1
\$\begingroup\$

This question already has an answer here:

This is blowing my mind.

Let's say I've a bunch of 100 Ohm resistors, all identical.

How is it possible that if I combine them in parallel, I can get something that has less resistance?

Check this out:

   ----100OHM---
---|           |---
   ----100OHM---

This is effectively a 50 Ohm resistor!

It was already hard for current to get through one resistor (at 100Ohms), but I combined two resistors together, and now it became easier for current to go through!

If I take ten 100 Ohm resistors in parallel:

Check this out:

   ----100OHM---
   ----100OHM---
   ----100OHM---
   ----100OHM---
   ----100OHM---
---|---100OHM--|---
   ----100OHM---
   ----100OHM---
   ----100OHM---
   ----100OHM---

This is now 5 Ohms!!!!!!!!!!!!!!!!!!!!!!! Electricity is no longer having almost any resistance, yet each of resistors is 100 Ohms. If I added even more 100 Ohm resistors, resistance would go close to 0, and current would just flow through with no resistance.

How is that even possible? If I think logically, then it could only have gotten HARDER for current to go through two resistors, but somehow it became easier.

Real life analogy:

Let's say there's a 500 pound bouncer at club entrance. It's hard to get past him right? He's strong and will resist you getting past him. Now if the club adds another bouncer at 500 pounds, it suddenly becomes EASIER to get past both of them??

There's no way it's possible!! Can anyone explain this in simple terms?

\$\endgroup\$

marked as duplicate by Alfred Centauri, Fizz, PeterJ, Daniel Grillo, Tom Carpenter Nov 23 '15 at 22:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    \$\begingroup\$ Let's say I have a bunch of narrow pipes. How is it possible that if I use several of them side by side, I can get more water to go through the combination at a given pressure than I could get to go through just one? \$\endgroup\$ – Chris Stratton Nov 22 '15 at 0:25
  • 2
    \$\begingroup\$ i won't add another answer to what is, a very fundamental question. but i will say this; replace "resistance" with "conductance" and 100 with its reciprocal, and then ask the same question. \$\endgroup\$ – robert bristow-johnson Nov 22 '15 at 0:30
  • 1
    \$\begingroup\$ think of these little two-wire devices as offering conductance, a partially free path for the electrons to flow from point A to point B. to put it in terms of current news affairs, think of the swarm of electrons in the same manner as a swarm of refugees: if there is a choice of two 0.01-mho gates letting them into a country, that would be about the same as if there were a single 0.02-mho gate (which is twice as wide as either of the 0.01-mho gates). \$\endgroup\$ – robert bristow-johnson Nov 22 '15 at 0:43
  • 1
    \$\begingroup\$ oooooooh another entrance and not another bouncer!! \$\endgroup\$ – bodacydo Nov 22 '15 at 8:34
  • 1
    \$\begingroup\$ More lanes on a motorway means less resistance to traffic. \$\endgroup\$ – Andy aka Nov 22 '15 at 20:41
2
\$\begingroup\$

When you add resistors in parallel, you add additional paths for the current to flow, so it is easier for current to flow through the two resistors in parallel, than it is for it to flow through a single resistor.

Do you know Ohm's Law? E = I x R, where E is voltage (electromotive force)(V is often used insead of E, and some languages use U), I is current in Amps, and R is resistance in Ohms.

schematic

simulate this circuit – Schematic created using CircuitLab

In the above circuit, Ohm's Law says 10 mA will flow through each resistor, so the total current produced bythe power supply will be 20 mA - twice the current for a single resistor.

Again using Ohm's Law, turned around to R = E/I = 10/.02 = 500 ohms. so the two 1000 ohm resistors in parallel act like a single 500 Ohm resistor.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your answer. But isn't resistance a property of how difficult it is for electricity to flow through material? If all resistors are at 100 Ohms, isn't that the lower limit of how hard it will be flow through this circuit? Resistance is how hard it will be for electrons to flow through material. How come that by combining resistors of the same material, we somehow made it easier for electrons to flow through? It's still the same material? I think this is my main concern. \$\endgroup\$ – bodacydo Nov 22 '15 at 0:39
  • \$\begingroup\$ You should consider each branch of a parallel circuit as a separate circuit - in my circuit above, if we only have R1 across the voltage source, we will have 10 mA flowing through the circuit. Similarly, if we only have R2 across the voltage source, we will also have 10 mA through the circuit. If we have both R1 and R2, each resistor will pass 10 mA as before, so the total current will be 20 mA. You might consider highway traffic - a two-lane road should be able to carry twice as many cars as a single lane road - the parallel resistors are similar to the two-lane road. \$\endgroup\$ – Peter Bennett Nov 22 '15 at 0:48
  • \$\begingroup\$ "conductance" is a measure of how wide the path is for the electricity to flow. "resistance" is the reciprocal of conductance. so resistance is a measure of how narrow the path is. \$\endgroup\$ – robert bristow-johnson Nov 22 '15 at 0:49
2
\$\begingroup\$

I want to touch on something that you have been placing in the comments on the other answers:

But isn't resistance a property of how difficult it is for electricity to flow through material?

