源相同的电流。

Connecting 2 different Current sources in series is a violation of the Kirchhoff’s current law.
将两个不同的电流源串联连接违反了基尔霍夫电流定律。

Again, you don’t want to be messing with Kirchhoff!!
再说一次,你可不想和基尔霍夫打交道!!

The problem with connecting 2 unequal current sources in series is that, you are asking the small current source to supply more than hat it is capable of.
将两个不等的电流源串联连接的问题在于,你要求小电流源提供超过其能力的电流。

Intuitively, this means one source is trying to push more charge than the other source is capable of accepting.
直观上,这意味着一个电源试图推送的电荷超过了另一个电源能够接受的量。

If two current sources are connected in parallel, the effective current output of the combination is their algebraic sum.
如果两个电流源并联连接,则组合的有效电流输出是它们的代数和。

If the sources are in opposite direction, then the single equivalent source will produce current in the direction of the larger current source.
如果源头方向相反,则单一等效源将产生朝向较大电流源的电流。

3.3 SOURCE TRANSFORMATION
3.3 源转换

In some circuits, you will encounter the presence of both current and voltage sources.
在某些电路中,您会遇到电流源和电压源的存在。

This makes things a little trickier.
这让事情变得有点棘手。

Lucky for us, it is possible to convert one type of source to other type and it’s pretty straightforward.
幸运的是,我们可以将一种类型的源转换为另一种类型,而且这非常简单。

Consider a voltage source having an internal resistance R se R se  R_("se ")R_{\text {se }} connected to a load resistor R L R L R_(L)\mathrm{R}_{\mathrm{L}}. Now consider a current source having an internal resistance R sh R sh R_(sh)\mathrm{R}_{\mathrm{sh}} supplying the same load.
考虑一个内部电阻为 R se R se  R_("se ")R_{\text {se }} 的电压源连接到负载电阻 R L R L R_(L)\mathrm{R}_{\mathrm{L}} 。现在考虑一个内部电阻为 R sh R sh R_(sh)\mathrm{R}_{\mathrm{sh}} 的电流源为相同的负载供电。

If the two supplies were to be equivalent, then the load current (or voltage) should be the same in both cases.
如果这两个电源是等效的,那么在两种情况下,负载电流(或电压)应该是相同的。

The current delivered by the voltage source is given by,
由电压源提供的电流为,

And the current delivered by the current source (applying current division rule) is given by,
由电流源提供的电流(应用电流分配规则)为:
I = I 1 × R s h R s h + R L I = I 1 × R s h R s h + R L I=I_(1)xx(R_(sh))/(R_(sh)+R_(L))\mathrm{I}=\mathrm{I}_{1} \times \frac{R_{s h}}{R_{s h}+R_{L}}
Equating both equations,
将两个方程等同,
V 1 R s e + R L = I 1 × R s h R s h + R L V 1 R s e + R L = I 1 × R s h R s h + R L (V_(1))/(R_(se)+R_(L))=I_(1)xx(R_(sh))/(R_(sh)+R_(L))\frac{V_{1}}{R_{s e}+R_{L}}=I_{1} \times \frac{R_{s h}}{R_{s h}+R_{L}}
Now if we equate the numerators and denominators separately, we get,
现在如果我们分别将分子和分母相等,我们得到,

R se = R sh R se  = R sh  R_("se ")=R_("sh ")R_{\text {se }}=R_{\text {sh }}
&
V 1 = I 1 R sh V 1 = I 1 R sh  V_(1)=I_(1)R_("sh ")V_{1}=I_{1} R_{\text {sh }}

Once the sources are transformed into same kind, they can be easily combined in series or parallel, as we did in the previous section.
一旦源被转化为相同类型,它们就可以像我们在前一节中所做的那样,轻松地串联或并联。

3.4 MESH ANALYSIS
3.4 网格分析

Using Circuit analysis techniques, we are essentially trying to find the voltage across or current through a component in a circuit.
使用电路分析技术,我们基本上是在试图找到电路中一个元件的电压或电流。

Two of the most popular and basic analysis techniques are the Node and the Mesh analysis.
两种最受欢迎和基本的分析技术是节点分析和网格分析。

These techniques were developed as an extension to the KVL and KCL.
这些技术是作为对基尔霍夫电压定律和基尔霍夫电流定律的扩展而开发的。

We’ll learn about Mesh analysis in this section & Node analysis in the next.
我们将在本节学习网格分析,在下一节学习节点分析。
In mesh analysis, we are dividing the circuit into areas or loops called Meshes and assigning them a Mesh current.
在网格分析中,我们将电路划分为称为网格的区域或回路,并为它们分配一个网格电流。

Consider the circuit below, just from observation, we can identify 3 loops or meshes.
考虑下面的电路,仅从观察中,我们可以识别出 3 个回路或网格。

Do note that, these loops have
请注意,这些循环有

some common components.
一些常见组件。

Now assume a loop current to flow in each of these loops and give them a random direction (although normally we assume clockwise direction as in the figure).
现在假设每个环路中都有一个环流,并给它们一个随机方向(尽管通常我们假设如图所示的顺时针方向)。

At first glance, this may seem like extra work, but it’s worth it, because reduces the no.
乍一看,这可能看起来像是额外的工作,但这是值得的,因为它减少了数量。

of equations significantly, making calculation very easy. Now let’s try out an example.
方程式显著地简化了计算,使得计算变得非常简单。现在让我们试一个例子。

Consider the circuit below, it has 2 voltage sources and a bunch of resistors.
考虑下面的电路,它有两个电压源和一堆电阻。

Simply through observation, we can identify 3 meshes. Let’s assume currents I A , I B , I C I A , I B , I C I_(A),I_(B),I_(C)I_{A}, I_{B}, I_{C} flow through the 3 meshes respectively.
通过观察,我们可以识别出 3 个网格。假设电流 I A , I B , I C I A , I B , I C I_(A),I_(B),I_(C)I_{A}, I_{B}, I_{C} 分别流过这 3 个网格。

Now let’s consider each mesh separately and form equations using KVL. Do note that the 5 Ω 5 Ω 5Omega5 \Omega resistor is common to both meshes A A AA and B B BB, so the current through it is the difference of the two mesh currents (because the currents are in opposite direction w.r.t 5 Ω 5 Ω 5Omega5 \Omega resistor.)
现在让我们分别考虑每个网格,并使用基尔霍夫电压定律形成方程。请注意, 5 Ω 5 Ω 5Omega5 \Omega 电阻器是两个网格 A A AA B B BB 的共同部分,因此通过它的电流是两个网格电流的差(因为电流相对于 5 Ω 5 Ω 5Omega5 \Omega 电阻器的方向相反)。

I A + 5 ( I A I B ) = 10 6 I A 5 I B = 10 I A + 5 I A I B = 10 6 I A 5 I B = 10 {:[I_(A)+5(I_(A)-I_(B))=10],[=>6I_(A)-5I_(B)=10]:}\begin{aligned} & I_{A}+5\left(I_{A}-I_{B}\right)=10 \\ & \Rightarrow 6 I_{A}-5 I_{B}=10 \end{aligned}
Similarly we form equation for the other too meshes.
同样,我们为其他两个网格形成方程。

5 I B + 2 ( I B I C ) + 5 ( I B I A ) = 0 12 I B 5 I A 2 I C = 0 5 I B + 2 I B I C + 5 I B I A = 0 12 I B 5 I A 2 I C = 0 {:[5I_(B)+2(I_(B)-I_(C))+5(I_(B)-I_(A))=0],[=>12I_(B)-5I_(A)-2I_(C)=0]:}\begin{aligned} & 5 I_{B}+2\left(I_{B}-I_{C}\right)+5\left(I_{B}-I_{A}\right)=0 \\ & \Rightarrow 12 I_{B}-5 I_{A}-2 I_{C}=0 \end{aligned}
2 I C + 2 ( I C I B ) = 5 4 I C 2 I B = 5 2 I C + 2 I C I B = 5 4 I C 2 I B = 5 {:[2I_(C)+2(I_(C)-I_(B))=-5],[=>4I_(C)-2I_(B)=-5]:}\begin{aligned} & 2 I_{C}+2\left(I_{C}-I_{B}\right)=-5 \\ & \Rightarrow 4 I_{C}-2 I_{B}=-5 \end{aligned}
Now we have 3 unknown variables I A , I B , I C I A , I B , I C I_(A),I_(B),I_(C)I_{A}, I_{B}, I_{C} and 3 equations.
现在我们有 3 个未知变量 I A , I B , I C I A , I B , I C I_(A),I_(B),I_(C)I_{A}, I_{B}, I_{C} 和 3 个方程。

This can be easily solved using the Cramer’s rule or by substitution.
这可以通过克拉默法则或代入法轻松解决。

3.5 SUPER MESH
3.5 超级网格

Mesh analysis is all well and good, but what if a current source is present in the circuit??
网格分析很好,但如果电路中存在电流源怎么办?

We could assign an unknown voltage across the current source, apply KVL around each mesh as before, and then relate the source current to the assigned mesh currents.
我们可以在电流源上施加一个未知电压,像之前一样对每个网格应用基尔霍夫电压定律,然后将源电流与分配的网格电流相关联。

This is generally the more difficult approach.
这通常是更困难的方法。

The easier method is to create something called the Super Mesh. Super Mesh is basically a
更简单的方法是创建一种叫做超级网格的东西。超级网格基本上是一个

mesh formed by combining 2 adjacent meshes, ignoring the branch which contains the current source.
通过组合两个相邻的网格形成的网格,忽略包含电流源的分支。

For example, in the circuit below, we create a Super Mesh by combining meshes A A AA and B B BB.
例如,在下面的电路中,我们通过结合网格 A A AA B B BB 来创建一个超级网格。

The Super Mesh equation can be obtained by applying KVL to the super mesh, ignoring the common branch (that contains the current source).
超级网方程可以通过对超级网应用基尔霍夫电压定律(KVL)来获得,忽略公共支路(包含电流源)。

I A + 5 I B + 2 I B = 10 I A + 7 I B = 10 I A + 5 I B + 2 I B = 10 I A + 7 I B = 10 {:[I_(A)+5I_(B)+2I_(B)=10],[=>I_(A)+7I_(B)=10]:}\begin{aligned} & I_{A}+5 I_{B}+2 I_{B}=10 \\ & \Rightarrow I_{A}+7 I_{B}=10 \end{aligned}
The second equation relating the 2 mesh currents can be obtained by applying KCL to the common branch.
第二个与两个网格电流相关的方程可以通过对公共支路应用基尔霍夫电流定律(KCL)来获得。

In our example, it is,
在我们的例子中,它是,
I B I A = 2 I B I A = 2 I_(B)-I_(A)=2I_{B}-I_{A}=2

3.6 NODAL ANALYSIS
3.6 节点分析

Much like the Mesh analysis, Nodal analysis is another commonly used circuit analysis technique.
与网分析类似,节点分析是另一种常用的电路分析技术。

The Nodal analysis is based on KCL, whereas Mesh analysis is based on KVL.
节点分析基于基尔霍夫电流定律,而网格分析基于基尔霍夫电压定律。

Before we go any further, we need to define a node.
在我们进一步讨论之前,我们需要定义一个节点。

A Node is simply a point where two or more circuit elements meet.
节点只是两个或多个电路元件相遇的点。

Let’s try using Nodal analysis in practice.
让我们在实践中尝试使用节点分析。

We’ll use the same circuit we used in Mesh analysis example to get a better understanding between the similarities and differences between the two techniques.
我们将使用在网格分析示例中使用的相同电路,以更好地理解这两种技术之间的相似性和差异。

The first task in Nodal analysis is to identify the nodes in the circuit.
在节点分析中的第一个任务是识别电路中的节点。

Do note that, in Nodal analysis, we are only interested in nodes where 3 or more components meet.
请注意,在节点分析中,我们只对三个或更多组件相交的节点感兴趣。

If we were to consider all the nodes, the method will still work, but the number of steps will increase.
如果我们考虑所有节点,这个方法仍然有效,但步骤的数量会增加。

In our example, we can identify 3 such nodes.
在我们的例子中,我们可以识别出 3 个这样的节点。

The next step is to assume one of those nodes as a reference node (usually the bottom one is chosen).
下一步是将其中一个节点假设为参考节点(通常选择底部的那个)。

The idea is assume zero voltage/potential at a point (Reference Node) in the circuit, so that we can
假设电路中某一点(参考节点)的电压/电位为零,这样我们就可以

measure/calculate voltage at different points with respect to this reference point.
在这个参考点相对于不同点测量/计算电压。

Once the Reference node is fixed, assume voltages at the other nodes ( V 1 , V 2 , V 3 V 1 , V 2 , V 3 V_(1),V_(2),V_(3)\mathrm{V}_{1}, \mathrm{~V}_{2}, \mathrm{~V}_{3} etc.). Once these things are taken care of, it’s time to look at the nodes separately and form node equations.
一旦参考节点固定,假设其他节点的电压( V 1 , V 2 , V 3 V 1 , V 2 , V 3 V_(1),V_(2),V_(3)\mathrm{V}_{1}, \mathrm{~V}_{2}, \mathrm{~V}_{3} 等)。一旦这些问题解决,就可以单独查看节点并形成节点方程。

Applying KCL at Node 1,
在节点 1 应用基尔霍夫电流定律,
V 1 10 1 + V 1 5 + V 1 V 2 5 = 0 6 V 1 V 2 50 = 0 V 1 10 1 + V 1 5 + V 1 V 2 5 = 0 6 V 1 V 2 50 = 0 {:[(V_(1)-10)/(1)+(V_(1))/(5)+(V_(1)-V_(2))/(5)=0],[=>6V_(1)-V_(2)-50=0]:}\begin{gathered} \frac{V_{1}-10}{1}+\frac{V_{1}}{5}+\frac{V_{1}-V_{2}}{5}=0 \\ \Rightarrow 6 V_{1}-V_{2}-50=0 \end{gathered}
Similarly applying KCL at Node 2,
同样在节点 2 应用基尔霍夫电流定律,

Solving these equations, we can obtain the node voltages and the rest of the parameters.
通过求解这些方程,我们可以获得节点电压和其他参数。

3.7 SUPER NODE
3.7 超级节点

In some circuits, a voltage source maybe present between 2 nodes.
在某些电路中,两个节点之间可能存在一个电压源。

To deal with such circuits it’s best to use the Super Node analysis.
处理此类电路最好使用超级节点分析。

The first step is the same, to identify nodes and assign nodal voltages.
第一步是相同的,识别节点并分配节点电压。

Once that is done, we need to create something called the super node, by combining the 2 nodes ignoring the voltage source in between them.
完成后,我们需要创建一个称为超级节点的东西,通过组合这两个节点,忽略它们之间的电压源。

Then to obtain the super node equation, KCL is applied to both the
然后为了获得超级节点方程,对两个节点应用基尔霍夫电流定律

nodes at the same time.
同时的节点。

The current through the common branch can be ignored, because the current exiting node 1 and the current entering node 2 are the same and hence they cancel out when taking the combined KCL equation.
可以忽略公共支路的电流,因为流出节点 1 的电流和流入节点 2 的电流是相同的,因此在进行组合的基尔霍夫电流定律方程时它们会相互抵消。

The second equation connecting the 2 nodes can be obtained by equating the difference between the 2 node voltages to the voltage of the source i.e. V 1 V 2 = V x V 1 V 2 = V x V_(1)-V_(2)=V_(x)\mathrm{V}_{1}-\mathrm{V}_{2}=\mathrm{V}_{\mathrm{x}}. All the other nodes can be treated as before and corresponding node equations can be found.
连接这两个节点的第二个方程可以通过将两个节点电压之间的差等于源电压即 V 1 V 2 = V x V 1 V 2 = V x V_(1)-V_(2)=V_(x)\mathrm{V}_{1}-\mathrm{V}_{2}=\mathrm{V}_{\mathrm{x}} 来获得。所有其他节点可以像之前一样处理,并可以找到相应的节点方程。

4. NETWORK THEOREMS
4. 网络定理

While the circuit analysis techniques discussed so far, are very handy for simple circuits.
虽然到目前为止讨论的电路分析技术对于简单电路非常方便。

They aren’t are the preferred choice for more complex circuits.
它们不是更复杂电路的首选。

For that we need the help of some theorems. The idea is to use one or more of these theorems to convert the complex circuit into a simple equivalent, which can be easily analyzed using our familiar basic analysis techniques.
为此,我们需要一些定理的帮助。这个想法是使用一个或多个这些定理将复杂电路转换为简单的等效电路,这样就可以使用我们熟悉的基本分析技术轻松分析。

Let’s look at these Theorems one by one in detail.
让我们逐一详细查看这些定理。

4.1 SUPERPOSITION THEOREM
4.1 叠加定理

Analysis of circuits having multiple energy sources is not the easiest of tasks, but Superposition theorem provides an easy solution to this.
对具有多个能源源的电路进行分析并不是一项简单的任务,但叠加定理为此提供了简单的解决方案。

According to the Superposition theorem, the effect or response in a component when 2 or more energy sources (voltage or current sources) are applied together is equal to the sum of effect/responses when the sources are applied individually.
根据叠加定理,当同时施加两个或更多能量源(电压源或电流源)时,组件中的效果或响应等于单独施加这些源时效果/响应的总和。

This may seem complicated, but that’s just the statement, the application is very easy.
这可能看起来很复杂,但这只是声明,应用程序非常简单。
What the Superposition theorem really does, is to convert a circuit with n energy sources into n circuits with a single energy source acting individually, so that they can be analyzed individually and the results can be added up.
叠加定理真正的作用是将一个具有 n 个能量源的电路转换为 n 个仅有一个能量源单独作用的电路,以便可以单独分析它们并将结果相加。

To study the effects of one particular energy source on the circuit, the other sources need to be eliminated.
为了研究某一特定能源对电路的影响,需要消除其他能源。

This can be done by Short Circuiting the Voltage sources and Open
这可以通过短路电压源和打开来完成

circuiting the Current sources, which are not under consideration.
电流源的电路,这些电流源不在考虑之内。

Now let’s try and use the Superposition theorem in practice with the help of an example.
现在让我们尝试在实践中使用叠加定理,借助一个例子。

In the circuit shown below there are 2 energy sources, one current and one voltage source and suppose we need to find the voltage across resistance R 2 R 2 R_(2)\mathrm{R}_{2}.
在下面所示的电路中,有两个能量源,一个电流源和一个电压源,假设我们需要找到电阻 R 2 R 2 R_(2)\mathrm{R}_{2} 上的电压。

  1. First thing to do is to split up the circuit into 2 circuits with a single energy source, as shown above.
    第一件事是将电路分成两个只有一个能源源的电路,如上所示。
  2. In the first circuit, as the current source is open circuited, the branch containing resistance R 3 R 3 R_(3)R_{3} is no longer relevant. By using the voltage divider rule, the voltage across Resistor R 2 R 2 R_(2)\mathrm{R}_{2} can be found as,
    在第一个电路中,由于电流源开路,包含电阻 R 3 R 3 R_(3)R_{3} 的支路不再相关。通过使用电压分压规则,可以找到电阻 R 2 R 2 R_(2)\mathrm{R}_{2} 上的电压,如下所示,
V R 2 = V R 1 + R 2 R 2 V R 2 = V R 1 + R 2 R 2 V_(R2)^(')=(V)/(R_(1)+R_(2))R_(2)\mathrm{V}_{\mathrm{R} 2}^{\prime}=\frac{\mathrm{V}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \mathrm{R}_{2}
  1. In the second circuit, the voltage across R 2 R 2 R_(2)R_{2} can be determined with the help of the current division rule.
    在第二个电路中,可以借助电流分配规则确定 R 2 R 2 R_(2)R_{2} 上的电压。
V R 2 = R 1 R 2 R 1 + R 2 V R 2 = R 1 R 2 R 1 + R 2 V_(R2)^('')=(R_(1)R_(2))/(R_(1)+R_(2))\mathrm{V}_{\mathrm{R} 2}^{\prime \prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}
  1. Once these results have been calculated, all you to do is to combine these results together, to find the voltage across the resistor R 2 R 2 R_(2)\mathrm{R}_{2} due to the both sources acting simultaneously.
    一旦计算出这些结果,您所要做的就是将这些结果结合在一起,以找到由于两个源同时作用而在电阻器 R 2 R 2 R_(2)\mathrm{R}_{2} 上的电压。
V R 2 = V R 2 + V R 2 V R 2 = V R 2 + V R 2 V_(R2)=V_(R2)^(')+V_(R2)^('')\mathrm{V}_{\mathrm{R} 2}=\mathrm{V}_{\mathrm{R} 2}^{\prime}+\mathrm{V}_{\mathrm{R} 2}^{\prime \prime}

4.2 THEVENIN'S THEOREM
4.2 泰夫宁定理

In circuit analysis, we often encounter large circuits and most times we are interested only in a portion of the circuit, and not the circuit as a whole.
在电路分析中,我们经常遇到大型电路,而大多数时候我们只对电路的一部分感兴趣,而不是整个电路。

In such cases, the analysis is cumbersome and the possibility of making errors is very high.
在这种情况下,分析繁琐,出错的可能性非常高。

Lucky for us, French engineer Léon Charles Thevenin found a solution.
幸运的是,法国工程师莱昂·查尔斯·特文宁找到了一个解决方案。

It’s what’s known as the Thevenin’s Theorem.
这被称为特文宁定理。

According to the Thevenin’s theorem, any two-terminal, dc network can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.
根据特文宁定理,任何两个端子的直流网络都可以用一个由电压源和串联电阻组成的等效电路来替代。

Thevenin Equivalent
特文宁等效

V t h V t h V_(th)V_{t h} is called the Thevenin equivalent voltage and R t h R t h R_(th)R_{t h} is called the Thevenin equivalent resistance.
V t h V t h V_(th)V_{t h} 被称为特文宁等效电压, R t h R t h R_(th)R_{t h} 被称为特文宁等效电阻。

The Thevenin’s theorem enables us to replace a large part of a circuit, often a complicated and uninteresting part, with a very simple equivalent.
泰文定理使我们能够用一个非常简单的等效电路替代电路中的大部分,通常是复杂且不有趣的部分。

With the help of an example, let’s see the Thevenin’s theorem in action.
在一个例子的帮助下,让我们看看泰文宁定理的实际应用。

(We have used a simple circuit for better understanding.)
(我们使用了一个简单的电路以便更好地理解。)

In the circuit shown above, let’s try to find the current through the 10 Ω 10 Ω 10 Omega10 \Omega resistor.
在上面的电路中,让我们尝试找出通过 10 Ω 10 Ω 10 Omega10 \Omega 电阻的电流。
  1. Firstly identify the part of the circuit, whose equivalent you want to determine.
    首先确定您想要确定其等效的电路部分。

    In this case it’s everything except the 10 Ω 10 Ω 10 Omega10 \Omega resistor.
    在这种情况下,除了 10 Ω 10 Ω 10 Omega10 \Omega 电阻器之外的所有东西。
  2. Then temporarily remove the load resistor (10 resistor from the circuit.
    然后暂时从电路中移除负载电阻(10 电阻)。
  3. To find the Thevenin equivalent Resistance ( R T H ) R T H (R_(TH))\left(R_{T H}\right), remove all the energy sources in the circuit.
    要找到特文宁等效电阻 ( R T H ) R T H (R_(TH))\left(R_{T H}\right) ,请移除电路中的所有能量源。

    This can be done by short circuiting the voltage sources and open circuiting the current sources.
    这可以通过短路电压源和开路电流源来实现。

    In our example, there is one voltage source, short it out.
    在我们的例子中,有一个电压源,将其短路。

  4. Now find the equivalent resistance between the terminals i.e. as if were looking from the terminals.
    现在找到端子之间的等效电阻,即就像从端子看一样。

    This will give the value of R T H R T H R_(TH)\mathrm{R}_{T H}. In our example, 3 Ω 3 Ω 3Omega3 \Omega resistor is in parallel with 6 Ω 6 Ω 6Omega6 \Omega resistor, which are in series with 6 Ω 6 Ω 6Omega6 \Omega. Therefore R T H = 8 Ω R T H = 8 Ω R_(TH)=8Omega\mathrm{R}_{T H}=8 \Omega (do the math.)
    这将给出 R T H R T H R_(TH)\mathrm{R}_{T H} 的值。在我们的例子中, 3 Ω 3 Ω 3Omega3 \Omega 电阻与 6 Ω 6 Ω 6Omega6 \Omega 电阻并联,这些电阻与 6 Ω 6 Ω 6Omega6 \Omega 串联。因此 R T H = 8 Ω R T H = 8 Ω R_(TH)=8Omega\mathrm{R}_{T H}=8 \Omega (做数学运算。)
  5. To find the Thevenin equivalent Voltage ( V TH ) V TH (V_(TH))\left(\mathrm{V}_{\mathrm{TH}}\right), return the energy sources to the way it was before, then
    要找到特文宁等效电压 ( V TH ) V TH (V_(TH))\left(\mathrm{V}_{\mathrm{TH}}\right) ,将能量源恢复到之前的状态,然后

    determine the open circuit voltage across the terminals.
    确定端子之间的开路电压。

Do note that current cannot flow through the 6 Ω 6 Ω 6Omega6 \Omega resistor (highlighted) because the load resistance is open circuited and hence no voltage drop across it.
请注意,电流无法通过 6 Ω 6 Ω 6Omega6 \Omega 电阻(高亮显示)流动,因为负载电阻是开路的,因此在其上没有电压降。

Therefore using Voltage Division rule, V TH = 6 × 9 / ( 3 + 6 ) V TH = 6 × 9 / ( 3 + 6 ) V_(TH)=6xx9//(3+6)\mathrm{V}_{\mathrm{TH}}=6 \times 9 /(3+6) = 6 V = 6 V =6V=6 \mathrm{~V}.
因此使用电压分配规则, V TH = 6 × 9 / ( 3 + 6 ) V TH = 6 × 9 / ( 3 + 6 ) V_(TH)=6xx9//(3+6)\mathrm{V}_{\mathrm{TH}}=6 \times 9 /(3+6) = 6 V = 6 V =6V=6 \mathrm{~V}

6. Now that we obtained both R T H R T H R_(TH)\mathrm{R}_{T H} and V T H V T H V_(TH)\mathrm{V}_{T H} values, we are ready to put the load resistance back in its place and obtain the Thevenin equivalent circuit.
6. 现在我们已经获得了 R T H R T H R_(TH)\mathrm{R}_{T H} V T H V T H V_(TH)\mathrm{V}_{T H} 的值,我们准备将负载电阻放回原位并获得泰文宁等效电路。


7. Now solving this circuit is a piece of cake. (Current through 10 I resistor is 0.33 Amperes)
7. 现在解决这个电路非常简单。(通过 10 欧姆电阻的电流是 0.33 安培)

4.3 NORTON'S THEOREM
4.3 诺顿定理

In the previous chapter we saw that it is possible to replace Voltage source by a Current source and vice versa.
在前一章中,我们看到可以用电流源替代电压源,反之亦然。

American engineer E.L. Norton made good use of this idea and theorized a corollary to the Thevenin’s theorem called the Norton’s theorem.
美国工程师 E.L.诺顿很好地利用了这个想法,并提出了一个与特文宁定理相关的推论,称为诺顿定理。

Norton’s theorem states that, any two-terminal, dc network can be replaced by an equivalent circuit consisting of a current source and a parallel resistor.
诺顿定理指出,任何两个端子的直流网络都可以被一个由电流源和并联电阻组成的等效电路所替代。

Now let’s try the Norton’s theorem on our example from previous section. Steps are as follows:
现在让我们在前一节的例子中尝试诺顿定理。步骤如下:
  1. Firstly identify the part of the circuit, whose equivalent you want to determine and then temporarily remove the load resistor (10 ) resistor from the circuit.
    首先确定电路中要确定其等效的部分,然后暂时从电路中移除负载电阻(10Ω)电阻。
  2. The Norton equivalent Resistance is the same as the Thevenin equivalent Resistance.
    诺顿等效电阻与特文宁等效电阻相同。

    So the procedure is the same, remove all the energy sources in the circuit and find the resistance between the load terminals.
    所以程序是一样的,去掉电路中的所有能源源,并找到负载端子之间的电阻。

    ( R N = R N = (R_(N)=:}\left(R_{N}=\right. 8 Ω 8 Ω 8Omega8 \Omega ).
  3. To find the Norton equivalent Current ( I N ) I N (I_(N))\left(I_{N}\right), return the energy sources to the way it was before, then determine the short circuit current through the terminals.
    要找到诺顿等效电流 ( I N ) I N (I_(N))\left(I_{N}\right) ,将能量源恢复到之前的状态,然后确定通过端子的短路电流。

    ( I N = 0.25 A ) I N = 0.25 A (I_(N)=0.25(A))\left(I_{N}=0.25 \mathrm{~A}\right)
  4. So the Norton’s equivalent circuit is:
    所以诺顿等效电路是:

  5. If you solve the circuit and the current through the 10 resistor would be 0.333A, exactly same as obtained from Thevenin’s theorem method.
    如果你解这个电路,10 欧姆电阻上的电流将是 0.333A,正好与通过泰文宁定理方法得到的结果相同。
We can easily switch between the two equivalent circuits simply by doing source transformation.
我们可以通过源变换轻松地在两个等效电路之间切换。

In doing so we can also come up a relation between the 3 quantities R T H , I N R T H , I N R_(TH),I_(N)R_{T H}, I_{N} and V T H V T H V_(TH)V_{T H}.
通过这样做,我们还可以建立这三个量 R T H , I N R T H , I N R_(TH),I_(N)R_{T H}, I_{N} V T H V T H V_(TH)V_{T H} 之间的关系。

4.4 MAXIMUM POWER TRANSFER THEOREM
4.4 最大功率传输定理

When we connect a load to a circuit, say a speaker to an amplifier circuit, it’s only sensible that the maximum power should be delivered to the load.
当我们将负载连接到电路时,比如将扬声器连接到放大器电路,最大功率应该传递给负载是合乎逻辑的。

So how do we go about this??
那么我们该如何进行呢?

