It's actually very similar to what you have in mesh analysis. It's actually the same principle. So the principle is identical to mesh analysis. When you look at the fourth page,

实际上，这和你现有的非常相似。就是在网格分析中。其实原理是一样的。所以，这个原理与网格分析完全相同。当你看到第四页时，

they are identical to mesh analysis.

它们与网格分析是一样的。

A mesh is a loop. A mesh is the smallest loop. And there's no loop inside of it.

所谓“网孔”，就是一个回路。网孔是最小的回路。而且里面没有其他回路。

For the mesh analysis, we first define the mesh. And then we assign mesh current. Mesh current is usually defined as clockwise direction.

对于网孔分析，我们首先要定义网孔。然后，我们给每个网孔分配一个网孔电流。网孔电流通常被定义为顺时针方向。

And then you apply Kirchhoff's voltage law around each mesh. to get the unknown mesh current.

接下来，你在每个网孔周围应用基尔霍夫电压定律。从而求出未知的网孔电流。

And here's an example.

这里有一个例子。

Here is the mesh current. the small i's. The small i is mesh current, the big I is the branch current. And mesh current has a clockwise direction. The small i1, i2 are mesh currents. the big i1, i2 and i3 are branch currents.

这就是网孔电流。就是那些小写的i。小i代表网孔电流，大I代表支路电流。而且网孔电流的方向是顺时针的。小i1和i2就是网孔电流。而大I1、I2和I3则是支路电流。

And for each mesh, we apply Kirchhoff's voltage law. So here we have two meshes. Let me assume the mesh current is clockwise direction. so small i1 and i2 as clockwise direction.

对于每个网孔，我们都要应用基尔霍夫电压定律。这里有两个网孔。我假设网孔电流是顺时针方向的。所以小i1和i2也是顺时针方向。

And let me apply the Kirchhoff's voltage law. for each mesh.

现在，我来应用基尔霍夫电压定律。对每个网孔都适用。

And here is important. because here we have mesh currents, right? small i1 and small i2 are mesh currents.

这里很重要。因为这里有网孔电流，对吧？小i1和小i2就是网孔电流。

And here is important. because here we have mesh currents, right? small i1 and small i2 are mesh currents.

这里很重要。因为这里有网孔电流，对吧？小i1和小i2就是网孔电流。

so the mesh current for resistor one is also i1. and for i3, for r3 is different. for r3, you have to apply kirchhoff's current law.

所以，流过电阻R1的网孔电流也是i1。而对于i3，流过R3的情况就不同了。对于R3，你必须应用基尔霍夫电流定律。

i found not many students has understood what we have told. so that's just a bit news to you.

我发现没有多少学生理解我们所说的内容。所以，这对你们来说可能有点新。

so for the mesh analysis, right, first let me define the mesh current. and here we have two mesh. the mesh direction is clockwise direction.

那么，对于网孔分析，首先让我来定义一下网孔电流。这里有两个网孔。网孔的方向是顺时针方向。

here are two meshes.

这里有两个网孔。

and we have i1 and i2. they are mesh current.

我们有I1和I2。它们是网孔电流。

and for each branch, these are the branches, right? so here are branch currents. they are mesh current. and the branch current, for this, it is identical to the mesh current. it's i1.

对于每一条支路，这些就是支路，对吧？所以，这里是支路电流。它们是网孔电流。对于这个支路来说，它的电流与网孔电流是相同的。就是I1。

small i1. here is small i1. and for this resistor, we have to apply the kirchhoff current law.

小写的i1。这里也是小写的i1。而对于这个电阻，我们有需要应用基尔霍夫电流定律。

so the current of this resistor should be i1 minus i2.

所以，这个电阻的电流应该是I1减去I2。

downwards. and we apply the kirchhoff voltage law for each mesh. to get the equation.

方向是向下的。然后，我们对每个网孔应用基尔霍夫电压定律。从而得到方程。

here's the i1.

这是I1。

here's i2. Here's i1 minus i2.

这是I2。这是I1减去I2。

and also for this circuit,

对于这个电路，

we also have two meshes. the mesh direction is clockwise direction. and i1 and i2 are mesh currents.

我们也有两个网孔。网孔的方向是顺时针方向。I1和I2是网孔电流。

and for this branch,

对于这个支路，

the branch current is also i1.

支路电流也是I1。

and for this current, the branch current is also i2. and for the branch current in the middle,

对于这个电流，支路电流也是I2。而对于中间的支路电流，

pay attention to the direction. direction of the i1. it's clockwise direction. and i2 is also clockwise direction. so the branch current flowing in this circuit is i1 minus i2.

要注意方向。就是I1的方向。它是顺时针方向。I2也是顺时针方向。所以，流过这个支路的电流是I1减去I2。

and we use the big i, big i is also we use. This big i is i1, i2 and this i1. i1 minus i2.

我们这里用大写的I，也就是用这个I。这个I就是I1、I2，还有这个I1。I1减去I2。

let me apply kirchhoff's voltage law for each mesh.

让我对每个网孔应用基尔霍夫电压定律。

but how can we design the sign? so what did you solve, max? if the voltage source has the same direction as the mesh current direction, then it's a positive. okay, this is also clockwise direction. so the 10 volts is positive. okay, the branch current for this resistor is

但是，我们该如何确定符号呢？那么，你解出来了什么，Max？如果电压源的方向与网孔电流的方向一致，那就是正的。好的，这也是顺时针方向。所以这10伏特是正的。好的，这个电阻的支路电流是

it's an upward branch. the branch current is upward. the voltage current is downward. so it's from low potential to high potential. so the voltage direction is not clockwise direction. so it's a minus. i1 times 10.

向上的支路电流。支路电流是向上的。而电压电流是向下的。所以它是从低电位指向高电位的。因此，电压的方向不是顺时针方向。所以是负的。I1乘以10。

this is equal to the mesh minus. because the voltage direction is not clockwise direction, it's the opposite direction as the mesh.

这等于网孔电压的负值。因为电压方向不是顺时针方向，而是与网孔方向相反。

and for this resistor, we know the branch current is i1 minus i2. and it's downward. so the voltage is upward. so it also has the opposite direction as the mesh. so it's the minus. i1 minus i2.

对于这个电阻，我们知道支路电流是I1减去I2。而且是向下的。所以电压是向上的。所以它的方向也与网孔方向相反。所以是负号。I1减去I2。

and for this resistor, we know the branch current is i1 minus i2. and it's downward. so the voltage is upward. so it also has the opposite direction as the mesh. so it's the minus. i1 minus i2.

对于这个电阻，我们知道支路电流是I1减去I2。而且是向下的。所以电压是向上的。所以它的方向也与网孔方向相反。所以是负号。I1减去I2。

times what is the resistor? 180.

乘以电阻是多少？180。

okay, this is my first equation.

好的，这是我的第一个方程。

and for the second mesh, for the second loop,

对于第二个网孔，对于第二个回路，

the voltage is upward. so it has the same direction as the mesh. also clockwise. so for this resistor, we should get positive. so it also has the opposite direction as the mesh.

电压是向上的。所以它的方向与网孔方向相同。也是顺时针方向。所以对于这个电阻，我们应该得到正号。所以它的方向与网孔方向相反。

and then for this resistor, the current is downwards. so the voltage is upward.

对于这个电阻，电流是向下的。所以电压是向上的。

it has the opposite direction as the mesh. so it's minus.

它的方向与网孔方向相反。所以是负号。

minus two times two ohm.

减去2乘以2欧姆。

and for the voltage source, it also has the opposite direction as the mesh. so it's minus. twenty.

对于电压源，它的方向也与网孔方向相反。所以也是负号。20。

is zero.

等于零。

okay, it's clear.

好的，明白了。

okay, a new definition. node analysis.

好的，一个新的定义。节点分析法。

so i've been introduced before. mesh analysis.

这个之前已经介绍过了。还有网孔分析法。

especially when you have many voltage sources.

特别是当有很多电压源的时候。

if you have many voltage sources in your circuit, we apply mesh analysis.

如果你的电路中有许多电压源，我们就使用网孔分析法。

because you will get less equations. you will get fewer.

因为这样得到的方程会更少。方程数量更少。

but if there are many current sources,

但是，如果有许多电流源，

in your circuit, then you should apply node analysis.

在电路中，你就应该使用节点分析法。

okay, this is the difference. If there are many voltage sources in your circuit, you should apply mesh analysis. but if there are more many current sources, we should use node analysis. but what is a node analysis?

好的，这就是区别。如果电路中有许多电压源，就应该用网孔分析法。但如果电流源更多，就应该用节点分析法。那么，什么是节点分析法呢？

we have introduced the definition of node. at the beginning. The node is defined as a junction of two or more branches. For example, here is a branch.

我们之前已经介绍过节点的定义了。在开头的时候。节点被定义为两个或更多支路的连接点。比如，这里就是一条支路。

and here is a second branch. Is the branch, right? branch one is the branch two. The junction between two branches.

