Why do resistors in series add-up? Is there any visualization equivalent of that? I have in mind electrons as cars driving on the highway, now a resistor could be a narrow spot (R1) where cars are moving slower, if I will "add" another congestion point R2 (resistor), after the first one, the cars that had already past by R1 will continue on R2, in case R2 is narrower they will slow down even more, in case R2 is wider it will be no problem since the cars will move freely, in case R2 is the same again the cars will move freely after the first congestion point R1

Can someone explain in that way why should the total R is equal to R1+R2?

|Many thanks in advance!

  • \$\begingroup\$ Many thanks to everyone helping me out to understand the analogy! \$\endgroup\$ – Electro Jo Nov 16 '14 at 11:33

Your car analogy is almost there, but not quite.

Instead of a single length of road, imagine instead a racetrack.

That racetrack is packed with cars, end to end, wall to wall. No space between them.

Now there are some narrow points on the racetrack. Each narrow point is a resistor. Each car is an electron.

The cars have to queue up to get through a narrow point. Not just because it's a narrow point, but because there are cars already in there, and cars filling the next section of road queueing to get into the next narrow point. That's the crucial difference with your analogy - you're assuming the "wire" after the resistor, and before the next resistor, is empty, but it isn't, it's full.

So the more "resistors" you have the more cars will be queuing, and the bigger the tailbacks will be.

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  • \$\begingroup\$ many thanks Majenko. I will try to understand the analogy better. so if the cars have to queue in order to pass through the narrow point R1, this means they are in line and let's say they pass something like 3 cars per second, now, those cars when they meet the second narrow point R2 which is wider and allows to pass 5 cars per second, why they should be waiting on queue? the way I am seeing it is, since the supply of electrons has already been constricted once and it was constricted by a number how can they be constricted even further if the second constriction is wider than the first? \$\endgroup\$ – Electro Jo Nov 16 '14 at 6:58
  • \$\begingroup\$ Because the cars can't get out of the first constriction as fast as they would be otherwise, because of the cars waiting to go through the second constriction. Remember, there is no space for them. \$\endgroup\$ – Majenko Nov 16 '14 at 10:51
  • \$\begingroup\$ thank you, for your description. what I was not taking into consideration was the always on "traffic" on the circuit. so in my mind I had an "empty" road where the first car/electron was the one pushed through the resistor by the EMF, so before that, the road was empty, but in reality there are lots of electrons already there which they move through both of the resistors simultaneously. could that be a correct understanding? \$\endgroup\$ – Electro Jo Nov 16 '14 at 11:22
  • \$\begingroup\$ Yep, you've got it now. \$\endgroup\$ – Majenko Nov 16 '14 at 11:23

The cars analogy doesn't model what's going on.

A better analogy is water in a pipe. Let's say you have a pump (like voltage source) that keeps a constant pressure from one end of a pipe to the other. With just a section of open pipe, the flow rate (like electric current) will be quite high.

Now imagine installing a constriction (like electric resistor) somewhere in the pipe. The flow rate will be less. Now install a second restriction further down the same pipe. The flow rate will be even less.

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  • \$\begingroup\$ thanks Olin for your reply. in the analogy of water current I am still having the question related to the already "passed" water out of the first constriction. For instance if my first constriction R1 is let's say 220Ohm that means an X amount of water will pass, now if my second constriction R2 is 110Ohm, which in my understanding is a "wider" constriction than the R1, the water (current) that already has passed R1 will not be constricted further, because during R1 it had to adjust to a tighter space. \$\endgroup\$ – Electro Jo Nov 16 '14 at 6:50
  • \$\begingroup\$ @ElectroJo the second restriction will restrict the flow still, no matter how small. Think of an extreme case when you have a 100k ohm resistor in series with a 1 ohm resistor. Sure, the total resistance is technically 100.001k, but you probably won't notice the difference and call it near enough 100k. Almost all of the resistance is due to the 100k resistor. This matches hooking a pipette to a 2" pipe: one has a much higher restriction than the other, but that doesn't mean the larger section has 0 restriction. \$\endgroup\$ – helloworld922 Nov 16 '14 at 7:33
  • \$\begingroup\$ in the water example if we assume we put a part of really narrow tube, equivalent of the 100K Ohm resistor, which let's say "allows" one drop of water to pass by and after that I have a much wider piece of tube of let's 1 Ohm, I cannot imagine how the wider piece will restrict the one drop that R1 allowed to pass prior. The only way I can imagine somehow that this work is only due to the fact that a conductor is already full of electrons in some kind of move so then we have what @Majenko described in his example of the packed race track. \$\endgroup\$ – Electro Jo Nov 16 '14 at 8:39
  • \$\begingroup\$ @ElectroJo The key point is that the same amount of water (by number of molecules or weight) moves past any point in the pipe in a given period. Otherwise water would accumulate somewhere. In a narrow section, the stream has to be faster than in a wide section. The same would be true in your traffic analogy. \$\endgroup\$ – ntoskrnl Nov 16 '14 at 11:13