Change the circuit to suit the load or change the load to suit the circuit??
更改电路以适应负载,还是更改负载以适应电路?

Both are possible, but we can’t randomly keep altering the components, we need to figure this out on paper.
两者都是可能的,但我们不能随意更改组件,我们需要在纸上弄清楚这一点。

This is exactly what the Maximum Power Transfer theorem is there for.
这正是最大功率传输定理的目的所在。
According to the Maximum Power Transfer theorem, the maximum power is delivered to the load, when the
根据最大功率传输定理,当负载时,最大功率被传递

load resistance is equal to the Thevenin equivalent resistance of the circuit.
负载电阻等于电路的泰文宁等效电阻。

First let’s make sense of this intuitively, before we go into the mathematical proof.
首先,让我们直观地理解这一点,然后再进行数学证明。

We know that power is the product of Voltage and Current, so for maximum power, both quantities need to be high.
我们知道功率是电压和电流的乘积,因此为了获得最大功率,这两个量都需要很高。

Say the load resistance is low, then the Current will be very high, but the Voltage will be equally low.
假设负载电阻很低,那么电流将会非常高,但电压也会相应很低。

Similarly, if the load resistance is high, then the Voltage will be high, but the Current though it will be very low.
类似地,如果负载电阻很高,那么电压将很高,但电流虽然会非常低。

So clearly the extremes are not the way to go. At R L = R T H R L = R T H R_(L)=R_(TH)R_{L}=R_{T H}, both voltage and current will high enough to deliver maximum power.
所以显然极端的方式不是可行的。在 R L = R T H R L = R T H R_(L)=R_(TH)R_{L}=R_{T H} ,电压和电流都会足够高,以提供最大功率。

It’s promising, but to confirm we need to use math.
这很有前途,但为了确认我们需要使用数学。
I = E TH R TH + R L P = I 2 R L P L = ( E TH R TH + R L ) 2 R L I = E TH R TH + R L P = I 2 R L P L = E TH R TH + R L 2 R L {:[I=(E_(TH))/(R_(TH)+R_(L))],[P=I^(2)R_(L)],[:.P_(L)=((E_(TH))/(R_(TH)+R_(L)))^(2)R_(L)]:}\begin{aligned} & \mathrm{I}=\frac{\mathrm{E}_{\mathrm{TH}}}{\mathrm{R}_{\mathrm{TH}}+\mathrm{R}_{\mathrm{L}}} \\ & \mathrm{P}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{L}} \\ & \therefore \mathrm{P}_{\mathrm{L}}=\left(\frac{\mathrm{E}_{\mathrm{TH}}}{\mathrm{R}_{\mathrm{TH}}+\mathrm{R}_{\mathrm{L}}}\right)^{2} \mathrm{R}_{\mathrm{L}} \end{aligned}
To find the R L R L R_(L)R_{L} corresponding to maximum power, P L P L P_(L)P_{L} is differentiated with respect to R L R L R_(L)R_{L}, keeping R T H R T H R_(TH)R_{T H} constant and equated to zero. If you actually bother to do the math, you can obtain the relation R L = R T H R L = R T H R_(L)=R_(TH)R_{L}=R_{T H}.
要找到与最大功率对应的 R L R L R_(L)R_{L} ,对 P L P L P_(L)P_{L} 关于 R L R L R_(L)R_{L} 进行微分,同时保持 R T H R T H R_(TH)R_{T H} 不变,并将其等于零。如果你真的愿意做数学运算,你可以得到关系 R L = R T H R L = R T H R_(L)=R_(TH)R_{L}=R_{T H}
The Maximum Power Transfer theorem isn’t a circuit analysis technique as such, but rather a practical application of the Thevenin and Norton theorems.
最大功率传输定理并不是一种电路分析技术,而是特夫南定理和诺顿定理的实际应用。

4.5 SUBSTITUTION THEOREM
4.5 替代定理

According to the Substitution theorem, any branch of a dc network can be replaced by a different combination of elements as long as the new combination of elements will maintain the same voltage across and current through, as the original branch.
根据替代定理,直流网络的任何支路都可以被不同的元件组合替代,只要新的元件组合能够保持与原支路相同的电压和电流。
For example, consider this particular of a network, it has a voltage of 10 V across it and a current of 2 A flowing through it.
例如,考虑这个特定的网络,它的电压为 10 伏,电流为 2 安培。

This branch can be replaced by any combination of elements as long as the voltage and the current remains the same.
这个分支可以被任何元素的组合替代,只要电压和电流保持不变。

Shown below are some of the possible replacement combinations.
下面是一些可能的替换组合。

Substitution theorem gives you the ability to replace complicated branches of a circuit with convenient components to make circuit analysis simpler.
替代定理使您能够用方便的元件替换电路中复杂的分支,从而简化电路分析。

4.6 RECIPROCITY THEOREM
4.6 互惠定理

This is one of those nasty theorems, a bit hard to understand, and has limited application.
这是一个令人讨厌的定理,有点难以理解,并且应用有限。

But we are engineers and we don’t have a choice but to learn.
但我们是工程师,别无选择,只能学习。

According to the Reciprocity theorem, if a voltage source in a circuit causes a current in some other part of the circuit, then the positions of the voltage source and the resulting current can be interchanged without a change in the current.
根据互易定理,如果电路中的电压源在电路的某个其他部分引起电流,则电压源和产生的电流的位置可以互换,而电流不变。

There’s no way that you can understand this theorem without the help of an example.
没有例子的帮助,你无法理解这个定理。

Let’s go step by step.
我们一步一步来。

Consider this example, here the 10 V voltage source causes a current I to flow throw the 4 Ω 4 Ω 4Omega4 \Omega resistor. If you do the math, you will get the magnitude of I as 0.45 A .
考虑这个例子,这里 10 V 电压源使电流 I 流过 4 Ω 4 Ω 4Omega4 \Omega 电阻。如果你计算一下,你会得到 I 的大小为 0.45 A。

Now if you interchange the positions of the voltage source and the resultant current, you will get this circuit, shown below.
现在,如果你交换电压源和结果电流的位置,你将得到下面所示的电路。

The 10V source will produce a current I’ in its new position. Let’s calculate it.
10V 源将在其新位置产生电流 I'。让我们来计算一下。

We’ve used mesh analysis (but you are free to use anything) and we got 2 equations: 4 I 2 I 1 = 0 4 I 2 I 1 = 0 4I^(')-2I_(1)=04 \mathrm{I}^{\prime}-2 \mathrm{I}_{1}=0 and 2 I + 12 I 1 = 10 2 I + 12 I 1 = 10 -2I^(')+12I_(1)=10-2 I^{\prime}+12 I_{1}=10. Solving them, we get I I I^(')I^{\prime} as 0.45 A , which is the same as before. This is what the Reciprocity Theorem is. The ratio V / I V / I V//I\mathrm{V} / \mathrm{I} is known as the transfer impedance.
我们使用了网路分析(但你可以自由使用任何方法),得到了两个方程: 4 I 2 I 1 = 0 4 I 2 I 1 = 0 4I^(')-2I_(1)=04 \mathrm{I}^{\prime}-2 \mathrm{I}_{1}=0 2 I + 12 I 1 = 10 2 I + 12 I 1 = 10 -2I^(')+12I_(1)=10-2 I^{\prime}+12 I_{1}=10 。解它们,我们得到 I I I^(')I^{\prime} 为 0.45 A,这与之前相同。这就是互易定理。比率 V / I V / I V//I\mathrm{V} / \mathrm{I} 被称为传输阻抗。

Do keep in mind that the reciprocity theorem’s use is strictly limited to single source circuits.
请记住,互易定理的使用严格限于单源电路。

5. CAPACITANCE
5. 电容

5.1 CAPACITORS
5.1 电容器

A capacitor is an electrical device that is used to store electrical energy.
电容器是一种用于储存电能的电气设备。

Isn’t that what batteries are for??
这不就是电池的用途吗??

Yes…In a way, a capacitor is like a battery, they both store electrical energy.
是的……在某种程度上,电容器就像电池,它们都储存电能。

But the difference is in how they store energy and hence their applications differ.
但它们存储能量的方式不同,因此它们的应用也不同。

In a battery, chemical reactions produce electrons at one terminal and absorb electrons at the other terminal.
在电池中,化学反应在一个端子产生电子,在另一个端子吸收电子。

Whereas, a capacitor is much simpler, it cannot produce new electrons, it only stores them.
然而,电容器要简单得多,它不能产生新的电子,它只能存储电子。

Next to the resistor, the capacitor is the most commonly encountered component in electrical circuits.
在电路中,电容器是仅次于电阻器的最常见组件。

A capacitor is constructed out of two metal plates, separated by an insulating material called dielectric.
电容器由两块金属板构成,中间隔着一种叫做介电材料的绝缘材料。

The plates are conductive and they are usually made of aluminum, tantalum or other metals, while the dielectric can be made out of any kind of insulating material such as paper, glass, ceramic or anything that obstructs the flow of the current.
这些板是导电的,通常由铝、钽或其他金属制成,而介电材料可以由任何种类的绝缘材料制成,例如纸、玻璃、陶瓷或任何阻碍电流流动的材料。

In fact, you can make a simple capacitor can be made from two strips of aluminum foil separated by two thin layers of wax paper (Check out this instructable: http://www.instructables.com/id/Aluminum-Foil-PlateCapacitor/).
实际上,你可以用两条铝箔条和两层薄蜡纸制作一个简单的电容器(查看这个说明书:http://www.instructables.com/id/Aluminum-Foil-PlateCapacitor/)。

Of course, our homemade capacitor won’t work very well, but it shows capacitor like behavior nonetheless.
当然,我们自制的电容器效果不是很好,但它仍然表现出电容器的特性。

Since the plates are made of metal, they contain a huge no. of free electrons.
由于这些板是由金属制成的,因此它们含有大量自由电子。

In their normal state, the plates are neutral, as there is no excess or deficiency of electrons.
在它们的正常状态下,电极是中性的,因为没有多余或缺乏电子。

But when we connect a power source to the metal plates of the capacitor, a current will try to flow i.e. the electrons from the plate connected to the positive lead of the battery will start moving to the plate connected to the negative lead of the battery.
但是当我们将电源连接到电容器的金属板时,电流会试图流动,即从连接到电池正极的板上的电子将开始移动到连接到电池负极的板上。

However, because of the dielectric between the plates, the electrons won’t be able to pass through the capacitor, so they will start accumulating on the plate.
然而,由于电容器板之间的介电质,电子将无法通过电容器,因此它们将开始在板上积累。

After a certain number of electrons accumulated on the plate, the battery will not have the sufficient energy to push any new electrons.
在一定数量的电子积累在电极上之后,电池将没有足够的能量来推动任何新的电子。

This leaves the top plate with a deficiency of electrons
这使得上板缺乏电子

(i.e. positive charge) and the bottom plate with an excess of electrons (i.e. negative charge).
(即正电荷)和底板上多余的电子(即负电荷)。

In this state, the capacitor is said to be charged.
在这种状态下,电容器被认为是充电的。

This state will remain even after the battery is removed and the Capacitor will only discharge once a load is connected across it.
即使在电池被移除后,该状态仍将保持,电容器只有在其两端连接负载时才会放电。


The ability of a capacitor to store an electric charge is referred to as its capacitance.
电容器存储电荷的能力称为其电容。

The capacitance C is the ratio of charge stored Q to the potential difference V between the conductors.
电容 C 是存储电荷 Q 与导体之间电势差 V 的比率。

Mathematically,
数学上,
C = Q V C = Q V C=(Q)/(V)\mathrm{C}=\frac{Q}{V}
So a better capacitor would be the one able to store more charge for a particular voltage applied.
因此,更好的电容器是能够在施加特定电压时存储更多电荷的电容器。

Capacitance is measured in farads.
电容以法拉为单位。

This is a very large unit and hence most capacitors are rated in microfarads or less.
这是一个非常大的单位,因此大多数电容器的容量以微法拉或更小的单位来表示。

The commonly used symbols for Capacitors are:
电容器常用的符号有:


Polarized capacitor Variable capacitor
电解电容器 可变电容器

5.2 HYDRAULIC ANALOGY
5.2 液压类比

A better understanding of how Capacitors store charge can be gained with the help of a hydraulic analogy.
通过液压类比,可以更好地理解电容器如何储存电荷。

Consider the arrangement shown below, it consists of a water tank separated by a diaphragm D in the middle and a piston P to force water into either side of the diaphragm.
考虑下面所示的装置,它由一个水箱组成,中间有一个隔膜 D,活塞 P 用于将水强制进入隔膜的任一侧。

Under normal circumstance, when the piston is left untouched, the diaphragm is flat as shown by the dotted line.
在正常情况下,当活塞未被触碰时,隔膜是平坦的,如虚线所示。

It’s similar to an uncharged capacitor, it has no energy.
它类似于一个未充电的电容器,没有能量。

But if the piston is pushed towards the left, water is drawn from the right side of the diaphragm and at the same time water is being forced into the left side.
但是如果活塞向左推,水会从隔膜的右侧被抽出,同时水被迫进入左侧。

Under this condition the diaphragm is no longer flat, as shown by the full line.
在这种情况下,隔膜不再是平的,如实线所示。

Greater the force applied to the piston, more water is displaced, and hence the diaphragm is under greater stress.
施加在活塞上的力越大,排出的水就越多,因此隔膜承受的压力也越大。

The force applied to the piston is analogous to the EMF applied, and the water displaced to the charge displaced, in case of a capacitor.
施加在活塞上的力类似于施加的电动势,而排开的水则类似于电容器中排开的电荷。

Just like the diagram separates the two halves of the tank and doesn’t allow water from either side to mix, the dielectric separate the charge in a capacitor.
就像图表将水箱的两个半部分分开,不允许任何一侧的水混合,电介质在电容器中分隔电荷。

If we now remove the force on the piston, the diaphragm will try to release its stress (energy) by becoming flat, hence pushing the piston back to its original position.
如果我们现在去掉施加在活塞上的力,隔膜将试图通过变平来释放其应力(能量),从而将活塞推回到原来的位置。

This is exactly what happens when a charged capacitor is connected to a load resistance.
当一个带电的电容器连接到负载电阻时,正是发生了这种情况。

A current rushes through the
电流涌过

resistance till the energy stored is released.
抵抗直到储存的能量释放。

The rate of flow of water is dependent on the resistance offered by the pipes, much like the rate of flow of charge (current) is dependent on the resistance offered by the wires.
水流的速率取决于管道提供的阻力,就像电流的流动速率取决于电线提供的阻力一样。

The diaphragm will rupture if sufficient enough force is applied on the piston, just as the Capacitor will breakdown under excess voltage.
如果在活塞上施加足够的力,隔膜将会破裂,就像电容器在过高电压下会击穿一样。

5.3 CAPACITORS IN PARALLEL
5.3 并联电容器

Like with resistors, capacitors can also be connected in series or parallel combination and to analyze such circuits, we can find equivalent capacitance for these combinations.
与电阻器一样,电容器也可以串联或并联连接,分析此类电路时,我们可以找到这些组合的等效电容。

When a set of capacitors are connected in parallel, the total equivalent capacitance is the sum of individual capacitances.
当一组电容器并联连接时,总的等效电容是各个电容的总和。

Suppose two capacitors, having capacitances C 1 C 1 C_(1)\mathrm{C}_{1} and C 2 C 2 C_(2)\mathrm{C}_{2} farads are connected in parallel across a potential difference
假设两个电容器,电容分别为 C 1 C 1 C_(1)\mathrm{C}_{1} C 2 C 2 C_(2)\mathrm{C}_{2} 法拉,连接在一个电压差的并联电路中

of V volts. Let the charge on C 1 C 1 C_(1)\mathrm{C}_{1} be Q 1 Q 1 Q_(1)\mathrm{Q}_{1} coulombs and that on C 2 C 2 C_(2)\mathrm{C}_{2} be Q 2 Q 2 Q_(2)\mathrm{Q}_{2} coulombs, where.
V 伏特。设 C 1 C 1 C_(1)\mathrm{C}_{1} 上的电荷为 Q 1 Q 1 Q_(1)\mathrm{Q}_{1} 库仑, C 2 C 2 C_(2)\mathrm{C}_{2} 上的电荷为 Q 2 Q 2 Q_(2)\mathrm{Q}_{2} 库仑,其中。
Q 1 = C 1 V , Q 2 = C 2 V Q 1 = C 1 V , Q 2 = C 2 V {:[Q_(1)=C_(1)V","],[Q_(2)=C_(2)V]:}\begin{aligned} & \mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}, \\ & \mathrm{Q}_{2}=\mathrm{C}_{2} \mathrm{~V} \end{aligned}
If we were to replace the capacitors by a single equivalent capacitor C , then a charge Q = Q 1 + Q 2 Q = Q 1 + Q 2 Q=Q_(1)+Q_(2)\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2} would be produced by the same potential difference.
如果我们用一个等效电容器 C 替换电容器,那么将会由相同的电压差产生一个电荷 Q = Q 1 + Q 2 Q = Q 1 + Q 2 Q=Q_(1)+Q_(2)\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}
Q = Q 1 + Q 2 CV = C 1 V + C 2 V C = C 1 + C 2 Q = Q 1 + Q 2 CV = C 1 V + C 2 V C = C 1 + C 2 {:[Q=Q_(1)+Q_(2)],[=>CV=C_(1)V+C_(2)V],[:.C=C_(1)+C_(2)]:}\begin{aligned} & \mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2} \\ \Rightarrow & \mathrm{CV}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V} \\ \therefore & \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2} \end{aligned}
This result can be extended to any no. of capacitors connected in parallel. For ’ n n nn ’ capacitors in parallel,
此结果可以扩展到任何数量的并联电容器。对于' n n nn '个并联电容器,
C e q = C 1 + C 2 + . + C n C e q = C 1 + C 2 + . + C n C_(eq)=C_(1)+C_(2)+dots.+C_(n)C_{e q}=C_{1}+C_{2}+\ldots .+C_{n}

5.4 CAPACITORS IN SERIES
5.4 串联电容器

For a series combination of capacitors, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.
对于电容器的串联组合,等效电容的倒数是各个电容的倒数之和。

Suppose two capacitors, having capacitances C 1 C 1 C_(1)\mathrm{C}_{1} and C 2 C 2 C_(2)\mathrm{C}_{2} farads are connected in series across a potential difference of V volts. Let he voltages across C 1 C 1 C_(1)\mathrm{C}_{1} and C 2 C 2 C_(2)\mathrm{C}_{2} be V 1 V 1 V_(1)\mathrm{V}_{1} and V 2 V 2 V_(2)\mathrm{V}_{2} volts respectively.
假设两个电容器,电容分别为 C 1 C 1 C_(1)\mathrm{C}_{1} C 2 C 2 C_(2)\mathrm{C}_{2} 法拉,串联连接在电压为 V 伏特的电源上。设 C 1 C 1 C_(1)\mathrm{C}_{1} C 2 C 2 C_(2)\mathrm{C}_{2} 上的电压分别为 V 1 V 1 V_(1)\mathrm{V}_{1} V 2 V 2 V_(2)\mathrm{V}_{2} 伏特。

Obviously, because it’s a series connection, the currents and hence the charge flowing through the capacitors are the same.
显然,由于这是串联连接,流经电容器的电流和电荷是相同的。
Q = C 1 V 1 = C 2 V 2 V 1 = Q C 1 & V 2 = Q C 2 Q = C 1 V 1 = C 2 V 2 V 1 = Q C 1 & V 2 = Q C 2 {:[:.Q=C_(1)V_(1)=C_(2)V_(2)],[=>V_(1)=(Q)/(C_(1))&V_(2)=(Q)/(C_(2))]:}\begin{aligned} & \therefore \mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2} \\ & \Rightarrow \mathrm{~V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}} \& \mathrm{~V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}} \end{aligned}
Now if we were to replace the 2 capacitors with an equivalent capacitor of capacitance C C CC, then it would have the same charge Q , when connected across the voltage V .
现在如果我们用一个电容值为 C C CC 的等效电容器替换这两个电容器,那么当它连接在电压 V 上时,它将具有相同的电荷 Q。

Also from KVL, we know that V = V 1 + V 2 V = V 1 + V 2 V=V_(1)+V_(2)\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2}. Therefore,
也来自 KVL,我们知道 V = V 1 + V 2 V = V 1 + V 2 V=V_(1)+V_(2)\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2} 。因此,
V = Q C = Q C 1 + Q C 2 1 C = 1 C 1 + 1 C 2 V = Q C = Q C 1 + Q C 2 1 C = 1 C 1 + 1 C 2 {:[V=(Q)/(C)=(Q)/(C_(1))+(Q)/(C_(2))],[=>(1)/(C)=(1)/(C_(1))+(1)/(C_(2))]:}\begin{aligned} & \mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}+\frac{\mathrm{Q}}{\mathrm{C}_{2}} \\ & \Rightarrow \frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}} \end{aligned}
This result can be extended to any no. of capacitors connected in series. For ’ n n nn ’ capacitors in series,
此结果可以扩展到任何数量的串联电容器。对于’ n n nn ’个串联电容器,

Do note that the expression for capacitors in series and resistors in parallel are the same and similarly the expression for capacitors in parallel and resistors in series are also the same.
请注意,串联电容器和并联电阻器的表达式是相同的,类似地,串联电阻器和并联电容器的表达式也是相同的。

5.5 CHARGING & DISCHARGING OF A CAPACITOR
5.5 电容器的充电与放电

A Capacitor doesn’t charge up all of a sudden, when connected to a voltage source.
电容器在连接到电压源时不会突然充电。

It takes some definite time for the capacitor to become fully charged and it does so in an exponential manner.
电容器需要一定的时间才能完全充电,并且这个过程是以指数方式进行的。

When an uncharged capacitor is
当一个未充电的电容器是

connected to a DC source, the voltage across is zero, as if there is a short circuit, then the voltage exponentially increases to the applied voltage after a while.
连接到直流电源时,电压为零,就像短路一样,然后电压在一段时间后指数增加到施加电压。

On the contrary, as soon the source is connected, the max current rushes to the capacitor.
相反,一旦电源连接,最大电流就会冲向电容器。

Later as the time passes, the current decreases to zero and acts like an open circuit.
随着时间的推移,电流减少到零,表现得像一个开路。

How fast the capacitor charges up depends on any resistance present in the circuit.
电容器充电的速度取决于电路中存在的任何电阻。
Charging of a Capacitor
电容器的充电


A fully charged capacitor will discharge in exactly the reverse manner, the voltage drops and the current picks up exponentially.
一个充满电的电容器将以完全相反的方式放电,电压下降,电流呈指数增长。
Discharging of a Capacitor
电容器的放电


We’ll study this in greater detail with the help of Laplace transform in chapter 12.
我们将在第 12 章中借助拉普拉斯变换更详细地研究这个问题。

5.6 ENERGY STORED BY CAPACITORS
5.6 电容器储存的能量

The energy stored in a Capacitor is basically the energy the battery expended in moving electrons from the positive plate to the negative plate of the capacitor against their natural tendency.
电容器中储存的能量基本上是电池在将电子从电容器的正极移动到负极时所消耗的能量,这一过程是与电子的自然倾向相反的。

Suppose a voltage V is applied to capacitor terminals, then the work done transferring an infinitesimal amount of charge d q d q dqd q from the negative to the positive plate is simply,
假设在电容器端子上施加了电压 V,那么将一个无穷小的电荷 d q d q dqd q 从负极转移到正极所做的功就是,
d W = V d q d W = V d q dW=Vdqd W=V d q
The work done is a variable quantity, because as the charge accumulates, more work needs to be done in moving the electrons.
所做的工作是一个变量,因为随着电荷的积累,移动电子需要做更多的工作。

Similarly, Voltage is also a function of charge. Hence the incremental work is given by,
类似地,电压也是电荷的一个函数。因此,增量功由以下公式给出,
V = q C d W = q d q C V = q C d W = q d q C {:[V=(q)/(C)],[:.dW=(qdq)/(C)]:}\begin{aligned} & \mathrm{V}=\frac{q}{\mathrm{C}} \\ & \therefore d W=\frac{q d q}{\mathrm{C}} \end{aligned}
To find the total work done, we need to integrate this quantity from 0 to the maximum charge Q .
要找到总的功,我们需要将这个量从 0 积分到最大电荷 Q。
W = d W = 0 Q q d q C W = Q 2 2 C W = d W = 0 Q q d q C W = Q 2 2 C {:[W=int dW=int_(0)^(Q)(qdq)/(C)],[:.W=(Q^(2))/(2C)]:}\begin{gathered} \mathrm{W}=\int d W=\int_{0}^{Q} \frac{q d q}{\mathrm{C}} \\ \therefore \mathrm{~W}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}} \end{gathered}
This expression has different forms, based on the quantities you choose:
这个表达式有不同的形式,取决于你选择的数量:
W = Q 2 2 C = C V 2 2 = QV 2 W = Q 2 2 C = C V 2 2 = QV 2 W=(Q^(2))/(2C)=(CV^(2))/(2)=(QV)/(2)\mathrm{W}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}=\frac{C \mathrm{~V}^{2}}{2}=\frac{\mathrm{QV}}{2}