这里还有第二条支路。这是一条支路，对吧？这是支路一，这是支路二。这两个支路之间的连接点，就是节点。

So, the junction between two branches is a node.

所以，两个支路之间的交点就是节点。

and if we define any one node as a reference node, the reference node is usually has zero potential. or the ground. And the remaining nodes, of the network will have fixed potential. we have a fixed potential relative to this reference.

如果我们把其中一个节点定义为参考节点，那么参考节点通常是电位为零的。或者说是接地。而剩下的节点，就具有固定的电位。这个电位是相对于参考节点而言的。

Okay, this means, for example, we have a current source.

好的，这意味着，举个例子，我们有一个电流源。

and we have a resistor here.

这里还有一个电阻。

another resistor.

还有另一个电阻。

Okay, if we have a circuit like this. and here is the ground.

好的，如果我们有这样一个电路。这里就是接地的。

here is the ground.

这里是接地。

and we use this node as reference.

我们把这个节点作为参考节点。

and here is another node, right?

这里还有另一个节点，对吧？

and here is another node, right?

这里还有另一个节点，对吧？

here's also another node.

这里也是另一个节点。

so the potential or the voltage here

所以，这里的电位或电压，

of this node is reference, is relative. relative to this node, relative to the reference node.

是相对于参考点的。也就是相对于这个节点，相对于参考节点。

and any node can be selected as reference node. so you might also select this node as reference node. but usually we select the node is zero potential as reference node. because that one is zero.

任何节点都可以被选为参考节点。所以，你也可以选择这个节点作为参考节点。但通常我们会选择电位为零的节点作为参考节点。因为那个点的电位是零。

so this is much simpler. but usually we select the node is zero potential as reference node.

这样会简单得多。通常，我们会选择电位为零的节点作为参考节点。

And if we have a electrical network of n nodes, there will be exist n minus one nodes. with a potential. We have to assign reference node. Here we have three nodes. So we have two nodes.

如果我们有一个包含n个节点的电路网络，那么就会存在n减1个节点。它们的电位各不相同。我们必须指定一个参考节点。这里我们有三个节点。所以，我们有两个节点。

and this is reference node. so we have two nodes with potential.

而这个是参考节点。所以，我们有两个具有电位的节点。

We have to assign reference node. and we have this potential could be, for example, here. so we have fixed the potential. we have fixed the potential relative to the assigned reference node.

我们需要指定一个参考节点。这个电位可以是，比如，这里的这个值。所以我们已经确定了电位。它是相对于所指定的参考节点而言的。

the node voltage is the voltage at a given node.

节点电压就是某个特定节点上的电压。

it's respect to the reference node.

它是相对于参考节点而言的。

Here is the node voltage. Node voltage is relative to the reference node.

这就是节点电压。节点电压是相对于参考节点的。

and now we apply the kcl. different from mesh analysis, here we apply kcl.

现在，我们应用基尔霍夫电流定律。与网孔分析不同，这里我们应用的是KCL。

for the n minus one node. so for these two nodes. we apply kcl.

对于n减1个节点。所以对于这两个节点。我们应用KCL。

so we have two equations. and solve the equations. get the current.

这样我们就得到了两个方程。然后解这两个方程。从而求出电流。

okay, here's the difference. for the mesh analysis, we apply kvl. right? here's voltage node. And we get the voltage.

好的，这就是两者的区别。对于网孔分析，我们应用的是KVL。对吧？这里是节点电压。我们得到的是电压。

and for the node analysis, node analysis, what we define the currents. and we have the remaining n minus one nodes. we apply kcl.

而对于节点分析，我们定义的是电流。剩下的还有n减1个节点。我们应用的是KCL。

and we will get the current.

这样就能得到电流。

for the branch. so this is the difference.

对于支路来说。所以，这就是两者的区别。

we apply kcl for the remaining nodes.

我们对剩余的节点应用KCL。

any questions?

有什么问题吗？

okay, here's the summary. how to use node analysis. first, we determine the number of nodes. determine the number of nodes in a network. and in the network, we have three nodes.

好的，这里做个总结。如何使用节点分析。首先，我们要确定节点的数量。也就是网络中的节点数。在这个网络中，有三个节点。

and it's the reference node.

其中一个是参考节点。

and label all other nodes. This is the reference node, and we label the other nodes, this is node 1 and node 2.

然后给其他所有节点编号。这是参考节点，我们给其他节点编号，这是节点1和节点2。

and then we assume the all node voltages. we assume the node voltage, here is v1, here is v2.

然后我们假设所有的节点电压。也就是这里的V1和V2。

with respect to the reference node.

这是相对于参考节点的电压。

and then we write KCL for each node. because here is the voltage v1. and here is the voltage v2.

然后我们对每个节点写出KCL方程。因为这里的电压是V1。这里是V2。

and we apply kirchhoff's current law.

我们应用基尔霍夫电流定律。

for each node. for example, for this node, we can apply the kirchhoff's current law. how can you write down the equation?

对每个节点都适用。例如，对于这个节点，我们可以应用基尔霍夫电流定律。那么，你怎么写出这个方程呢？

if this, for example, is 10 ohms, this is 20 ohms.

比如说，这个是10欧姆，这个是20欧姆。

v1 minus v2.

那就是V1减去V2。

here is v1 voltage, here is the node voltage v1, here is node voltage v2. v1 minus v2 is the voltage of this resistor.

这里是V1电压，也就是节点电压V1；这里是节点电压V2。V1减去V2就是这个电阻两端的电压。

divided by 10 ohms is the current. for this resistor. so we get the current here.

除以10欧姆，就是电流。对于这个电阻来说。所以我们在这里得到了电流。

and

而且，

for this branch,

对于这个支路来说，

we know here is the voltage. and the resistor is 10 ohms.

我们知道这里的电压。电阻是10欧姆。

so v2 divided by 10 is the current.

所以，V2除以10就是电流。

because here is zero. v2 minus zero, actually minus zero. divided by 10 is the current here. and how do we get the current here? here is zero, here is zero, zero.

因为这里是零。其实是V2减去0。再除以10，就是这里的电流。那我们怎么得到这里的电流呢？这里是零，这里也是零，零。

so v2 divided by 20 is the current here.

所以，V2除以20就是这里的电流。

so we get the three current. and then we apply the kirchhoff's current law. the current entering this node is equivalent to the current leaving the node.

这样我们就得到了三个电流。然后，我们应用基尔霍夫电流定律。流入这个节点的电流等于流出这个节点的电流。

okay, here is the entering, right? only one, this one is entering. the other two are leaving.

好的，这里是流入的，对吧？只有一个，就是这个流入。另外两个是流出的。

so you can get one equation.

这样你就能得到一个方程。

is it correct?

对吗？

and so for each node, we write the KCL. okay, this is the KCL applied for this node. and we should write KCL for each node. for each of the remaining node. and here's another node. so we should also write KCL.

所以，对于每个节点，我们都写出基尔霍夫电流定律（KCL）。好的，这就是应用于这个节点的KCL方程。我们应该为每个节点都写出KCL方程。对于剩下的每个节点都是如此。这里还有一个节点。所以我们也应该写出它的KCL方程。

for this node.

对于这个节点。

this is the current that we assume is 1 ampere. okay.

这是我们假设为1安培的电流。好的。

and here is the current flowing, is entering this node. and current is leaving this node. So current entering this node is 1 ampere.

这里有一个电流流入这个节点。还有一个电流流出这个节点。所以，流入这个节点的电流是1安培。

The current leaving the node is v1 minus v2. v1 minus v2 is the potential difference. divided by 10.

流出这个节点的电流是V1减去V2。V1减去V2就是电位差。除以10。

is the current here.

这就是这里的电流。

So the current entering the node is equivalent to the current leaving the node. so for each node, we can write KCL here. by solving the two equations, we will get v1 and v2.

所以，流入节点的电流等于流出节点的电流。因此，对于每个节点，我们都可以写出KCL方程。通过解这两个方程，我们可以得到V1和V2的值。

the current flowing from high potential to lower potential

电流是从高电位流向低电位的。

okay. if we assume here is zero, and here has voltage. v2. so the current flowing from high potential to low potential. The current is downwards.

好的。如果我们假设这里为零，那么这里就有电压。就是V2。所以，电流是从高电位流向低电位的。电流方向是向下的。

Any questions?

有什么问题吗？

If no questions, let's have an example.

如果没有问题的话，我们来看一个例子。

We find all of those, right? And here is one of those. And I found another node.

我们找到了所有这些，对吧？这里就是其中之一。我还发现了另一个节点。

Actually, there's all the nodes. Actually, there's also a node here. Because the definition of node is a junction. It's a junction of two branches. Here is a branch, here is another branch. so there's also a node here.

实际上，这里包含了所有的节点。其实，这里还有一个节点。因为节点的定义就是一个交汇点。它是两个分支的交汇点。这里是一个分支，这里是另一个分支。所以这里也有一个节点。

Okay, second step is you define one node. as reference node. We define one node as reference node. So you can select this node as reference node. And also select this one as reference. And keep going. You can select this one. The one at bottom. The one at bottom as reference node.