I don't know if some sophisticated math could explain the situation also with cars. An often used analogy for electric current is water flow.

Think about a big pipeline, with low pressure you can push a lot of water through it. If you replace part of this pipeline with a small tube the water flow will be strongly reduced. Adding another such stage the amount of water flowing through will be about halved.

Putting the second tube in parallel to the first one you will get the water flow doubled.

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  • \$\begingroup\$ hi kitana, i would prefer a more visual analogy than sophisticated math is true :) how can it be reduced even further since the water that already passed let's say 5litres/second is the same that has the potential to pass from the second part of smaller tube which also allows 5litres/second? so, from the first small piece of tube we can pass 5litres/second which is the same as the second piece of smaller tube a bit further down the circuit. and what if the second is wider than the first? \$\endgroup\$ – Electro Jo Nov 16 '14 at 7:02
  • \$\begingroup\$ @ElectroJo - Not realy because with a smal tube you have a pressure loss. After the first tube you will have a lower pressure (voltage) than before. With this lower pressure you're not able to push the same amount of water/time (current) through the second pipe. \$\endgroup\$ – Kitana Nov 16 '14 at 8:45
  • \$\begingroup\$ The flow (current) will necessarily be the same in both segments, unless you have a leaky pipe. What matters here is pressure - the water will create a pressure difference as the flow is forced through the constriction; higher flow will result in a larger pressure difference. If you put two narrow segments in series, then the pressure difference across both of them will be the sum of the pressure difference across each one individually, but the flow through both will be the same. \$\endgroup\$ – alex.forencich Nov 16 '14 at 10:31
  • \$\begingroup\$ thanks @alex.forencich, and this is because the pipe is always filled with water regardless if it moves or not \$\endgroup\$ – Electro Jo Nov 16 '14 at 11:29

Why do resistors in series add-up?

No fancy analogies, just ohms law. The voltage across a resistor is the current through the resistor multiplied by resistance i.e. V = IR.

Now, if two resistors were put in series (with the same current flowing) the voltage has to become twice as big because you have individual "V=IR"s adding up. so now you have the equation

2V = IR + IR = I*(2R).

2V = I*(2R)

I think that should be clear but then again I've known and understood ohms law for a million years and it's easy to forget what it's like for someone starting out on these things.

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  • \$\begingroup\$ thanks @Andy aka, I have read the Ohm's Law again and again, I am trying to understand the concept in more "everyday" anologies :) \$\endgroup\$ – Electro Jo Nov 16 '14 at 11:31
  • \$\begingroup\$ @ElectroJo Although I understand the need for visual analogies from a newbie, I don't quite favor them. They could be useful to understand very simple situations, but they rapidly get out of hands. I've seen so many students studying more complex circuits trying to force their behavior to a water analogy and completely missing the point. Thus I'm with Andy: it's worthwhile to make the effort of understanding the basic laws of electricity (Ohm's, Kirchhoff's, etc.) even if they are not "visual" at all. With practice they become intuitive and this greatly simplify further learning IMHO. \$\endgroup\$ – Lorenzo Donati -- Codidact.com Nov 16 '14 at 12:15
  • \$\begingroup\$ thanks for the advice @LorenzoDonati I just have a feeling that is necessary to have an understanding in physical world before get deeper into maths \$\endgroup\$ – Electro Jo Nov 16 '14 at 15:04
  • \$\begingroup\$ @ElectroJo - imagine a thin 1m long water pipe. To get water to flow at 10 litres per minute (current flow) might require a pressure (potential-difference aka voltage) of 10 kPa. If two of these pipes were put in series, to get the same flow would require twice the pressure. This is because the resistance to water flow has doubled. \$\endgroup\$ – Andy aka Nov 16 '14 at 16:47

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