Why doesn't current pass through a resistance if there is another path without resistance? How does it know there is resistance on that path ?

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    $\begingroup$ Related: physics.stackexchange.com/q/33621/2451 , physics.stackexchange.com/q/188371/2451 $\endgroup$ – Qmechanic Oct 28 '18 at 12:40
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    $\begingroup$ RE your edit. I see that you've added three additional questions to you original post after quite a few answers were written. This is considered bad form here. I recommend that you roll back the edits and then start a new post that references this one, explains what isn't clear about the answers here, and ask the follow-up questions there $\endgroup$ – Alfred Centauri Oct 29 '18 at 12:34
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    $\begingroup$ @AlfredCentauri is right; accordingly I've rolled back the edits that added additional questions. $\endgroup$ – David Z Oct 29 '18 at 20:08
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    $\begingroup$ Can you explain what you mean by "no resistance?" That's a red flag that suggests you are simplifying things, because everything has resistance (except superconductors in some sense). There's a good chance your confusion is coming from the simplification you used. If we know what you are thinking, we can help better. $\endgroup$ – Cort Ammon Oct 30 '18 at 0:02
  • $\begingroup$ I agree with Cort's comment. The question is not well posed because in a realistic setup where there is a high resistance path and a low resistance path, current does pass on both paths, so asking why it current does not pass on one path is contrary to observation of the physical world. The question "why does an impossible thing happen?" is "it doesn't". $\endgroup$ – Eric Lippert Oct 31 '18 at 15:36

13 Answers 13


The basic circuit theory "rules" you imply, are high level simplifications applicable at a large scale and at slow speeds.

If you look at it close and fast enough, you could say that a current really starts to go into the obstructed path, but the electric field in front of the obstruction would build up gradually and current will start to repartition into the free path where it can start to flow. Naively you could say that the electric field will "sniff out" the paths. Actually in reality the current will also bounce off the obstructions, reflect and go back and forth etc. This is a real mess in practical electrical engineering at high frequencies.

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    $\begingroup$ @ten1o Because electron density builds up (transiently) in front of the obstructed path, for the same reason that car density builds up in front of a bottleneck on a highway. $\endgroup$ – probably_someone Oct 29 '18 at 12:59
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    $\begingroup$ perhaps another comparison might be water through pipes of different diameters. Water at too high a pressure can't transition from one pipe-diameter to another effectively because water is incompressible and it backs up. Meanwhile the pipe that isn't getting narrower continues at its normal pressure and flows faster for it. $\endgroup$ – Ruadhan2300 Oct 29 '18 at 13:22
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    $\begingroup$ @ten1o Yes, this kind of thing will happen for a very short time whenever there's a sufficiently fast change in the current. In DC and low-frequency AC circuits, this only happens when the current or voltage source is suddenly turned on, and since we're only interested in the behavior of the system on longer timescales, we tend to ignore these transient effects. For high-frequency AC current, though, these transients are important, as they happen on the same timescale as the oscillations in the current. $\endgroup$ – probably_someone Oct 29 '18 at 13:51
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    $\begingroup$ @ten1o, would you please clarify for the community if the context of your questions is steady state or transient? $\endgroup$ – Alfred Centauri Oct 29 '18 at 15:39
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    $\begingroup$ At high frequency it's customary to use impedance matched connections and constant-width (and impedance) tracks, because those reflections and current loops occur when there is a change in impedance in the path of a conductor. This is why you will see very high frequency circuits having round tracks (no corners) with weird layouts, and unnecessarily long tracks. They are designed that way so that two differential signals will have the same track length and the exact same impedance, to limit reflections, noise and radiation/interference. $\endgroup$ – Drunken Code Monkey Oct 31 '18 at 7:21

I'll try to offer a simpler analogy of how that works.

Camp A on the side of a mountain is full of hikers. There is another empty campsite B on the other side of the mountain. And there are two possible paths between A and B - over the mountain or straight through a tunnel.

You order (apply voltage) the hikers (electrons) to go to camp B. While most are still packing, some hikers have their packs ready almost instantly and head out. A few of them go to the path leading to tunnel, a few go towards the mountain pass.