好的，第二步是定义一个节点。作为参考节点。我们定义一个节点为参考节点。所以你可以选择这个节点作为参考节点。也可以选择这个作为参考节点。然后继续。你可以选择这个。就是底部的那个。把底部的那个作为参考节点。

The reference node has zero voltage. And then we assume the voltage here is V1.

参考节点的电压为零。然后，我们假设，这里的电压是V1。

The voltage here is V2. We assume voltage for each node.

这里的电压是V2。我们为每个节点都假设了电压。

And then we apply KCL for each node.

接下来，我们对每个节点应用基尔霍夫电流定律。

always remember, the current flowing from high potential to low potential. so here has voltage V1. here is zero. So the current here flowing from high potential to low potential.

一定要记住，电流是从高电位流向低电位的。所以这里有一个电压V1。这里是零。所以这里的电流是从高电位流向低电位的。

and here the current for this branch is already given. it's 1 ampere. this direction 1 ampere.

而这条支路的电流已经给出了。是1安培。这个方向是1安培。

and what is the current here? for this branch, because we have to apply KCL for this node. so we can obtain the current here.

那这里的电流是多少呢？对于这条支路，因为我们必须对这个节点应用基尔霍夫电流定律。这样我们就能求出这里的电流。

here is zero. here is 24.

这里是零。这里是24。

so v2 is actually 24.

所以V2实际上是24。

the voltage of this node is 20 times 2 directly. but here is zero. and here is 24, it's a high potential. so v is 24.

这个节点的电压直接就是20乘以2。但这里是零。而这里是24，这是一个高电位。所以V就是24。

so, and here, for the current direction, we have no idea. it can be either flowing from left to right.

那么，在这里，对于电流的方向，我们并不知道。它可能是从左向右流动的。

or it can be flowing from right to left.

也可能是从右向左流动的。

the both direction are okay. if you assume the direction of the current,

这两种方向都是可以的。如果你假设了电流的方向，

it's flowing from left to right. then you should use 24. minus v1. here is 24. here is v1. v1 is 6. it's the voltage in here. it's the current here.

那就是从左向右流动。那么你就应该用24。减去V1。这里是24。这里是V1。V1等于6。这是这里的电压。这是这里的电流。

and the current is flowing from high potential to low potential. so it should be 24. minus v1. but if you assume the current is going from right, from left to right,

而且，电流是从高电位流向低电位的。所以应该是24。减去V1。但是，如果你假设电流是从右向左流动的，

if you assume the voltage or the current is flowing from left, from right to left. because the current is going from high potential to low potential, that means

也就是假设电流从左向右流动，那么就是从右向左。因为电流是从高电位流向低电位的，这意味着

for this direction, it is v1 minus 24. is v1 is 6. it's very important difference. You can assume different direction, but you should change the expression. based on your assumed.

对于这种方向，应该是V1减去24。而V1等于6。这是一个非常重要的区别。你可以假设不同的方向，但你必须相应地改变表达式。这要根据你的假设来调整。

here's two options, but let's just keep one. you can assume, for example,

这里有两个选项，但我们只保留一个。比如，你可以假设

left to right. then it should be 24 minus v1. v1 is 6. it's the current flowing from left to right. for this node, we apply kirchhoff current law. all the current entering the node is equivalent to all the current leaving the node. here's entering. The one ampere is also entering. and this current is leaving.

从左到右。那么表达式应该是24减去V1。V1等于6。这就是从左到右流动的电流。对于这个节点，我们应用基尔霍夫电流定律。所有流入节点的电流，等于所有流出节点的电流。这是流入的。那一安培也是流入的。而这个电流是流出的。

so we can get the expression like this.

所以我们可以得到这样的表达式。

that's all the problem.

问题就解决了。

okay, new line is 20.

好的，下一行是20。

so it's all the problem. any problems? any questions?

这就是全部的问题了。有什么问题吗？还有问题吗？

the solution is a little bit complex, but actually similar. you define one node as reference, and you assign the same voltage for each node. the remaining nodes. and you assign Vcfr for those one.

这个解法有点复杂，但其实思路是相似的。你定义一个节点作为参考节点，然后给每个节点分配相同的电压。剩下的节点。然后给那些节点分配Vcfr。

and you solve the kcl equation.

然后解KCL方程。

and you must be know the voltage here. once you know the voltage here, you can get the current. on each branch. for example, for the current here, it is 24 volt.

而且你必须知道这里的电压。一旦知道了这里的电压，就能算出电流。在每一条支路上。比如，这里的电流是24伏。

so the current here is 24 divided by 12. so it is 2 ampere.

所以这里的电流就是24除以12。结果是2安培。

the current here. Oh, it is 10 volt, sorry. 10 divided by 12.

这里的电流。哦，是10伏，抱歉。10除以12。

and current here is here 24, here 20. so the current here is 4 volt.

这里的电流是24，这里是20。所以这里的电流是4伏。

and for the solution on the screen, we have negative sign. negative sign for the current, cathode gate. that means the actual current, the actual direction of the current

至于屏幕上的解，有一个负号。这个负号代表电流是从阴极流向栅极的。这意味着实际的电流方向，也就是电流的实际流向，

is the other direction of the assumed. here, the solution assumes the current direction is flowing from right to left, right? but here, the value is negative. negative value, which means the actual current is flowing from left to right. that's basically it gives the positive value direction. it don't give negative. If it gives a negative value, which means the actual current is flowing from left to right. opposite the side of what you have assumed.

与假设的方向相反。在这里，解法假设电流是从右向左流动的，对吧？但是这里的数值是负的。负值意味着，实际的电流是从左向右流动的。这基本上就给出了正确的方向。它没有给出负值。如果给出负值，那就意味着实际电流是从左向右流动的。与你假设的方向相反。

is it clear?

明白了吗？

and here's a short summary. we use KVL. we use KVL to find the current.

这里有一个简短的总结。我们使用了基尔霍夫电压定律。用KVL来求解电流。

we can find mesh currents. and we can solve the equation. get some mesh currents.

我们可以求出网孔电流。然后我们可以解这个方程。得到网孔电流的值。

and here are the steps to do the mesh analysis. and one mesh is a loop.

以下是进行网孔分析的步骤。所谓网孔，就是一个回路。

doesn't have to be any other loop inside it. so it is the smallest loop.

它内部不能包含其他回路。所以它是最小的回路。

and we assign mesh current to each mesh. apply KVL.

我们为每个网孔分配一个网孔电流。然后应用基尔霍夫电压定律。

finally we can solve it. solving the equations, we can get the mesh currents.

最后，我们就可以求解了。通过解方程，我们就能得到网孔电流。

and for the node analysis, we select one node as reference. and we assign voltages. for the remaining nodes. and then we apply KCL. current law for the remaining nodes. and solving the equations, we can get the node voltage.

至于节点分析，我们选择一个节点作为参考节点。然后给其他节点分配电压。对于剩余的节点。接下来，我们应用基尔霍夫电流定律。对剩余的节点应用电流定律。然后解方程，就能得到节点电压。

and current is flowing from high potential to low potential.

电流是从高电位流向低电位的。

okay, that's all for mesh analysis and node analysis.

好的，关于网孔分析和节点分析的内容就到这里。

we take time.

我们休息一下。

it's still have five minutes.

还有五分钟。

and here are some general guidance.

这里有一些通用的指导原则。

in mesh analysis,

在网孔分析中，

for a circuit, if your mesh is known, if the number of the meshes is fewer, it is known. because you can get less equations, right? You can get fewer equations.

对于一个电路，如果某个网孔是已知的，那么网孔的数量较少，就更容易处理。因为这样可以得到更少的方程。对吧？方程数量减少了。

for each mesh is an equation, for each node is an equation. if the number of the meshes is fewer, then nodes. then you should use mesh analysis. because the final equations, the number of the equations will be fewer.

每个网孔对应一个方程，每个节点也对应一个方程。如果网孔的数量比节点少，那就更好了。那么你应该使用网孔分析。因为最终的方程数量会更少。

and if the circuit has fewer nodes, than meshes, then you should choose the nodal analysis. because the number of the equations will be fewer.

如果电路的节点较少，而不是网孔，那么你应该选择节点分析。因为这样方程的数量会更少。

and second guidance is if the network contain many series connected elements or voltage sources, then you should choose the mesh analysis.

第二条建议是，如果电路中包含许多串联的元件或电压源，那么你应该选择网孔分析。

because the number of the equations will be fewer. because if the voltage source is given, we can use the voltage sources. we don't have to compute the voltages.

因为这样方程的数量会更少。因为如果已知电压源，我们就可以利用电压源。这样就不需要计算电压了。

and the elements connected in series, which means we can get voltage easily. because all elements have the same current.

而且这些元件是串联的，这意味着我们可以很容易地得到电压。因为所有元件中的电流都是相同的。

and if the network has many current sources, then you should choose mesh analysis. because the current is given.

如果电路中有许多电流源，那么你应该选择网孔分析。因为电流是已知的。

and the elements are connected in parallel, so we can get the current easily. they have the same voltage.