No. Well, partly. But that is a very over simplistic way of looking at things.

A given material has a Resistivity associated with it. What this is a measure of is how much resistance per metre a material will present. Actually it is related to resistance by the equation:

$$R=\frac{\rho L}{A}$$

Where \$R\$ is resistance, \$\rho\$ is resistivity, \$L\$ is the length of the piece of material, and \$A\$ is its cross-sectional area.

If you take a \$100 \Omega\$ resistor, it will be made of a material with a known resistivity, formed into a device where \$A\$ and \$L\$ are selected to get \$100 \Omega\$.

If you take two of these, and place them in parallel, you haven't changed the resistivity of the material, or its length, but what you have done is increased the cross-sectional area - doubled it to be precise. If you increase \$A\$ in the equation above, you can see that \$R\$ goes down.


Perhaps an analogy would help. Electricity is the flow of electrons, so lets make the analogy that it is like moving water. Bear with me.

Imagine taking a glass of water and putting a drinking straw in it. You suck on the straw and water moves through. Now if you suck harder, you get more water through. You are behaving like a voltage source. You are trying to move the water through the straw which is offering some resistance. You can imagine this resistance if you compare the case of using a straw, to simply taking a big gulp directly from the glass without the straw.

Now try to drink with two straws instead of one (in parallel with each other) - it gets easier doesn't it? You can drink more water with the same amount of effort. This is because the two straws in parallel have a larger cross sectional area to let water through.

Same with putting two resistors in parallel.


For completeness, what happens if you put two resistors in series?

You aren't putting them in parallel, so the area doesn't change - all the electricity has to flow through one resistor to get to the other as opposed to flowing through both at once.

But you have made the length longer. To get from one end of the series resistor pair, the electricity has to flow through the first resistor and then through the second (over simplification). You've doubled the length.

If the length gets longer, we can see from the equation above that if the length increases, so does the resistance.



As to your analogy of the bouncers at the club. Well, you can't fight both at the same time, you would have to take on one, then on the other (or alternating between, whatever). So essentially by putting a second bouncer at the door, you have doubled the length of the path through the bouncers into the club, hence you have made it harder.

It's difficult to actually make that analogy into one where the bouncers are in parallel. It would be almost like if there were now two doors but still only one bouncer - it's easier to get.

\$\endgroup\$
  • \$\begingroup\$ The parallel bouncer idea could work if you assume a typical bouncer loses 1 fight per hour. Then with two parallel bouncers each facing the same crowd pressure, there could be 2 lost fights per hour, resulting in a higher bad customer flow per hour. \$\endgroup\$ – Nedd Nov 22 '15 at 2:18
1
\$\begingroup\$

Basically a current divider works as if you add more water pipes in parallel. More pipes, more flow. Resistors in parallel form a current divider.

Regarding

But doesn't electricity still have At Least 100 Ohm resistance, no matter what path it takes? Let's say all electrons choose to go through upper resistors.

No, they can't make that kind of choice. Statistically they get distributed equally through equal resistors in parallel.


How come that by combining resistors of the same material, we somehow made it easier for electrons to flow through? It's still the same material?

It's the same material but there's more of it in terms of cross-sectional area, which reduces resistance. For more details (here) see: Role of Resistor in circuits mainly in analog electronic.


And finally

Let's say there's a 500 pound bouncer at club entrance. It's hard to get past him right? He's strong and will resist you getting past him. Now if the club adds another bouncer at 500 pounds, it suddenly becomes EASIER to get past both of them??

There's no way it's possible!!

You bouncer analogy is flawed. When two bouncers are working as sidekicks on the same door they are "in series" as far as opposition is concerned, you'd have to get past both of them to get in (and they can help each other so it's a non-linear effect). But if they are working different doors that are far apart, then to get in you'd only have to get past the dumbest/weakest of them indeed. Furthermore your analogy is also flawed because current is (ignoring quantum mechanics for now) infinitely divisible, so [about] half can try one door and half try the other, something that a man can't do simultaneously, but if we consider sequential attempts by the same guy on all doors [or by a group that splits up], then it is a fair comparison when multiple doors far apart are considered.

Think what is the probability of rolling a six in one dice throw. And then what is the probability of rolling at least one 6 using two dice throws [I'm not talking about adding the values rolled, just any of them falling with its 6-face up]. What about using three throws? Beware however that he probability analogy is only qualitative; the actual math formulas differ between parallel resistance and the probability of success on repeated attempts. Conduction doesn't work by electrons "slipping by" some opposition, so that's why thinking of it in such terms doesn't get one very far (quantitatively) even when you do get a qualitative analogy somewhat right.