When the next batch is ready to go, once again few will go towards the tunnel and few will choose the mountain way. However, the latter group will get stuck as the previous mountain guys will be seriously slow trying to get up. So a queue will start to form.

When the next batch is ready to go, they will see that there is a queue on one of the paths and will (almost) all choose the easy way where none of the previous hikers got stuck.

Similarly, the electrons don't in some magical way feel that the path will be harder. They are simply stuck between a bunch of previous electrons that have hard time going that way so in the juncture they redirect to the route without the traffic jam.

The main difference between electrons in electrical paths and hikers on hiking paths is that all electrical paths are initially already full of electrons so the next electrons will instantly observe which path has trouble moving forward.

  • $\begingroup$ In the analogy, I believe the hikers are electrons. Do the electrons change their paths according to the electric field created by the crowd of electrons or by their change in potential energy as they enter and exit the resistance? If so, how is an electric field created within a resistor? @BjornW $\endgroup$ – ten1o Oct 29 '18 at 9:37
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    $\begingroup$ Both. "Change in potential energy" and "electric field" are two tools for looking at almost the same thing. Integrate the electric field along the path of particle and you'll get voltage. Multiply voltage by the charge of the particle and that's the change of her potential energy. $\endgroup$ – Džuris Oct 29 '18 at 12:36
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    $\begingroup$ Electric field inside a resistor is created by the electrons stuck there. They repel the following electrons. $\endgroup$ – Džuris Oct 29 '18 at 12:38
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    $\begingroup$ Yes, exactly. Voltage is defined as the difference of potentials. Say you have potential of 15V at one end and 12V at another. Then the voltage between those points is 3V or -3V (depending on which way you look at it). Thus the voltage is obviously the same on any path between the points. $\endgroup$ – Džuris Oct 29 '18 at 14:41
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    $\begingroup$ @coniferous_smellerULPBG-W8ZgjR I don't think that really paints a more understandable picture for the OP. Either you see it as electrons repulsing other electrons, or holes attracting electrons - in reality, both are in play, but looking at just the electrons is sufficient to explain the phenomenon and lines up better with what the OP already thinks is true ("electrons travel through a conductor"). Electrons don't go this way, because there's already too many electrons there - instead, they go the way that doesn't have as many electrons is a fine simplification IMO. $\endgroup$ – Luaan Oct 31 '18 at 10:09

If there is a parallel path without resistance then the voltage across the terminals is zero. If the voltage is zero then, by Ohm’s law, the current through any branch with resistance is also zero.

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    $\begingroup$ I don't think this is the complete explanation OP was looking for. $\endgroup$ – Sam Spade Oct 28 '18 at 23:38
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    $\begingroup$ @SamSpade but it is a perfect Microsoft Answer. alunthomasevans.blogspot.com/2007/10/old-microsoft-joke.html $\endgroup$ – Carl Witthoft Oct 29 '18 at 13:31
  • $\begingroup$ I have a hunch that the current drop through a resistor in case of a short circuit is actually what the OP is wondering about. $\endgroup$ – Peter A. Schneider Oct 29 '18 at 16:09
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    $\begingroup$ I don’t know what you guys think is insufficient about it. It was a question about circuits answered clearly using the standard laws of circuit theory. The OP gave no indication in the question that an answer in the context of circuit theory was unwanted. The best policy (IMO) is to use the simplest theory available to answer a question unless specifically requested otherwise. $\endgroup$ – Dale Oct 29 '18 at 21:39
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    $\begingroup$ Actually, I think the phrase "How does it know there is resistance on that path?" strongly implies that OP is looking for something more than just Ohm's law. And anyway, I think the best policy is to try to figure out what the point of the question is, not just follow the letter of the law. Formulating a good question can be just as hard as finding the answer. $\endgroup$ – Javier Oct 30 '18 at 18:43

Why doesn't current pass through a resistance if there is another path without resistance?

Stipulate that there are two parallel connected resistors with resistance $R_1$ and $R_2$ respectively.

