By Paul Michaels

2011-02-22 18:51:37 8 Comments

Electricity takes the path of least resistance!

Is this statement correct?

If so, why is it the case? If there are two paths available, and one, for example, has a resistor, why would the current run through the other path only, and not both?


@A.H. van Herp 2012-07-18 11:03:29

Electricity takes the path of least resistance. Is this statement correct?


People often don't understand what the shortest path means. If I have a lightning rod connected to a wire which has a slight bend in it people still believe that the current will follow the wire to Earth and are astonished when the voltage jumps to a tree 15 feet away. We are talking here about millions of volts and hundreds of thousands of amperes. At these frequencies the slight bend will present an enormous impedance to the current and the tree is far more attractive in spite of the distance.

BTW: when a lightning rod gets struck it is not doing its job. It has a point and it is known that electrons will gather around this point and since the lightning is negative like will repulse like. That is the principle of the workings of a lightning rod.

@Vorac 2012-09-11 10:50:53

I ... do not agree about the lightning rods...

@V .Kiran Bharadwaj 2017-12-09 10:18:30

The statement is not true. Electricity passes through all possible paths whether the resistance is high or low. Just the difference is that the current is more in which resistance is less. This is a direct implication of Ohm's law.

@Anuj Tanwar 2017-12-09 07:53:25

Actually, the current flows in every wire connected to its path. There can be difference in amount of current flowing through different wires.

That's true for all case except when one wire connected in its path with no resistance or nothing (bulb, resistor) is connected to it. In this case current will flow only through this path leaving all others

@Yrogirg 2012-07-18 18:27:02

"Least resistance" can be interpreted as least heat generation. There might be such principle, at least I can show it for @Ted Bunn example so that the answer would be "yes". The largest difficulty in formulating extremal principles is specifying the constraints. I chose fixed current, because I don't see a way to fix voltage for the model in hand without fixing everything else.

In any case I think reformulating least resistance as least dissipation under certain constraints is a right direction.

What you have is two bulbs connected in parallel. Let's fix the overall current $I$ through the bulbs rather than voltage $U$. That is it is a case when you are to push a certain amount of electricity through the system. In this setting currents on the bulbs $I_1$ and $I_2$ would be to minimize heat dissipation:

$$ \begin{cases} I_1 + I_2 = I, \\ I_1^2 R_1 + I_2^2 R_2 \to \min \end{cases} $$

Using Lagrange multipliers:

$$ \begin{cases} I_1 + I_2 = I, \\ d \left[ I_1^2 R_1 + I_2^2 R_2 + \lambda (I_1 + I_2 - I) \right] = 0 \end{cases} $$

which leads to

$$I_1 R_1 - I_2 R_2 = 0$$

Thus having assumed the extremality of current distribution we arrived to the distribution which is in harmony with the Ohm's law. One can check that it correspond to the minimum of heat dissipation.

@Ron Maimon 2012-07-19 03:09:57

This is a nice answer, but least heat generation is not what people usually mean by the statement. They mean the wrong statement that others have interpreted it as.

@Corbs 2011-12-31 04:50:38

If you turn on the water at your sink it comes out the nozzle, not the pipe.(unless you have a leak) Or in the case of a rocket if you ignite the fuel it comes out of the opening. These all have the path of least resistance, if you have two different paths the flow of energy will go through both of them until one of the paths has too much resistance, then the flow of energy will go through only one path. The same basically applies to electrical circuits.

@Paul 2011-07-03 02:42:32

This statement is true and a direct consequence of the 5th Law of Thermodynamics, the Onsager Relations for which Lars Onsager of Yale received the Nobel Prize in 1968.

In an electrical circuit, for DC, current takes the path of least reisitance; For AC it takes the path of least inductance (impedance). So a pulse of voltage will cause the current distribution to be determined by path inductance and then finish with the distribution determined by resistance.

On a circuit board, this is critical as the ground plane insures the differences between these two distribution paths are minimal.

@Vorac 2012-09-11 10:47:22

-1 The statement is actually false and -1 inductance and impedance is not the same.

@Omega Centauri 2011-02-22 21:02:01

I suspect the statement was meant to be about electrical discharges via dielectric breakdowm. Such as lightening bolts etc. As such it has partial validity, in that say a tall tree is more likely to be hit than a short one. But the reality is that dielectric breakdown is a chaotic process, which is why lightening appears forked, rather than taking a straight pathway. Once you get ionization along a path, more current flows along it causing more ionization, and so on.

For simple circuits not relying on breakdown, its a simple matter of resistance/impedance, and the current will distribute itself among multiple paths as described above. But for breakdown situation, whatever path gets connected first often takes the entire current.

@Georg 2011-02-23 10:41:08

This expression "takes the path" depicts a process involving some selection (it only seemingly) eg electrical sparks. Your interpretation is th correct one, Omega +1

@Approximist 2011-02-22 20:09:42

No. The statement is not correct. Current will take any path that is available to it. Which means it can even take the path of leaking out of the wire into the surrounding air, which is seen as sparks when dielectric breakdown of air happens. What you intend to mean perhaps is why current distributes itself in the inverse ratio of resistances, given the same potential difference across different resistive elements.

Ohm's law $I=\frac{V}{R}$ would explain what you are asking. Given a common potential, the amount of current flowing through a resistive element is inversely proportional to the resistance. This would mean, and hopefully answer your question, that a path of lower resistance will have more current flowing through it and vice versa. (Normally the resistance of air is so high that current taking that path and leaking out of the cable is negligibly zero under normal circumstances.)

For a more thorough explanation, currents (and voltages) are distributed to minimize the total power dissipated as heat. This is a consequence of making the action of a disspative system stationary


Here W is the virtual work done by dissipative elements (resistance, capacitance, inductance etc) and L is the dissipation free dynamical system

For an alternative, this link explains how Ohm's Law corresponds to Fermat's Principle of least time.

@Ted Bunn 2011-02-22 19:02:44

It's not true. To see this, you can try an experiment with some batteries and light bulbs. Hook up two bulbs of different wattages (that is, with different resistances) in parallel with a single battery:

|                     |                   |
Battery              Bulb 1              Bulb 2
|                     |                   |

Both bulbs will light up, although with different brightnesses. That is, current is flowing through the one with more resistance as well as through the one with less resistance.

@oosterwal 2011-09-01 18:16:31

Can this be viewed by describing the bulb with less resistance as reaching a saturation point, after which the conductive paths become equally resistant? Perhaps "saturation point" is a misnomer and each bulb should be viewed as having infinitely many, infinitesimally small, saturation points but with different rates of saturation?

@dbrane 2011-02-22 19:02:32

The statement is correct if you interpret it to mean that there is a larger current in the path that has a lower resistance, when both paths have the same voltage across them. (This doesn't mean that the path with higher resistance has no current, just less current - as Ted Bunn's example shows)

You can understand this by thinking of the analagous situation of a a long pipe that diverges into two branches and re-converges back again. Suppose that the pipe is filled with water and there's a pressure difference (say using a pump) between the two extreme ends of the pipe. One of the branches is just like the rest of the pipe while the other branch is lined with, say, wheels that add to the resistance and make the water flow slower in that branch.

The pressure difference across both branches are the same (just like the voltage between two parallel electrical resistances is the same) but the water flows at a faster speed in the branch without the wheels, just like there's a larger current (rate of flow of electrons) in the path with lower resistance.

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