6. INDUCTANCE
6. 电感

6.1 ELECTROMAGNETISM
6.1 电磁学

Firstly, Electromagnetism is a huge topic and doesn’t really have a place in a circuit analysis text.
首先,电磁学是一个庞大的主题,并不真正适合电路分析教材。

But since we want to give our readers a proper introduction to inductance, we will quickly go through the fundamental ideas in electromagnetism without going into the minute details.
但由于我们想给读者一个关于电感的适当介绍,我们将快速浏览电磁学中的基本概念,而不深入细节。

The most fundamental idea in electromagnetism is that there is magnetic field surrounding every current carrying object.
电磁学中最基本的思想是,每个带电流的物体周围都有磁场。

These magnetic fields take the shape of concentric rings around a straight wire, called magnetic field lines.
这些磁场呈现出围绕直线电线的同心环形状,称为磁场线。

Larger the current flowing through wire, more the no. of magnetic field lines.
电线中流动的电流越大,磁场线的数量就越多。

These lines are not random, they have direction, which can be determined by using the Right hand thumb rule.
这些线不是随机的,它们有方向,可以通过使用右手拇指法则来确定。

It goes like this, if you point your thumb in the direction of the current, then the fingers curl in the direction of the field lines
如果你把拇指指向电流的方向,那么手指就会向场线的方向弯曲

Similarly, when current flows through a coil, a magnetic field is generated, such that coil acts like a magnet with a north and south polarity.
类似地,当电流通过线圈时,会产生磁场,使得线圈像一个具有南北极性的磁铁。

The pattern of field lines is as shown below.
场线的模式如下所示。

Do note that field lines are concentric if you consider a tiny portion of the coil, but these field lines add and cancel each other giving us this effective pattern.
请注意,如果考虑线圈的一个微小部分,场线是同心的,但这些场线相互叠加和抵消,形成了这种有效的模式。

By the way these sort of coils are called Solenoids.
顺便说一下,这种线圈被称为电磁铁。

Faraday’s Laws: Michael Faraday formulated 2 laws, which form the basis of Electromagnetic studies, called the Faraday’s Laws.
法拉第定律:迈克尔·法拉第提出了两个定律,这些定律构成了电磁学研究的基础,称为法拉第定律。

These laws introduces us to the phenomenon called Electromagnetic Induction.
这些法律让我们了解了一个叫做电磁感应的现象。

According to the Faraday’s first law, when a conductor is placed in a varying magnetic field, an EMF gets induced across the conductor and if the conductor offers a closed circuit then induced current flows through it.
根据法拉第的第一定律,当导体置于变化的磁场中时,导体上会感应出电动势,如果导体形成闭合电路,则感应电流会通过它流动。

And Faraday’s second law states that, the induced EMF is directly proportional to the rate of change of magnetic flux.
法拉第的第二定律指出,感应电动势与磁通量变化率成正比。

If you place a bar magnet near a wire, nothing happens, no voltage is induced.
如果你把一个条形磁铁放在一根电线附近,什么也不会发生,没有电压被感应。

But if you move the magnet such that some of the flux lines (imaginary) are cut by the wire, then a voltage is induced.
但是如果你移动磁铁,使得一些磁通线(想象中的)被导线切割,那么就会产生电压。

There are two ways to obtain varying magnetic field:
有两种方法可以获得变化的磁场:
  1. One is relative spatial movement that is, if the distance between the magnet and the conductor keeps changing, the magnetic field also keeps changing and induction is possible.
    一个是相对空间运动,也就是说,如果磁铁和导体之间的距离不断变化,磁场也会不断变化,从而可能产生感应。
  2. The other is to vary the magnetic field originating from the source itself.
    另一个是改变来自源本身的磁场。

    This is not possible with permanent magnets, but it’s easy to do with solenoid magnets we discussed earlier.
    这在永久磁铁中是不可能的,但在我们之前讨论的电磁铁中很容易做到。

    All you need to do is to vary the current through the coils, the magnetic field also varies as a result.
    您需要做的就是改变通过线圈的电流,磁场也会因此变化。
Guess what would happen if we placed 2 coils close to each other, one connected to a varying current source and the other to an ammeter?
如果我们将两个线圈放得很近,一个连接到变化电流源,另一个连接到电流表,你猜会发生什么?

Yes, the ammeter will show deflection, proving that a current has been induced in the second coil.
是的,电流表会显示偏转,证明第二个线圈中产生了电流。

So can we just place many coils in the proximity of a current carrying coil and induce current in all of them?
那么我们可以在一个带电流的线圈附近放置许多线圈,并在它们中感应电流吗?

Yes, that’s
是的,那是

possible. Wait! Did we just invent a new method to generate electricity??
可能。等一下!我们刚刚发明了一种新的发电方法吗??

Unfortunately not, there’s a catch in all this, called mutual induction.
不幸的是,这一切都有一个陷阱,叫做互感应。

When we induce a current in the secondary coil, this current will produce itself produce a flux in the secondary coil.
当我们在次级线圈中感应电流时,这个电流将自身在次级线圈中产生磁通。

This flux will link with primary coil, inducing an EMF. So this is a mutual process.
这个磁通将与初级线圈连接,感应出电动势。因此这是一个相互的过程。

To sum up, the primary induces a voltage, therefore a current in the secondary, which in turn will induce a voltage and a current back in the primary.
总之,初级绕组感应出电压,从而在次级绕组中产生电流,反过来又会在初级绕组中感应出电压和电流。

The catch is that the current induced back in the primary will be in the opposite direction as the original applied current in the primary, thus reducing the overall effect.
问题在于,感应回流的电流方向与原始施加在初级的电流方向相反,从而减少了整体效果。

This isn’t a wild theory or anything, it’s a direct consequence of the law of conservation of energy.
这不是一个疯狂的理论,而是能量守恒定律的直接结果。

In electromagnetics it’s called the Lenz’s law.
在电磁学中,这被称为伦茨定律。

Lenz’s Law ensures that the electrical energy of the primary coil is reduced by the same amount as the energy gained by the secondary coil.
楞次定律确保主线圈的电能减少的量与副线圈获得的能量相同。

In layman’s terms, an induced effect is always such as to oppose the cause that produced it.
用通俗的话来说,诱导效应总是与产生它的原因相对立。

Electromagnetic induction is the principle behind the working of devices like transformers, motors etc.
电磁感应是变压器、电动机等设备工作的原理。

Now there’s another type of Inductance called Self Inductance.
现在有另一种电感称为自感。

We’ll study about it in detail in the next section.
我们将在下一节中详细研究它。

6.2 INDUCTOR
6.2 电感器

Inductor is the final member of our amazing trio that includes the resistor and the capacitor.
电感是我们惊人三人组的最后一员,包括电阻器和电容器。

Like the other two components, the inductor is practically used everywhere.
与其他两个组件一样,电感器几乎在任何地方都被广泛使用。

Have you seen a copper coil in an electronic circuit??
你见过电子电路中的铜线圈吗?

That’s the inductor, that’s right it’s just a coil, nothing else.
那就是电感器,没错,它只是一个线圈,别无其他。

Inductor like the capacitor is an energy storing device, but it uses a completely different mechanism to do so.
电感器像电容器一样是一个储能装置,但它使用完全不同的机制来实现这一点。

While the capacitor stores energy in the form of electrostatic energy, the inductor stores its energy in the form of magnetic energy.
电容器以静电能的形式储存能量,而电感器以磁能的形式储存能量。

Despite this, Inductors aren’t primarily used as a storage devices, they are commonly used as filters and chokes.
尽管如此,电感器并不是主要用作储存设备,它们通常用作滤波器和扼流圈。

That’s because Inductors have the ability to suppress variation in current flowing through it.
这是因为电感器具有抑制流过它的电流变化的能力。
The inductors ability to resist variation in current can be attributed to a phenomenon called Self Induction.
电感器抵抗电流变化的能力可以归因于一种称为自感应的现象。

The phenomenon can be better understood with the help of the figure below.
这个现象可以通过下面的图更好地理解。

Consider just two loops of an inductor coil.
考虑仅两个电感线圈的回路。

When a current is passed through the inductor or more specifically the first loop of the inductor, it produces magnetic a field around it in a concentric manner (as with any other conductor).
当电流通过电感器,或更具体地说,通过电感器的第一个线圈时,它会在周围产生一个同心的磁场(与任何其他导体一样)。

This magnetic field created by the first loop also links with the second loop, because of their proximity.
这个由第一个线圈产生的磁场也与第二个线圈相连,因为它们彼此接近。

The natural response of the second loop to this magnetic field, is to produce a current (or a counter magnetic field as represented by the bottom ring) such as to oppose the original current, in accordance with the Lenz’s law.
第二个回路对这个磁场的自然反应是产生一个电流(或如底部环所示的反向磁场),以抵消原始电流,符合伦茨定律。

The direction of the current induced in the second loop due to the field generated by the first loop is show by the dotted arrow.
由于第一个线圈产生的场,第二个线圈中感应电流的方向由虚线箭头表示。

These currents will be generated whenever there is a variation in current in the inductor and it opposes the original inductor current.
这些电流将在电感器中的电流发生变化时产生,并且它们会抵消原来的电感器电流。

So this ability of an Inductor to oppose change in current is called the Self Inductance or simply Inductance.
因此,电感器抵抗电流变化的能力称为自感或简单称为电感。

It is denoted by the letter L L LL and its unit is Henry (H).
它用字母 L L LL 表示,单位是亨利(H)。
Our analysis was just with 2 loops, but the inductance will increase if the number of winds in the coil is increased since the magnetic field from one coil will have more coils to interact with.
我们的分析仅涉及 2 个线圈,但如果线圈中的绕组数量增加,电感将会增加,因为一个线圈产生的磁场将与更多的线圈相互作用。

So self-induction in a way, is the mutual induction between the loops of an inductor coil.
因此,自感在某种程度上是电感线圈中线圈之间的互感。

The commonly used symbol for an Inductor is,
电感器的常用符号是,

6.3 INDUCTORS IN SERIES
6.3 串联电感器

For inductors in series, the total inductance is simply the sum of individual inductances, just as with resistors in series.
对于串联的电感器,总电感就是各个电感的总和,就像串联的电阻一样。

L eq = L 1 + L 2 + . + L n L eq = L 1 + L 2 + . + L n L_(eq)=L_(1)+L_(2)+dots.+L_(n)\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}+\ldots .+\mathrm{L}_{n}
Do note that this result is under the assumption that the magnetic fields of the inductors do not interact with each other.
请注意,这个结果是在假设电感器的磁场彼此不相互作用的情况下得出的。

6.4 INDUCTORS IN PARALLEL
6.4 并联电感器

For a parallel combination of inductors, the reciprocal of the equivalent inductance is the sum of the reciprocals of the individual inductances, just as with resistors.
对于电感器的并联组合,等效电感的倒数是各个电感倒数之和,就像电阻一样。

Once again, this result is under the assumption that the magnetic fields of the inductors do not interact with each other.
再次强调,这个结果是在假设电感器的磁场彼此不相互作用的情况下得出的。

6.5 CHARGING & DISCHARGING OF A INDUCTOR
6.5 电感器的充电与放电

Like the capacitor, it takes some definite time for the inductor to become fully charged.
像电容器一样,电感器也需要一定的时间才能完全充电。

When an uncharged inductor is connected to a DC source, it acts as an open circuit and the voltage across it is equal to the applied voltage, then the voltage exponentially decreases to zero after a while.
当一个未充电的电感器连接到直流电源时,它表现为一个开路,电感器两端的电压等于施加的电压,然后电压会在一段时间后指数性地降低到零。

On the contrary, the current flowing the capacitor initially is zero.
相反,电容器初始时流过的电流为零。

Later as the time passes, the current builds to a maximum value and acts like a short circuit.
随着时间的推移,电流达到最大值并像短路一样。

How fast the capacitor charges up depends on any resistance present in the circuit.
电容器充电的速度取决于电路中存在的任何电阻。
Charging of a Inductor
电感器的充电


A fully charged inductor will discharge in exactly the reverse manner, the voltage picks up and the current drops exponentially.
一个充满电的电感器将以完全相反的方式放电,电压上升而电流指数下降。

Discharging of a Inductor
电感器的放电


6.6 ENERGY STORED BY AN INDUCTOR
6.6 电感器储存的能量

The EMF induced across the inductor due to variation in current (which leads to change in flux) is given by,
由于电流变化(导致磁通变化)而在电感器上感应的电动势由以下公式给出,
v = L d i d t v = L d i d t v=L(di)/(dt)\mathrm{v}=\mathrm{L} \frac{d i}{d t}
Therefore the instantaneous power which must be supplied to initiate the current in the inductor is,
因此,必须提供给电感器以启动电流的瞬时功率是,
P = vi = Li d i d t d W = Pdt = Li di P = vi = Li d i d t d W = Pdt = Li di {:[P=vi=Li(di)/(dt)],[=>dW=Pdt=Lidi]:}\begin{aligned} & \mathrm{P}=\mathrm{vi}=\mathrm{Li} \frac{d i}{d t} \\ & \Rightarrow d W=\mathrm{Pdt}=\mathrm{Li} \mathrm{di} \end{aligned}
To find the total work done, we need to integrate this quantity from 0 to the maximum charge I .
要找到总的工作量,我们需要将这个量从 0 积分到最大电荷 I。

7. AC FUNDAMENTALS
7. 交流电基础知识

7.1 INTRODUCTION TO AC
7.1 AC 简介

So far we have only discussed about DC circuits and its analysis.
到目前为止,我们只讨论了直流电路及其分析。

Now we’ll turn our attention to AC circuits. AC stands for Alternating current.
现在我们将注意力转向交流电路。交流电代表交流电流。

AC is of interest to us, because 90 % 90 % 90%90 \% of supply used for commercial purposes is AC.
我们对交流电感兴趣,因为用于商业目的的供应中有 90 % 90 % 90%90 \% 是交流电。

DC supply, we dealt with so far had constant magnitude and direction (positive to negative).
直流电源,我们迄今为止处理的具有恒定的大小和方向(从正到负)。

A DC source like your car battery will always have a constant magnitude between its terminals.
像汽车电池这样的直流电源在其端子之间始终具有恒定的大小。

Its positive and negative terminals will always remain as it is.
它的正负端子将始终保持不变。

On the contrary, for AC supply like your power outlet, both magnitude and direction changes periodically.
相反,对于像您的电源插座这样的交流电源,幅度和方向都会周期性变化。

The whole process takes place in 2 parts or 2 half cycles, Positive half cycle and the negative half cycle.
整个过程分为两个部分或两个半周期,正半周期和负半周期。

In the positive half cycle, the voltage (and therefore the current) will gradually increase from 0 to a max value, then starts decreasing back to zero.
在正半周期中,电压(因此电流)将逐渐从 0 增加到最大值,然后开始减少回到零。

The same thing happens in the negative half cycle, but in reverse direction. Reverse direction??
负半周期也会发生同样的事情,但方向相反。方向相反??

So does the current flows from negative to positive terminal in the negative half cycle??
那么电流在负半周期间是从负极流向正极吗?

No, it doesn’t happen that way. It’s the terminals that change its polarity.
不,这样是不会发生的。是端子改变了它的极性。

The terminal that would have been positive in the positive half cycle changes to negative in the negative half cycle and similarly for the other terminal.
在正半周期中为正的端子在负半周期中变为负,另一个端子也是如此。

This essentially means that there is no fixed Positive and Negative terminals for AC supply.
这基本上意味着交流电源没有固定的正极和负极。

A terminal can have one polarity in a half cycle and the opposite polarity in the other half cycle.
一个端子在半个周期内可以具有一种极性,而在另一个半个周期内则具有相反的极性。


DC

AC is complex, DC was straightforward. Why would we even bother generating AC?
交流电是复杂的,直流电则很简单。我们为什么还要费心去产生交流电呢?

That would be the obvious question on your mind at this point.
那将是你此刻脑海中显而易见的问题。

The answer is simple, it’s a way lot easier to generate, transmit and manipulate AC supply.
答案很简单,生成、传输和处理交流电源要容易得多。
Unfortunately for us, this simplicity in operation doesn’t translate into easier math.
不幸的是,这种操作的简单性并没有转化为更简单的数学。

So far we have discussed about variation of voltage in AC supply, but not about the pattern of this variation.
到目前为止,我们讨论了交流电源中电压的变化,但没有讨论这种变化的模式。

Does the voltage shoot up to a max value all of a sudden and fall back to zero again or does it follow a triangular pattern??
电压是否突然升高到最大值然后又降回零,还是遵循三角形模式?

Square Waveform
方波形

Triangular Waveform
三角波形

Sine Waveform
正弦波形
All these patterns are called waveforms. A waveform is basically a plot of a quantity (in our case voltage/current) against time.
所有这些模式称为波形。波形基本上是一个量(在我们的例子中是电压/电流)与时间的图表。

All these waveforms shown in the figure above and many more, are definite possibilities and many of them have real practical applications.
上面图中显示的所有这些波形以及更多,都是明确的可能性,其中许多具有实际应用。

But the pattern or waveform of our interest at least in this book, is the sine waveform.
但我们在本书中感兴趣的模式或波形至少是正弦波形。

For commercial AC supply pure sine wave is the most preferred waveform, because it’s easier to generate and mathematically simpler to analyze.
对于商业交流电源,纯正弦波是最受欢迎的波形,因为它更容易生成,数学分析也更简单。

7.2.1 Instantaneous value
7.2.1 瞬时值

The value or the magnitude of an alternating quantity at a particular instant of time is known as its instantaneous value.
交变量在特定时刻的数值或大小称为其瞬时值。

For example, in the Voltage-time waveform, the instantaneous values of voltage at instants t 1 , t 2 , t 3 t 1 , t 2 , t 3 t_(1),t_(2),t_(3)t_{1}, t_{2}, t_{3} are v 1 , v 2 v 1 , v 2 v_(1),v_(2)v_{1}, v_{2} and v 3 v 3 v_(3)v_{3} respectively. Instantaneous quantities are always denoted by small letters (v, e, i etc.)
例如,在电压-时间波形中,瞬时电压值在时刻 t 1 , t 2 , t 3 t 1 , t 2 , t 3 t_(1),t_(2),t_(3)t_{1}, t_{2}, t_{3} 分别为 v 1 , v 2 v 1 , v 2 v_(1),v_(2)v_{1}, v_{2} v 3 v 3 v_(3)v_{3} 。瞬时量总是用小写字母表示(v, e, i 等等)。
Voltage
电压

7.2.2 Cycle
7.2.2 循环

A Cycle is a portion of a waveform, which when repeated makes up the entire waveform.
一个周期是波形的一部分,当它重复时构成整个波形。

In the figure below, the shaded portion is the only unique part of the entire waveform, rest of the waveform is just repetitions of this portion.
在下图中,阴影部分是整个波形中唯一独特的部分,其余波形只是这一部分的重复。

A more formal definition would be: an alternating quantity is said to have completed a cycle when it goes through the entire range of positive and negative instantaneous values without reoccurrence.
一个更正式的定义是:当一个交替量经历整个正负瞬时值的范围而没有重复时,称其已完成一个周期。

Obviously it goes without mentioning that the concept of a cycle is only relevant to periodic waveforms like the sine waveform.
显然,不用提也知道,周期的概念仅与正弦波等周期性波形相关。

Do

note that a cycle needn’t start from zero value and end at zero value. It’s only for convenience.
请注意,一个周期不必从零值开始并以零值结束。这只是为了方便。

For example, V max V max  V_("max ")\mathrm{V}_{\text {max }} to the next V max V max  V_("max ")\mathrm{V}_{\text {max }} is also a cycle.
例如,从 V max V max  V_("max ")\mathrm{V}_{\text {max }} 到下一个 V max V max  V_("max ")\mathrm{V}_{\text {max }} 也是一个循环。

7.2.3 Time Period
7.2.3 时间段

The time period is the time taken by an alternating quantity to complete one cycle.
时间周期是交变量完成一个周期所需的时间。

In other words, a cycle of an alternating quantity repeats after every T seconds, where T T T\mathbf{T} denotes the Time period.
换句话说,交替量的周期在每 T 秒后重复,其中 T T T\mathbf{T} 表示时间周期。

7.2.4 Frequency
7.2.4 频率

The number of cycles completed by an alternating quantity in a second is known as its frequency.
一个交变量在一秒钟内完成的周期数称为其频率。

It’s measured in cycles per second or Hertz.
它以每秒周期或赫兹来测量。

So a 60 Hz supply means that the waveform complete 60 cycles in a second. It is denoted by f.
因此,60 赫兹的电源意味着波形在一秒钟内完成 60 个周期。它用 f 表示。

Did you notice something interesting??
你注意到什么有趣的事情了吗?

The definitions for Frequency and Time Period were kind of the reverse of each other.
频率和时间周期的定义有点相反。

One is the time taken for a cycle and other is the number of cycles per time.
一个是一个周期所需的时间,另一个是每单位时间的周期数。

That’s because Frequency and Time Period are inversely related quantities i.e.
这是因为频率和周期是成反比的量,即
T = 1 f T = 1 f T=(1)/(f)T=\frac{1}{f}
So as the frequency increases, time period decreases and vice versa.
因此,频率增加时,周期减少,反之亦然。

7.2.5 Amplitude
7.2.5 振幅

Amplitude is the maximum value (positive or negative) attained by an alternating quantity during its cycle.
幅度是交变量在其周期内达到的最大值(正值或负值)。

7.3 EQUATION
7.3 方程

Now that you have a basic idea about alternating quantities, let’s talk math.
既然你对交替数量有了基本的了解,我们来谈谈数学。

The general equation for an AC sinusoidal voltage is:
交流正弦电压的一般方程是:
v = V max sin ( ω t ) v = V max sin ( ω t ) v=V_(max)sin(omegat)\mathrm{v}=\mathrm{V}_{\max } \sin (\omega \mathrm{t})
This equation can be understood better, if we take a look at the working of a generator.
如果我们看看发电机的工作原理,这个方程就能更好地理解。

Inside a generator a coil is made to rotate with the help of external forces like water or steam or other form of energy.
在发电机内部,线圈在水、蒸汽或其他形式的能量等外部力量的帮助下旋转。

As the coil moves within a magnetic field, voltage is induced in the coil, this is the basic working.
当线圈在磁场中移动时,线圈中会感应出电压,这就是基本工作原理。

The voltage induced is a function of the sine of the angle ( Θ ) ( Θ ) (Theta)(\Theta) the coil makes with the center line.
感应电压是线圈与中心线所成角度 ( Θ ) ( Θ ) (Theta)(\Theta) 的正弦函数。

When the coil is along the center line, no voltage is induced and when the coil is at 90 degrees to the center line, max voltage is induced.
当线圈沿中心线时,没有感应电压;当线圈与中心线成 90 度时,感应电压达到最大。

It is better to represent the voltage as a function of time instead of the physical angle of the coil, so the term ??
将电压表示为时间的函数比表示为线圈的物理角度更好,因此术语??

t is used. It is usually measured in radians.
它被使用。它通常以弧度为单位测量。

Going back to the general equation, v represents the instantaneous value of the voltage and V max V max  V_("max ")\mathrm{V}_{\text {max }} represents the amplitude of the voltage waveform.
回到一般方程,v 代表电压的瞬时值,而 V max V max  V_("max ")\mathrm{V}_{\text {max }} 代表电压波形的幅度。

7.4 AVERAGE VALUE
7.4 平均值

Average value is a pretty common and useful concept in technical fields, yet its meaning is often misunderstood.
平均值是一个在技术领域中相当常见且有用的概念,但其含义常常被误解。

Imagine sand piled up in the form of a mountain over a certain distance, then the average value is that height obtained if the same distance is maintained while the sand is leveled off.
想象一下沙子在一定距离内堆成一座山,那么如果保持相同的距离将沙子抹平,得到的平均值就是那种高度。

From observation itself, it is pretty clear that the average value of the sine waveform over a full cycle is zero.
从观察本身来看,正弦波形在一个完整周期内的平均值显然为零。

So for symmetrical waveforms such as the sine waveform, the average value is calculated over a half cycle rather the full cycle.
因此,对于正弦波等对称波形,平均值是计算在半个周期而不是整个周期上。

Average value = Area under the curve Length of the base = 0 π V m sin ( ω t ) π 0 = V m [ cos ( ω t ) ] o π π = V m [ cos ( π ) + cos ( 0 ) ] π V avg = 2 V m π  Average value  =  Area under the curve   Length of the base  = 0 π V m sin ( ω t ) π 0 = V m [ cos ( ω t ) ] o π π = V m [ cos ( π ) + cos ( 0 ) ] π V avg = 2 V m π {:[" Average value "=(" Area under the curve ")/(" Length of the base ")=(int_(0)^(pi)V_(m)sin(omega t))/(pi-0)],[=(V_(m)[-cos(omega t)]_(o)^(pi))/(pi)=(V_(m)[-cos(pi)+cos(0)])/(pi)],[V_(avg)=(2V_(m))/(pi)]:}\begin{aligned} & \text { Average value }= \frac{\text { Area under the curve }}{\text { Length of the base }}=\frac{\int_{0}^{\pi} V_{m} \sin (\omega t)}{\pi-0} \\ &=\frac{\mathrm{V}_{\mathrm{m}}[-\cos (\omega t)]_{o}^{\pi}}{\pi}=\frac{\mathrm{V}_{\mathrm{m}}[-\cos (\pi)+\cos (0)]}{\pi} \\ & \mathrm{V}_{\mathrm{avg}}=\frac{2 \mathrm{~V}_{\mathrm{m}}}{\pi} \end{aligned}

7.5 RMS VALUE
7.5 RMS 值

For a long time, AC was thought to be a useless form of electricity, primarily because its average value is zero over the full cycle, but experiments showed otherwise.
长期以来,交流电被认为是一种无用的电力,主要是因为它在整个周期内的平均值为零,但实验结果显示并非如此。

When an AC current is passed through a wire, the wire gradually heated up, showing that power is being delivered.
当交流电流通过一根电线时,电线逐渐加热,表明正在传递功率。

How is that possible??
这怎么可能??