Since they are parallel connected, the current $I$ into the resistor network divides according to current division:

$$I_1 = I\frac{R_2}{R_1 + R_2}$$

$$I_2 = I\frac{R_1}{R_1 + R_2}$$

Now, let the resistance $R_2$ go to zero while holding $R_1$ fixed and see that, as $R_2$ gets smaller, the current through $R_1$ gets smaller and that, when $R_2 = 0$

$$I_1 = I \frac{0}{R_1 + 0} = 0$$

$$I_2 = I \frac{R_1}{R_1 + 0} = I$$

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    $\begingroup$ Why doesn't current pass through a resistance if there is another path without resistance!? Your answer explains how to calculate it but it doesn't explain why this is how it works. Granted though, this could be down to different interpretations of what is being asked for with that troublesome word 'why'. :) $\endgroup$ – RyanfaeScotland Oct 28 '18 at 20:02
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – RyanfaeScotland Oct 28 '18 at 21:21
  • $\begingroup$ This answer assumes a constant overall current $I$! Only then does $I_1$ shrink with shrinking $R_2$. Constant $I$ though implies a declining voltage because, after all, the overall resistance shrinks with shrinking $R_2$. Constant current was never assumed and actually needs a nice lab transformer. (With a constant voltage the current through $R_1$ wouldn't change a bit, obviously. $R_1$ does not care about remote parts of the universe.) $\endgroup$ – Peter A. Schneider Oct 29 '18 at 16:18
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Peter A. Schneider Oct 29 '18 at 17:31
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    $\begingroup$ Seeing as there are now two chat rooms made for the comments on this post, I've removed the comments except for (apparently) the first one in each discussion. $\endgroup$ – David Z Oct 29 '18 at 20:06

Current will flow through all possible paths no matter how high the resistance. The amount of current flowing through any given path will depend upon voltage and resistance. Given two parallel paths, one very high resistance and one very low, most of the current will flow through the low resistance path, but some will still flow through the high resistance path.

Even an electrical "short" will offer some small resistance. As current flows through a "short" there will still be a small voltage across it. So, if a high resistance is shorted and current flows through the short, there will be some small voltage across it, so some small amount of current will still flow through the high resistance.

In practical terms, we consider a short to pass all of the available current, but in truth, it is never all of the current; small, perhaps vanishingly and inconsequentially small amounts of current will still flow though other paths.


Greatly simplified, let's say we have some electrons and two paths:

enter image description here

Now we apply electric field to them and they move:

enter image description here

The ones in the low resistance path have moved quite a distance, but the ones in the high resistance path didn't manage to move at all. Also a new electron has arrived at the junction and needs to make a decision.

The absence of electrons is a positively charged hole. So now Coulomb force acts on that new electron, and it is more likely to choose a low resistance path.

enter image description here

So there will be not enough charge carriers at the beginning of a low resistance path, and too many of them at the beginning of a high resistance path. It will cause charge carriers at the junction to prefer the low resistance path.

  • $\begingroup$ Except that the electron-electron interaction is usually negligible, so this picture might possibly work as an analogy (that eventually breaks down), but is nowhere near a real description of what happens in reality. $\endgroup$ – thermomagnetic condensed boson Oct 30 '18 at 13:01
  • $\begingroup$ This picture is a simplified analogy of all answers here. Slow electrons "prevent" new ones from entering, while holes behind fast electrons attract new ones. Add some "misdirected" electrons and the interaction increases. $\endgroup$ – Andrew Svietlichnyy Oct 30 '18 at 15:17
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    $\begingroup$ @coniferous_smellerULPBG-W8ZgjR That's interesting, I would have thought that this is indeed the mechanism. What else if not an "electron backup" would prevent an electron from entering a wire leading to a resistor? The only possible reason is a weaker electrostatic field, and that in turn can only be caused by the other electrons. $\endgroup$ – Peter A. Schneider Oct 30 '18 at 15:58
  • $\begingroup$ @PeterA.Schneider The E field is basically setup "instantly" compared to the motion of the conductive electrons. The orders of magnitude differ by a factor 2 or so. All of these electrons are going to feel that same E field. If a region has a higher resistance, it means the electrons are going to be scattered more (for several possible reasons) than in a region with a lesser resistance. I do not know enough of solid state physics, but any solid state or condensed matter physicist should be able to set up a corresponding Boltzmann transport equation and explain what's going on, I believe. $\endgroup$ – thermomagnetic condensed boson Oct 31 '18 at 8:39
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    $\begingroup$ @coniferous_smellerULPBG-W8ZgjR The speed of charge carriers doesn't matter. 1 A is 1 C passing per second, kind of 1 C "entering" and 1 C "leaving". The path with higher current has higher amount of charge entering and leaving by definition of electric current. $\endgroup$ – Andrew Svietlichnyy Oct 31 '18 at 10:08

I'll give a partial answer because the real answer probably involves heavy math and is beyond my current knowledge. I wish a condensed matter or solid state physicist would take over and either demolish what I write or improve it.