而且这些元件是并联的，所以我们可以很容易地得到电流。因为它们两端的电压是相同的。

and circuit analysis is if branch or mesh current are required. then we should use mesh analysis directly. because mesh analysis will give you the branch and mesh current.

而节点分析则是在需要支路电流或网孔电流时使用的。那么我们就应该直接使用网孔分析。因为网孔分析可以直接给出支路电流和网孔电流。

and if you want to obtain the load voltage, the node analysis is most important.

如果你想获得负载电压，那么节点分析就最为重要。

because the node analysis will gives you the voltage directly. and solving the equation.

因为节点分析可以直接给出电压值。然后解方程即可。

any questions?

大家有什么问题吗？

okay, if there are no questions, we are be learn new knowledge, superposition.

好的，如果没有问题的话，我们将学习新的知识，那就是叠加定理。

and so first the definition of linear circuit. the linear circuit is means the output is the linear relation. for example, the voltage has a linear relation. or the current.

首先，我们先来了解一下线性电路的定义。线性电路的意思是，输出与输入之间是线性的。关系。例如，电压之间就存在线性关系。或者电流。

the voltage has a linear relation to the current. but the power is s square. so it's not linear. it's not linear.

电压与电流之间是线性关系。但是，功率却是平方关系。所以它不是线性的。它不是线性的。

and the superposition is only applicable for linear circuits.

而叠加原理仅适用于线性电路。

the definition is that the voltage across or current through an element. in a linear circuit, it's the algebraic sum of the voltages. due to each independent source acting alone.

其定义是，电压跨接在某个元件上的电压或电流。在线性电路中，它就是这些电压的代数和。这是由每个独立源单独作用时产生的结果。

so let's explain this long sentence.

那么，让我们来解释一下这句话。

if there's two sources, you have a voltage source and you have a current source.

如果有两个电源，一个是电压源，另一个是电流源。

and this is a linear circuit.

这是一个线性电路。

the total voltage or current, the total voltage for this resistor, or the current through this resistor.

这个电阻上的总电压或总电流，也就是流过这个电阻的电流。

is the algebraic sum. of each source.

就是它们的代数和。即每个电源单独作用时的叠加。

which means for the first case, we don't consider this current source.

这意味着，对于第一种情况，我们不考虑这个电流源。

If we don't consider this current source, the deal can be simplified. So this is the situation one.

如果我们不考虑这个电流源，电路就可以简化。这就是第一种情况。

for the second situation, we don't consider the voltage source.

对于第二种情况，我们不考虑电压源。

we don't consider the voltage source.

我们不考虑电压源。

this is the situation two.

这就是第二种情况。

and then for each situation, we can get the voltage. for this resistor and the current.

然后，对于每种情况，我们都可以得到电压。对于这个电阻和电流。

the superposition means you can add v1 plus v2. to get the voltage. for this original circuit. and you can also add i1 and i2 to get the the final current. for the resistor. so this is superposition. For each situation, you can set only one source. and then you add them together. it's the algebraic sum.

叠加原理的意思是，你可以把V1和V2相加。从而得到总电压。也就是这个原始电路的电压。你也可以把I1和I2相加，得到最终的电流。对于那个电阻来说。这就是叠加原理。对于每种情况，你只需要考虑一个电源。然后把它们加起来。这就是代数相加。

and here is important. for the first situation, for the current source, we set it to open circuit. so there's no current anymore. And for the second situation,

这里很重要。对于第一种情况，对于电流源，我们设置它为开路。这样就没有电流了。而对于第二种情况，

we set the voltage source as short circuit. so there's no voltage drop.

我们将电压源设为短路。这样就没有电压降了。

And for each situation, we just can use the voltage. and current. and then we add them together. it's the algebraic sum. get the final, the original final voltage or current.

对于每种情况，我们都可以使用电压。和电流。然后把它们加起来。这也是代数相加。这样就能得到最终的、原始的电压或电流。

Is it clear?

清楚了吗？

here is important. for each situation, we have situation one and situation two, right? you can set only one. independent source.

这里很重要。对于每种情况，我们有情况一和情况二，对吧？你只能设置一个。也就是独立电源。

so you can set only one independent source. We remove all the other.

所以，你只能保留一个独立电源，其他的都要移除。

but how can you remove? for voltage source, we set it to short circuit. so there's no voltage. and for the current source, we set it open circuit. so there's no current. that means we remove all the other independent source.

但是，你怎么移除呢？对于电压源，我们要把它短路。这样就没有电压了。而对于电流源，我们要把它开路。这样就没有电流了。这意味着我们要移除所有其他的独立电源。

and then we find the unknown quantity from the active source.

然后，我们从有源电源那里求出未知的量。

we find the voltage and current for this resistor. when this voltage source is active.

我们求出这个电阻的电压和电流。当这个电压源处于激活状态时。

we find the unknown quantity from the active source. This is the active one. It just here, this one. So that means this source, this is the active. for the other are removed.

我们从有源电源那里求出未知量。这就是那个有源的电源。就是这里，这个。所以这意味着，这个电源是有源的。其他的都被移除了。

we find the unknown quantity, we find the voltage and current.

我们求出了未知量，也就是电压和电流。

and then we repeat.

然后我们重复这个过程。

repeat this step. because for this case, we just keep this one and remove all the other. and for the second case, we remove this one and keep the current source. and repeat the process. for each of the independent source.

重复这一步。因为在这种情况下，我们只保留这一个，而移除所有其他的。而对于第二种情况，我们移除这个，保留电流源。然后重复这个过程。对于每一个独立电源。

and finally, we add them together. as really all contributions.

最后，我们将它们相加。也就是所有的贡献之和。

and for this independent source, we get the voltage and current. and for the second, we also get the voltage and current. then we add them together. obtain the final voltage or current.

对于第一个独立电源，我们得到了电压和电流。而对于第二个，我们也得到了电压和电流。然后我们将它们相加。从而得到最终的电压或电流。

but here is a very important note. independent source should be kept.

但是，这里有一个非常重要的注意事项。独立电源必须保留。

because here we have an independent source, right? so we have to remove it. but if there's a dependent source, dependent source we use this symbol, right? this is the dependent.

因为这里有一个独立电源，对吧？所以我们必须把它去掉。但是，如果存在受控源，我们用的是这个符号，对吧？这就是受控源。

if there's a dependent source in a circuit,

如果电路中存在受控源，

you should keep it.

就必须保留它。

you should not remove it.

不能把它去掉。

because they are controlled by the circuit variables. you know, the output of a dependent source is dependent on the parameter of circuit. for example, the voltage is five times i. which means the output is dependent on the current. so dependent source should not be removed. you should always keep it in a circuit. when consider each situation,

因为它们是由电路变量控制的。大家知道，受控源的输出是依赖于电路参数的。比如，电压是电流的5倍。这意味着输出依赖于电流。所以，受控源是不能被移除的。你应该始终把它保留在电路中。在考虑每种情况时，

and a superposition is only applicable for linear circuits. it's not applicable for non-linear circuits. for example, you want to calculate power. you cannot calculate power together. for v1 and i1, you can add them together. so if you want to calculate power, you also have p1. and p2. when you add p1 and p2, you will get wrong answer. for the final power, because power is non-linear. so we cannot use the superposition to calculate the power.

叠加原理只适用于线性电路。它不适用于非线性电路。例如，如果你想计算功率。你不能直接用叠加原理来计算功率。但对于V1和I1，你可以把它们相加。所以，如果你想计算功率，你还有P1。还有P2。当你把P1和P2相加，得到的结果是错误的。因为功率是非线性的。所以我们不能用叠加原理来计算功率。

this is incorrect. You should not use that method to calculate the final power.

这是错误的。你不应该用这种方法来计算最终的功率。

The superposition can help simplify a circuit. because for this circuit, we simplify, we just remove those sources. so the circuit becomes simpler. And for the second case, the voltage source is removed. it also becomes simpler.

叠加原理可以帮助简化电路。因为对于这个电路，我们可以通过移除那些电源来简化它。这样电路就变得简单了。对于第二种情况，电压源被移除了。这样电路也变得简单了。

So it can be used to simplify a circuit.

所以，它可以用来简化电路。

And what you saw is instead of a short circuit, no voltage drop. and for the current source, we said open circuit. so there is no current.

大家看到的是，短路时没有电压降。而对于电流源，我们说的是开路。所以没有电流。

Okay, that's the theory of superposition.

好的，这就是叠加原理。

Any questions?

有什么问题吗？

Any other questions?

还有其他问题吗？

if this is a short-circuited, the voltage and the current is zero. you add them together. should be v1, v2, zero.

如果这是短路，那么电压和电流都是零。然后把它们加起来。应该是V1、V2和零。

okay, superposition.

好的，叠加原理。

use the superposition to solve this example. okay, time is.

用叠加原理来解决这个例子。好的，时间到了。

if one source, for example, if just keep v1,

如果只有一个源，比如只保留V1，

you have to remove the

你就必须把

all the remaining independent source. and the voltage source is代替ed to open circuit, short circuit. so there is no voltage drop.