It’s possible because both Voltage and Current are changing direction simultaneously and power being the product of these 2 quantities, power is always delivered.
这是可能的,因为电压和电流同时改变方向,而功率是这两个量的乘积,因此功率始终被传递。

Consider this ridiculous example, say someone punched in your face, then he decides to you punch on the back of your head, but if you turn around at the exact moment, you’ll once again be punched in the face.
考虑这个荒谬的例子,比如有人打了你的脸,然后他决定在你后脑勺上打你,但如果你在那个确切的时刻转过身,你又会再次被打到脸。

So as long as both you and the attacker moves simultaneously, all the punches are delivered at the same place, your face (Ouchh!).
只要你和攻击者同时移动,所有的拳头都会打在同一个地方,你的脸上(哎哟!)。

Similarly as long as both current and voltage have same direction, their product is always positive, hence the power is always delivered.
同样,只要电流和电压的方向相同,它们的乘积总是正的,因此功率总是被传递。
The electrons are forced to switch direction ever so quickly that they practically remain still and yet power is being
电子被迫以极快的速度改变方向,以至于它们几乎保持静止,但仍在提供能量

delivered by them.
由他们交付。

Getting an intuitive feel of how AC power is delivered is not the easiest of tasks, but a water analogy might help.
了解交流电如何传输并不是一件容易的事情,但水的类比可能会有所帮助。

When you throw a rock into a pond, the ripples formed will travel throughout the pond causing leaves and other debris to oscillate on the water’s surface.
当你把一块石头扔进池塘时,形成的涟漪会在池塘中传播,导致水面上的叶子和其他杂物摇晃。

This means that energy has been transferred from the rock to the floating leaves, even though no single water molecule has actually travelled all the way from the rock’s impact point to the floating debris.
这意味着能量已经从岩石转移到漂浮的叶子上,即使没有单个水分子实际上从岩石的冲击点一路移动到漂浮的碎片上。

The energy is carried by the waves formed on the water’s surface, in which chains of water molecules push and pull on each other in succession, transferring energy without actually moving anyone around.
能量通过水面形成的波浪传递,其中水分子链相互推拉,依次传递能量,而实际上并没有移动任何物体。
By now it should be pretty clear that average value is not the most effective parameter to measure AC.
到现在为止,平均值显然不是测量交流电的最有效参数。

So we need a better parameter to quantity AC, it is called the RMS or Root Mean Square value.
因此,我们需要一个更好的参数来量化交流电,它被称为均方根值或 RMS 值。

It is developed by comparing the heating effect caused by DC and AC sources.
它是通过比较直流和交流源造成的加热效果来开发的。

The RMS value of AC current is the magnitude of DC current which need to be passed through a resistor, so as to produce same heat as the AC, for the same duration of time.
交流电流的有效值是通过电阻器所需的直流电流的大小,以便在相同的时间内产生与交流电相同的热量。

Say we pass an AC current through a resistor for 1 minute and measure its temperature and it’s found to be say 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}.
假设我们通过一个电阻器通入交流电流 1 分钟,并测量其温度,发现温度为 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}

Now if we connect a DC source to the same resistor for the same duration of 1 minute and the temperature is raised to 100 C 100 C 100^(@)C100^{\circ} \mathrm{C}. Then, that value of DC current gives the RMS value of the AC current.
现在如果我们将一个直流电源连接到同一个电阻上,持续 1 分钟,并将温度提高到 100 C 100 C 100^(@)C100^{\circ} \mathrm{C} 。那么,该直流电流的值给出了交流电流的有效值。
P DC = P AC (avg) I RMS 2 R = I m 2 R 2 I RMS = I m 2 Similarly, V RMS = V m 2 P DC = P AC  (avg)  I RMS 2 R = I m 2 R 2 I RMS = I m 2  Similarly,  V RMS = V m 2 {:[P_(DC)=P_(AC" (avg) ")],[=>I_(RMS)^(2)R=(I_(m)^(2)R)/(2)],[:.I_(RMS)=(I_(m))/(sqrt2)],[" Similarly, "V_(RMS)=(V_(m))/(sqrt2)]:}\begin{aligned} & \mathrm{P}_{\mathrm{DC}}=\mathrm{P}_{\mathrm{AC} \text { (avg) }} \\ \Rightarrow & \mathrm{I}_{\mathrm{RMS}}{ }^{2} \mathrm{R}=\frac{\mathrm{I}_{\mathrm{m}}{ }^{2} \mathrm{R}}{2} \\ \therefore & \mathrm{I}_{\mathrm{RMS}}=\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}} \\ \text { Similarly, } & \mathrm{V}_{\mathrm{RMS}}=\frac{\mathrm{V}_{\mathrm{m}}}{\sqrt{2}} \end{aligned}
Hence the RMS value of AC is 1 / 2 1 / 2 1//sqrt21 / \sqrt{2} or 0.707 times the maximum value.
因此,交流的有效值为 1 / 2 1 / 2 1//sqrt21 / \sqrt{2} 或最大值的 0.707 倍。

When you measure the voltage of your power socket, the reading indicates the RMS value.
当你测量电源插座的电压时,读数表示有效值。

Unless specifically mentioned, all values related to AC voltages and currents are RMS values.
除非特别提及,所有与交流电压和电流相关的值均为有效值。

7.6 PHASE
7.6 阶段

In our general equation, we have only considered sinusoids having zero value at ?? t = 0 , π t = 0 , π t=0,pi\mathrm{t}=0, \pi and maximum value at ? ? ?? ? t = t = t=\mathrm{t}= π / 2 , 3 π / 2 π / 2 , 3 π / 2 pi//2,3pi//2\pi / 2,3 \pi / 2.
在我们的通用方程中,我们只考虑了在 t = 0 , π t = 0 , π t=0,pi\mathrm{t}=0, \pi 时值为零,在 ? ? ?? 时值为最大值的正弦波 ? t = t = t=\mathrm{t}= π / 2 , 3 π / 2 π / 2 , 3 π / 2 pi//2,3pi//2\pi / 2,3 \pi / 2

But this needn’t be the case always, sinusoids can be shifted to the left or the right as shown below.
但这并不总是如此,正弦波可以向左或向右移动,如下所示。




The waveforms are identical in all aspects, but the second waveform starts earlier than the first, and the third waveform has a delayed start than the first. In other words, the second wave leads the first wave and the third wave lags the first. The lead or lag of a waveform is denoted by ϕ ϕ phi\phi known as the phase angle.
波形在各个方面都是相同的,但第二个波形的开始时间比第一个波形早,而第三个波形的开始时间比第一个波形晚。换句话说,第二个波浪领先于第一个波浪,第三个波浪滞后于第一个波浪。波形的领先或滞后用 ϕ ϕ phi\phi 表示,称为相位角。

Considering this concept of phase angle, we can modify the general equation for an AC sinusoidal voltage as,
考虑到相位角的概念,我们可以将交流正弦电压的一般方程修改为,
v = V max sin ( ω t + ϕ ) v = V max sin ( ω t + ϕ ) v=V_(max)sin(omega t+phi)v=V_{\max } \sin (\omega t+\phi)
The difference between phase angles of 2 sinusoids is called the phase difference.
两个正弦波的相位角之差称为相位差。

7.7 COMPLEX NUMBERS
7.7 复数

DC circuit analysis we did so far were pretty easy, because voltages and currents could be added or subtracted directly, but now that we introduced the demon called phase, things are about to get a little trickier.
到目前为止,我们所做的直流电路分析都很简单,因为电压和电流可以直接相加或相减,但现在我们引入了一个叫做相位的恶魔,事情将变得有些棘手。

To tackle this problem, we need the help of a mathematical tool called complex numbers.
为了解决这个问题,我们需要一种叫做复数的数学工具的帮助。

Complex numbers can be represented in 2 forms:
复数可以用两种形式表示:
  1. Rectangular Form: It’s the most commonly used representation for complex numbers.
    矩形形式:这是复数最常用的表示方式。

    Note that in electrical engineering, letter j j j\mathbf{j} is used to denote the imaginary part, instead of i i i\mathbf{i}, to not get mixed up with symbol for current.
    请注意,在电气工程中,字母 j j j\mathbf{j} 用于表示虚部,而不是 i i i\mathbf{i} ,以免与电流符号混淆。


  2. Polar Form: In polar form, a quantity is denoted in terms of its magnitude and the angle it makes with the positive x -axis.
    极坐标形式:在极坐标形式中,一个量用其大小和与正 x 轴的夹角表示。

Magnitude Angle
幅度角

Imaginary axis
虚轴

Converting between the two forms is very easy and will come in handy later.
在这两种形式之间转换非常简单,稍后会派上用场。

Rectangular to Polar Form
矩形到极坐标形式

Z = X 2 + Y 2 θ = tan 1 Y X Z = X 2 + Y 2 θ = tan 1 Y X {:[Z=sqrt(X^(2)+Y^(2))],[theta=tan^(-1)((Y)/(X))]:}\begin{aligned} \mathrm{Z} & =\sqrt{X^{2}+Y^{2}} \\ \theta & =\tan ^{-1} \frac{Y}{X} \end{aligned}

Polar to Rectangular Form
极坐标到直角坐标形式

X = Z cos θ Y = Z sin θ X = Z cos θ Y = Z sin θ {:[X=Zcos theta],[Y=Zsin theta]:}\begin{aligned} & \mathrm{X}=\mathrm{Z} \cos \theta \\ & \mathrm{Y}=\mathrm{Z} \sin \theta \end{aligned}

7.8 OPERATIONS USING COMPLEX NUMBERS
7.8 使用复数的运算

7.8.1 Addition/Subtraction:
7.8.1 加法/减法:

Complex Addition/ Subtraction is as easy as they some.
复杂的加法/减法和他们一样简单。

To add two complex number’s, simply add the real and imaginary parts separately.
要将两个复数相加,只需分别将实部和虚部相加。

Similarly, to subtract two complex number’s, simply subtract the real and imaginary parts separately.
同样,要减去两个复数,只需分别减去实部和虚部。
If, C 1 = X 1 + j Y 1 and C 2 = X 2 + j Y 2  If,  C 1 = X 1 + j Y 1  and  C 2 = X 2 + j Y 2 " If, "C_(1)=X_(1)+jY_(1)" and "C_(2)=X_(2)+jY_(2)\text { If, } C_{1}=X_{1}+j Y_{1} \text { and } C_{2}=X_{2}+j Y_{2}
Then,
然后,
C 1 ± C 2 = ( X 1 ± X 2 ) + j ( Y 1 ± Y 2 ) C 1 ± C 2 = X 1 ± X 2 + j Y 1 ± Y 2 C_(1)+-C_(2)=(X_(1)+-X_(2))+j(Y_(1)+-Y_(2))C_{1} \pm C_{2}=\left(X_{1} \pm X_{2}\right)+j\left(Y_{1} \pm Y_{2}\right)

7.8.2 Multiplication:
7.8.2 乘法:

To multiply two Complex numbers in rectangular form, each term of the first complex number is multiplied separately by each term of the second complex number.
要将两个矩形形式的复数相乘,第一个复数的每一项分别与第二个复数的每一项相乘。

Then the real parts and the imaginary parts are separated out to obtain the product complex number.
然后将实部和虚部分开,以获得乘积复数。
C 1 C 2 = ( X 1 + j Y 1 ) ( X 2 + j Y 2 ) = X 1 X 2 + j X 1 Y 2 + j Y 1 X 2 + j Y 1 j Y 2 = ( X 1 X 2 Y 1 Y 2 ) + j ( X 1 Y 2 + X 2 Y 1 ) C 1 C 2 = X 1 + j Y 1 X 2 + j Y 2 = X 1 X 2 + j X 1 Y 2 + j Y 1 X 2 + j Y 1 j Y 2 = X 1 X 2 Y 1 Y 2 + j X 1 Y 2 + X 2 Y 1 {:[C_(1)C_(2)=(X_(1)+jY_(1))*(X_(2)+jY_(2))],[=X_(1)X_(2)+jX_(1)Y_(2)+jY_(1)X_(2)+jY_(1)jY_(2)],[=(X_(1)X_(2)-Y_(1)Y_(2))+j(X_(1)Y_(2)+X_(2)Y_(1))]:}\begin{aligned} C_{1} C_{2} & =\left(X_{1}+j Y_{1}\right) \cdot\left(X_{2}+j Y_{2}\right) \\ & =X_{1} X_{2}+j X_{1} Y_{2}+j Y_{1} X_{2}+j Y_{1} j Y_{2} \\ & =\left(X_{1} X_{2}-Y_{1} Y_{2}\right)+j\left(X_{1} Y_{2}+X_{2} Y_{1}\right) \end{aligned}
Complex multiplication is a lot easier in polar form, the magnitudes are multiplied and the angles added algebraically.
在极坐标形式下,复数乘法要简单得多,模长相乘,角度代数相加。
C 1 C 2 = Z 1 θ 1 Z 2 θ 1 = Z 1 Z 2 ( θ 1 + θ 2 ) C 1 C 2 = Z 1 θ 1 Z 2 θ 1 = Z 1 Z 2 θ 1 + θ 2 {:[C_(1)C_(2)=Z_(1)/_theta_(1)*Z_(2)/_theta_(1)],[=Z_(1)Z_(2)/_(theta_(1)+theta_(2))]:}\begin{aligned} \mathrm{C}_{1} \mathrm{C}_{2} & =\mathrm{Z}_{1} \angle \theta_{1} \cdot \mathrm{Z}_{2} \angle \theta_{1} \\ & =\mathrm{Z}_{1} \mathrm{Z}_{2} \angle\left(\theta_{1}+\theta_{2}\right) \end{aligned}

7.8.2 Division:
7.8.2 部门:

In rectangular form, Complex multiplication is done by multiplying both the numerator and denominator with the denominator of the denominator and separating out the real and imaginary parts.
在矩形形式中,复数乘法是通过将分子和分母都乘以分母的分母,并分离出实部和虚部来完成的。
C 1 C 2 = ( X 1 + j Y 1 ) ( X 2 j Y 2 ) ( X 2 + j Y 2 ) ( X 2 j Y 2 ) C 1 C 2 = X 1 + j Y 1 X 2 j Y 2 X 2 + j Y 2 X 2 j Y 2 (C_(1))/(C_(2))=((X_(1)+jY_(1))(X_(2)-jY_(2)))/((X_(2)+jY_(2))(X_(2)-jY_(2)))\frac{C_{1}}{C_{2}}=\frac{\left(X_{1}+j Y_{1}\right)\left(X_{2}-j Y_{2}\right)}{\left(X_{2}+j Y_{2}\right)\left(X_{2}-j Y_{2}\right)}
In polar form, the magnitudes of the numerator is divided by the magnitude or the denominator and the angle is subtracted from the other.
在极坐标形式中,分子的大小除以分母的大小,角度从另一个角度中减去。
C 1 C 2 = Z 1 θ 1 Z 2 θ 2 = Z 1 Z 2 ( θ 1 θ 2 ) C 1 C 2 = Z 1 θ 1 Z 2 θ 2 = Z 1 Z 2 θ 1 θ 2 {:[(C_(1))/(C_(2))=(Z_(1)/_theta_(1))/(Z_(2)/_theta_(2))],[=(Z_(1))/(Z_(2))/_(theta_(1)-theta_(2))]:}\begin{aligned} \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}} & =\frac{\mathrm{Z}_{1} \angle \theta_{1}}{\mathrm{Z}_{2} \angle \theta_{2}} \\ & =\frac{\mathrm{Z}_{1}}{\mathrm{Z}_{2}} \angle\left(\theta_{1}-\theta_{2}\right) \end{aligned}
In the next chapter, we’ll put all that we learnt in this chapter in understanding AC circuit further.
在下一章中,我们将把本章所学的内容应用于进一步理解交流电路。

8. AC CIRCUITS
8. 交流电路

8.1 AC THROUGH RESISTANCE
8.1 交流通过电阻

Before we delve into the deep end, we will study the behavior of AC when passed through our amazing trio of resistors, capacitors, and inductors and build from there.
在深入探讨之前,我们将研究交流电通过我们惊人的三重电阻、电容和电感时的行为,并从那里开始。

Resistance is the easiest component to analyze in AC circuits, because it behaves the same way for DC as well as AC.
在交流电路中,电阻是最容易分析的元件,因为它在直流和交流中表现相同。
Consider the circuit shown below, where an AC voltage v = v = v=\mathrm{v}= V m sin ( t ) V m sin t V_(m)sin(◻_(t))\mathrm{V}_{\mathrm{m}} \sin \left(\square_{\mathrm{t}}\right) is applied across a resistor R .
考虑下面所示的电路,其中交流电压 v = v = v=\mathrm{v}= V m sin ( t ) V m sin t V_(m)sin(◻_(t))\mathrm{V}_{\mathrm{m}} \sin \left(\square_{\mathrm{t}}\right) 施加在电阻 R 上。

Obviously a current will flow through the circuit, which as per the ohms law is given by,
显然,电流将通过电路流动,根据欧姆定律,电流由以下公式给出,
i = v R = V m sin ( ω t ) R = ( V m R ) sin ( ω t ) i = v R = V m sin ( ω t ) R = V m R sin ( ω t ) {:[i=(v)/(R)=(V_(m)sin(omega t))/(R)],[=((V_(m))/(R))sin(omega t)]:}\begin{aligned} i=\frac{v}{R} & =\frac{V_{m} \sin (\omega t)}{R} \\ & =\left(\frac{V_{m}}{R}\right) \sin (\omega t) \end{aligned}
If you compare this equation with the equation for the applied voltage, you can identify that the current and applied voltage are in phase (phase shift ϕ = 0 ϕ = 0 phi=0\phi=0 ) and also the maximum values are related as, I m = V m / R I m = V m / R I_(m)=V_(m)//R\mathrm{I}_{\mathrm{m}}=\mathrm{V}_{\mathrm{m}} / \mathrm{R}.
如果将这个方程与施加电压的方程进行比较,可以发现电流和施加电压是同相的(相位差 ϕ = 0 ϕ = 0 phi=0\phi=0 ),并且最大值之间的关系为 I m = V m / R I m = V m / R I_(m)=V_(m)//R\mathrm{I}_{\mathrm{m}}=\mathrm{V}_{\mathrm{m}} / \mathrm{R}

Both the current and the voltage waveforms are exactly the same and only difference is that the voltage waveform is R times bigger than the current waveform, as shown below.
电流和电压波形完全相同,唯一的区别是电压波形比电流波形大 R 倍,如下所示。


8.2 AC THROUGH INDUCTOR
8.2 交流通过电感器

We have learnt that, an inductor is a component that resists change in current, due to its self-inductance property.
我们了解到,电感器是一个由于其自感特性而抵抗电流变化的元件。

When an Inductor is connected to an AC source, the current will repeatedly change in the magnitude and direction.
当电感器连接到交流电源时,电流的大小和方向会反复变化。

The
inductor will try to oppose this change by inducing a voltage across it, which limits the current in the circuit.
电感器将通过在其上感应电压来试图抵抗这种变化,从而限制电路中的电流。

This opposition due to the inductance is called inductive reactance.
由于电感引起的这种反对称为电感抗。

Inductive reactance is denoted by the symbol X L X L X_(L)X_{L} and is measured in ohms.
感抗用符号 X L X L X_(L)X_{L} 表示,单位为欧姆。

Inductive reactance is dependent on the frequency of the applied AC voltage, as given by the relation,
感抗依赖于施加的交流电压的频率,如下关系所示,
X L = ω L = 2 π f L X L = ω L = 2 π f L X_(L)=omega L=2pi fLX_{L}=\omega L=2 \pi f L
As the frequency of the applied voltage increases, the Inductive reactance increases and hence the voltage drop across it also increases.
随着施加电压的频率增加,感抗增加,因此其上的电压降也增加。

The inductor can be thought of as a variable resistor, whose opposition to the current is controlled by the frequency of the supply voltage.
电感器可以被视为一个可变电阻,其对电流的阻抗由供电电压的频率控制。

Consider an AC voltage applied to a pure inductor (the coil offers no resistance) of inductance L L LL, as shown in the circuit below.
考虑施加在纯电感(线圈没有电阻)的电路中的交流电压,如下图所示。

The current flowing through the circuit can be calculated as follows:
通过电路的电流可以按如下方式计算:
v L = L d i d t v L = L d i d t v_(L)=-L((d)i)/((d)t)\mathrm{v}_{\mathrm{L}}=-\mathrm{L} \frac{\mathrm{~d} i}{\mathrm{~d} t}
Since the applied emf and the induced emf oppose each other,
由于施加的电动势和感应电动势相互抵消,
V = V L V m sin ω t = L d i d t i = V m sin ω t L d t = V m ω L cos ω t i = V m X L sin ( ω t π 2 ) V = V L V m sin ω t = L d i d t i = V m sin ω t L d t = V m ω L cos ω t i = V m X L sin ω t π 2 {:[V=-V_(L)],[=>V_(m)sin omega t=L((d)i)/((d)t)],[i=int(V_(m)sin omega t)/((L))dt=(-V_(m))/(omega(L))cos omega t],[i=(V_(m))/(X_(L))sin(omega t-(pi)/(2))]:}\begin{aligned} & \mathrm{V}=-\mathrm{V}_{\mathrm{L}} \\ & \Rightarrow \mathrm{~V}_{\mathrm{m}} \sin \omega t=\mathrm{L} \frac{\mathrm{~d} i}{\mathrm{~d} t} \\ & i=\int \frac{\mathrm{V}_{\mathrm{m}} \sin \omega t}{\mathrm{~L}} \mathrm{~d} t=\frac{-\mathrm{V}_{\mathrm{m}}}{\omega \mathrm{~L}} \cos \omega t \\ & i=\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{X}_{\mathrm{L}}} \sin \left(\omega t-\frac{\pi}{2}\right) \end{aligned}
This derivation itself isn’t very important, but what is important are these conclusions than can be made from it.
这个推导本身并不是很重要,但从中得出的这些结论才是重要的。
  1. Comparing the equation with that of supply voltage, it is obvious that the current has a phase angle of π / 2 π / 2 -pi//2-\pi / 2 or 90 90 -90^(@)-90^{\circ} i.e.the current lags the applied voltage by π / 2 π / 2 pi//2\pi / 2 or 90 90 90^(@)90^{\circ}.
    将该方程与供电电压的方程进行比较,很明显电流的相位角为 π / 2 π / 2 -pi//2-\pi / 2 90 90 -90^(@)-90^{\circ} ,即电流滞后施加电压 π / 2 π / 2 pi//2\pi / 2 90 90 90^(@)90^{\circ}
  2. The maximum magnitude of current is related to the maximum magnitude of the applied voltage as,
    电流的最大幅度与施加电压的最大幅度有关,如下所示:
I m = V m X L I m = V m X L I_(m)=(V_(m))/(X_(L))I_{m}=\frac{V_{m}}{X_{L}}

8.3 AC THROUGH CAPACITOR
8.3 交流通过电容器

When an AC voltage is applied to a capacitor, a voltage is developed across its plates, as the electrostatic charge is built up.
当交流电压施加到电容器上时,电容器的极板之间会产生电压,因为静电荷在积累。

This impressed voltage opposes the applied voltage and limits the flow of current in the circuit.
这个感应电压与施加电压相反,并限制电路中的电流流动。

This opposition caused by capacitance is called capacitive reactance ( X c X c X_(c)\mathrm{X}_{\mathrm{c}} ) and is measured in ohms. It is similar to Inductive reactance in a lot of ways, but the key difference is that the capacitive reactance opposes the change in voltage, whereas the inductive reactance opposes the change in current caused as a result of the applied voltage.
这种由电容引起的反对称为电容反应( X c X c X_(c)\mathrm{X}_{\mathrm{c}} ),以欧姆为单位进行测量。它在很多方面与电感反应相似,但关键区别在于电容反应反对电压的变化,而电感反应则反对由于施加电压而引起的电流变化。

Capacitive reactance is also frequency dependent, as given by the relation,
电容性反应也依赖于频率,如下关系所示,
X C = 1 ω C = 1 2 π f C X C = 1 ω C = 1 2 π f C X_(C)=(1)/(omega C)=(1)/(2pi fC)X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}
Obviously as you can see, unlike inductive reactance, the capacitive reactance is inversely proportional to the frequency of the supply voltage.
显然,如你所见,与感抗不同,电容抗是与供电电压的频率成反比的。

Why it is so, is beyond the scope of this book.
这背后的原因超出了本书的范围。

Think of it this way, as voltage changes faster, lesser the time for charge to accumulate, hence lesser capacitive reactance.
这样想吧,电压变化得越快,电荷积累的时间就越少,因此电容性反应就越小。

The current flowing in the circuit, shown above can be determined as follows:
上面所示电路中的电流可以通过以下方式确定:
i = d q d t = C d v d t i = C d ( V m sin ω t ) d t = V m C d ( sin ω t ) d t = V m ω C cos ω t i = V m X C sin ( ω t + π 2 ) i = d q d t = C d v d t i = C d V m sin ω t d t = V m C d ( sin ω t ) d t = V m ω C cos ω t i = V m X C sin ω t + π 2 {:[i=(dq)/((d)t)=C((d)v)/((d)t)],[i=C((d)(V_(m)sin omega t))/(dt)],[=V_(m)C((d)(sin omega t))/(dt)=V_(m)omegaCcos omega t],[i=(V_(m))/(X_(C))sin(omega t+(pi)/(2))]:}\begin{aligned} i & =\frac{\mathrm{d} q}{\mathrm{~d} t}=\mathrm{C} \frac{\mathrm{~d} v}{\mathrm{~d} t} \\ i & =\mathrm{C} \frac{\mathrm{~d}\left(\mathrm{~V}_{\mathrm{m}} \sin \omega t\right)}{\mathrm{d} t} \\ & =\mathrm{V}_{\mathrm{m}} \mathrm{C} \frac{\mathrm{~d}(\sin \omega t)}{\mathrm{d} t}=\mathrm{V}_{\mathrm{m}} \omega \mathrm{C} \cos \omega t \\ i & =\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{X}_{\mathrm{C}}} \sin \left(\omega t+\frac{\pi}{2}\right) \end{aligned}
Again this derivation itself isn’t very important, but some inferences can be made from it.
这个推导本身并不是很重要,但可以从中得出一些推论。
  1. Comparing the equation with that of supply voltage, it is obvious that the current has a phase angle of π / 2 π / 2 pi//2\pi / 2 or 90 90 90^(@)90^{\circ} i.e.the current leads the applied voltage by π / 2 π / 2 pi//2\pi / 2 or 90 90 90^(@)90^{\circ}.
    将该方程与供电电压的方程进行比较,很明显电流的相位角为 π / 2 π / 2 pi//2\pi / 2 90 90 90^(@)90^{\circ} ,即电流超前施加电压 π / 2 π / 2 pi//2\pi / 2 90 90 90^(@)90^{\circ}
  2. The maximum magnitude of current is related to the maximum magnitude of the applied voltage as,
    电流的最大幅度与施加电压的最大幅度有关,如下所示:
I m = V m X C I m = V m X C I_(m)=(V_(m))/(X_(C))I_{m}=\frac{V_{m}}{X_{C}}

8.4 IMPEDANCE
8.4 阻抗

Impedance is defined as the opposition to the flow of alternating current in a circuit.
阻抗被定义为电路中对交流电流流动的阻碍。

As we have seen, in a pure inductive circuit, the opposition was the inductive reactance, in a pure capacitive circuit, it was the capacitive reactance and in a resistive circuit it was the resistance.
正如我们所看到的,在纯电感电路中,阻抗是电感抗,在纯电容电路中,阻抗是电容抗,而在电阻电路中,阻抗是电阻。

Similarly, in a circuit with one or more of these elements, in any combination, the impedance is the total current limiting element in the circuit.
类似地,在一个包含一个或多个这些元件的电路中,无论组合如何,阻抗都是电路中总的电流限制元件。

It is denoted by Z and its unit is obviously Ohm.
它用 Z 表示,单位显然是欧姆。

As seen in the previous sections, the Inductive reactance introduces a phase shift of 90 90 -90^(@)-90^{\circ} to the current and capacitive reactance introduces a phase shift of + 90 + 90 +90^(@)+90^{\circ} to the current. Whereas the resistance doesn’t cause any phase shift.
如前面各节所述,感抗对电流引入了 90 90 -90^(@)-90^{\circ} 的相位偏移,而容抗对电流引入了 + 90 + 90 +90^(@)+90^{\circ} 的相位偏移。 而电阻则不会引起任何相位偏移。

