Series, Parallel, Kirchhoff

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Introduction: Series, Parallel, Kirchhoff

A network of resistors behaves as a single resistor: the current through the network is proportional to the potential difference applied. Calculating the equivalent resistance of a network starts for many at high school but may come back later with more complex cases also at university. For success in science and engineering it is crucial to be skilled in these calculations but also to develop an intuition on how to make quick approximate solutions. Being an expert in handling electrical resistance networks will come into great use for other types of networks that have similar behaviour: magnetic circuits, fluid dynamics, heat conduction, traffic flow, just to name a few.

To hear is to forget, to see is to remember, and to do is to understand, so let’s practice resistor networks hands-on!

This tutorial is great for the class-room, since it can be performed by students individually or in small groups: the equipment is cheap and common (a multimeter and some crocodile clips) and the consumables are very cheap (7 resistors per experiment - they cost less than a cent each when bought in 100-packs) - The students can keep the final resistor network as a souvenir.

There are two levels: for high-school students do all the series and parallel exercises, and skip the final assignment that requires Kirchhoff law’s. For bachelor students, the series and parallel networks provide a nice warm-up to the real work, which is to solve a multi-loop network with Kirchhoff’s laws.

用品:

  • A multimeter - sold at hardware stores for as little as 5EUR
  • It’s helpful to have crocodile clips at the end of the measuring leads - I did this by replacing the multimeter cables with banana cables and attaching crocodile clips. An alternative is to attach crocodile leads to the end of the multimeter leads.
  • 2 more Crocodile leads - sold online for as little as 1 or 2 EUR for a pack of 10
  • 7 identical resistors. 100Ohms or 1kOhm are fine. Common 0.25W 1% types will do. We’ll call the resistance value ‘R’ from now on, to get into the right habit of doing math with symbols.

Step 1: Measure the Resistances

Put the multimeter in the resistance setting, with the maximum just above the value of R. So set it to ‘200’ if R=100Ohm. Check that all 7 resistors are good!

How does a multimeter measure resistance? Internally, it has a current source, which will force a very small current through anything connected to the leads. The potential difference that results from it is proportional to the resistance: V=iR. Yes, that’s called Ohm’s law. The value displayed on the multimeter is simply the voltage, divided by the value of the current.

Step 2: Attach Two Resistors

In electronic equipment, components are connected by soldering them to a printed circuit board (PCB). Here we use a simpler method: we wind the leads around each other. Cross the leads of two resistors and twist them around each other. Keep twisting until there are at least 10 turns and they are tightly connected. This is a quick way to make a connection that is both mechanically electrically strong.

This is our first network: A current that passes through both resistors will incur a voltage drop in both resistors. For the same current, the voltage will be the sum of the voltage on both resistors, and the equivalent resistance is the sum of all resistance values in series: Req=R1+R2. Since both resistors are equal we expect 2R as the equivalent resistance.

Go ahead and measure the resistance of the two resistors in series. Is the value that you find consistent with the calculation?

Step 3: Keep Adding Resistors

重复上一步,直到有6个电阻串。从一端到另一端的总阻力是多少?对,6R。将万用表的引线夹在不同的位置,检查是否可以得到1R、2R、3R、4R、5R和6R的所有值。

第四步:做一个戒指

进行最终连接,使6个电阻器形成一个环。万用表导线现在可以连接到6个不同的位置,我们称它们为a、b、c、d、e和f。

当测量a和d之间的电阻时,你期望有什么电阻?我们现在有两个并联的支路,每个支路的电阻为3R。并联电阻的等效电阻计算为逆电阻之和:1/Req=1/R1+1/R2+1/R3+。对于两个电阻器,这简化为乘积除以和:Req=(R1*R2)/(R1+R2)。对于n个相同的电阻,并联电阻就是R/n,所以我们期望3R/2=1,5R,这就是你们测量的吗?

Of course, the multimeter leads can be connected at other points too: what about the resistance between point a and c? We then have one branch with a resistance of 4R and one branch with a resistance of 2R. So in parallel, this gives Req=(4R*2R)/(4R+2R)=(4/3)R or 1,33R. And between a and b? Req=(5R*R)/(5R+R)=(⅚)R=0,83R

Step 5: Make a Short-cut

让我们回到a和d之间的电阻。电阻是1,5R。我们将用c和f之间的鳄鱼形导线做一个额外的连接。在测量或计算之前,争论一下你所期望的:电阻会保持在1,5R,还是会变高或变低?想想等效的方法:在一个管网络中,你再加一个零电阻管。或者在一个街道狭窄的城市,你开了一条新的宽马路,交通会增加还是减少?正确,额外的0欧姆连接将使电流更容易流动,因此网络的电阻只能下降。

这可能不是很明显,但网络可以重新绘制,这样更明显的是,我们现在有两个子电路串联:每个子电路包含一个R和一个2R并联。所以我们得到Req=2*(R*2R)/(R+2R)=(4/3)R=1,33R。事实上,计算证实了我们的直觉,捷径把电阻从1,5R降低到了1,33R。如果一切连接良好,测量应该证实这一点

第六步:硬电路

And now for the genius of the class: Let’s replace the shortcut between point c and f by a resistor between point c and f. We use one side of each crocodile clip to clamp the resistor in position, or, as in he picture, we wrap the leads of the resistor around point c and point f. It is no longer possible to calculate the equivalent resistance simply by finding combinations of series and parallel resistors! What then?