I think most of the answers (not all) are wrong in that they assume that the electron-electron interaction is the responsible to prevent electrons to pass through a more resistive path than a less resistive path. This is wrong because the $e^--e^-$ interaction is "usually" negligible, and in any case do not account for the observed phenomena.

Instead I think the answer should come out of setting up a Boltzmann transport equation for the (quasi)electrons, considering the transient period of time. In other words, the density of electrons $f$ satisfy an equation of the type $\frac{df}{d t}=\frac{\partial f}{\partial t} \big|_{\text{scattering}} + \frac{\partial f}{\partial t} \big|_{\text{drift}} + \frac{\partial f}{\partial t} \big|_{\vec E \text{ field}}$.

$f$ depends on the position, time and is satisfied for each state $\vec k$. In the transient period of time, $\frac{df}{dt} \neq 0$, but after a short time, when the steady-state is reached, it is worth $0$.

To solve the equation and give an accurate answer, several assumptions have to be made. The first it to make clear whether we're dealing with a metal or a semiconductor. Then, some assumptions that reduces the range of validity of the analysis, such as making the relaxation time approximation that greatly simplifies the scattering (or collision) term. See the book of Ziman "Principles of the theory of solids", around page 215 for such a treatment.

An important and relevant point to note is that in metals, current is not due to slowly (drift velocity of order of $1\mathrm{cm}/\mathrm{s}$) moving electrons (this arises from the now obsolete Drude's model that many, many, many people still take way too seriously and would defend to death). Instead, current is mainly caused by the few electrons that have a speed near Fermi velocity.

So my current unfortunately not rigorous answer is that the electrons are taking all possible path they can, but the electrons responsible for the current (the few ones at speeds roughly equal to the Fermi speed) are getting scattered by impurities, grain boundaries, physical boundaries, phonons (and not so much with other electrons). This yields what we observe as resistance. So it is not that the electrons are avoiding the path with high resistances, it's that they do take it but they get affected in such a way that the resulting current is small. I emphasize once more: these electrons are few, move very fast (Fermi speed, i.e. about $10^6 \mathrm{m}/\mathrm{s}$ and for the most part, do not interact significantly with each other. Screening is a thing that many people here have forgotten.

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    $\begingroup$ The solid state physics may explain how resistance in general works; but since resistance of the wire leading to a resitor in a typical circuit is low (as low as the wires leading to other paths from an assumed junction in the circuit), this answer does not explain why electrons are taking a different route at the wire junction. I still think it's basically a capacitor effect of backed-up electrons. All wires are little capacitors (noticable at high frequencies) and can take only so much charge before the resulting electric field annihilates the field created by the potential difference. $\endgroup$ – Peter A. Schneider Nov 1 '18 at 10:30
  • $\begingroup$ @PeterA.Schneider I think this should come out of the BTE, which describes the electron density everywhere, at the junction included. By analyzing what happens with varying every terms of that equation, one should see the impact on the electron distribution. I am not really convinced that this would still miss why the electrons are taking a particular path or not. I think this should clear things up. $\endgroup$ – thermomagnetic condensed boson Nov 1 '18 at 11:32

An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current. This is what the basic idea of 'Electric Currents in Conductors' is and this is apparently known to you. In nature, free charged particles do exist like in upper strata of atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules a gram of water, for example, contains approximately $ 10^{22} $ molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied. In other materials, notably metals, according to the Drude-Lorentz Electron-sea theory, some electrons are practically free to move within the bulk material.