Hence the inductive part of the circuit leads the resistive part of the circuit by 90 90 90^(@)90^{\circ} and similarly, the capacitive part of the circuit lags the resistive part of the circuit by 90 90 90^(@)90^{\circ}. For this reason, Impedance is a complex quantity that has a magnitude and a phase.
因此,电路的感性部分领先于电路的阻性部分 90 90 90^(@)90^{\circ} ,而电路的容性部分滞后于电路的阻性部分 90 90 90^(@)90^{\circ} 。因此,阻抗是一个具有大小和相位的复数量。
Here’s an example of a phasor diagram for a circuit containing all these elements.
这是一个包含所有这些元件的电路相量图的示例。


X L X L X_(L)X_{L} and X C X C X_(C)X_{C} lie on the Imaginary axis of the complex plane. Therefore to represent them, X L X L X_(L)\mathbf{X}_{\mathbf{L}} is multiplied by j j j\mathbf{j} and X c X c X_(c)\mathbf{X}_{\mathbf{c}} is multiplied -j. Here are some examples on how to calculate Impedance of a circuit.
X L X L X_(L)X_{L} X C X C X_(C)X_{C} 位于复平面的虚轴上。因此,为了表示它们, X L X L X_(L)\mathbf{X}_{\mathbf{L}} 乘以 j j j\mathbf{j} X c X c X_(c)\mathbf{X}_{\mathbf{c}} 乘以 -j。以下是一些计算电路阻抗的示例。

Z = 5 j Z = 5 j Z=-5jZ=-5 j
  • R X C = 3 Ω X C = 4 Ω Z = 2 7 j R X C = 3 Ω X C = 4 Ω Z = 2 7 j ^(R)X_(C)=3OmegaquadX_(C)=4OmegaquadZ=2-7j^{\mathrm{R}} \mathrm{X}_{\mathrm{C}}=3 \Omega \quad \mathrm{X}_{\mathrm{C}}=4 \Omega \quad \mathrm{Z}=2-7 \mathrm{j}
By the way, impedances in series/parallel are calculated in the same way as resistances in series/parallel.
顺便提一下,串联/并联的阻抗计算方式与串联/并联的电阻相同。
Impedance in series: Z = Z 1 + Z 2 + Z 3 + Z = Z 1 + Z 2 + Z 3 + Z=Z_(1)+Z_(2)+Z_(3)+dots\mathrm{Z}=\mathrm{Z}_{1}+\mathrm{Z}_{2}+\mathrm{Z}_{3}+\ldots
串联中的阻抗: Z = Z 1 + Z 2 + Z 3 + Z = Z 1 + Z 2 + Z 3 + Z=Z_(1)+Z_(2)+Z_(3)+dots\mathrm{Z}=\mathrm{Z}_{1}+\mathrm{Z}_{2}+\mathrm{Z}_{3}+\ldots

Impedance in parallel: 1 Z = 1 Z 1 + 1 Z 2 + 1 Z 3 + 1 Z = 1 Z 1 + 1 Z 2 + 1 Z 3 + (1)/(Z)=(1)/(Z_(1))+(1)/(Z_(2))+(1)/(Z_(3))+dots\frac{1}{\mathrm{Z}}=\frac{1}{\mathrm{Z}_{1}}+\frac{1}{\mathrm{Z}_{2}}+\frac{1}{\mathrm{Z}_{3}}+\ldots
并联阻抗: 1 Z = 1 Z 1 + 1 Z 2 + 1 Z 3 + 1 Z = 1 Z 1 + 1 Z 2 + 1 Z 3 + (1)/(Z)=(1)/(Z_(1))+(1)/(Z_(2))+(1)/(Z_(3))+dots\frac{1}{\mathrm{Z}}=\frac{1}{\mathrm{Z}_{1}}+\frac{1}{\mathrm{Z}_{2}}+\frac{1}{\mathrm{Z}_{3}}+\ldots
Here are some more examples:
这里还有一些例子:

8.5 POWER & POWER FACTOR
8.5 功率与功率因数

As mentioned earlier in the book, electrical power is the product of voltage and current.
如书中早前提到的,电功率是电压和电流的乘积。

But how does this translate to AC circuits, where both voltage and current both vary sinusoidally?
但这如何转化为交流电路,其中电压和电流都呈正弦波变化?

Does the power also vary?
功率也会变化吗?
Generally if,
一般来说,如果,
v = V m sin ( ω t + θ v ) & i = I m sin ( ω t + θ i ) v = V m sin ω t + θ v & i = I m sin ω t + θ i {:[v=V_(m)sin(omega t+theta_(v))&],[i=I_(m)sin(omega t+theta_(i))]:}\begin{aligned} & \mathrm{v}=\mathrm{V}_{\mathrm{m}} \sin \left(\omega t+\theta_{\mathrm{v}}\right) \& \\ & \mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \left(\omega t+\theta_{\mathrm{i}}\right) \end{aligned}
Then,
然后,
P avg = V m I m 2 cos ( ϕ ) P avg = V m I m 2 cos ( ϕ ) P_(avg)=(V_(m)I_(m))/(2)cos(phi)P_{\mathrm{avg}}=\frac{\mathrm{V}_{\mathrm{m}} \mathrm{I}_{\mathrm{m}}}{2} \cos (\phi)
Where, ϕ = θ v θ i ϕ = θ v θ i phi=theta_(v)-theta_(i)\phi=\theta_{\mathrm{v}}-\theta_{\mathrm{i}}
哪里, ϕ = θ v θ i ϕ = θ v θ i phi=theta_(v)-theta_(i)\phi=\theta_{\mathrm{v}}-\theta_{\mathrm{i}}
The term cos ( ϕ ) cos ( ϕ ) cos(phi)\cos (\phi) is called the Power factor of the circuit.
该术语 cos ( ϕ ) cos ( ϕ ) cos(phi)\cos (\phi) 称为电路的功率因数。

Resistive circuit:
电阻电路:

In the earlier section we saw that in a purely resistive circuit, the voltage and the current are in phase.
在前面的部分,我们看到在一个纯电阻电路中,电压和电流是同相的。

In the first half cycle, both voltage and current are positive, therefore power is the product of these two quantities is also positive in the first half cycle.
在第一个半周期中,电压和电流都是正值,因此这两个量的乘积也在第一个半周期中为正。

Similarly, in the second half cycle, both the voltage and current are negative, therefore the power is positive in this half cycle too.
同样,在下半周期中,电压和电流都是负值,因此在这个半周期中功率也是正值。

Hence the average power is always positive in a pure resistive circuit.
因此,在纯电阻电路中,平均功率始终为正。

In a purely resistive circuit, phase difference is zero, hence the power factor is equal to 1 .
在纯电阻电路中,相位差为零,因此功率因数等于 1。

Inductive circuit:
感应电路:

In a purely inductive circuit, the current lags the voltage by 90 90 90^(@)90^{\circ}. Therefore the power factor is zero and consequently the average power is also zero.
在纯电感电路中,电流滞后于电压 90 90 90^(@)90^{\circ} 。因此,功率因数为零,因此平均功率也为零。

In the interval A, both the voltage and current are positive, therefore the power is also positive.
在区间 A 中,电压和电流都是正值,因此功率也是正值。

During this interval, the power is absorbed by the inductor to set up a magnetic field.
在此期间,电感器吸收电能以建立磁场。

In the interval B, the current is positive, but the voltage is negative, therefore the power is negative.
在区间 B 中,电流为正,但电压为负,因此功率为负。

Negative power means the power is being returned back to the source, as the magnetic field collapses.
负功意味着功率正在返回到源头,因为磁场正在崩溃。

This process continues for the next 2 intervals C C CC and D D DD as well.
这个过程在接下来的两个时间间隔 C C CC D D DD 中也会继续。

So it is very evident that over a full cycle, the average power absorbed by the inductor is zero.
因此,很明显,在一个完整的周期内,电感器吸收的平均功率为零。

To sum up, a pure inductor doesn’t dissipate energy like a resistor, it only stores energy in the form of magnetic field for a while, then releases it back.
总之,纯电感器不像电阻器那样消耗能量,它只是以磁场的形式存储能量一段时间,然后再释放出来。

Capacitive circuit:
电容电路:

In a purely capacitive circuit, the current leads the voltage by 90 90 90^(@)90^{\circ}.
在纯电容电路中,电流超前电压 90 90 90^(@)90^{\circ}

Therefore, similar to a purely inductive circuit, here too the power factor and the average power are zero.
因此,与纯电感电路类似,这里的功率因数和平均功率都是零。

In the interval A, both the voltage and current are positive, therefore the power is also positive.
在区间 A 中,电压和电流都是正值,因此功率也是正值。

During this interval, the power is absorbed by the capacitor to build up charge and increase electrostatic energy.
在此期间,电容器吸收电能以积累电荷并增加静电能。

In the interval B, the voltage is positive, but the current is negative, therefore the power is negative.
在区间 B 中,电压为正,但电流为负,因此功率为负。

Now the capacitor starts discharging and the returns gathered electrostatic energy back to the source.
现在电容器开始放电,并将收集到的静电能量返回给源头。

This process continues for the next 2 intervals C C CC and D D DD as well. So over the full cycle, the average power absorbed by the capacitor is zero.
这个过程在接下来的两个时间间隔 C C CC D D DD 中也会继续。因此,在整个周期内,电容器吸收的平均功率为零。

By now, you may have figured out that, resistor is the only component that absorbs and dissipates energy, whereas inductors and capacitors can only store energy for a while.
到现在为止,你可能已经明白,电阻器是唯一一个吸收和耗散能量的元件,而电感器和电容器只能暂时储存能量。
Returning to our earlier discussion, the power factor of a circuit tells us the amount of resistance contributing to the total impedance of the circuit.
回到我们之前的讨论,电路的功率因数告诉我们对电路总阻抗贡献的电阻量。

Mathematically,
数学上,
cos ϕ = R | Z | cos ϕ = R | Z | cos phi=(R)/(|Z|)\cos \phi=\frac{\mathrm{R}}{|\mathrm{Z}|}
To sum up, the Electrical power in an AC circuit, depends on three factors: Voltage, Current and the Power factor.
总之,交流电路中的电功率取决于三个因素:电压、电流和功率因数。

8.6 SERIES R-L CIRCUIT
8.6 系列 R-L 电路

Before we close out this chapter, we’ll take a look at few circuits with combinations of these components and their response to AC, to solidify what we’ve learnt so far.
在我们结束这一章之前,我们将看看几个电路,这些电路结合了这些组件及其对交流电的响应,以巩固我们迄今为止所学的内容。

First stop is the RL series circuit, shown below.
第一个停靠点是下面所示的 RL 系列电路。

Since you are more or less familiar with complex number math at this stage, we’ll go that route.
既然你在这个阶段或多或少熟悉复数数学,我们就走这条路。

The total impedance
总阻抗

of this circuit is Z = 5 + j 7 Z = 5 + j 7 Z=5+j7Z=5+j 7 (Always remember to multiply X L X L X_(L)X_{L} by j) and the current and the power factor can be calculated as,
该电路的 Z = 5 + j 7 Z = 5 + j 7 Z=5+j7Z=5+j 7 (始终记得将 X L X L X_(L)X_{L} 乘以 j),电流和功率因数可以计算为,
I = V Z = 12 5 + j 7 = 0.81 j 1.13 cos ϕ = R | Z | = 0.58 I = V Z = 12 5 + j 7 = 0.81 j 1.13 cos ϕ = R | Z | = 0.58 {:[I=(V)/(Z)=(12)/(5+j7)],[=0.81-j 1.13],[cos phi=(R)/(|Z|)=0.58]:}\begin{aligned} & I=\frac{V}{Z}=\frac{12}{5+j 7} \\ &=0.81-j 1.13 \\ & \cos \phi=\frac{R}{|Z|}=0.58 \end{aligned}
Generally, in an RL series circuit, the current lags the applied voltage by an angle less than 90 90 90^(@)90^{\circ}.
通常,在 RL 串联电路中,电流滞后于施加的电压,滞后角小于 90 90 90^(@)90^{\circ}

If the resistance is very high compared to inductive reactance, then the phase difference will be closer to zero and if resistance is
如果电阻与电感抗相比非常高,那么相位差将更接近于零,如果电阻是

negligible, then the phase difference will be close to 90 90 90^(@)90^{\circ}. In our example, the current lags the voltage by 54.54 54.54 54.54^(@)54.54^{\circ}.
微不足道,那么相位差将接近 90 90 90^(@)90^{\circ} 。在我们的例子中,电流滞后电压 54.54 54.54 54.54^(@)54.54^{\circ}

8.7 SERIES R-C CIRCUIT
8.7 系列 R-C 电路

Consider the example of an RC series circuit shown below.
考虑下面所示的 RC 串联电路的例子。

The total impedance of this circuit is Z = 2 j 4 Z = 2 j 4 Z=2-j4\mathrm{Z}=2-\mathrm{j} 4 (Always remember to multiply X c X c X_(c)\mathrm{X}_{\mathrm{c}} by -j ) and the current and the power factor can be calculated as,
该电路的总阻抗为 Z = 2 j 4 Z = 2 j 4 Z=2-j4\mathrm{Z}=2-\mathrm{j} 4 (始终记得将 X c X c X_(c)\mathrm{X}_{\mathrm{c}} 乘以 -j )电流和功率因数可以计算为,
I = V Z = 10 2 j 4 = 1 + j 2 cos ϕ = R | Z | = 0.44 I = V Z = 10 2 j 4 = 1 + j 2 cos ϕ = R | Z | = 0.44 {:[I=(V)/(Z)=(10)/(2-j4)],[=1+j2],[cos phi=(R)/(|Z|)=0.44]:}\begin{aligned} & I=\frac{V}{Z}=\frac{10}{2-j 4} \\ &=1+j 2 \\ & \cos \phi=\frac{R}{|Z|}=0.44 \end{aligned}
Generally, in an RC series circuit, the current leads the applied voltage by an angle less than 90 90 90^(@)90^{\circ}.
一般来说,在 RC 串联电路中,电流相对于施加电压的相位角小于 90 90 90^(@)90^{\circ}

If the resistance is very high compared to capacitive reactance, then the phase difference will be closer to zero and if resistance is negligible, then the phase difference will be close to 90 90 90^(@)90^{\circ}. In this example, the current leads the voltage by 63.89 63.89 63.89^(@)63.89^{\circ}.
如果电阻与电容性反应相比非常高,则相位差将更接近于零;如果电阻可以忽略不计,则相位差将接近 90 90 90^(@)90^{\circ} 。在这个例子中,电流超前电压 63.89 63.89 63.89^(@)63.89^{\circ}

8.8 SERIES RLC CIRCUIT
8.8 系列 RLC 电路

The total impedance of the RLC circuit shown above is Z = Z = Z=Z= 2 j 2 j 2-j2-\mathrm{j}. Hence the power and the power factor can be calculated as:
上述 RLC 电路的总阻抗为 Z = Z = Z=Z= 2 j 2 j 2-j2-\mathrm{j} 。因此,功率和功率因数可以计算为:
I = V Z = 10 2 j = 4 + j 2 cos ϕ = R | Z | = 0.89 I = V Z = 10 2 j = 4 + j 2 cos ϕ = R | Z | = 0.89 {:[I=(V)/(Z)=(10)/(2-j)],[=4+j2],[cos phi=(R)/(|Z|)=0.89]:}\begin{aligned} & I=\frac{V}{Z}=\frac{10}{2-j} \\ &=4+\mathrm{j} 2 \\ & \cos \phi=\frac{R}{|Z|}=0.89 \end{aligned}
In the RLC circuit, the Inductive reactance and the Capacitive reactance oppose each other.
在 RLC 电路中,电感抗和电容抗相互对立。

In our example, the capacitive reactance is more than the inductive reactance, hence the current leads the voltage by an angle.
在我们的例子中,电容性反应比电感性反应大,因此电流超前电压一个角度。

At a certain frequency called the resonance frequency, the inductive reactance and the capacitive reactance become equal.
在一个称为共振频率的特定频率下,感抗和容抗变得相等。

Then the circuit becomes a purely resistive circuit with capacitor-inductor combination acting as a short.
然后电路变成一个纯电阻电路,电容-电感组合作为短路。

At resonance frequency, the capacitor and the inductor exchanges energy back and forth, without effect the rest of the circuit.
在共振频率下,电容器和电感器相互交换能量,而不影响电路的其余部分。

9. ANALYSIS TECHNIQUES (FOR AC)
9. 分析技术(用于交流电)

9.1 VOLTAGE DIVIDER RULE
9.1 电压分压规则

A lot of the laws and theorems used in this chapter and the next chapter are very similar to what we learnt for DC circuits, but there are some differences as well.
本章和下一章中使用的许多定律和定理与我们在直流电路中学到的非常相似,但也有一些不同之处。

So we’ll go through each of these with the help of examples, rather than repeating the theory.
所以我们将通过示例逐一讲解这些内容,而不是重复理论。

This way you can get better at handling complex number math.
通过这种方式,你可以更好地处理复杂的数字数学。
In a series AC circuit, the voltage will be divided amongst the components according to their impedance values, and to find the exact values, we need to use the Voltage divider rule.
在串联交流电路中,电压将根据各个元件的阻抗值进行分配,要找到确切的值,我们需要使用电压分压规则。

Consider the RL series circuit shown below.
考虑下面所示的 RL 串联电路。

In this circuit the voltage applied by the source is divided amongst the resistor R R RR and an inductor L L LL. The resistor offers an impedance Z 1 = R Z 1 = R Z_(1)=RZ_{1}=R, and the inductor L L LL offers an impedance of Z 2 = j X L Z 2 = j X L Z_(2)=jX_(L)Z_{2}=j X_{L} (in rectangular form), where X L = X L = X_(L)=X_{L}= ?? L L LL.
在这个电路中,源施加的电压在电阻器 R R RR 和电感器 L L LL 之间分配。电阻器提供了阻抗 Z 1 = R Z 1 = R Z_(1)=RZ_{1}=R ,而电感器 L L LL 提供了阻抗 Z 2 = j X L Z 2 = j X L Z_(2)=jX_(L)Z_{2}=j X_{L} (以矩形形式),其中 X L = X L = X_(L)=X_{L}= ?? L L LL

I = V ( Z 1 + Z 2 ) V R = VZ 1 Z T , V L = VZ 2 Z T I = V Z 1 + Z 2 V R = VZ 1 Z T , V L = VZ 2 Z T {:[I=(V)/((Z_(1)+Z_(2)))],[:.V_(R)=(VZ_(1))/(Z_(T))","V_(L)=(VZ_(2))/(Z_(T))]:}\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\left(\mathrm{Z}_{1}+\mathrm{Z}_{2}\right)} \\ \therefore \mathrm{V}_{\mathrm{R}}=\frac{\mathrm{VZ}_{1}}{\mathrm{Z}_{T}}, \mathrm{~V}_{\mathrm{L}}=\frac{\mathrm{VZ}_{2}}{\mathrm{Z}_{T}} \end{gathered}
In general the Voltage divider rule for AC circuits is,
一般来说,交流电路的电压分压规则是,
V x = VZ x Z T V x = VZ x Z T V_(x)=(VZ_(x))/(Z_(T))\mathrm{V}_{\mathrm{x}}=\frac{\mathrm{VZ}_{\mathrm{x}}}{\mathrm{Z}_{T}}
Were x x xx is the component whose voltage we want to find out.
x x xx 是我们想要找出电压的组件。

9.2 CURRENT DIVIDER RULE
9.2 当前分流器规则

As you already know, the current divider rule is to find the Current division between components in a parallel circuit.
正如您所知,当前的分流规则是找到并联电路中组件之间的电流分配。
This time we’ll use a parallel RL circuit example to derive the result.
这次我们将使用一个并联 RL 电路的例子来推导结果。

V = I Z T = I Z 1 Z 2 ( Z 1 + Z 2 ) V = I Z T = I Z 1 Z 2 Z 1 + Z 2 V=IZ_(T)=(IZ_(1)Z_(2))/((Z_(1)+Z_(2)))V=I Z_{T}=\frac{I Z_{1} Z_{2}}{\left(Z_{1}+Z_{2}\right)}
I R = IZ T Z 1 = IZ 2 ( Z 1 + Z 2 ) , I L = IZ 1 ( Z 1 + Z 2 ) I R = IZ T Z 1 = IZ 2 Z 1 + Z 2 , I L = IZ 1 Z 1 + Z 2 :.I_(R)=(IZ_(T))/(Z_(1))=(IZ_(2))/((Z_(1)+Z_(2))),I_(L)=(IZ_(1))/((Z_(1)+Z_(2)))\therefore \mathrm{I}_{\mathrm{R}}=\frac{\mathrm{IZ}_{T}}{\mathrm{Z}_{1}}=\frac{\mathrm{IZ}_{2}}{\left(\mathrm{Z}_{1}+\mathrm{Z}_{2}\right)}, \mathrm{I}_{\mathrm{L}}=\frac{\mathrm{IZ}_{1}}{\left(\mathrm{Z}_{1}+\mathrm{Z}_{2}\right)}
In general the Current divider rule for A C A C ACA C is,
一般来说, A C A C ACA C 的电流分配规则是,
I x = I I T Z X I x = I I T Z X I_(x)=(II_(T))/(Z_(X))\mathrm{I}_{\mathrm{x}}=\frac{\mathrm{I} \mathrm{I}_{\mathrm{T}}}{\mathrm{Z}_{\mathrm{X}}}

9.3 SOURCE CONVERSION
9.3 源转换

Just like DC sources, AC voltage sources and AC current sources can also be converted to one another.
就像直流电源一样,交流电压源和交流电流源也可以相互转换。

The process
过程

is the same, only difference is that we need to use phasors in this case.
是一样的,唯一的区别是我们在这种情况下需要使用相量。

9.4 MESH ANALYSIS
9.4 网格分析

Like we mentioned before, most of these techniques and theorems are same as for DC and only different in math part, so will go through examples for each of them than repeat the theory.
正如我们之前提到的,这些技术和定理大多数与直流相同,仅在数学部分有所不同,因此我们将逐个例子进行讲解,而不是重复理论。

Consider the following example:
考虑以下示例:

It’s easy to see that there are 3 loops or meshes in the circuit.
很容易看出电路中有 3 个回路或网格。

Let’s assign a mesh current to each of them and obtain 3 mesh equations using KVL.
让我们为它们每一个分配一个网格电流,并使用基尔霍夫电压定律获得三个网格方程。

Notice how there is a polarity given to the voltage source.
注意电压源的极性。

Well, technically AC doesn’t have a direction, we know that, but for analysis a direction, which denotes to the direction in the positive half cycle, is often given.
好吧,从技术上讲,交流电没有方向,我们知道这一点,但为了分析,通常会给出一个方向,表示正半周期的方向。
Mesh 1: ( 8 2 j + 4 j ) I 1 4 j I 2 = 0 ( 8 + 2 j ) I 1 4 j I 2 = 0  Mesh 1:  ( 8 2 j + 4 j ) I 1 4 j I 2 = 0 ( 8 + 2 j ) I 1 4 j I 2 = 0 {:[" Mesh 1: "(8-2j+4j)I_(1)-4jI_(2)=0],[(8+2j)I_(1)-4jI_(2)=0]:}\begin{gathered} \text { Mesh 1: }(8-2 j+4 j) I_{1}-4 j I_{2}=0 \\ (8+2 j) I_{1}-4 j I_{2}=0 \end{gathered}
Mesh 2: 4 j I 1 + ( 4 j + 6 ) I 2 6 I 3 = 50 < 30 4 j I 1 + ( 4 j + 6 ) I 2 6 I 3 = 50 < 30 -4jI_(1)+(4j+6)I_(2)-6I_(3)=-50 < 30-4 j I_{1}+(4 j+6) I_{2}-6 I_{3}=-50<30
Mesh 3: I 3 = 10 0 I 3 = 10 0 I_(3)=-10/_0I_{3}=-10 \angle 0
In mesh 1, current I 1 I 1 I_(1)I_{1} passes through all the components and current I 2 I 2 I_(2)I_{2} passes through the inductor 4 j 4 j 4j4 j, and because I 2 I 2 I_(2)I_{2} is in opposite direction to I 1 I 1 I_(1)\mathrm{I}_{1}, the voltage 4 jl 2 4 jl 2 4jl_(2)4 \mathrm{jl}_{2} has to be subtracted from the equation. Similarly currents I 1 I 1 I_(1)I_{1} and I 3 I 3 I_(3)I_{3} has to be considered in mesh 2 along with its mesh current I 2 I 2 I_(2)I_{2}.
在网格 1 中,电流 I 1 I 1 I_(1)I_{1} 通过所有组件,电流 I 2 I 2 I_(2)I_{2} 通过电感 4 j 4 j 4j4 j ,因为 I 2 I 2 I_(2)I_{2} I 1 I 1 I_(1)\mathrm{I}_{1} 方向相反,所以电压 4 jl 2 4 jl 2 4jl_(2)4 \mathrm{jl}_{2} 必须从方程中减去。同样,电流 I 1 I 1 I_(1)I_{1} I 3 I 3 I_(3)I_{3} 也必须在网格 2 中考虑,以及它的网格电流 I 2 I 2 I_(2)I_{2}

Also the current direction we assumed is entering into the positive terminal of the voltage source, hence the negative sign for the voltage source in the equation.
我们假设的当前方向是进入电压源的正极,因此方程中电压源的负号。

Mesh 3 equation is easy because it has a current source, all we need to do is equate I 3 I 3 I_(3)I_{3} to it (they are in opposite direction though).
网格 3 方程很简单,因为它有一个电流源,我们需要做的就是将 I 3 I 3 I_(3)I_{3} 与之相等(不过它们的方向相反)。

It is not necessary to assume mesh currents in clockwise direction, it can be chosen as per your wish, only thing is the equation should be formed accordingly, the results will be the same.
不必假设网格电流为顺时针方向,可以根据你的意愿选择,唯一需要注意的是方程应相应地形成,结果将是相同的。

Now the equations can be solved easily using the Cramer’s rule, to find the currents.
现在可以使用克拉默法则轻松求解方程,以找到电流。
Converting the quantities to polar form
将数量转换为极坐标形式
50 30 = 43.3 25 j 10 0 = 10 50 30 = 43.3 25 j 10 0 = 10 {:[-50/_30=-43.3-25 j],[-10/_0=-10]:}\begin{aligned} & -50 \angle 30=-43.3-25 j \\ & -10 \angle 0=-10 \end{aligned}
I 1 = Δ 1 Δ = [ 0 4 j 0 43.3 25 j 4 j + 6 6 10 0 1 ] [ 8 + 2 j 4 j 0 4 j 4 j + 6 6 0 0 1 ] = 5.969 65.45 I 1 = Δ 1 Δ = 0 4 j 0 43.3 25 j 4 j + 6 6 10 0 1 8 + 2 j 4 j 0 4 j 4 j + 6 6 0 0 1 = 5.969 65.45 {:[I_(1)=(Delta_(1))/(Delta)=([[0,-4j,0],[-43.3-25 j,4j+6,-6],[-10,0,1]])/([[8+2j,-4j,0],[-4j,4j+6,-6],[0,0,1]])],[=-5.969/_65.45^(@)]:}\begin{aligned} \mathrm{I}_{1}=\frac{\Delta_{1}}{\Delta} & =\frac{\left[\begin{array}{ccc} 0 & -4 j & 0 \\ -43.3-25 j & 4 j+6 & -6 \\ -10 & 0 & 1 \end{array}\right]}{\left[\begin{array}{ccc} 8+2 j & -4 j & 0 \\ -4 j & 4 j+6 & -6 \\ 0 & 0 & 1 \end{array}\right]} \\ & =-5.969 \angle 65.45^{\circ} \end{aligned}
Do note that this circuit can be solved in multiple ways, for instance, converting the current source and the σ Ω σ Ω sigma^(Omega)\sigma^{\Omega} resistor to a voltage source, would reduce the circuit to a 2 meshes, which is significantly easier to solve (Sorry it’s better to learn the hard way).Concepts like Super Mesh analysis are applicable for AC too.
请注意,这个电路可以通过多种方式解决,例如,将电流源和 σ Ω σ Ω sigma^(Omega)\sigma^{\Omega} 电阻转换为电压源,可以将电路简化为两个网,这样解决起来要容易得多(抱歉,最好还是通过艰苦的方式学习)。像超级网分析这样的概念也适用于交流电。