剩下的都去掉。独立源都去掉。电压源要替换成开路，电流源要替换成短路。这样就没有电压降了。

so this is the first case. for the second case, you remove v1.

所以这是第一种情况。对于第二种情况，你要去掉V1。

instead you will short circuit.

把它替换成短路。

and here is the important thing. for the first case, the current is flowing this direction.

这里有一点很重要。在第一种情况下，电流是沿着这个方向流动的。

and for the second case, the current

而在第二种情况下，电流

is going the other direction, opposite direction. so we have to minus them together.

是朝着另一个方向，也就是相反的方向。所以我们需要把它们相减。

for the first one, the current is i1. divided by r1 plus r2. and second is e1, e2 divided by r1.

对于第一个情况，电流是I1。除以R1加R2。第二个情况是E1减去E2，再除以R1。

and i have to minus them together because the current at different direction. current is always flowing from high potential. here is high potential, here is low potential.

我需要把它们相减，因为电流的方向是不同的。电流总是从高电位流向低电位。这里是高电位，这里是低电位。

so it's minus.

所以是减号。

when you look at the solution,

大家看这个解法，

very identical to what we have written on the screen. but pay attention to the sign.

和我们写在屏幕上的内容完全一致。但是要注意符号。

okay, summary. so in this course, we have introduced many new knowledge, right? we have introduced the measurement, this, total analysis and superposition. measurement is the most important. and for the measurement analysis, measurement is the most important, and we assume measurement are perfect. in each measurement, and current is clockwise. and let me determine the branch current. branch current, you think?

好的，总结一下。在这门课程中，我们介绍了很多新的知识，对吧？我们介绍了测量、对偶分析以及叠加原理。其中，测量是最重要的。对于测量分析来说，测量是最关键的，而且我们假设测量是完美的。在每次测量中，电流都是顺时针方向的。现在，让我来确定一下支路电流。支路电流，你觉得呢？

apply kirchhoff's voltage law for each mesh.

对每个回路应用基尔霍夫电压定律。

for the other meshes, we define a reference direction. and then assign voltage for each loop. and we apply kcl for each node.

对于其他回路，我们定义一个参考方向。然后为每个回路分配电压。接着，我们对每个节点应用基尔霍夫电流定律。

for the superposition,

至于叠加原理，

for each time, we only can screw one independent source. and remove all the others.

每次我们只能保留一个独立电源。而将其他电源移除。

For voltage source, we set it with short circuit. and for the current source, we set it with open circuit.

对于电压源，我们将其视为短路。而对于电流源，我们将其视为开路。

and then we add all the contributions algebraically.

最后，我们将所有贡献进行代数相加。

Do you have any questions?

大家有什么问题吗？

Okay, if there's no questions, we will see you all in the lab.

好的，如果没有问题的话，我们实验室见。

The four states are very similar to what happens in mesh analysis. Mesh analysis follows the same principle. Kirchhoff's law is identical to mesh analysis when you look at the four states. A mesh is the smallest loop, and there is no loop inside it.

四个状态与网孔分析中发生的情况非常相似。网孔分析遵循相同的原理。当你观察这四个状态时，基尔霍夫定律与网孔分析是等价的。网孔是最小的回路，其内部不包含任何其他回路。

For the mesh analysis, we first find a mesh, and then find the mesh parent. The mesh parent is the only one defined not by direction. Unless you apply...

在进行网格分析时，我们首先找到一个网格，然后找到该网格的父网格。父网格是唯一一个不由方向定义的。除非你应用……

Here is an example. Here is the matrix part, and here is the matrix part. Small i denotes the matrix part; large i denotes the branch part. The matrix part has a clockwise direction: small i₁, i₂ of the matrix part. The branch part also has a clockwise direction: large i, i₁, i₂, i₃ of the branch part.

这里是一个例子。这里是矩阵部分，这里是矩阵部分。小写 i 表示矩阵部分；大写 i 表示支路部分。矩阵部分具有顺时针方向：矩阵部分的小 i₁、i₂。支路部分也具有顺时针方向：支路部分的大 i、i₁、i₂、i₃。

First, let me define a mesh. Here we have two meshes. We assume the mesh currents I₁ and I₂ flow in the clockwise direction. Now, let me apply Kirchhoff's Voltage Law to each mesh.

首先，让我定义网孔。这里有两个网孔。我们假设网孔电流 I₁ 和 I₂ 沿顺时针方向流动。现在，让我对每个网孔应用基尔霍夫电压定律。

This is important because here we have mesh currents. I1 and I2 are mesh currents. So the mesh current for resistor 1 is also I1. But for R3 it is different. For R3, both currents flow through it for mesh B.

这一点很重要，因为这里涉及网孔电流。I1和I2都是网孔电流。因此，流过电阻1的网孔电流就是I1。但对于R3则不同，对于R3，两个电流都流过了它，即网孔B中的电流。

For mesh analysis, first let me define the mesh. Here we have two meshes. The mesh direction is clockwise.

对于网孔分析，首先让我定义网孔。这里有两个网孔。网孔方向为顺时针。

We have I1 and I2. These are mesh currents. Now, for each branch—this is a branch—here we have branch currents. A branch current equals the sum of the mesh currents passing through it. However, for this particular resistor, the branch current is simply equal to one mesh current, namely I1.

我们有 I1 和 I2，它们是网孔电流。对于每一条支路——这是一条支路——这里存在支路电流。支路电流等于流过该支路的各网孔电流之和。然而，对于这个特定电阻，其支路电流仅等于一个网孔电流，即 I1。

For this resistor, we have to apply Kirchhoff's Current Law at the node to get the current flowing through this resistor. It should be I₁ minus I₂, directed downward. And then we apply Kirchhoff's Voltage Law for each branch to get the equations.

对于这个电阻，我们需要在节点处应用基尔霍夫电流定律来求得流过该电阻的电流。它应为 I₁ 减去 I₂，方向向下。然后我们对每条支路应用基尔霍夫电压定律，得到方程组。

For this circuit, we also have two meshes. We also have two mesh currents. The mesh current is different from the branch current. And I₁ and I₂ are mesh currents.

对于这个电路，我们同样有两个网孔，也有两个网孔电流。网孔电流与支路电流是不同的。I₁和I₂就是网孔电流。

And for this branch, the branch current is also I1. And for this branch, the branch current is also I2. But for the branch current in the middle, the direction of I1 is clockwise, and I2 is counterclockwise. So the branch current flowing in this branch is I1 minus I2.

对于这条支路，支路电流也是 I1。对于另一条支路，支路电流也是 I2。但对于中间这条支路的电流，I1 的方向是顺时针，I2 的方向是逆时针。因此，流过这条支路的支路电流为 I1 减去 I2。

Now we apply KVL to find the loop current. We define the loop current as big I, where big I equals I1 minus I2. Then we apply KVL here for the voltage drop across the resistors. How do we determine the sign? If the voltage source has the same direction as the mesh current direction, then it is positive. This is clockwise direction, so the term is positive. The branch current for this resistor is upward.

现在我们应用基尔霍夫电压定律来求回路电流。我们定义回路电流为大 I，其中大 I 等于 I1 减去 I2。然后我们在这里对电阻上的电压降应用 KVL。如何判断符号？如果电压源的方向与网孔电流方向相同，则为正。这是顺时针方向，所以该项为正。该电阻的支路电流方向向上。

The branch current is upward, so the voltage drop is downward, flowing from low potential to high potential. Since the voltage direction is not clockwise, we assign a minus sign: −I₁ × R₁. This is important—the voltage direction is opposite to the mesh direction. For this second resistor, the branch current is I₁ − I₂, and it flows downward, so the voltage is upward. This also has the opposite direction as the mesh, so we assign a minus sign: −(I₁ − I₂) × R₂.

支路电流方向向上，因此电压降方向向下，从低电位流向高电位。由于电压方向不是顺时针方向，我们赋予负号：−I₁ × R₁。这一点很重要——电压方向与网孔方向相反。对于第二个电阻，支路电流为 I₁ − I₂，方向向下，因此电压方向向上。这同样与网孔方向相反，所以我们赋予负号：−(I₁ − I₂) × R₂。

This is the first equation. For the second mesh, for the second loop, the voltage is upward, so it has the same direction as the mesh, also clockwise. So for this resistor, it should get positive voltage. And for this resistor, the current is downward, so the voltage is upward. It has opposite direction.

这是第一个方程。对于第二个网孔，即第二个回路，电压方向向上，与网孔绕行方向相同，也是顺时针方向。因此对于这个电阻，应取正电压。而对于这个电阻，电流方向向下，电压方向向上，二者方向相反。

For the voltage source, it also has the opposite direction — from minus to plus. Is it clear?

对于电压源，它的方向也是相反的——从负极到正极。清楚了吗？

A new definition: null analysis. As we introduced before, patient analysis and null analysis.