Resistance to electrical flow is due to the fact that when charge is given to a resistance, it remains stationary. In case of conductor, it is delocalised so it gets displaced and spread evenly on the surface, so note this carefully: in a conductor, charge flows mostly on the surface itself. This requires, undoubtedly some potential difference across the ends of the conductor but very less in magnitude. So, another fundamental rule/observation of universe is that "any dynamical process occurs in the path that requires least energy expense".

The most general and fundamental formula for Joule heating is: $$ {\displaystyle P=(V_{A}-V_{B})I} $$


$P$ is the power (energy per unit time) converted from electrical energy to thermal energy,

$I$ is the current travelling through the resistor or other element,

${\displaystyle V_{A}-V_{B}}$ is the voltage drop across the element.

The explanation of this formula (P=VI) is:

(Energy dissipated per unit time) = (energy dissipated per charge passing through resistor) × (charge passing through resistor per unit time)

When Ohm's law is also applicable, the formula can be written in other equivalent forms: $${\displaystyle P=IV=I^{2}R=V^{2}/R}$$

When current varies, as it does in AC circuits,

$${\displaystyle P(t)=U(t)I(t)}$$

where $t$ is time and $P$ is the instantaneous power being converted from electrical energy to heat. Far more often, the average power is of more interest than the instantaneous power:

$${\displaystyle P_{avg}=U_{\text{rms}}I_{\text{rms}}=I_{\text{rms}}^{2}R=U_{\text{rms}}^{2}/R}$$

where "avg" denotes average (mean) over one or more cycles, and "rms" denotes root mean square.

These formulas are valid for an ideal resistor, with zero reactance. If the reactance is nonzero, the formulas are modified:$${\displaystyle P_{avg}=U_{\text{rms}}I_{\text{rms}}\cos \phi =I_{\text{rms}}^{2}\operatorname {Re} (Z)=U_{\text{rms}}^{2}\operatorname {Re} (Y^{*})}$$

where $\phi$ is the phase difference between current and voltage,$Re$ means real part, $Z$ is the complex impedance, and Y* is the complex conjugate of the admittance (equal to $1/Z*$).

So, this shows how energy inefficient is electric flow through a resistance under an applied potential.

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    $\begingroup$ Please don't use all-caps for emphasis. Use italics by surrounding your content with _ characters instead. $\endgroup$ – user191954 Oct 28 '18 at 16:48

"Current flows through a path with no resistance" or "current flows through the path with least resistance" is a common misconception in electronics. In reality current flows though all paths, and the current in each path is proportional to that path's conductance.

If you apply a voltage V to a resistance R, the current I=V/R will flow through it, regardless of other available paths. In reality, you will have a hard time providing a path with strictly no resistance, or applying any significant voltage across a path those resistance is very low. In the end however, you will end up applying some voltage, at which point the Ohm's law will define the current in each path.

  • $\begingroup$ This is the proper answer. The question is most likey based on a misonception which is best dealt with by carefully laying out and systematically applying the basic principles. $\endgroup$ – Peter A. Schneider Nov 1 '18 at 11:10

Okay, so we know that if a voltage is applied over a resistor with resistance R, then V/R amps with pass through the resistor. The problem is, what happens when R is zero? We have infinite current?

for the purposes of this example, assume that, when I say short circuit I mean "extrmely low resistance path." When I say infinite currrent, I mean extremely high current, and when I say no current, I mean basically no current.

Basically, yes. In a perfect world, if you shorted out a resistor that was connected to a perfect power supply, nothing would happen. The voltage across the perfect power supply (and therefore the resistor) would be unchanged, and a truly ludicrous amount of current would flow through the short circuit, while a normal amount of current would flow through the resistor. However, we do not live in a perfect world, and any real power supply will have a limited amount of current.

As the power supply loses its ability to supply the current the system is demanding (infinite), the voltage across the resistor will no longer be constant and will decrease to ~0. Since the voltage has dropped to zero, no current will pass through the resistor.

To put this another way, perhaps more clearly, there is no arbitrary rule that says a shorted resistor cannot have current pass through it, but the voltage across a resistor is proportional to the current, and the voltage across a short circuit is defined to be zero. Trying to apply a voltage to a short circuit will do nothing, it will simply short out whatever it touches.