9.5 NODAL ANALYSIS
9.5 节点分析

Now let’s try to solve this circuit using nodal analysis, the procedure is same as with DC.
现在让我们尝试使用节点分析来解决这个电路,过程与直流相同。

The first step is to identify the nodes and to select a reference node (Remember that node is a point where 2 or more components meet).
第一步是识别节点并选择一个参考节点(请记住,节点是两个或多个组件相遇的点)。

Reference Node
参考节点

Unlike our last example, the impedance values are not in the complex form, so we to convert them before proceeding further.
与我们上一个例子不同,阻抗值不是复数形式,因此我们需要在进一步处理之前将其转换。

It’s simple, just add j before inductive impedance and add -j before capacitive impedance and leave resistance as such.
很简单,只需在感抗前加上 j,在容抗前加上 -j,电阻保持不变。

Then assign voltages to the nodes and use KCL to form equations for each node.
然后为节点分配电压,并使用基尔霍夫电流定律(KCL)为每个节点形成方程。

Node 1: ( V 1 12 0 ) 0.5 + V 1 j 10 + ( V 1 V 2 ) 2 = 0 V 1 12 0 0.5 + V 1 j 10 + V 1 V 2 2 = 0 ((V_(1)-12/_0))/(0.5)+(V_(1))/(j 10)+((V_(1)-V_(2)))/(2)=0\frac{\left(V_{1}-12 \angle 0\right)}{0.5}+\frac{V_{1}}{j 10}+\frac{\left(V_{1}-V_{2}\right)}{2}=0
( 25 j ) V 1 5 V 2 = 240 ( 25 j ) V 1 5 V 2 = 240 (25-j)V_(1)-5V_(2)=240(25-j) V_{1}-5 V_{2}=240
Node 2 : ( V 2 V 1 ) 2 + V 2 j 6 + 4 = 0 2 : V 2 V 1 2 + V 2 j 6 + 4 = 0 2:((V_(2)-V_(1)))/(2)+(V_(2))/(-j6)+4=02: \frac{\left(V_{2}-V_{1}\right)}{2}+\frac{V_{2}}{-j 6}+4=0
3 V 1 + ( 3 + j ) V 2 = 24 3 V 1 + ( 3 + j ) V 2 = 24 -3V_(1)+(3+j)V_(2)=-24-3 V_{1}+(3+j) V_{2}=-24
When forming the equations, assume that all currents flow away from the node.
在形成方程时,假设所有电流都从节点流出。

After the equations are obtained, proceed as we did in the last example using the crammers rule.
得到方程后,按照我们在上一个例子中使用克拉默法则的方式进行。

10. NETWORK THEOREMS (FOR AC)
10. 网络定理(交流电)

10.1 SUPERPOSITION THEOREM
10.1 叠加定理

If you recall correctly, we used the superposition theorem to convert circuits with multiple sources into circuits with single sources.
如果你记得正确的话,我们使用叠加定理将多个源的电路转换为单个源的电路。

Here too we are doing the same, we are eliminating the other sources in the circuit and analyzing the circuit and repeating the same for the other sources and finally adding up the results.
在这里我们也在做同样的事情,我们在电路中消除其他源,分析电路,并对其他源重复相同的操作,最后将结果相加。

Consider this example, it has two voltage sources and suppose we are required to find the current through inductor.
考虑这个例子,它有两个电压源,假设我们需要找出通过电感的电流。

This is an ideal situation to use the superposition theorem, even though this circuit can be solved in plenty ways, including the techniques we learnt so far and the one’s we are going to.
这是一个使用叠加定理的理想情况,尽管这个电路可以通过多种方式解决,包括我们到目前为止学到的技术和我们即将学习的技术。

The first task while using the superposition theorem is to remove the energy sources than the one under consideration.
使用叠加定理的第一个任务是去除除正在考虑的能量源之外的其他能量源。

This can be done by shorting the
这可以通过缩短来完成

voltage sources and opening the current sources.
电压源和打开电流源。

In this example, we can obtain 2 circuits, as there are 2 sources.
在这个例子中,我们可以获得 2 个电路,因为有 2 个源。

For a while let’s consider only the 10 0 V 10 0 V 10◻0V10 \square 0 \mathrm{~V} source and we get this circuit.
暂时只考虑 10 0 V 10 0 V 10◻0V10 \square 0 \mathrm{~V} 源,我们得到这个电路。

Let’s use mesh analysis to solve this circuit.
让我们使用网分析来解决这个电路。
Mesh 1: ( 2 + j 10 ) I 1 j 10 I 2 = 10 0 Mesh 2: j 10 I 1 + ( 5 j 2 ) I 2 = 0  Mesh 1:  ( 2 + j 10 ) I 1 j 10 I 2 = 10 0  Mesh 2:  j 10 I 1 + ( 5 j 2 ) I 2 = 0 {:[" Mesh 1: "(2+j 10)I_(1)-j 10I_(2)=10/_0],[" Mesh 2: "-j 10I_(1)+(5-j2)I_(2)=0]:}\begin{aligned} & \text { Mesh 1: }(2+j 10) I_{1}-j 10 I_{2}=10 \angle 0 \\ & \text { Mesh 2: }-j 10 I_{1}+(5-j 2) I_{2}=0 \end{aligned}
Solving the equations we get, I 1 = 0.29 j 0.25 I 1 = 0.29 j 0.25 I_(1)=0.29-j0.25\mathrm{I}_{1}=0.29-\mathrm{j} 0.25 and I 2 = 0.24 + I 2 = 0.24 + I_(2)=0.24+\mathrm{I}_{2}=0.24+ j0.68. Therefore,
解方程我们得到, I 1 = 0.29 j 0.25 I 1 = 0.29 j 0.25 I_(1)=0.29-j0.25\mathrm{I}_{1}=0.29-\mathrm{j} 0.25 I 2 = 0.24 + I 2 = 0.24 + I_(2)=0.24+\mathrm{I}_{2}=0.24+ j0.68。因此,
I x 1 = I 1 I 2 = 0.04 j 0.93 I x 1 = I 1 I 2 = 0.04 j 0.93 I_(x1)=I_(1)-I_(2)=0.04-j 0.93I_{x 1}=I_{1}-I_{2}=0.04-j 0.93
Now let’s shift our focus to the second voltage source.
现在让我们将注意力转向第二个电压源。

To solve this circuit, let’s use ohm’s law and KVL.
要解决这个电路,我们使用欧姆定律和基尔霍夫电压定律。
Z T = ( j 10 ) ( 2 ) 2 + j 10 + 5 j 12 = 6.9 j 11.6 Ω I = 6 + j 8 6.9 j 11.6 I x 2 = 6 + j 8 5 + j 12 j 10 = 1 + j 20 j 10 = 2 j 0.1 Z T = ( j 10 ) ( 2 ) 2 + j 10 + 5 j 12 = 6.9 j 11.6 Ω I = 6 + j 8 6.9 j 11.6 I x 2 = 6 + j 8 5 + j 12 j 10 = 1 + j 20 j 10 = 2 j 0.1 {:[Z_(T)=((j 10)(2))/(2+j 10)+5-j 12],[=6.9-j 11.6 Omega],[I=(6+j8)/(6.9-j 11.6)],[I_(x2)=(6+j8-5+j 12)/(j 10)=(1+j 20)/(j 10)],[=2-j0.1]:}\begin{aligned} \mathrm{Z}_{\mathrm{T}} & =\frac{(j 10)(2)}{2+j 10}+5-j 12 \\ & =6.9-j 11.6 \Omega \\ \mathrm{I} & =\frac{6+j 8}{6.9-j 11.6} \\ \mathrm{I}_{\mathrm{x} 2} & =\frac{6+j 8-5+j 12}{j 10}=\frac{1+j 20}{j 10} \\ & =2-\mathrm{j} 0.1 \end{aligned}
Now that we have analyzed the circuits separately, let’s combine the results.
现在我们已经分别分析了电路,让我们将结果结合起来。

So the current through the inductor is,
所以电感器中的电流是,
I x = I x 1 + I x 2 = 2.04 j 1.03 I x = I x 1 + I x 2 = 2.04 j 1.03 I_(x)=I_(x1)+I_(x2)=2.04-j 1.03I_{x}=I_{x 1}+I_{x 2}=2.04-j 1.03
The superposition theorem may seem a little complex and cumbersome, and that’s probably true, but nonetheless it’s a handy tool to use in analyzing certain types of circuits.
叠加定理可能看起来有点复杂和繁琐,这可能是事实,但尽管如此,它仍然是分析某些类型电路的一个方便工具。

10.2 THEVENIN THEOREM
10.2 特文定理

Thevenin’s theorem states that, any two-terminal, ac network can be replaced by an equivalent circuit consisting of a voltage source and a series impedance.
特文宁定理指出,任何两个端子的交流网络都可以用一个由电压源和串联阻抗组成的等效电路来替代。

This is essentially the same as Thevenin’s theorem for DC, except that we use Thevenin impedance here.
这本质上与直流的特文宁定理相同,只是我们在这里使用特文宁阻抗。
Consider the circuit shown below, now let’s try to find the current through the 4 Ω 4 Ω 4Omega4 \Omega resistor. The steps are the same as for DC, but we’ll go through them once again.
考虑下面所示的电路,现在让我们尝试找出通过 4 Ω 4 Ω 4Omega4 \Omega 电阻的电流。步骤与直流相同,但我们将再次进行一遍。

  1. Identify the part of the circuit whose equivalent you need to find and then temporarily open circuit the load impedance ( 4 Ω 4 Ω 4Omega4 \Omega resistor in our case).
    确定您需要找到其等效的电路部分,然后暂时断开负载阻抗(在我们的例子中是 4 Ω 4 Ω 4Omega4 \Omega 电阻)。

  2. To find the Thevenin equivalent Impedance ( Z T H ) Z T H (Z_(TH))\left(Z_{T H}\right), remove all the energy sources in the circuit.
    要找到特文宁等效阻抗 ( Z T H ) Z T H (Z_(TH))\left(Z_{T H}\right) ,请移除电路中的所有能量源。

    This can be done by short circuiting the voltage sources and open circuiting the current sources.
    这可以通过短路电压源和开路电流源来实现。

  3. Now the equivalent impedance between the terminals will give us the Thevenin equivalent impedance.
    现在端子之间的等效阻抗将给我们提供特文宁等效阻抗。

    Here
    这里

    the inductor and the capacitor are in parallel, therefore Z TH = ( j 8 ) ( j 2 ) / ( j 6 ) = 2.67 j Z TH  = ( j 8 ) ( j 2 ) / ( j 6 ) = 2.67 j Z_("TH ")=(j8)(-j2)//(j6)=-2.67 jZ_{\text {TH }}=(j 8)(-j 2) /(j 6)=-2.67 j
    电感器和电容器是并联的,因此 Z TH = ( j 8 ) ( j 2 ) / ( j 6 ) = 2.67 j Z TH  = ( j 8 ) ( j 2 ) / ( j 6 ) = 2.67 j Z_("TH ")=(j8)(-j2)//(j6)=-2.67 jZ_{\text {TH }}=(j 8)(-j 2) /(j 6)=-2.67 j
  4. To find the Thevenin equivalent voltage ( V th ) V th (V_(th))\left(\mathrm{V}_{\mathrm{th}}\right), energy sources are returned to their original position and then the open circuit voltage across the terminals is determined ( V th = 3.33 V ) V th = 3.33 V (V_(th)=-3.33(V))\left(\mathrm{V}_{\mathrm{th}}=-3.33 \mathrm{~V}\right).
    要找到特文宁等效电压 ( V th ) V th (V_(th))\left(\mathrm{V}_{\mathrm{th}}\right) ,能量源被返回到其原始位置,然后确定端子之间的开路电压 ( V th = 3.33 V ) V th = 3.33 V (V_(th)=-3.33(V))\left(\mathrm{V}_{\mathrm{th}}=-3.33 \mathrm{~V}\right)
  5. Finally put the resistor back in its place and we are ready to draw the equivalent circuit.
    最后将电阻放回原位,我们准备绘制等效电路。

Finding the current through the resistor in this circuit is now a piece of cake ( I = 0.57 j 0.38 A ) ( I = 0.57 j 0.38 A ) (I=-0.57-j 0.38A)(I=-0.57-j 0.38 \mathrm{~A}).
在这个电路中找到电阻器中的电流现在轻而易举。

10.3 NORTON'S THEOREM
10.3 诺顿定理

Norton’s equivalent circuit is essentially the source transformed version of the Thevenin’s equivalent circuit.
诺顿的等效电路本质上是西门子的等效电路的源变换版本。

Using Thevenin’s theorem, we could replace a complex portion of a circuit by a voltage source in series with an equivalent impedance, whereas using Norton’s theorem, we
使用戴维南定理,我们可以用一个电压源串联一个等效阻抗来替代电路的复杂部分,而使用诺顿定理,我们

could replace the circuit by a current source in parallel with an equivalent impedance.
可以用一个与等效阻抗并联的电流源替换电路。

Let’s use the Norton’s theorem on our last example and spot the similarities between the two theorems.
让我们在最后一个例子中使用诺顿定理,并找出这两个定理之间的相似之处。
  1. Repeat all the steps and find Z t h Z t h Z_(th)Z_{t h}. Norton’s equivalent impedance is same as the Thevenin’s equivalent impedance.
    重复所有步骤并找到 Z t h Z t h Z_(th)Z_{t h} 。诺顿的等效阻抗与特文宁的等效阻抗相同。
  2. To find the Norton equivalent current ( I N ) I N (I_(N))\left(I_{N}\right), energy sources are returned to their original position and then the closed circuit current through terminals is determined.
    要找到诺顿等效电流 ( I N ) I N (I_(N))\left(I_{N}\right) ,能量源被返回到其原始位置,然后通过端子确定闭合电路电流。

When the load is shorted, the capacitor is also shorted out.
当负载短路时,电容器也被短路。

Hence the Norton equivalent current is given by I N I N I_(N)I_{N} = 10 / j 8 = 1.25 j = 10 / j 8 = 1.25 j =10//j8=-1.25j=10 / \mathrm{j} 8=-1.25 \mathrm{j}
因此,诺顿等效电流由 I N I N I_(N)I_{N} = 10 / j 8 = 1.25 j = 10 / j 8 = 1.25 j =10//j8=-1.25j=10 / \mathrm{j} 8=-1.25 \mathrm{j} 给出

3. Finally put the resistor back in its place and we are ready to draw the equivalent circuit.
最后将电阻放回原位,我们准备绘制等效电路。

The current through the resistor can be found using the Current division rule. I = ( 1.25 j ) ( 2.67 j ) / ( 4 2.67 j ) = 0.57 I = ( 1.25 j ) ( 2.67 j ) / ( 4 2.67 j ) = 0.57 I=(-1.25j)(-2.67j)//(4-2.67j)=-0.57-I=(-1.25 \mathrm{j})(-2.67 \mathrm{j}) /(4-2.67 \mathrm{j})=-0.57- j0.38 A, which is exactly the value we obtained using the Thevenin’s theorem.
通过电阻器的电流可以使用电流分配规则找到。 I = ( 1.25 j ) ( 2.67 j ) / ( 4 2.67 j ) = 0.57 I = ( 1.25 j ) ( 2.67 j ) / ( 4 2.67 j ) = 0.57 I=(-1.25j)(-2.67j)//(4-2.67j)=-0.57-I=(-1.25 \mathrm{j})(-2.67 \mathrm{j}) /(4-2.67 \mathrm{j})=-0.57- j0.38 A,这正是我们使用特文宁定理得到的值。

10.4 MAXIMUM POWER TRANSFER THEOREM
10.4 最大功率传输定理

All the theorems stated so far were pretty much the same as for DC, but this theorem is slightly different, the idea is the same, the math part is significantly different.
到目前为止所述的所有定理与直流电的定理基本相同,但这个定理略有不同,思想是相同的,数学部分则有显著不同。
Maximum Power Transfer Theorem for AC states that, maximum power will be delivered to a load when the load impedance is the conjugate of the Thevenin impedance across its terminals.
交流最大功率传输定理指出,当负载阻抗是其端子上的泰文宁阻抗的共轭时,最大功率将传递给负载。

The statement is all sorts of confusing, where did the conjugate term come from?
这个陈述非常令人困惑,那个共轭项是从哪里来的?

Let’s try to prove this theorem our self to get a better understanding.
让我们尝试自己证明这个定理,以便更好地理解。

Suppose we reduced a random circuit to its Thevenin’s equivalent and connected a load impedance across it, we
假设我们将一个随机电路简化为其特文宁等效电路,并在其上连接一个负载阻抗,我们

get a circuit like this.
得到一个这样的电路。

The current through this circuit will be,
该电路中的电流将是,
I = V T H Z T H + Z L = V T H R L + R T H + j ( X L + X T H ) I = V T H Z T H + Z L = V T H R L + R T H + j X L + X T H I=(V_(TH))/(Z_(TH)+Z_(L))=(V_(TH))/(R_(L)+R_(TH)+j(X_(L)+X_(TH)))I=\frac{V_{T H}}{Z_{T H}+Z_{L}}=\frac{V_{T H}}{R_{L}+R_{T H}+j\left(X_{L}+X_{T H}\right)}
Hence the power delivered to the load impedance is given by,
因此,施加到负载阻抗上的功率由下式给出,
P L = V T H 2 R L ( R L + R T H ) 2 + ( X L + X T H ) 2 P L = V T H 2 R L R L + R T H 2 + X L + X T H 2 P_(L)=(V_(TH)^(2)R_(L))/((R_(L)+R_(TH))^(2)+(X_(L)+X_(TH))^(2))P_{L}=\frac{V_{T H}^{2} R_{L}}{\left(R_{L}+R_{T H}\right)^{2}+\left(X_{L}+X_{T H}\right)^{2}}
Notice how there is only R L R L R_(L)R_{L} term in the numerator and the X L X L X_(L)X_{L} term is missing, that’s because the reactive part of the impedance doesn’t consume any power over the full cycle.
注意到分子中只有 R L R L R_(L)R_{L} 项,而 X L X L X_(L)X_{L} 项缺失,这是因为阻抗的反应部分在整个周期内不消耗任何功率。

To get the condition for max power, we need to differentiate the P L P L P_(L)P_{L} with respect to X L X L X_(L)X_{L} (I’m not going to, but you should) and equate it to zero. Then we get the condition X T H + X L = 0 X T H + X L = 0 X_(TH)+X_(L)=0X_{T H}+X_{L}=0
要获得最大功率的条件,我们需要对 X L X L X_(L)X_{L} P L P L P_(L)P_{L} 求导(我不打算这样做,但你应该这样做)并将其等于零。然后我们得到条件 X T H + X L = 0 X T H + X L = 0 X_(TH)+X_(L)=0X_{T H}+X_{L}=0

i.e. X L = X T H X L = X T H X_(L)=-X_(TH)X_{L}=-X_{T H}. Substituting this relation in the power equation, we obtain a simpler expression.
X L = X T H X L = X T H X_(L)=-X_(TH)X_{L}=-X_{T H} 。将此关系代入功率方程,我们得到一个更简单的表达式。
P L = V T H 2 R L ( R L + R T H ) 2 P L = V T H 2 R L R L + R T H 2 P_(L)=(V_(TH)^(2)R_(L))/((R_(L)+R_(TH))^(2))P_{L}=\frac{V_{T H}^{2} R_{L}}{\left(R_{L}+R_{T H}\right)^{2}}
To get the next condition for maximum power transfer, differentiate the P P P_(llcorner)P_{\llcorner }once again, this time with respect to R L R L R_(L)R_{L} and equate it to zero, it’s much easier this time. This time we get the condition R L + R T H = 2 R L R L + R T H = 2 R L R_(L)+R_(TH)=2R_(L)R_{L}+R_{T H}=2 R_{L} i.e. R L = R T H R L = R T H R_(L)=R_(TH)R_{L}=R_{T H}.
为了获得最大功率传输的下一个条件,再次对 P P P_(llcorner)P_{\llcorner } 进行一次微分,这次是关于 R L R L R_(L)R_{L} 的,并将其等于零,这次要简单得多。这次我们得到了条件 R L + R T H = 2 R L R L + R T H = 2 R L R_(L)+R_(TH)=2R_(L)R_{L}+R_{T H}=2 R_{L} ,即 R L = R T H R L = R T H R_(L)=R_(TH)R_{L}=R_{T H}

So the two conditions for Maximum power transfer in AC circuits is,
因此,交流电路中最大功率传输的两个条件是,
R L = R TH & X L = X TH R L = R TH & X L = X TH R_(L)=R_(TH)&X_(L)=-X_(TH)\mathrm{R}_{\mathrm{L}}=\mathrm{R}_{\mathrm{TH}} \& \mathrm{X}_{\mathrm{L}}=-\mathrm{X}_{\mathrm{TH}}
Combining the 2 we get,
结合这两者,我们得到,
Z L = R L + j X L = R T H j X T H Z L = Z T H Z L = R L + j X L = R T H j X T H Z L = Z T H {:[Z_(L)=R_(L)+jX_(L)=R_(TH)-jX_(TH)],[:.Z_(L)=Z_(TH)^(**)]:}\begin{aligned} Z_{L}= & R_{L}+j X_{L}=R_{T H}-j X_{T H} \\ & \therefore \mathrm{Z}_{\mathrm{L}}=\mathrm{Z}_{T H}^{*} \end{aligned}
To sum up, maximum power can be transferred from source to the load in an AC circuit, if the resistive part of the source and the load are the same and the reactive parts cancel each other out.
总之,在交流电路中,如果源和负载的电阻部分相同且反应部分相互抵消,则可以从源到负载传输最大功率。

11. LAPLACE TRANSFORM
拉普拉斯变换

11.1 INTRODUCTION
11.1 引言

So far we dealt with DC circuits and sinusoidal AC circuits in steady state (more on this in the next chapter).
到目前为止,我们处理了直流电路和稳态下的正弦交流电路(下一章将对此进行更多讨论)。

But in most real life circuits, the sources may not always be sinusoidal and quantities of interest in these circuits may be in transient state etc. So the math we used so far will prove inadequate to deal with these circuits.
但在大多数实际电路中,电源可能并不总是正弦波,电路中感兴趣的量可能处于瞬态状态等。因此,我们迄今为止使用的数学将不足以处理这些电路。

The way to deal with such circuits is to model them with the help of differential equations.
处理此类电路的方法是借助微分方程对其进行建模。

Perhaps an example will make things more clear.
也许一个例子会使事情更清楚。

Consider this simple RLC circuit with a voltage source v s ( t ) v s ( t ) v_(s)(t)\mathrm{v}_{\mathrm{s}}(\mathrm{t}) and suppose the current through the circuit is the quantity of our interest. Using KVL, v s ( t ) = v R + v L + v C v s ( t ) = v R + v L + v C v_(s)(t)=v_(R)+v_(L)+v_(C)\mathrm{v}_{\mathrm{s}}(\mathrm{t})=\mathrm{v}_{\mathrm{R}}+\mathrm{v}_{\mathrm{L}}+\mathrm{v}_{\mathrm{C}}. The voltage across each components can be replaced by the relations
考虑这个简单的 RLC 电路,电压源为 v s ( t ) v s ( t ) v_(s)(t)\mathrm{v}_{\mathrm{s}}(\mathrm{t}) ,假设电路中的电流是我们关注的量。使用基尔霍夫电压定律, v s ( t ) = v R + v L + v C v s ( t ) = v R + v L + v C v_(s)(t)=v_(R)+v_(L)+v_(C)\mathrm{v}_{\mathrm{s}}(\mathrm{t})=\mathrm{v}_{\mathrm{R}}+\mathrm{v}_{\mathrm{L}}+\mathrm{v}_{\mathrm{C}} 。每个元件上的电压可以用以下关系替代

shown in the table below (memorize this table).
如下表所示(记住这个表)。

This is done to make each term a function of current i ( t ) i ( t ) i(t)i(t), which is common to all components, as this is a series circuit.
这是为了使每个术语成为当前 i ( t ) i ( t ) i(t)i(t) 的函数,这在所有组件中是共同的,因为这是一个串联电路。
Component
组件
Voltage across the component
组件上的电压
Current through the component
通过组件的电流
Resistor
电阻器
V R = i R R V R = i R R V_(R)=i_(R)R\mathrm{V}_{\mathrm{R}}=\mathrm{i}_{\mathrm{R}} \mathrm{R} i R = V R R i R = V R R i_(R)=(V_(R))/(R)\mathrm{i}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}}
Inductor
电感器
v L = L d i L dt v L = L d i L dt v_(L)=L(di_(L))/(dt)\mathrm{v}_{\mathrm{L}}=\mathrm{L} \frac{\mathrm{d} \mathrm{i}_{\mathrm{L}}}{\mathrm{dt}} i L = 1 L v L dt i L = 1 L v L dt i_(L)=(1)/((L))intv_(L)dt\mathrm{i}_{\mathrm{L}}=\frac{1}{\mathrm{~L}} \int \mathrm{v}_{\mathrm{L}} \mathrm{dt}
Capacitor
电容器
v C = 1 C i L dt v C = 1 C i L dt v_(C)=(1)/(C)inti_(L)dt\mathrm{v}_{\mathrm{C}}=\frac{1}{\mathrm{C}} \int \mathrm{i}_{\mathrm{L}} \mathrm{dt} i C = C d v C dt i C = C d v C dt i_(C)=C(dv_(C))/(dt)\mathrm{i}_{\mathrm{C}}=\mathrm{C} \frac{\mathrm{d} \mathrm{v}_{\mathrm{C}}}{\mathrm{dt}}
Component Voltage across the component Current through the component Resistor V_(R)=i_(R)R i_(R)=(V_(R))/(R) Inductor v_(L)=L(di_(L))/(dt) i_(L)=(1)/((L))intv_(L)dt Capacitor v_(C)=(1)/(C)inti_(L)dt i_(C)=C(dv_(C))/(dt)| Component | Voltage across the component | Current through the component | | :---: | :---: | :---: | | Resistor | $\mathrm{V}_{\mathrm{R}}=\mathrm{i}_{\mathrm{R}} \mathrm{R}$ | $\mathrm{i}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}}$ | | Inductor | $\mathrm{v}_{\mathrm{L}}=\mathrm{L} \frac{\mathrm{d} \mathrm{i}_{\mathrm{L}}}{\mathrm{dt}}$ | $\mathrm{i}_{\mathrm{L}}=\frac{1}{\mathrm{~L}} \int \mathrm{v}_{\mathrm{L}} \mathrm{dt}$ | | Capacitor | $\mathrm{v}_{\mathrm{C}}=\frac{1}{\mathrm{C}} \int \mathrm{i}_{\mathrm{L}} \mathrm{dt}$ | $\mathrm{i}_{\mathrm{C}}=\mathrm{C} \frac{\mathrm{d} \mathrm{v}_{\mathrm{C}}}{\mathrm{dt}}$ |
Now the KVL equation becomes:
现在 KVL 方程变为:
v s ( t ) = R i ( t ) + L d i ( t ) dt + 1 C i ( t ) dt To remove integral, differentiate both sides, d s ( t ) dt = R d i ( t ) dt + L d 2 i ( t ) dt t 2 + 1 C i ( t ) v s ( t ) = R i ( t ) + L d i ( t ) dt + 1 C i ( t ) dt  To remove integral, differentiate both sides,  d s ( t ) dt = R d i ( t ) dt + L d 2 i ( t ) dt t 2 + 1 C i ( t ) {:[v_(s)(t)=Ri(t)+L((d)i(t))/(dt)+(1)/(C)inti(t)dt],[" To remove integral, differentiate both sides, "],[(d_(s)(t))/(dt)=R((d)i(t))/(dt)+L(d^(2)i(t))/(dtt^(2))+(1)/(C)i(t)]:}\begin{aligned} & v_{s}(\mathrm{t})=\mathrm{R} \mathrm{i}(\mathrm{t})+\mathrm{L} \frac{\mathrm{~d} \mathrm{i}(\mathrm{t})}{\mathrm{dt}}+\frac{1}{\mathrm{C}} \int \mathrm{i}(\mathrm{t}) \mathrm{dt} \\ & \text { To remove integral, differentiate both sides, } \\ & \frac{\mathrm{d}_{\mathrm{s}}(\mathrm{t})}{\mathrm{dt}}=\mathrm{R} \frac{\mathrm{~d} i(\mathrm{t})}{\mathrm{dt}}+\mathrm{L} \frac{\mathrm{~d}^{2} \mathrm{i}(\mathrm{t})}{\mathrm{dt} \mathrm{t}^{2}}+\frac{1}{\mathrm{C}} \mathrm{i}(\mathrm{t}) \end{aligned}
This is the differential equation for this particular circuit.
这是该特定电路的微分方程。