From the previous calculations and measurements we know the range in which the resistance must be: the present circuit cannot have a resistance more than 1,5R, because that’s what we get without the resistance between c and f. For the same reason it cannot have a resistance less than 1,33R, because that’s what we get when we short between c and f. So we’re somewhere in between 1,33R and 1,5R. Let’s measure: 1,4R, in range with expectation, but how would you calculate it?

Kirchhoff’s laws come to rescue: the first says that for every node the sum of currents must be zero and the second says that for every loop the sum of voltage differences must be zero. So let’s give names to the currents, make sure they add up to zero in all nodes and evaluate the voltages over some loops.

First of all, to calculate the behaviour of the circuit, we see what happens when we apply a potential difference V across it. There will be a current and we determine Req=V/i.

In point a, the current splits into two branches, we indicate these currents as i1 in the branch with resistance 2R, and i2 for the branch with resistance R. The network is perfectly symmetric, so also on the right side, the current through the 2R branch is i1 and the current through the R branch i2. In the central branch we’ll have i2-i1.

现在我们把基尔霍夫定律应用到一些回路上。让我们从包含电池和一个电阻分支的大回路开始。我们发现V-i1*2R-i2*R=0。

现在我们需要一个小环,包括没有电池but does include the central branch: -i2*R-(i2-i1)R+i1*2R=0. This equation gives us the ratio of i1 and i2: i2=(3/2)i1. We substitute it in the first equation and get V=3,5Ri1, or i1=(2/7)V/R. Then, i2=(3/2)i1=(3/7)V/R and i=i1+i2=(5/7)V/R. As the last step we find Req=V/i=(7/5)R=1.4R. Exactly like the measurement!

Step 7: Conclusions

This exercise teaches the students a lot of skills in a single lesson, with hands-on practice and with a lot of fun: handling resistors, connecting cables, using a multimeter, making electrical connections, performing measurements. The abstract academic methods of calculations come to life in the student’s hands. The students are confronted with a difficult problem for which it turns out a solution does exist, although it requires non-trivial math: finding the solution to a linear system of two equations with two unknowns.

And, if you are ready for more, check out NerdSnipe's fun video about way more complex resistor networks:https://www.youtube.com/watch?v=nHUv80RsQns

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    5 Comments

    0
    书呆子

    3天前Step 7

    哈,看来你比我更擅长算出数学捷径。不久前,我做了一个视频,里面有这个精确的电路,但是在视频结束时,我把它构建成了一个16x17的正方形网格。在这里查看:https://www.youtube.com/watch?v=nHUv80RsQns

    0
    rgco

    Reply 2 days ago

    Awesome! I really enjoyed your video and added a link at the end! Indeed, I used symmetry arguments to reduce the number od unknowns and simplify the math. I have a special connection to this circuit: I solved it at a high-school physics olympiad, which helped me get into the national finals. Later, in grad school, the equivalent of a 1xn network came up for a problem of gas distribution in a straw-tube chamber for a neutrino detector. My supervisor assumed that the flow (current) would be equal through all branches but I could show that the flow through the central branches is much less than that of the lateral branches, and we had to redo the design to assure equal flow.
    两周前我民用建筑ht the resistors and a multimeter to a class, since I was annoyed that this circuit was not in the student's textbook. I could not make them do it themselves because of limited equipment, limited time and COVID restrictions, but I did a demo projected from a document camera. They seemed to like it, so I figured I might as well put it on the web. Good luck with your channel, I'm following!

    0
    书呆子

    Reply 2 days ago

    Yeah, I missed doing this stuff. Been working 10 years, and haven't touched any of it. I'm ashamed to say it took me 3 tries just to get kirkhoff's laws right.
    Just checked your stuff out too. It looks like we have a lot of similar things we're working on. I was debating turning the pico into a rudementary oscilloscope. I saw you already turned it into a waveform generator.

    Keep up the good work! I'll be following you too!

    0
    banman11

    6 days ago

    伟大的导师!我对电子学是个初学者,讲解很有帮助。
    Thank you.