  • $\begingroup$ I wonder if such a situation could actually happen in practice in a superconductor circuit (at least to the limit of what current that superconductor can take); there might be extremely high currents on the "short-circuit" without damaging either the wire or the power supply, but then again, the short-circuit could have literally zero resistance (until the current rises too high). But I suppose there's little point in applying usual circuit logic ("everything has some resistance") to a superconductor :P $\endgroup$ – Luaan Oct 31 '18 at 10:17

Let's assume that the resistor and the wire around the resistor are part of a circuit with a battery and a switch.

Before the switch is closed, all battery voltage drops on the switch and all electric field is concentrated between the terminals of the switch, i.e., there is no electric field anywhere else in the circuit. The field across the switch is created by opposite charges on the switch terminals, which represent a small capacitor.

So, when the switch is closed, the initial voltage across the resistor is zero. As the the capacitance of the closed switch is discharged, the voltage and electric field across the switch decrease, while the voltage and electric field across the rest of the circuit increase, causing the current to flow.

Given a uniform initial field distribution, the current will flow faster where the resistance is smaller and slower where the resistance is greater. As a result, there will be a build-ups of opposite charges around the sections of the circuit with high resistance. This build-ups will cause redistribution of the initially uniform field, so that the field is concentrated in the sections with higher resistance, which will speed up the current through those sections, equalizing it with the current through the sections with low resistance.

Since the resistor in question has the low resistance path around it, there won't be any significant charge build-up and no significant field or voltage across the resistor, so the current through the resistor, according the Ohm's law, will be small in comparison with the current through the wire around it.

In summary, the current does not flow through the resistor with an alternative low resistance path, because there is no voltage across the resistor to push it through.


Let's suppose that a single battery is connected with a wire, which does not have resistance. Electrons will start to flow , in reality, with a wire with resistance, a potential difference would be generated across it. The current would build up until the potential difference is equal to the voltage of the battery. In the case in which potential difference is not created by the wire because there is no resistivity, the potential difference across will immediately become equal to that of the battery.


I think conceiving of it as the current choosing to flow along the path of lesser resistance rather than the path of greater resistance is a misleading & complicating way of conceiving of it. If a resistor of such & such a value is placed across a source of EMF a certain current given by V/R will flow. If a different resistor is placed across the EMF, a different current will flow. If both the resistors are placed across the EMF simultaneously, then each resistor will simply conduct the current it would have done had the other been absent.

This argument assumes a perfect EMF source for simplicity; but it doesn't matter, because the effect of the being real rather than theoretically perfect of the EMF source is that the voltage across the resistors will fall slightly; but the situation is exactly the same as were you simply considering a perfect EMF source at the new lower voltage.

If the total resistance loading the real EMF source is very much less than it's internal resistance, to the degree that the source is supplying very nearly its closed-circuit current, then the voltage across the parallel load resistors will be a tiny fraction of the source's open-circuit voltage; but it's still the same as were you considering a perfect EMF source at _that _ tiny voltage: each resistor has the current flowing through it that it would have were it alone, at that voltage.

  • $\begingroup$ What happens when the EMF can't supply enough current to get a full voltage drop on each resistor? $\endgroup$ – JMac Oct 29 '18 at 11:01
  • $\begingroup$ This answer ignores superconductors which are a reality nowadays. The 1st paragraph is wrong because it does not apply to superconductors. $\endgroup$ – thermomagnetic condensed boson Oct 30 '18 at 13:06
  • $\begingroup$ I believe this is the correct answer to the main question. The current is just a function of resistance and potential difference and does not change due to other paths in the circuit. (If the current through other paths, e.g. short circuits, overloads the voltage source then the voltage drops which is the single reason for the current through our original resisitor to drop as well.) Constant voltage + constant resistance -> constant current. $\endgroup$ – Peter A. Schneider Oct 30 '18 at 16:03
  • $\begingroup$ The second question is vague ("how does the current know..."). One may understand it as asking for the mechanism of electric resistance, but it is completely unrelated to the misconception underlying the first question. The misconception is the assumption "current [doesn't] pass through a resistance if there is another path without resistance": It does continue to pass, completely unfazed by other paths (unless you take the voltage away)! $\endgroup$ – Peter A. Schneider Oct 30 '18 at 16:07

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