There are lots of advantages to modelling circuits this way.
以这种方式建模电路有很多优点。

For one, this equation is a general one, it is applicable to any kind of source voltage, DC or sinusoidal AC or any other waveform.
首先,这个方程是一个通用方程,适用于任何类型的源电压,无论是直流电还是正弦交流电或其他任何波形。

Also a lot of inferences can be made just from the nature of the differential equation.
从微分方程的性质中也可以得出很多推论。

For instance, the equation in our example is a second order differential
例如,我们例子中的方程是一个二阶微分方程

equation and that is enough information to predict the general behavior of this circuit to various inputs.
方程,这足以预测该电路对各种输入的总体行为。

The only problem with this method is that, solving differential equations isn’t the easiest of tasks and not everyone’s an expert in calculus.
这种方法唯一的问题是,解微分方程并不是最简单的任务,并不是每个人都是微积分专家。

But luckily there’s an easier way to solve differential equations, using Laplace transform.
但幸运的是,有一种更简单的方法可以解决微分方程,即使用拉普拉斯变换。

11.2 LAPLACE TRANSFORM
11.2 拉普拉斯变换

The Laplace transform is a well-established mathematical technique for solving differential equations.
拉普拉斯变换是一种成熟的数学技术,用于求解微分方程。

It is named in honor of the great French mathematician, Pierre Simon De Laplace.
它是为了纪念伟大的法国数学家皮埃尔·西蒙·拉普拉斯而命名的。

Like all transforms, the Laplace transform changes a mathematical function into another according to some fixed set of rules or equations.
像所有变换一样,拉普拉斯变换根据一些固定的规则或方程将一个数学函数转换为另一个函数。

Before we dig into Laplace transform, let’s look into transforms in general.
在我们深入研究拉普拉斯变换之前,先来看看变换的一般概念。

So what is a transform? Why do we need them?
那么什么是变换?我们为什么需要它们?
Let’s begin by considering a simple computational problem: compute the value of x = 3 . 4 2 . 4 x = 3 . 4 2 . 4 x=3.4^(2.4)\mathbf{x}=\mathbf{3 . 4 ^ { 2 . 4 }}. It is not easy to get the exact value using straightforward methods.
让我们先考虑一个简单的计算问题:计算 x = 3 . 4 2 . 4 x = 3 . 4 2 . 4 x=3.4^(2.4)\mathbf{x}=\mathbf{3 . 4 ^ { 2 . 4 }} 的值。使用直接的方法很难得到确切的值。

What we can do to make this problem solvable is to take natural log on both sides: now the equation becomes l n ( x ) = 2 . 4 I n ( 3 . 4 ) l n ( x ) = 2 . 4 I n ( 3 . 4 ) ln(x)=2.4In(3.4)\boldsymbol{\operatorname { l n } ( \mathbf { x } ) = \mathbf { 2 } . 4} \mathbf{I n}(\mathbf{3 . 4}). Now the value of ln ( x ) ln ( x ) ln(x)\ln (x) can be easily obtained from a log table. And to obtain the value of x x xx, all we have to do is to take the antilog of the value obtained.
我们可以做的使这个问题可解的方法是对两边取自然对数:现在方程变为 l n ( x ) = 2 . 4 I n ( 3 . 4 ) l n ( x ) = 2 . 4 I n ( 3 . 4 ) ln(x)=2.4In(3.4)\boldsymbol{\operatorname { l n } ( \mathbf { x } ) = \mathbf { 2 } . 4} \mathbf{I n}(\mathbf{3 . 4}) 。现在 ln ( x ) ln ( x ) ln(x)\ln (x) 的值可以很容易地从对数表中获得。要获得 x x xx 的值,我们所要做的就是取所获得值的反对数。

What we did was to take the hard problem, convert it into an
我们所做的是将这个难题转化为一个

easier equivalent problem.
更简单的等价问题。

This is the very idea behind transforms. The concept of transformation can be illustrated with the simple diagram below:
这正是变换背后的理念。变换的概念可以用下面的简单图示来说明:

What kind of transformation might we use with ODEs?
我们可以使用什么样的变换来处理常微分方程(ODE)?

Based on our experience with logarithms, the dream would be a transformation, it would be useful if some transformation allowed us to replace the operation of differentiation by some easier operation, perhaps something similar to multiplication.
基于我们对对数的经验,这个梦想将是一个变换,如果某种变换能够让我们用一些更简单的操作来替代微分的操作,那将是有用的,也许类似于乘法的操作。

This is exactly what the Laplace transform is used for.
这正是拉普拉斯变换的用途。

The Laplace transform, transforms the differential equations into algebraic equations which are easier to manipulate and solve.
拉普拉斯变换将微分方程转化为代数方程,这些方程更容易处理和求解。

Once the solution in the Laplace transform domain is obtained, the inverse Laplace transform is used to obtain the solution to the differential equation.
一旦在拉普拉斯变换域中获得了解,便使用逆拉普拉斯变换来获得微分方程的解。
The Laplace transform of a function f ( t ) f ( t ) f(t)\mathbf{f}(\mathbf{t}), denoted as F ( s ) F ( s ) F(s)\mathbf{F}(\mathbf{s}), is defined as:
函数 f ( t ) f ( t ) f(t)\mathbf{f}(\mathbf{t}) 的拉普拉斯变换,记作 F ( s ) F ( s ) F(s)\mathbf{F}(\mathbf{s}) ,定义为:
F ( s ) = 0 f ( t ) e s t d t F ( s ) = 0 f ( t ) e s t d t F(s)=int_(0)^(oo)f(t)e^(-st)dtF(s)=\int_{0}^{\infty} f(t) e^{-s t} d t
This equation looks menacing at first glance.
这个方程乍一看显得很可怕。

But fortunately, most times you don’t need to use this equation, you can easily get away with knowing some standard results and some properties.
但幸运的是,大多数时候你不需要使用这个方程,你可以轻松地通过了解一些标准结果和一些性质来应对。

11.3 PROPERTIES OF LAPLACE TRANSFORM
11.3 拉普拉斯变换的性质

Some of the basic properties of Laplace transform are listed here,
这里列出了一些拉普拉斯变换的基本性质,
Property
财产
Operation in time domain
时域中的操作
Operation in s domain
s 域中的运算
Linearity
线性
a 1 x 1 ( t ) + a 2 x 2 ( t ) a 1 x 1 ( t ) + a 2 x 2 ( t ) a_(1)x_(1)(t)+a_(2)x_(2)(t)\mathrm{a}_{1} \mathrm{x}_{1}(\mathrm{t})+\mathrm{a}_{2} \mathrm{x}_{2}(\mathrm{t}) a 1 X 1 ( s ) + a 2 X 2 ( s ) a 1 X 1 ( s ) + a 2 X 2 ( s ) a_(1)X_(1)(s)+a_(2)X_(2)(s)\mathrm{a}_{1} \mathrm{X}_{1}(\mathrm{~s})+\mathrm{a}_{2} \mathrm{X}_{2}(\mathrm{~s})
Differentiation
微分
d n x ( t ) d t n d n x ( t ) d t n (d^(n)x(t))/(dt^(n))\frac{d^{n} x(t)}{d t^{n}} s n X ( s ) s n 1 x ( 0 ) x n 1 ( 0 ) s n X ( s ) s n 1 x 0 x n 1 0 s^(n)X(s)-s^(n-1)x(0^(-))dots-x^(n-1)(0^(-))\mathrm{~s}^{\mathrm{n}} \mathrm{X}(\mathrm{s})-\mathrm{s}^{\mathrm{n}-1} \mathrm{x}\left(0^{-}\right) \ldots-\mathrm{x}^{\mathrm{n}-1}\left(0^{-}\right)
Integration
集成
t x ( τ ) d τ t x ( τ ) d τ int_(-oo)^(t)x(tau)d tau\int_{-\infty}^{t} x(\tau) d \tau X ( s ) s + x n 1 ( 0 ) s X ( s ) s + x n 1 0 s (X(s))/(s)+(x^(n-1)(0^(-)))/(s)\frac{\mathrm{X}(\mathrm{s})}{s}+\frac{\mathrm{x}^{n-1}\left(0^{-}\right)}{s}

初值定理
Initial value
theorem
Initial value theorem| Initial value | | :--- | | theorem |
x ( 0 ) = lim t 0 x ( t ) x ( 0 ) = lim t 0 x ( t ) x(0)=lim_(t rarr0)x(t)\mathrm{x}(0)=\lim _{t \rightarrow 0} x(t) x ( 0 ) = lim s sX ( s ) x ( 0 ) = lim s sX ( s ) x(0)=lim_(s rarr oo)sX(s)\mathrm{x}(0)=\lim _{s \rightarrow \infty} \mathrm{sX}(\mathrm{s})
Final value theorem
最终值定理
x ( ) = lim t x ( t ) x ( ) = lim t x ( t ) x(oo)=lim_(t rarr oo)x(t)\mathrm{x}(\infty)=\lim _{t \rightarrow \infty} x(t) x ( 0 ) = lim s 0 sX ( s ) x ( 0 ) = lim s 0 sX ( s ) x(0)=lim_(s rarr0)sX(s)\mathrm{x}(0)=\lim _{s \rightarrow 0} \mathrm{sX}(\mathrm{s})
Time scaling
时间缩放
x ( at ) x ( at ) x(at)\mathrm{x}(\mathrm{at}) a 1 X ( s a ) a 1 X s a a^(-1)X((s)/(a))\mathrm{a}^{-1} \mathrm{X}\left(\frac{s}{a}\right)
Property Operation in time domain Operation in s domain Linearity a_(1)x_(1)(t)+a_(2)x_(2)(t) a_(1)X_(1)(s)+a_(2)X_(2)(s) Differentiation (d^(n)x(t))/(dt^(n)) s^(n)X(s)-s^(n-1)x(0^(-))dots-x^(n-1)(0^(-)) Integration int_(-oo)^(t)x(tau)d tau (X(s))/(s)+(x^(n-1)(0^(-)))/(s) "Initial value theorem" x(0)=lim_(t rarr0)x(t) x(0)=lim_(s rarr oo)sX(s) Final value theorem x(oo)=lim_(t rarr oo)x(t) x(0)=lim_(s rarr0)sX(s) Time scaling x(at) a^(-1)X((s)/(a))| Property | Operation in time domain | Operation in s domain | | :--- | :--- | :--- | | Linearity | $\mathrm{a}_{1} \mathrm{x}_{1}(\mathrm{t})+\mathrm{a}_{2} \mathrm{x}_{2}(\mathrm{t})$ | $\mathrm{a}_{1} \mathrm{X}_{1}(\mathrm{~s})+\mathrm{a}_{2} \mathrm{X}_{2}(\mathrm{~s})$ | | Differentiation | $\frac{d^{n} x(t)}{d t^{n}}$ | $\mathrm{~s}^{\mathrm{n}} \mathrm{X}(\mathrm{s})-\mathrm{s}^{\mathrm{n}-1} \mathrm{x}\left(0^{-}\right) \ldots-\mathrm{x}^{\mathrm{n}-1}\left(0^{-}\right)$ | | Integration | $\int_{-\infty}^{t} x(\tau) d \tau$ | $\frac{\mathrm{X}(\mathrm{s})}{s}+\frac{\mathrm{x}^{n-1}\left(0^{-}\right)}{s}$ | | Initial value <br> theorem | $\mathrm{x}(0)=\lim _{t \rightarrow 0} x(t)$ | $\mathrm{x}(0)=\lim _{s \rightarrow \infty} \mathrm{sX}(\mathrm{s})$ | | Final value theorem | $\mathrm{x}(\infty)=\lim _{t \rightarrow \infty} x(t)$ | $\mathrm{x}(0)=\lim _{s \rightarrow 0} \mathrm{sX}(\mathrm{s})$ | | Time scaling | $\mathrm{x}(\mathrm{at})$ | $\mathrm{a}^{-1} \mathrm{X}\left(\frac{s}{a}\right)$ |

11.4 STANDARD LAPLACE TRANSFORM PAIRS
11.4 标准拉普拉斯变换对

f(t) F(s)
1 1 s 1 s (1)/(s)\frac{1}{s}
Constant K
常数 K
K
t 1 s 2 1 s 2 (1)/(s^(2))\frac{1}{s^{2}}
t n t n t^(n)t^{n} n ! s n + 1 n ! s n + 1 (n!)/(s^(n+1))\frac{n!}{s^{n+1}}
e a t e a t e^(-at)e^{-a t} 1 s + a 1 s + a (1)/(s+a)\frac{1}{s+a}
e at e at  e^("at ")e^{\text {at }} 1 s a 1 s a (1)/(s-a)\frac{1}{s-a}
e a t t n e a t t n e^(-at)t^(n)e^{-a t} t^{n} n ! ( s + a ) n + 1 n ! ( s + a ) n + 1 (n!)/((s+a)^(n+1))\frac{n!}{(s+a)^{n+1}}
sin ω t sin ω t sin omega t\sin \omega t ω s 2 + ω 2 ω s 2 + ω 2 (omega)/(s^(2)+omega^(2))\frac{\omega}{s^{2}+\omega^{2}}
cos ω t cos ω t cos omega t\cos \omega t s s 2 + ω 2 s s 2 + ω 2 (s)/(s^(2)+omega^(2))\frac{s}{s^{2}+\omega^{2}}
e a t sin ω t e a t sin ω t e^(-at)sin omega te^{-a t} \sin \omega t ω ( s + a ) 2 + ω 2 ω ( s + a ) 2 + ω 2 (omega)/((s+a)^(2)+omega^(2))\frac{\omega}{(s+a)^{2}+\omega^{2}}
e a t cos ω t e a t cos ω t e^(-at)cos omega te^{-a t} \cos \omega t ( s + a ) ( s + a ) 2 + ω 2 ( s + a ) ( s + a ) 2 + ω 2 ((s+a))/((s+a)^(2)+omega^(2))\frac{(s+a)}{(s+a)^{2}+\omega^{2}}
f(t) F(s) 1 (1)/(s) Constant K K t (1)/(s^(2)) t^(n) (n!)/(s^(n+1)) e^(-at) (1)/(s+a) e^("at ") (1)/(s-a) e^(-at)t^(n) (n!)/((s+a)^(n+1)) sin omega t (omega)/(s^(2)+omega^(2)) cos omega t (s)/(s^(2)+omega^(2)) e^(-at)sin omega t (omega)/((s+a)^(2)+omega^(2)) e^(-at)cos omega t ((s+a))/((s+a)^(2)+omega^(2))| f(t) | F(s) | | :---: | :---: | | 1 | $\frac{1}{s}$ | | Constant K | K | | t | $\frac{1}{s^{2}}$ | | $t^{n}$ | $\frac{n!}{s^{n+1}}$ | | $e^{-a t}$ | $\frac{1}{s+a}$ | | $e^{\text {at }}$ | $\frac{1}{s-a}$ | | $e^{-a t} t^{n}$ | $\frac{n!}{(s+a)^{n+1}}$ | | $\sin \omega t$ | $\frac{\omega}{s^{2}+\omega^{2}}$ | | $\cos \omega t$ | $\frac{s}{s^{2}+\omega^{2}}$ | | $e^{-a t} \sin \omega t$ | $\frac{\omega}{(s+a)^{2}+\omega^{2}}$ | | $e^{-a t} \cos \omega t$ | $\frac{(s+a)}{(s+a)^{2}+\omega^{2}}$ |

11.5 INVERSE LAPLACE TRANSFORM
11.5 反拉普拉斯变换

Finding the Inverse Laplace transforms of functions isn’t terribly difficult.
求函数的逆拉普拉斯变换并不是特别困难。

Most times Inverse Laplace transforms of functions can be figured out by inspection.
大多数情况下,函数的逆拉普拉斯变换可以通过观察来确定。

The general method to find the Inverse Laplace transforms of functions is to express them as partial fractions and then make it into a convenient form and figure out which function’s Laplace each term is.
找到函数的逆拉普拉斯变换的一般方法是将它们表示为部分分式,然后将其转化为方便的形式,并找出每个项对应的拉普拉斯变换函数。

Keeping the various properties of Laplace transform is very handy.
保持拉普拉斯变换的各种性质是非常方便的。

Here are some examples on finding Laplace Inverse: Link
这里有一些关于寻找拉普拉斯逆变换的例子:链接

11.6 SOLVING DIFFERENTIAL EQUATIONS
11.6 解微分方程

We started with Laplace transform as an easier method to solve differential equations.
我们从拉普拉斯变换开始,作为解决微分方程的一个更简单的方法。

The procedure is best illustrated with an example.
这个过程最好通过一个例子来说明。

Example:
示例:

f ( t ) + 3 f ( t ) + 2 f ( t ) = e t f ( t ) + 3 f ( t ) + 2 f ( t ) = e t f^('')(t)+3f^(')(t)+2f(t)=e^(-t)\mathrm{f}^{\prime \prime}(\mathrm{t})+3 \mathrm{f}^{\prime}(\mathrm{t})+2 \mathrm{f}(\mathrm{t})=\mathrm{e}^{-\mathrm{t}}, with the initial conditions f ( 0 ) = f ( 0 ) = 0 f ( 0 ) = f ( 0 ) = 0 f(0)=f^(')(0)=0f(0)=f^{\prime}(0)=0
f ( t ) + 3 f ( t ) + 2 f ( t ) = e t f ( t ) + 3 f ( t ) + 2 f ( t ) = e t f^('')(t)+3f^(')(t)+2f(t)=e^(-t)\mathrm{f}^{\prime \prime}(\mathrm{t})+3 \mathrm{f}^{\prime}(\mathrm{t})+2 \mathrm{f}(\mathrm{t})=\mathrm{e}^{-\mathrm{t}} ,初始条件为 f ( 0 ) = f ( 0 ) = 0 f ( 0 ) = f ( 0 ) = 0 f(0)=f^(')(0)=0f(0)=f^{\prime}(0)=0

f ( t ) + 3 f ( t ) + 2 f ( t ) = e t f ( t ) + 3 f ( t ) + 2 f ( t ) = e t f^('')(t)+3f^(')(t)+2f(t)=e^(-t)f^{\prime \prime}(t)+3 f^{\prime}(t)+2 f(t)=e^{-t}
s 2 F ( s ) + 3 s F ( s ) + 2 F ( s ) = 1 s + 1 s 2 F ( s ) + 3 s F ( s ) + 2 F ( s ) = 1 s + 1 s^(2)F(s)+3sF(s)+2F(s)=(1)/(s+1)s^{2} F(s)+3 s F(s)+2 F(s)=\frac{1}{s+1}
F ( s ) = 1 s + 1 1 s 2 + 3 s + 2 F ( s ) = 1 s + 1 1 s 2 + 3 s + 2 F(s)=(1)/(s+1)(1)/(s^(2)+3s+2)\mathrm{F}(\mathrm{s})=\frac{1}{s+1} \frac{1}{s^{2}+3 s+2}
Taking Laplace
进行拉普拉斯变换

transform on both sides
双侧变换
Decomposing into partial fractions,
分解成部分分式,

F ( s ) = 1 s + 2 1 s + 1 + 1 ( s + 1 ) 2 F ( s ) = 1 s + 2 1 s + 1 + 1 ( s + 1 ) 2 F(s)=(1)/(s+2)-(1)/(s+1)+(1)/((s+1)^(2))\mathrm{F}(\mathrm{s})=\frac{1}{s+2}-\frac{1}{s+1}+\frac{1}{(s+1)^{2}}
f ( t ) = f ( t ) = f(t)=f(\mathrm{t})= Taking Inverse Laplace
f ( t ) = f ( t ) = f(t)=f(\mathrm{t})= 进行反拉普拉斯变换

f ( t ) = e 2 t e t + t e t f ( t ) = e 2 t e t + t e t f(t)=e^(-2t)-e^(-t)+te^(-t)f(t)=e^{-2 t}-e^{-t}+t e^{-t}
transform on both
在两者上变换

sides
边缘
At first glance, this may not seem any better than differential equations, but trust us, using Laplace transform is very easy with some practice.
乍一看,这可能看起来并不比微分方程好,但相信我们,经过一些练习,使用拉普拉斯变换是非常简单的。

Here are more examples to practice: Link
这里有更多练习的例子:链接

11.7 MODELLING CIRCUITS IN SDOMAIN
11.7 在 SDOMAIN 中建模电路

Once you gain enough confidence with Laplace transform, you don’t have to find the differential equations for the circuits, then convert it into Laplace transform.
一旦你对拉普拉斯变换有足够的信心,就不必为电路找到微分方程,然后将其转换为拉普拉斯变换。

Instead you can form algebraic equations in the Laplace domain or the s-domain, directly by inspection.
相反,您可以通过观察直接在拉普拉斯域或 s 域中形成代数方程。

Voltage or current in an element in the circuit can be represented as given in the table below.
电路中元件的电压或电流可以如下面表格所示表示。

These are just the Laplace transforms of the relations from the earlier table.
这些只是早期表格中关系的拉普拉斯变换。
With the differentiation and integration gone, the relations look easier already.
去掉微分和积分后,关系看起来简单多了。

We have ignored the initial states of the components in these relations (That’s for the next chapter).
我们在这些关系中忽略了组件的初始状态(这将在下一章讨论)。
Component
组件
Voltage across the component
组件上的电压
Current through the component
通过组件的电流
Resistor
电阻器
V ( s ) = I ( s ) R V ( s ) = I ( s ) R V(s)=I(s)R\mathrm{V}(\mathrm{s})=\mathrm{I}(\mathrm{s}) \mathrm{R} I ( s ) = V ( s ) R I ( s ) = V ( s ) R I(s)=(V(s))/(R)\mathrm{I}(\mathrm{s})=\frac{\mathrm{V}(\mathrm{s})}{\mathrm{R}}
Inductor
电感器
V ( s ) = sL I ( s ) V ( s ) = sL I ( s ) V(s)=sLI(s)\mathrm{V}(\mathrm{s})=\mathrm{sL} \mathrm{I}(\mathrm{s}) I ( s ) = V ( s ) sL I ( s ) = V ( s ) sL I(s)=(V(s))/(sL)\mathrm{I}(\mathrm{s})=\frac{\mathrm{V}(\mathrm{s})}{\mathrm{sL}}
Capacitor
电容器
V ( s ) = I ( s ) sC V ( s ) = I ( s ) sC V(s)=(I(s))/(sC)\mathrm{V}(\mathrm{s})=\frac{\mathrm{I}(\mathrm{s})}{\mathrm{sC}} V ( s ) = sC I ( s ) V ( s ) = sC I ( s ) V(s)=sCI(s)\mathrm{V}(\mathrm{s})=\mathrm{sC} \mathrm{I}(\mathrm{s})
Component Voltage across the component Current through the component Resistor V(s)=I(s)R I(s)=(V(s))/(R) Inductor V(s)=sLI(s) I(s)=(V(s))/(sL) Capacitor V(s)=(I(s))/(sC) V(s)=sCI(s)| Component | Voltage across the component | Current through the component | | :---: | :---: | :---: | | Resistor | $\mathrm{V}(\mathrm{s})=\mathrm{I}(\mathrm{s}) \mathrm{R}$ | $\mathrm{I}(\mathrm{s})=\frac{\mathrm{V}(\mathrm{s})}{\mathrm{R}}$ | | Inductor | $\mathrm{V}(\mathrm{s})=\mathrm{sL} \mathrm{I}(\mathrm{s})$ | $\mathrm{I}(\mathrm{s})=\frac{\mathrm{V}(\mathrm{s})}{\mathrm{sL}}$ | | Capacitor | $\mathrm{V}(\mathrm{s})=\frac{\mathrm{I}(\mathrm{s})}{\mathrm{sC}}$ | $\mathrm{V}(\mathrm{s})=\mathrm{sC} \mathrm{I}(\mathrm{s})$ |
Now we’ll try to model a circuit in the Laplace domain directly.
现在我们将直接尝试在拉普拉斯域中建模一个电路。

What better than our circuit from earlier, to try this out.
还有什么比我们之前的电路更好的呢,来试试这个。