To actually understand conductivity at quantum-mechanics level requires models like the free or nearly free electron model which frankly even most EEs probably heaven't heard of, so don't feel bad if that flies over your head until you get to seriously study physics.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your answer. But isn't resistance a property of how difficult it is for electricity to flow through material? If all resistors are at 100 Ohms, isn't that the lower limit of how hard it will be flow through this circuit? Resistance is how hard it will be for electrons to flow through material. How come that by combining resistors of the same material, we somehow made it easier for electrons to flow through? It's still the same material? I think this is my main concern. I understand pipe analogy now but I still don't understand it from this other perspective. \$\endgroup\$ – bodacydo Nov 22 '15 at 0:39
  • \$\begingroup\$ You are confusing the flow in one resistor to the total flow in the whole circuit. The current flow is just as hard for "each" single resistor. But now the total circuit flow is the combination of all the currents flowing in each resistor. Using this total current you can determine an equivalent total resistance for the circuit, which decreases with each added parallel resistor. \$\endgroup\$ – Nedd Nov 22 '15 at 1:59
  • \$\begingroup\$ @Nedd: I'm guessing you're addressing Boda. \$\endgroup\$ – Fizz Nov 22 '15 at 2:35
  • \$\begingroup\$ @bodacydo: "Resistance is how hard it will be for electrons to flow through material." that's not a very good explanation and you should banish that thought from your head. \$\endgroup\$ – whatsisname Nov 22 '15 at 4:56
  • 1
    \$\begingroup\$ @whatsisname I learned it's resistivity not resistance. \$\endgroup\$ – bodacydo Nov 22 '15 at 8:33
0
\$\begingroup\$

You are thinking about this the wrong way. Although you are putting in more resistance into the circuit, you are also giving another path for the current to go through. In your analogy, think about this like there are two body guards, but instead of the body gaurds both blocking one path (resistors in series), each of them have to individually block two separate paths (resistors in parallel).

\$\endgroup\$
  • \$\begingroup\$ Thanks for your answer. But isn't resistance a property of how difficult it is for electricity to flow through material? If all resistors are at 100 Ohms, isn't that the lower limit of how hard it will be flow through this circuit? Resistance is how hard it will be for electrons to flow through material. How come that by combining resistors of the same material, we somehow made it easier for electrons to flow through? It's still the same material? I think this is my main concern. I understand pipe analogy now but I still don't understand it from this other perspective. \$\endgroup\$ – bodacydo Nov 22 '15 at 0:40
  • \$\begingroup\$ Whenever you are learning anything you should find the method that works best for you. If the pipe analogy worked better for you then feel free to ignore mine. To elaborate on my answer more: Imagine a hall with a filter and that many people are trying to get to the other side. This filter only will let specific people in and not others. If you are to put two filters in the same hall, then you will get less people out on the other side. But now lets say you build another hall with a filter so you have two halls that people can go through. Now you will have more people on the other side. \$\endgroup\$ – user92289 Nov 22 '15 at 1:06
0
\$\begingroup\$

Remember what current is. Current is the amount of charge that flows in some time, the charge carrier of which is the electron, thus current is a rate. Resistance resists the flow, so it reduces the rate at THAT path, if you add another path of the same resistance, you get two paths of the same rate... Without sounding insulting, if the rate of both paths are 1 and you now have two paths, 1+1=2, so you new rate is higher. Said another way, your resistance is now the reciprocal of that, or 1/2, thus 50 ohms. Though this parallel trick will only reduce the resistance so far, resistance is indeed an innate property of matter (electron bands and phonon properties), and you'll probably find infinite parallel resistors have a vertical asymptote at zero ohms.

\$\endgroup\$
0
\$\begingroup\$

How about this simple explanation:

Using an electrical current flow perspective, assume 100V is across a single 100 ohm resistor. The current flow in this one resistor will be 1A, (by ohms law, the current I = V/R, or 100V/100 ohms).

Now add another similar resistor in parallel with the first. Nothing changes in the first resistor, (there is still the 100V across it and it still passes the same 1A), but now the second resistor also has the 100V across it and it also passes 1A. Now the total current passed in the revised circuit will be 2A, (1A in "each" resistor).

With this new total current you can use the ohms law calculation in another way to find the "equivalent" resistance in the circuit. The resistance R = V/I = 100V/2A = 50 ohms. The increase in total current would continue for each additional resistor placed in parallel. So the equivalent resistance in the circuit continues to be divided down with a simple formula of R/n, (where n is the number of resistors in parallel).

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.