一个新定义：零分析。正如我们之前介绍的，患者分析与零分析。

If a circuit has many voltage sources, we apply mesh analysis because you will get fewer equations. But if there are many current sources...

如果一个电路中有许多电压源，我们采用网孔分析法，因为这样得到的方程更少。但如果电路中有许多电流源……

In some cases, we should apply node analysis. If there are many voltage sources, we should apply mesh analysis; but if there are many current sources, we should use node analysis. What is node analysis? We have introduced the definition of node at the beginning. A node is defined as the junction of two or more branches. Here is a branch, and here is a second branch.

在某些情况下，我们应该应用节点分析法。如果电路中存在大量电压源，则应使用网孔分析法；但如果存在大量电流源，则应使用节点分析法。什么是节点分析？我们在课程开始时已经引入了节点的定义。节点被定义为两条或更多条支路的交汇点。这里是一条支路，这里是第二条支路。

If the ground wires—there is one ground wire, there are two ground wires—the junction between two grounds is a node. And if we define one node as the reference node, the reference node has zero potential. The ground has zero potential, and the remaining nodes of the network will have a fixed potential relative to this reference.

如果有接地线——一根接地线，两根接地线——两个接地点之间的连接点就是一个节点。如果我们定义一个节点为参考节点，则参考节点的电势为零。地的电势为零，网络中其余节点相对于该参考节点将具有固定的电势。

This means, for example, we have a current source here, and we have a resistor here, another resistor. If we have a circuit like this, and here is the ground, we use this node as the reference node.

这意味着，例如，我们这里有一个电流源，这里有一个电阻，还有另一个电阻。如果我们有这样一个电路，这里是地，我们就把这个节点用作参考节点。

Here is the reference node. And here is another node. Here is also another node. The potential, or the voltage, of this node is relative to the reference node.

这里是参考节点。这里是另一个节点。这里也是另一个节点。该节点的电位，或者说电压，是相对于参考节点而言的。

Any node can be selected as a reference node. You can also select this node as the reference node. Usually, we select the node with zero potential as the reference node, because that is why we define it as zero — it is simply zero. This makes things much simpler. So usually, we select the node with zero potential as the reference node, but of course, we can also select other nodes as the reference node.

任何节点都可以被选为参考节点。你也可以选择这个节点作为参考节点。通常，我们选择电位为零的节点作为参考节点，因为这就是我们将其定义为零的原因——它本身就是零。这样会简单得多。所以通常我们选择电位为零的节点作为参考节点，但当然，我们也可以选择其他节点作为参考节点。

If we have a network of n nodes, they are partitioned into n minus one nodes, and a potential of zero is assigned to the reference node. Here we have three nodes, so we have two nodes. This node is grounded, and this is the reference node. So we have two nodes, with zero potential for the reference node.

如果我们有一个由 n 个节点组成的网络，它们被划分为 n 减一个节点，并为参考节点分配零电位。这里我们有三个节点，所以有两个节点。这个节点是接地的，这就是参考节点。所以我们有两个节点，参考节点的电位为零。

We have two nodes: a reference node, and we have fixed the potential relative to the reference node, which is 20 volts here. The node voltage is the voltage at a given node of the circuit with respect to the reference node.

我们有两个节点：一个参考节点，并且我们已经固定了相对于参考节点的电位，这里是20伏。节点电压是电路中给定节点相对于参考节点的电压。

The bottom node is related to the reference node. Now we apply Kirchhoff's Current Law. Different from mesh analysis, here we apply Kirchhoff's Current Law for the n minus one nodes. For each two nodes, we apply Kirchhoff's Current Law, so we have two equations. Solve the equations to get the current.

底部节点与参考节点相关。现在我们应用基尔霍夫电流定律。与网孔分析不同，这里我们对n减一个节点应用基尔霍夫电流定律。对每两个节点分别列写基尔霍夫电流方程，因此我们得到两个方程。求解方程即可得到电流。

Here is the difference. For mesh analysis, we apply Kirchhoff's voltage law, and we get the voltages. For node analysis, once we define the reference node, and we have the remaining n minus one nodes, we apply Kirchhoff's current law.

这就是两者的区别。对于网孔分析，我们应用基尔霍夫电压定律，从而求得各电压。对于节点分析，一旦确定了参考节点，剩余 n 减一个节点上我们应用基尔霍夫电流定律。

We get the current for the branch. So this is the difference. Here is the current of the remaining nodes. Any questions? Here is a summary of how to use node analysis. First step: determine the number of nodes in your network. Here is the network; we have three nodes.

我们求得支路电流。这就是区别所在。这里是剩余节点的电流。有没有问题？下面总结一下如何使用节点分析法。第一步：确定网络中的节点数。这是该网络，我们有三个节点。

We select a reference node and label all other nodes. This is the reference node, and we label the other nodes—node one, node two. Then we assume the voltage at each node with respect to the reference node: V₁, V₂. We assume the voltage with respect to the reference node.

我们选择一个参考节点，并对所有其他节点进行标注。这是参考节点，我们标注其余节点——节点一、节点二。然后我们假设每个节点相对于参考节点的电压：V₁、V₂。我们假设的是相对于参考节点的电压。

Kirchhoff's Current Law applies to each node. Here we have velocity v₁ and here we have velocity v₂. We apply Kirchhoff's Current Law to each node. For example, for this node, we can apply Kirchhoff's Current Law. How can you write down the equation?

基尔霍夫电流定律适用于每一个节点。这里我们有速度 v₁，这里有速度 v₂。我们对每个节点应用基尔霍夫电流定律。例如，对于这个节点，我们可以应用基尔霍夫电流定律。你如何写出方程？

This is a 10-ohm resistor. Here is V1, and here is V2. V1 is the load voltage here; V2 is the load voltage here. V1 minus V2 is the voltage across this resistor. Divided by 10 ohms, that is the current through this resistor. So we get the current here.

这是一个10欧姆电阻。这里是V1，这里是V2。V1是此处的负载电压，V2是此处的负载电压。V1减去V2就是该电阻两端的电压。除以10欧姆，即为通过该电阻的电流。因此我们得到此处的电流。

For this branch, we know the voltage is V2 and the resistor is 10 ohms. So V2 divided by 10 is the current here. Because the other end is at zero, V2 minus zero divided by 10 gives us the current. And how did we get the current...

对于这条支路，我们知道电压为V2，电阻为10欧姆。因此V2除以10就是这里的电流。因为另一端电位为零，V2减去零再除以10就得到电流。那么我们是如何得到这个电流的……

There is zero here, so C2 divided by 20 is the current here. So we get three currents. Then we apply Kirchhoff's Current Law: the current entering this node is equivalent to the current leaving the node. Here is the entering — only one, this one is entering. The other two are leaving. So you can get one equation.

这里为零，所以C2除以20就是这里的电流。这样我们得到三个电流。然后应用基尔霍夫电流定律：流入该节点的电流等于流出该节点的电流。这里流入的只有一个，就是这一个。另外两个是流出的。因此可以得到一个方程。

Is it correct? Yes. Poison will be attracted.

对吗？是的。毒物会被吸引。

This is the KCL applied to this node. You should write KCL for each node, for each remaining node. And here is another node, so you should also write KCL for this node. This is the current value: one ampere here, one ampere here, one ampere here. And here is the current flowing and entering...

这是对该节点应用的基尔霍夫电流定律（KCL）。你应该对每个节点、每个剩余节点都写出KCL方程。这里是另一个节点，所以你也应该对它写出KCL方程。这是电流值：这里一安培，这里一安培，这里一安培。而这里的电流正在流入……

The current entering this load is one ampere. The current leaving the load is V1 minus V2. V1 minus V2 is the potential difference divided by ten ohms. Divided by ten ohms is the current here. So the current entering the load is equivalent to the current leaving the load. So the equation of V1...

进入该负载的电流为一安培。离开该负载的电流为V1减去V2。V1减去V2是电位差除以十欧姆。除以十欧姆即为此处的电流。因此，进入负载的电流等于离开负载的电流。所以V1的方程……

For each loop, we have x here. By solving the two equations directly, v1 and v2. And here, the current flows from high potential to lower potential.

对于每个回路，这里有x。通过直接求解两个方程，v1和v2。而在这里，电流从高电位流向低电位。

Current flows from high potential to low potential. If we assume a equals zero, then it has to be true. So the current is flowing from high potential to low potential; the current is downward. Any questions? If no questions, let's put an example.

电流从高电势流向低电势。如果我们假设 a 等于零，那么它必然成立。因此电流从高电势流向低电势，方向向下。有问题吗？如果没有问题，我们来看一个例子。

We define all nodes. Here is one node, and here is another node.

我们定义所有节点。这里是一个节点，这里是另一个节点。

Actually, there is also a node here, because the definition of a node is a junction of two branches. Here is one branch, here is another branch, so there is also a node here. The second statement is: you define one node as the reference node. You can select this one as reference, or this one as reference. Typically, we select the one at the bottom as the reference node.

