I was always taught that electricity flows through the path of least resistance.

The problem that I'm having is that I've seen/done this: Take 3 wires have one wire break off into the 2 other wire (like a fork in the road) and put electricity through the first single wire.

Now I take a multi-meter and test the end of each of the wires that forked from the original wire. Both wires show current/energy.

How can this be possible, this happens even if I use 2 wires with different conductivity.

The ending output might differ between the wires, but both wires are getting energy, even though one of them is more resistant.

Can someone please explain this to me?

  • 1
    $\begingroup$ You have always been taught wrongly. That statement is often made and has NEVER been correct. It is a misstatement of the true* position that, where there is more than one path current will flow in each path in INVERSE-PROPORTION to its resistance. So - for two paths A & B, if the resistance of B is 1/3rd of A then 3 times as much current flows in B compared to A. So, MORE current flow in paths of lower resistance, but not all current. $\endgroup$ May 11 '15 at 0:15
  • $\begingroup$ is there anyway to force the electricity down the path of less resistance without a switch or relay? $\endgroup$
    – huddie96
    May 11 '15 at 0:32
  • $\begingroup$ Essentially no, and yiu MUST get what is happening clear in your mind. IF the source is an ideal one (and many sources are close to ideal in this case) then the source does not "know" if there is 1 wire or 2 or 20 or 200. If the source is able to provide the total demand (asn it usually is in eg power supplies, mains electricity etc) then EVERY load circuit attached is independent. The "least resistance" circuit takes what it "needs" regardless of what else is connected. I = V/R for every path, regardless of others. Smaller R = larger I. The problems occur when people try to insist on ... $\endgroup$ May 11 '15 at 2:59
  • $\begingroup$ ... "models" which do not match this simple relationship. For given fixed source V and 3 paths. I1 = V/R1. I2 = V/R2. I3 = V/R3. There is NO interaction between paths if V is fixed. IF a heavy load vauses V to drop then all paths are affected BUT the ratio of I's is the same. || The answer to your question re "forcing" current is NO because "forcing" IS what switching does. IF something stops current flowing we call it a switch. A relay is a controlled switch. A transistor or SCR or TRIAC of IGBT or ... is a controlled switch. ... $\endgroup$ May 11 '15 at 3:04
  • $\begingroup$ .... | Remember - trying to force wrong 'models' onto the simple formula causes errors. | PER BRANCH I = V/R. | IF the branch is non linear then extra laws and rules may apply. Just as if I tell you how to get on a bus and what arrives is a motorcycle, or a tank, or a unicycle, if you have other devices then other rules apply. But for resistive loads I=V/R per branch. $\endgroup$ May 11 '15 at 3:06

Ohm's law states that current through a resistor is voltage over resistance. Resistors in parallel will see an equal voltage. This means that each resistor will get some current through them. The one with lower resistance will get more. But they will both conduct current and the total current through the circuit will be the sum of the current through each.

Given that you can work out that formula for resistance of 2 different resistors in parallel is:

$$\frac{1}{R_{tot}} = \frac1{R_1}+\frac{1}{R_2}$$

Otherwise to get a 5 $\Omega$ resistance from 2 10 $\Omega$ resistors would be much more finicky as under your assumption even the slightest change in resistance would cut off the slightly stronger resistor.

  • $\begingroup$ So if I had a metal rod and I attached the + of a battery to one side of it and a - to the other, there would be no way to divert the electricity from the middle of the rod to somewhere else leaving no electricity after the diversion point $\endgroup$
    – huddie96
    May 10 '15 at 22:21
  • $\begingroup$ @user3080828 Not quite: If you connected a wire to the middle of the rod, via a light bulb, and back to the negative terminal, you would still get some current through the light bulb. However, because (presumably) the resistance of the rod is so low, most of the current would still go through the rod, but a small amount of current would go through the light bulb. Using the formula above in ratchet freak's answer, you can see that if R1 is the resistance of the rod and R2 that of the light bulb, with R1 being so low compared to R2, it would get the majority of the current. $\endgroup$
    – jhabbott
    May 10 '15 at 22:59
  • $\begingroup$ Lets say I wrap copper wire (more conductive than metal) around the rod and connect the copper wire to rubber or something. Will more electricity travel to the rubber or the end of the rod $\endgroup$
    – huddie96
    May 10 '15 at 23:06
  • $\begingroup$ Or if I don't connect the copper to anything lets say $\endgroup$
    – huddie96
    May 10 '15 at 23:07
  • $\begingroup$ Comments aren't ideal for teaching this, watch this video to understand: youtube.com/watch?v=8UP4TLSFU9Q - I think what you are asking is how to make the current go one way - and the way to do that is to reduce the resistance of that path where you want current to flow and increase the resistance of the other paths. A great example of this is the hot-suits worn by cable repair technicians: youtube.com/watch?v=9tzga6qAaBA $\endgroup$
    – jhabbott
    May 11 '15 at 0:51

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