U sing K V L , V s ( s ) = RI ( s ) + sLI ( s ) + I ( s ) sC sC V s ( s ) = I ( s ) [ sRC + s 2 LC + 1 ] I ( s ) = sC LCs 2 + R C s + 1 V s ( s ) U sing K V L , V s ( s ) = RI ( s ) + sLI ( s ) + I ( s ) sC sC V s ( s ) = I ( s ) sRC + s 2 LC + 1 I ( s ) = sC LCs 2 + R C s + 1 V s ( s ) {:[U sing KVL","],[V_(s)(s)=RI(s)+sLI(s)+(I((s)))/(sC)],[=>sCV_(s)(s)=I(s)[sRC+s^(2)LC+1]],[:.I(s)=(sC)/(LCs^(2)+RCs+1V_(s)((s)))]:}\begin{aligned} & U \operatorname{sing} K V L, \\ & \mathrm{~V}_{\mathrm{s}}(\mathrm{~s})=\mathrm{RI}(\mathrm{~s})+\mathrm{sLI}(\mathrm{~s})+\frac{\mathrm{I}(\mathrm{~s})}{\mathrm{sC}} \\ & \Rightarrow \mathrm{sC} \mathrm{~V}_{\mathrm{s}}(\mathrm{~s})=\mathrm{I}(\mathrm{~s})\left[\mathrm{sRC}+\mathrm{s}^{2} \mathrm{LC}+1\right] \\ & \therefore \mathrm{I}(\mathrm{~s})=\frac{\mathrm{sC}}{\mathrm{LCs}^{2}+R C s+1 \mathrm{~V}_{\mathrm{s}}(\mathrm{~s})} \end{aligned}
Modelling circuits in the s-domain has lots of advantages, it’s easier to study stability, natural response, frequency response etc., but that’s more of a control systems terro and we are not going into it.
在 s 域中建模电路有很多优点,研究稳定性、自然响应、频率响应等更为简单,但这更多是控制系统的领域,我们不打算深入探讨。

You can check out our control systems book if you are interested: Control Systems for Complete Idiots
如果您感兴趣,可以查看我们的控制系统书籍:《完全傻瓜的控制系统》

12. TRANSIENT ANALYSIS
12. 瞬态分析

12.1 INTRODUCTION
12.1 引言

A circuit whose circuit parameters or conditions remain constant, is said to be in a steady state.
一个电路的电路参数或条件保持不变时,称为稳态。

But a circuit isn’t always in steady state, when a circuit or a portion of the circuit is switched on or off, there is a sudden change in circuit parameters (like amplitude, frequency etc.). A certain amount of time is taken for these changes to take place, this duration is called the Transient period and this phenomenon is known as Transient.
但电路并不总是处于稳态,当电路或电路的一部分被打开或关闭时,电路参数(如幅度、频率等)会发生突然变化。这些变化需要一定的时间才能发生,这段时间称为瞬态期,这种现象被称为瞬态。

Once the transient period is over, the circuit settles down and attains the steady state, if not disturbed further.
一旦瞬态期结束,电路就会稳定下来并达到稳态,前提是没有进一步的干扰。
So when you switch on a circuit, there are 2 responses; one is the transient response or the natural response and the other is the steady state response or the forced response.
所以当你打开一个电路时,会有两种响应;一种是瞬态响应或自然响应,另一种是稳态响应或强迫响应。

All the circuit analysis we did till now was to find the steady state response, we ignored the transient response.
到目前为止,我们所做的所有电路分析都是为了找到稳态响应,我们忽略了瞬态响应。

Transients are due to the presence of energy storing elements (Capacitors and Inductors) in a circuit.
瞬态是由于电路中存在储能元件(电容器和电感器)所致。

These elements don’t respond instantly to change in circuit conditions.
这些元件对电路条件的变化不会立即作出反应。

12.2 TRANSIENT RESPONSE OF AN RL CIRCUIT TO DC EXCITATION
12.2 RL 电路对直流激励的瞬态响应

Consider this initially uncharged inductor in series with a resistor. At t = 0 t = 0 t=0t=0, the switch S is closed.
考虑这个最初未充电的电感器与电阻器串联。在 t = 0 t = 0 t=0t=0 时,开关 S 关闭。

Being an initially uncharged inductor, the current before the instant of closing, i 0 ) i 0 {:i^(-)0^(-))\left.\mathrm{i}^{-} 0^{-}\right)is zero.
作为一个最初未充电的电感器,在闭合瞬间之前, i 0 ) i 0 {:i^(-)0^(-))\left.\mathrm{i}^{-} 0^{-}\right) 的电流为零。

But as the inductor cannot quickly respond to the change in current, the current at the instant right after the closing of the switch, i ( 0 + ) i 0 + i(0^(+))i\left(0^{+}\right)is also zero i.e.
但由于电感器无法快速响应电流的变化,因此在开关关闭的瞬间, i ( 0 + ) i 0 + i(0^(+))i\left(0^{+}\right) 也为零,即
i ( 0 ) = i ( 0 + ) = 0 i 0 = i 0 + = 0 i(0^(-))=i(0^(+))=0i\left(0^{-}\right)=i\left(0^{+}\right)=0
Then the current that flows through the circuit can be found using differential equation. Using KVL,
然后可以使用微分方程找到流经电路的电流。使用基尔霍夫电压定律,
V = i R + L d i d t V = i R + L d i d t V=iR+L(di)/(dt)V=i R+L \frac{d i}{d t}
This is a simple first order differential equation and can be solved easily, but we’ll go with the Laplace transform approach.
这是一个简单的一阶微分方程,可以很容易地解决,但我们将采用拉普拉斯变换的方法。

Do not forget to include the initial value terms in the Laplace transform of the differential.
不要忘记在微分的拉普拉斯变换中包含初始值项。
Taking Laplace on both sides, V s = RI ( s ) + L [ sI ( s ) i ( 0 ) ] I ( s ) = V s [ R + Ls ]  Taking Laplace on both sides,  V s = RI ( s ) + L sI ( s ) i 0 I ( s ) = V s [ R + Ls ] {:[" Taking Laplace on both sides, "],[(V)/((s))=RI(s)+L[sI((s))-i(0^(-))]],[=>I(s)=(V)/((s)[R+Ls])]:}\begin{aligned} & \text { Taking Laplace on both sides, } \\ & \frac{\mathrm{V}}{\mathrm{~s}}=\mathrm{RI}(\mathrm{~s})+\mathrm{L}\left[\mathrm{sI}(\mathrm{~s})-\mathrm{i}\left(0^{-}\right)\right] \\ & \Rightarrow \mathrm{I}(\mathrm{~s})=\frac{\mathrm{V}}{\mathrm{~s}[\mathrm{R}+\mathrm{Ls}]} \end{aligned}
Writing the RHS as partial fraction, I ( s ) = V R [ 1 s 1 s + R / L ] I ( s ) = V R 1 s 1 s + R / L I(s)=(V)/(R)[(1)/(s)-(1)/(s+R//L)]I(s)=\frac{V}{R}\left[\frac{1}{s}-\frac{1}{s+R / L}\right]
将右侧写成部分分式, I ( s ) = V R [ 1 s 1 s + R / L ] I ( s ) = V R 1 s 1 s + R / L I(s)=(V)/(R)[(1)/(s)-(1)/(s+R//L)]I(s)=\frac{V}{R}\left[\frac{1}{s}-\frac{1}{s+R / L}\right]
i ( t ) = V R V R e R t / L i ( t ) = V R V R e R t / L :.i(t)=(V)/(R)-(V)/(R)e^(-Rt//L)\therefore \mathrm{i}(\mathrm{t})=\frac{\mathrm{V}}{\mathrm{R}}-\frac{\mathrm{V}}{\mathrm{R}} e^{-\mathrm{R} t / \mathrm{L}}
Steady state response
稳态响应
Transient
瞬态

response
响应
The equation just validates our discussion, that circuits have 2 responses, steady state and transient.
这个方程验证了我们的讨论,即电路有两种响应,稳态和瞬态。

As t increases the transient response term decreases exponentially and nullifies, leaving only the steady state response.
随着 t 的增加,瞬态响应项指数下降并消失,只留下稳态响应。

If you go by the methods used prior to this chapter, current value by ohms law will give the result I = V/R, which is our steady state response.
如果你按照本章之前使用的方法,使用欧姆定律计算当前值将得到结果 I = V/R,这就是我们的稳态响应。

The graph of the current response of this circuit will look like this:
该电路的当前响应图将如下所示:

The constant τ = L / R τ = L / R tau=L//R\tau=L / R is called the time constant of the circuit.
常数 τ = L / R τ = L / R tau=L//R\tau=L / R 被称为电路的时间常数。

This value decides how fast this circuit will reach steady state.
这个值决定了电路达到稳态的速度。

Typically current will reach steady state after t = 5 τ = 5 τ =5tau=5 \tau.
通常电流将在 t = 5 τ = 5 τ =5tau=5 \tau 后达到稳态。

The important thing to note is that after the brief transient period, Inductor acts as a short circuit (just like a normal wire) in a DC circuit.
需要注意的重要一点是,在短暂的瞬态期间之后,电感在直流电路中表现得像短路(就像普通导线一样)。

12.3 TRANSIENT RESPONSE OF AN RC CIRCUIT TO DC EXCITATION
12.3 RC 电路对直流激励的瞬态响应

This time consider an initially uncharged capacitor in series with a resistor. At t = 0 t = 0 t=0t=0, the switch S S SS is closed.
这次考虑一个最初未充电的电容器与电阻器串联。在 t = 0 t = 0 t=0t=0 时,开关 S S SS 被关闭。

As the capacitor cannot quickly respond to the change in voltage, the voltage before and right after the closing of the switch are the same i.e.
由于电容器无法快速响应电压的变化,因此开关关闭前后的电压是相同的,即

Current through the circuit, i ( t ) = C dv C dt I ( s ) = C [ sV ( s ) v C ( 0 ) ] = Cs V C ( s ) . . (1)  Current through the circuit,  i ( t ) = C dv C dt I ( s ) = C sV ( s ) v C 0 = Cs V C ( s ) . .  (1)  {:[" Current through the circuit, "],[{:[i(t)=C(dv_(C))/(dt)],[{:[I(s)=C[sV((s))-v_(C)(0^(-))]],[=CsV_(C)(s)dots dots..]:}]:}{:" (1) ":}]:}\begin{aligned} & \text { Current through the circuit, } \\ & \begin{aligned} & \mathrm{i}(\mathrm{t})=\mathrm{C} \frac{\mathrm{dv}_{\mathrm{C}}}{\mathrm{dt}} \\ & \begin{aligned} \mathrm{I}(\mathrm{~s}) & =\mathrm{C}\left[\mathrm{sV}(\mathrm{~s})-\mathrm{v}_{\mathrm{C}}\left(0^{-}\right)\right] \\ & =\mathrm{Cs} \mathrm{~V}_{\mathrm{C}}(\mathrm{~s}) \ldots \ldots . . \end{aligned} \end{aligned} \begin{array}{l} \text { (1) } \end{array} \end{aligned}

Using KVL,
使用基尔霍夫电压定律,

V s = R I ( s ) + V C ( s ) . V s = R I ( s ) + V C ( s ) . (V)/(s)=RI(s)+V_(C)(s)dots dots.\frac{V}{s}=R I(s)+V_{C}(s) \ldots \ldots .. (2)
Substituting (2) in (1)
在(1)中代入(2)
V s = R C s V C ( s ) + V C ( s ) V s = R C s V C ( s ) + V C ( s ) (V)/(s)=RCsV_(C)(s)+V_(C)(s)\frac{V}{s}=R C s V_{C}(s)+V_{C}(s)
V C ( s ) = V s [ RCs + 1 ] v C ( t ) = F Steady state response t / RC Transient response V C ( s ) = V s [ RCs + 1 ] v C ( t ) = F  Steady state   response  t / RC  Transient   response  {:[V_(C)(s)=(V)/((s)[RCs+1])],[:.v_(C)(t)=F_({:[" Steady state "],[" response "]:})ubrace(-t//RCubrace)_({:[" Transient "],[" response "]:})]:}\begin{aligned} & V_{\mathrm{C}}(\mathrm{~s})=\frac{\mathrm{V}}{\mathrm{~s}[\mathrm{RCs}+1]} \\ & \therefore \mathrm{v}_{\mathrm{C}}(\mathrm{t})=\mathrm{F}_{\begin{array}{c} \text { Steady state } \\ \text { response } \end{array}} \underbrace{-\mathrm{t} / \mathrm{RC}}_{\begin{array}{c} \text { Transient } \\ \text { response } \end{array}} \end{aligned}
This is the generalized expression for v c ( t ) v c ( t ) v_(c)(t)\mathrm{v}_{\mathrm{c}}(\mathrm{t}) and the corresponding graph looks like:
这是 v c ( t ) v c ( t ) v_(c)(t)\mathrm{v}_{\mathrm{c}}(\mathrm{t}) 的广义表达式,对应的图形如下:

The constant τ = R C τ = R C tau=RC\tau=R C is the time constant of this circuit.
该常数 τ = R C τ = R C tau=RC\tau=R C 是该电路的时间常数。

In a DC circuit, the capacitor acts as an open circuit in the steady state.
在直流电路中,电容器在稳态下表现为开路。

12.4 EXAMPLE
12.4 示例

Using a similar approach like in the last two cases, we can obtain the general response for any circuit.
使用与前两个案例类似的方法,我们可以获得任何电路的一般响应。

Consider this example, say the switch has been in position 1 for a long time and then it’s moved to position 2 at t = 0 t = 0 t=0t=0 and we are required to find the voltage across the capacitor.
考虑这个例子,假设开关在位置 1 上待了很长时间,然后在 t = 0 t = 0 t=0t=0 时移动到位置 2,我们需要找出电容器两端的电压。

Since the switch has been in position 1 for a long time it’s in the steady state, hence the initial voltage across the capacitor will be equal to the applied voltage.
由于开关在位置 1 上已经很长时间,因此它处于稳态,因此电容器两端的初始电压将等于施加的电压。
v C ( 0 ) = v C ( 0 + ) = 40 V v C 0 = v C 0 + = 40 V v_(C)(0^(-))=v_(C)(0^(+))=40Vv_{\mathrm{C}}\left(0^{-}\right)=v_{\mathrm{C}}\left(0^{+}\right)=40 \mathrm{~V}
In position 2,
在位置 2,
i ( t ) = C d v c dt = 0.1 dv v c dt i ( t ) = C d v c dt = 0.1 dv v c dt i(t)=C((d)v_(c))/(dt)=0.1(dvv_(c))/(dt)\mathrm{i}(\mathrm{t})=\mathrm{C} \frac{\mathrm{~d} \mathrm{v}_{\mathrm{c}}}{\mathrm{dt}}=0.1 \frac{\mathrm{dv} \mathrm{v}_{\mathrm{c}}}{\mathrm{dt}}
Using KVL,
使用基尔霍夫电压定律,
v C + i ( t ) [ R 1 + R 2 ] = 0 v C + i ( t ) R 1 + R 2 = 0 v_(C)+i(t)[R_(1)+R_(2)]=0v_{C}+i(t)\left[R_{1}+R_{2}\right]=0
Taking Laplace transform on both sides,
对两边进行拉普拉斯变换,
V C ( s ) + 10 × 0.1 [ s V C ( s ) V C ( 0 ) ] = 0 V C ( s ) + s V C ( s ) 40 = 0 V C ( s ) = 40 ( 1 + s ) V C ( s ) + 10 × 0.1 s V C ( s ) V C 0 = 0 V C ( s ) + s V C ( s ) 40 = 0 V C ( s ) = 40 ( 1 + s ) {:[V_(C)(s)+10 xx0.1[(s)V_(C)((s))-V_(C)(0^(-))]=0],[V_(C)(s)+sV_(C)(s)-40=0],[V_(C)(s)=(40)/((1+s))]:}\begin{aligned} & \mathrm{V}_{\mathrm{C}}(\mathrm{~s})+10 \times 0.1\left[\mathrm{~s} \mathrm{~V}_{\mathrm{C}}(\mathrm{~s})-\mathrm{V}_{\mathrm{C}}\left(0^{-}\right)\right]=0 \\ & \mathrm{~V}_{\mathrm{C}}(\mathrm{~s})+\mathrm{s} \mathrm{~V}_{\mathrm{C}}(\mathrm{~s})-40=0 \\ & \mathrm{~V}_{\mathrm{C}}(\mathrm{~s})=\frac{40}{(1+\mathrm{s})} \end{aligned}
v C ( t ) = 40 e t v C ( t ) = 40 e t :.v_(C)(t)=40e^(-t)\therefore \mathrm{v}_{\mathrm{C}}(\mathrm{t})=40 \mathrm{e}^{-\mathrm{t}}
Similarly, the general response for any type of circuits including AC circuits can be found.
类似地,任何类型电路的总体响应,包括交流电路,都可以找到。

13. 3-PHASE SYSTEMS
13. 三相系统

13.1 INTRODUCTION
13.1 引言

There are 2 popular kinds of electrical systems, Single phase and Three phase.
有两种流行的电力系统,单相和三相。

In a single phase system, there will be live wire and a neutral return path for the current to flow.
在单相系统中,将有火线和中性回路供电流流动。

In a three phase system, there will be 3 live wires and a common neutral return path for the current.
在三相系统中,将有 3 根火线和一个共同的中性回路。

There are several advantages to having 3 phase system over single phase; more power can be delivered, cheaper to generate, transmit etc.
三相系统相对于单相系统有几个优点;可以提供更多的电力,发电、传输等成本更低。

Three phase voltage is generated with the help of 3 coils separated by 120 120 120^(@)120^{\circ} inside a generator. Due to this arrangement, the voltage induced on each coil will lag the other by 120 120 120^(@)120^{\circ}. Mathematically,
三相电压是在发电机内部由三个相隔 120 120 120^(@)120^{\circ} 的线圈产生的。由于这种排列,每个线圈上感应的电压将滞后于其他线圈 120 120 120^(@)120^{\circ} 。在数学上,
V R = V m sin ( ω t ) V Y = V m sin ( ω t 120 ) V B = V m sin ( ω t 240 ) V R = V m sin ( ω t ) V Y = V m sin ω t 120 V B = V m sin ω t 240 {:[V_(R)=V_(m)sin(omega t)],[V_(Y)=V_(m)sin(omega t-120^(@))],[V_(B)=V_(m)sin(omega t-240^(@))]:}\begin{aligned} & V_{R}=V_{m} \sin (\omega t) \\ & V_{Y}=V_{m} \sin \left(\omega t-120^{\circ}\right) \\ & V_{B}=V_{m} \sin \left(\omega t-240^{\circ}\right) \end{aligned}

13.2 STAR CONNECTION
13.2 星形连接

In a single phase connection, 2 wires are sufficient for transmitting power to the load.
在单相连接中,2 根电线足以将电力传输到负载。

But in a 3 phase connection, 6 terminals (2 ends of each phase) are available to supply power to the loads.
但在三相连接中,有 6 个端子(每相的两个端)可用于向负载供电。

Using these 6 terminals individually, like in single phase connection will prove expensive and unnecessary.
单独使用这 6 个端子,就像在单相连接中一样,将会显得昂贵且不必要。

There are 2 better ways to connect three phase terminals to deliver power to the loads.
有两种更好的方法将三相端子连接以向负载供电。

First is the Star or the Wye Connection.
首先是星形连接或三角形连接。

In such a connection, one terminal of each coil is terminated at a common point called the neutral.
在这种连接中,每个线圈的一个端子连接到一个称为中性的公共点。

Loads can be connected either between the phases or between the phase and the neutral.
负载可以连接在相与相之间或相与中性线之间。

13.3 DELTA CONNECTION
13.3 三角洲连接

Another possible way to connect coils is the Delta Connection.
另一种连接线圈的可能方式是三角连接。

In such a connection, the ending terminal of a coil is connected to the starting terminal of the other coil, so as to form a closed loop as shown below.
在这种连接中,一个线圈的结束端连接到另一个线圈的起始端,从而形成一个闭合回路,如下所示。

In delta connection there is no common neutral point, so the only way to connect load is between the phases.
在三角连接中没有公共中性点,因此连接负载的唯一方法是在相之间。

13.4 LINE & PHASE VOLTAGE
13.4 线路与相电压

While studying 3 ϕ 3 ϕ 3phi3 \phi circuits, two types of voltages can be defined; line voltage and phase voltage (this applies for both connections).
在研究 3 ϕ 3 ϕ 3phi3 \phi 电路时,可以定义两种类型的电压:线路电压和相电压(这适用于两种连接方式)。

The potential difference or voltage between any two phases is defined as the line voltage.
任意两个相之间的电位差或电压被定义为线电压。

It is denoted as V L V L V_(L)\mathrm{V}_{\mathrm{L}}. And the potential difference between any one phase and neutral is called phase voltage.
它表示为 V L V L V_(L)\mathrm{V}_{\mathrm{L}} 。任何一个相与中性点之间的电位差称为相电压。

It is denoted as V ph V ph V_(ph)\mathrm{V}_{\mathrm{ph}}.
它表示为 V ph V ph V_(ph)\mathrm{V}_{\mathrm{ph}}

In a delta connection, there is no neutral point, hence the line voltage and the phase voltage are the same.
在三角连接中,没有中性点,因此线电压和相电压是相同的。

But in star connection, these are two different quantities, whose relation can be derived as follows:
但在星形连接中,这两个量是不同的,其关系可以推导如下:

Consider phases R & Y,
考虑相 R 和 Y,

V R N = V p h 0 = V p h V R N = V p h 0 = V p h V_(RN)=V_(ph)/_0=V_(ph)V_{R N}=V_{p h} \angle 0=V_{p h},
V Y N = V p h 120 = V p h 2 + j 3 V p h 2 V Y N = V p h 120 = V p h 2 + j 3 V p h 2 V_(YN)=V_(ph)/_-120=(-V_(ph))/(2)+j(-sqrt3V_(ph))/(2)V_{Y N}=V_{p h} \angle-120=\frac{-V_{p h}}{2}+j \frac{-\sqrt{3} V_{p h}}{2}
V L = V R N V Y N V L = V R N V Y N V_(L)=V_(RN)-V_(YN)V_{L}=V_{R N}-V_{Y N} = 3 V p h 2 + j 3 V p h 2 = 3 V p h 2 + j 3 V p h 2 =(3V_(ph))/(2)+j(-sqrt3V_(ph))/(2)=\frac{3 V_{p h}}{2}+j \frac{-\sqrt{3} V_{p h}}{2}
| V L | = 3 | V ph | V L = 3 V ph :.|V_(L)|=sqrt3|V_(ph)|\therefore\left|\mathrm{V}_{\mathrm{L}}\right|=\sqrt{3}\left|\mathrm{~V}_{\mathrm{ph}}\right|
Point to note is that, in a delta connection, the line voltage is higher than the phase voltage.
需要注意的是,在三角连接中,线电压高于相电压。

13.5 LINE & PHASE CURRENT
13.5 线电流和相电流

Similar to voltage, current can also be defined in 2 ways in a 3 ϕ 3 ϕ 3phi3 \phi circuit. Current flowing through the coil (or the load) is called as the phase current ( I p h ) I p h (I_(ph))\left(I_{p h}\right) and current flowing through any line is called line current ( I L ) I L (I_(L))\left(I_{L}\right).
类似于电压,电流在 3 ϕ 3 ϕ 3phi3 \phi 电路中也可以用两种方式定义。流过线圈(或负载)的电流称为相电流 ( I p h ) I p h (I_(ph))\left(I_{p h}\right) ,而流过任何线路的电流称为线路电流 ( I L ) I L (I_(L))\left(I_{L}\right)
In a star connection, the line current and the phase current are one and the same.
在星形连接中,线电流和相电流是相同的。

But in delta connection, these are two different quantities, whose relation can be derived as follows:
但在三角连接中,这两个量是不同的,其关系可以推导如下:

Consider the line R ,
考虑线 R,
I R = I p h 0 = I p h , I Y N = I p h 120 = I p h 2 + j 3 I p h 2 I R = I p h 0 = I p h , I Y N = I p h 120 = I p h 2 + j 3 I p h 2 {:[I_(R)=I_(ph)/_0=I_(ph)","],[I_(YN)=I_(ph)/_-120=(-I_(ph))/(2)+j(-sqrt3I_(ph))/(2)]:}\begin{aligned} & I_{R}=I_{p h} \angle 0=I_{p h}, \\ & I_{Y N}=I_{p h} \angle-120=\frac{-I_{p h}}{2}+j \frac{-\sqrt{3} I_{p h}}{2} \end{aligned}
I L = I R I B = 3 I ph 2 + j 3 I ph 2 | I L | = 3 | I ph | I L = I R I B = 3 I ph 2 + j 3 I ph 2 I L = 3 I ph {:[I_(L)=I_(R)-I_(B)],[=(3I_(ph))/(2)+j(-sqrt3I_(ph))/(2)],[quad:.|I_(L)|=sqrt3|I_(ph)|]:}\begin{aligned} & \mathrm{I}_{\mathrm{L}}=\mathrm{I}_{\mathrm{R}}-\mathrm{I}_{\mathrm{B}} \\ & =\frac{3 \mathrm{I}_{\mathrm{ph}}}{2}+\mathrm{j} \frac{-\sqrt{3} \mathrm{I}_{\mathrm{ph}}}{2} \\ & \quad \therefore\left|\mathrm{I}_{\mathrm{L}}\right|=\sqrt{3}\left|\mathrm{I}_{\mathrm{ph}}\right| \end{aligned}
Thus in a delta connection, the line current is higher than phase current.
因此,在三角连接中,线电流高于相电流。

13.6 LOAD CONNECTIONS
13.6 负载连接

Loads can also be connected in several ways in a 3 ϕ 3 ϕ 3phi3 \phi system as shown below (there are still more connections).
3 ϕ 3 ϕ 3phi3 \phi 系统中,负载也可以通过多种方式连接,如下所示(还有更多连接)。

The appropriate connection is chosen according the voltage and the current requirements of the load.
根据负载的电压和电流要求选择适当的连接。

Each connection has certain advantages and disadvantages.
每个连接都有其优缺点。

If the impedances or the loads are equally distributed among the 3 phases, such a load is called balanced load.
如果阻抗或负载在三个相之间均匀分布,则该负载称为平衡负载。


Z 2 Z 2 Z_(2)\mathrm{Z}_{2}
For Balanced Load,
对于平衡负载,
Z = Z 1 = Z 2 = Z 3 Z = Z 1 = Z 2 = Z 3 Z=Z_(1)=Z_(2)=Z_(3)\mathrm{Z}=\mathrm{Z}_{1}=\mathrm{Z}_{2}=\mathrm{Z}_{3}

13.7 POWER
13.7 功率

Three phase power in a circuit is given by:
三相电路中的功率为:
P = 3 V L I L cos ϕ P = 3 V ph I ph cos ϕ P = 3 V L I L cos ϕ P = 3 V ph I ph cos ϕ {:[P=sqrt3V_(L)I_(L)cos phi],[P=3V_(ph)I_(ph)cos phi]:}\begin{aligned} & \mathrm{P}=\sqrt{3} \mathrm{~V}_{\mathrm{L}} \mathrm{I}_{\mathrm{L}} \cos \phi \\ & \mathrm{P}=3 \mathrm{~V}_{\mathrm{ph}} \mathrm{I}_{\mathrm{ph}} \cos \phi \end{aligned}
These equations are applicable to both star and delta connections.
这些方程适用于星形和三角形连接。

REFERENCES
参考文献

  1. Introductory Circuit Analysis by Robert L. Boylestad
    罗伯特·L·博伊尔斯塔德的《电路分析导论》
  2. Engineering Circuit Analysis by Hayt, Kimmerly
    《海特与基默利的工程电路分析》
  3. Delmar’s Standard Textbook of Electricity by Stephen L. Herman
    德尔玛电力标准教科书 斯蒂芬·L·赫尔曼著
  4. Basic AC Circuits by John Clayton Rawlins
    《基本交流电路》 约翰·克莱顿·罗林斯
  5. Discovering the Laplace Transform in Undergraduate Differential Equations by Terrance J.
    发现本科生微分方程中的拉普拉斯变换 由特伦斯·J。

    Quinn and Sanjay Rai
    奎因和桑贾伊·赖

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