实际上，这里也是一个节点，因为节点的定义是两条支路的交汇点。这里是一条支路，这里是另一条支路，所以这里也是一个节点。第二点是：你定义一个节点作为参考节点。你可以选这个作为参考，也可以选这个作为参考。通常，我们选择最下面那个作为参考节点。

Replace components with zero voltage sources. Then we assume voltage gains V1 for each node. Next, we apply KCL for each node. Always remember: current is flowing from high potential.

将元件替换为零电压源。然后我们假设每个节点的电压增益为V1。接下来，对每个节点应用KCL。务必记住：电流从高电位流向低电位。

From high potential to low potential. So here has voltage V1, here is zero. The current is flowing from high potential to low potential. The current for this branch is already given: 1 ampere in this direction. Now, what is the current here for this branch? Because we have applied KCL at this node, we can obtain the current here.

从高电位流向低电位。这里电压为V1，这里为零。电流从高电位流向低电位。这条支路的电流已经给出：沿此方向为1安培。那么，这条支路的电流是多少？因为我们在该节点应用了基尔霍夫电流定律（KCL），所以可以求得这里的电流。

Here is zero, here is 24. So V₂ is minus 24. The voltage of this node is 24 plus 2R₂. Because here is zero and here is 24, there is a high potential and a low potential, so here is 24. Now, for the current direction, we have no idea. It can be either flowing from left to right, or it can be flowing from right to left.

这里是零，这里是24。所以V₂等于负24。这个节点上的电压是24加2R₂。因为这里是零，这里是24，存在高电位和低电位，所以这里是24。现在，对于电流方向，我们并不确定。它可能是从左向右流动，也可能是从右向左流动。

Both directions give almost the same result. If you assume the direction of the current is going from left to right, then you should use 24 minus E1. Here is 24, here is E1, divided by 6 gives the voltage here, and that is the current here. The current is flowing from high potential to low potential, so it should be 24 minus E1. But if you assume the current is flowing from right to left—

两个方向得到的结果几乎相同。如果你假设电流方向是从左到右，那么应该用24减去E1。这里是24，这里是E1，除以6得到这里的电压，也就是这里的电流。电流从高电位流向低电位，所以应该是24减去E1。但如果你假设电流是从右到左流动——

The voltage on the current is going from right to left, because the current is flowing from high potential to low potential. For this direction, you should use V₁ minus 24 divided by 6. This is a very important difference. You can assume a different direction, but you must change the expression accordingly based on your assumed direction.

电流上的电压方向是从右到左，因为电流从高电位流向低电位。对于这个方向，你应该使用 V₁ 减去 24 再除以 6。这是一个非常重要的区别。你可以假设不同的方向，但必须根据你所假设的方向相应地改变表达式。

There are two options for assigning current direction; let us pick one. For example, the current is flowing from left to right, labeled as i₁, given as 6 A. For this node, we apply Kirchhoff's Current Law: all current entering the node is equal to all current leaving the node. The 1 ampere is also entering, and this current is leaving. So we can get the equation.

电流方向有两种指定方式，我们选一种。例如，电流从左向右流动，标记为 i₁，给定为 6 A。对于这个节点，我们应用基尔霍夫电流定律：所有流入节点的电流等于所有流出节点的电流。1 安培也是流入的，而这个电流是流出的。因此我们可以列出方程。

396 divided by 20.

396 除以 20。

Stress. It's on the bottom there. Any problems? Any questions?

应力。它在底部那里。有问题吗？有疑问吗？

The solution is a little bit complex, but it is similar. You perform one node analysis and assign the same voltage for each node. For the remaining nodes, you assign KCL for the other one. You solve the KCL equations and you can get V1. Once you know the voltage here, you can get the current for each branch. For example, for the current here, it is 24 volts, so the current here is 24 divided by...

求解过程稍微复杂一些，但方法是类似的。你对一个节点进行节点分析，并为每个节点指定相同的电压。对于剩余节点，对其余节点列写KCL方程。求解KCL方程即可得到V1。一旦知道了这里的电压，就可以求出每条支路的电流。例如，对于这里的电流，它是24伏，所以这里的电流等于24除以……

I1 is 24, I2 is 20, so I2 is 4 minus 6. For the solution, we have a negative sign for the current. That is okay. That means the actual direction of the current is the opposite direction.

I1 等于 24，I2 等于 20，所以 I2 等于 4 减 6。在求解过程中，电流出现了负号，这是正常的。负号意味着电流的实际方向与假设方向相反。

Of the assumed direction, here the student assumes the current direction is going from right to left, right? But here he finds this negative value, which means the actual current is flowing from left to right. That's the reason: if you get a positive value, the direction is correct for the current; if you get a negative value, it means the actual current's actual direction is on the opposite side of what you have assumed.

关于假设方向，这里学生假设电流方向是从右到左，对吧？但这里他得到了一个负值，这意味着实际电流是从左到右流动的。原因就在于：如果你得到正值，说明方向假设是正确的；如果你得到负值，则意味着实际电流的真实方向与你所假设的方向相反。

For mesh analysis, we use KVL to find the current.

对于网孔分析，我们使用基尔霍夫电压定律来求电流。

We define mesh currents, and we can solve the equations to get the mesh currents. There are steps to perform mesh analysis. A mesh is a loop that does not contain any other loop inside it; it is the smallest loop. We assign a mesh current to each mesh and apply KVL to each mesh.

我们定义网孔电流，然后可以求解方程得到各网孔电流。进行网孔分析有若干步骤。网孔是指内部不包含任何其他回路的回路，即最小的回路。我们为每个网孔指定一个网孔电流，并对每个网孔应用基尔霍夫电压定律（KVL）。

For the mesh analysis, solve the equations simultaneously to get the mesh currents. For the nodal analysis, select one node as reference and assign voltages for the remaining nodes. Then apply KCL for the remaining nodes, and solve the equations simultaneously to get the node voltages.

对于网孔分析，联立求解方程组即可得到各网孔电流。对于节点分析，选取一个节点作为参考节点，并为其余节点指定电压。然后对其余节点列写KCL方程，联立求解即可得到各节点电压。

Currently, deploy from application to document. That is more from class analysis and not analysis. Here are some general guidance into class analysis.

目前，从应用程序部署到文档。这更多来自类分析而非分析。以下是一些关于类分析的一般指导。

If the number of meshes is fewer than the number of nodes, we should use mesh analysis. For each mesh we get one equation; for each node we get one equation. Therefore, when there are fewer meshes than nodes, mesh analysis yields fewer equations overall.

如果网孔数少于节点数，则应使用网孔分析法。每个网孔对应一个方程，每个节点也对应一个方程。因此，当网孔数少于节点数时，网孔分析法所得方程总数更少。

If the circuit has fewer nodes than meshes, then you should choose nodal analysis, because the number of equations will be fewer. The second guideline is: if the network contains many series-connected elements or voltage sources, then you should choose mesh analysis, as it is more suitable. This is because if the voltage source is given, you can use the value directly.

如果电路的节点数少于网孔数，则应选择节点分析法，因为方程的数量会更少。第二条准则是：如果网络包含大量串联元件或电压源，则应选择网孔分析法，因为它更为适用。这是因为如果电压源已知，可以直接使用其数值。

And the elements are connected in series, which means we can get voltage easily because all elements have the same current. And if the network has many current sources, then you should choose nodal analysis because the current is given.

各元件串联连接，这意味着我们可以方便地求得电压，因为所有元件具有相同的电流。如果网络中包含许多电流源，则应选择节点分析法，因为电流是已知的。

When elements are connected in parallel, they have the same voltage. If branch or mesh currents are required, use mesh analysis directly, because mesh analysis will give you the branch and mesh currents. If you want to obtain the load voltage, load analysis is most suitable.

当元件并联连接时，它们具有相同的电压。如果需要求支路电流或网孔电流，直接使用网孔分析法，因为网孔分析法会给出支路电流和网孔电流。如果想求负载电压，负载分析法最为合适。

The double-nested analysis gives you the voltage directly after solving the equations. Any questions?

双重嵌套分析在求解方程后可直接得到电压。有问题吗？

If there are no questions, we will learn new knowledge: the superposition theorem. First, the definition of a linear circuit. What is a linear circuit? A linear circuit means the output is linearly related to its inputs. For example, the voltage has a linear relation to the current.

如果没有问题，我们将学习新知识：叠加定理。首先是线性电路的定义。什么是线性电路？线性电路是指输出与其输入呈线性关系的电路。例如，电压与电流呈线性关系。

For power, however, there is a square relationship—power is proportional to s squared—so it is not linear. Superposition is therefore only applicable for linear circuits. The definition states that the voltage across or current through an element in a linear circuit is the algebraic sum of the voltage across or current through that element due to each independent source acting alone.

然而对于功率而言，存在平方关系——功率与s的平方成正比——因此它不是线性的。叠加定理仅适用于线性电路。其定义为：在线性电路中，某元件两端的电压或流过该元件的电流，等于各独立源单独作用时在该元件上产生的电压或电流的代数和。

When each independent source acts alone, the x values are non-standard. For example, if there are two sources — you have a voltage source and you have a current source — this is a linear circuit.

当每个独立源单独作用时，x值是非标准的。例如，如果有两个源——一个电压源和一个电流源——这是一个线性电路。

The total voltage across this resistor, or the current through this resistor, is the algebraic sum of each source acting alone. This means, for the first case, we cannot consider this current source.

该电阻两端的总电压，或通过该电阻的总电流，等于每个电源单独作用时的代数和。这意味着，在第一种情况下，我们不考虑这个电流源。

If we do not consider these internal sources, we can simplify this circuit. This is situation one. For situation two, we do not consider the voltage source. This is situation two. And then, for each situation...

如果不考虑这些内部电源，我们可以简化这个电路。这是第一种情况。第二种情况是，我们不考虑电压源。这是第二种情况。然后，对于每种情况……

For each excitation, you can get the voltage for each resistor and the current. The superposition means you can add V1 plus V2 to get the voltage for this original circuit, and you can also add I1 and I2 to get the final current through the resistor. So this is superposition: for each source, you consider only one source at a time, and then you add them together — it is the algebraic sum.

对于每个激励，你可以求出每个电阻上的电压和电流。叠加定理的含义是：你可以将V1加上V2得到原电路中的电压，也可以将I1加上I2得到流过电阻的最终电流。这就是叠加定理——每次只考虑一个电源，然后将它们相加，即代数求和。

Here is important. For the first situation, for the current source, we set the voltage source to open circuit, so there is no current anymore. And for the second situation, we set the voltage source as short circuit, so there is no voltage drop. And for each situation, we just calculate the voltage and current, and then we add them together to get the final original voltage or current.

这里很重要。对于第一种情况，即电流源的情况，我们将电压源设为开路，因此不再有电流流过。对于第二种情况，我们将电压源设为短路，因此没有电压降。对于每种情况，我们分别计算电压和电流，然后将它们叠加，得到最终的原始电压或电流。

Here it is important. For each expression, we have expression one and expression two. We consider only one independent source. We remove all the other independent sources.

这里很重要。对于每个表达式，我们有表达式一和表达式二。我们只考虑一个独立源，移除所有其他独立源。

A dependent source is not deactivated. For the other independent sources: a voltage source is replaced by a short circuit, so there is no voltage drop; a current source is replaced by an open circuit, so there is no current. That means we remove all the other independent sources. Then we find the unknown quantity from the active source. We find the voltage and the current for this resistor when this voltage source is active.

受控源不予去激活。对于其他独立源：电压源用短路替代，因此没有电压降；电流源用开路替代，因此没有电流。也就是说，我们移除所有其他独立源。然后从仍然激活的源中求出未知量。当该电压源激活时，我们求出该电阻上的电压和电流。

For inactive nodes, we keep this one—the active node—and remove all the others. They become unoccupied; they become what we call parents. Then we repeat this step. Because for this case, we just keep this one and remove all the others. For the second case, we remove this one and keep the parent nodes. We repeat the process for each of the inactive nodes. And finally—

对于非活跃节点，我们保留这一个——即活跃节点——并移除所有其他节点。它们变为未占用状态，成为我们所说的父节点。然后我们重复这一步骤。因为在这种情况下，我们只保留这一个并移除所有其他节点。对于第二种情况，我们移除这一个并保留父节点。我们对每一个非活跃节点重复该过程。最后——

We add them together to get all our contributions. We put them into the nodal equations. For the first independent source, we get the voltage and current; for the second, we also get the voltage and current. Then we add them together to obtain the final voltage or current. But here is a very important note: the dependent source should be kept, because we have...

我们将它们相加，得到所有的分量。然后将这些分量代入节点方程。对于第一个独立源，我们求得电压和电流；对于第二个独立源，同样求得电压和电流。最后将它们相加，得到最终的电压或电流。但这里有一个非常重要的注意事项：受控源应当保留，因为我们有……

For an independent source, we remove it and just keep one. But if there is a dependent source in the circuit — we use this symbol for a dependent source — we should keep it. We should not remove it.

对于独立源，我们将其移除，只保留一个。但如果电路中存在受控源——我们用这个符号表示受控源——则必须保留它，不能将其移除。

Dependent sources are controlled by circuit variables. The output of a dependent source depends on the parameters of the circuit. For example, the voltage is five times i, which means the output depends on the current of the circuit. Therefore, dependent sources should not be removed; they should always be kept in the circuit when considering these situations.

受控源由电路变量控制。受控源的输出取决于电路的参数。例如，电压等于五倍的i，这意味着输出取决于电路中的电流。因此，在考虑这些情况时，受控源不应被移除，而应始终保留在电路中。

Superposition is only applicable for linear circuits. It is not applicable for nonlinear circuits. For example, if you want to compute power, you cannot add the powers together. For V1 and I1, you can add them to get V2 and compute the power. Then you also have P1 and P2. When you add P1 and P2, you will get the wrong answer for the final power, because power is nonlinear. So you cannot use superposition to compute power.

叠加定理仅适用于线性电路，不适用于非线性电路。例如，如果你想计算功率，不能将各功率直接相加。对于V1和I1，你可以将它们叠加得到V2再计算功率，这样你会得到P1和P2。当你把P1和P2相加时，得到的最终功率是错误的，因为功率是非线性的。因此，不能使用叠加定理来计算功率。

In practice, superposition is an often-used method to obtain the final model. It produces simplified circuits. For this circuit, we simplify it by removing one source, so the circuit becomes simpler. In the second case, once a source is removed, the circuit also becomes simpler. This method can be used.

在实际应用中，叠加法是一种常用的获取最终模型的方法。它能够产生简化的电路。对于这个电路，我们通过移除一个电源来简化它，因此电路变得更简单。在第二种情况下，一旦移除一个电源，电路同样变得更简单。这种方法是可以使用的。

To simplify the circuit, a voltage source is replaced by a short circuit, so there is no voltage drop. A current source is replaced by an open circuit, so there is no current. That is the principle of superposition. Any questions? Any other questions?

为了简化电路，电压源用短路替代，因此没有电压降；电流源用开路替代，因此没有电流流过。这就是叠加原理。有问题吗？还有其他问题吗？

If this is a short circuit, the voltage and the current are zero. You add them together, it should be V1 equals zero.

如果这是短路，则电压和电流均为零。将它们相加，应得V1等于零。

For example, if you want to remove all the remaining conditions, you have removed all the remaining if-statements.

例如，如果你想移除所有剩余的条件，你就已经移除了所有剩余的 if 语句。

For an independent voltage source, we replace it with a short circuit. For the first case, we replace u1 with a short circuit. For the second case, we replace u2 with a short circuit. Here is the important point: for the first case, the current flows in this direction; for the second case, the current flows in the opposite direction.

对于独立电压源，我们用短路来替代它。第一种情况下，我们将u1替换为短路。第二种情况下，我们将u2替换为短路。这里有一个关键点：第一种情况下，电流沿这个方向流动；第二种情况下，电流沿相反方向流动。

We have to minus them together because the currents are in different directions. The first current is E1 divided by R1 plus R2. The second is E2 divided by R1 plus R2. We have to minus them together because the currents are in different directions. Current always flows from high potential to low potential.

我们必须将它们相减，因为电流方向不同。第一个电流是 E1 除以 R1 加 R2。第二个是 E2 除以 R1 加 R2。我们必须将它们相减，因为电流方向不同。电流总是从高电位流向低电位。

When you look at the solution, there are many identical permutations—one grouping. Then pay attention to the sign here. Pay attention to the sign.

当你观察这个解时，存在许多相同的排列——归为一组。然后注意这里的符号。注意符号。

Okay, summary. So in this course we have introduced many new objects. We have introduced mesh analysis, nodal analysis, and superposition theorem. For mesh analysis, mesh is the smallest loop, and we assign a mesh current in each mesh. The mesh current is clockwise. Then we determine the branch current using KVL.

好，总结一下。在本课程中，我们引入了许多新的对象。我们介绍了网孔分析法、节点分析法和叠加定理。对于网孔分析法，网孔是最小的回路，我们在每个网孔中指定一个网孔电流，网孔电流方向为顺时针。然后我们利用基尔霍夫电压定律确定支路电流。

We finally apply KVL for each mesh. For the other method, we find a reference node and then assign voltages to the remaining nodes. We apply KCL at each node. For the supernode position, each time we only consider one independent source and one independent node, and we remove all

我们最终对每个网孔应用基尔霍夫电压定律。对于另一种方法，我们先找一个参考节点，然后为其余节点分配电压。我们在每个节点应用基尔霍夫电流定律。对于超节点的位置，每次我们只考虑一个独立源和一个独立节点，并移除所有

For a voltage source, we set it to a short circuit. For a current source, we set it to an open circuit. That means all the contributions are additive, because they are independent sources. Right, any questions?

对于电压源，我们将其设为短路。对于电流源，我们将其设为开路。这意味着所有分量都是可叠加的，因为它们都是独立源。好，有什么问题吗？

Any questions? If there are no questions, we proceed.

有问题吗？如果没有问题，我们继续。