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How is it possible that in scenario 1 resistor R1 converts 5V of energy to heat and in scenario 2 the same resistor R1 converts 2.5V of energy to heat?

The resistor R1 in both scenarios are exacly the same, I don't seem to grasp what happens on the physical level to the electrons how they "know" when to dump all their energy at once in resistor R1 in scenario 1 or to divide the energy they have proportionally over the resistors R1 and R2 in scenario 2.

I have read an explenation here, but it's still not clear. How is it possible that the voltage is different at point A going through the same amount of resistance? I would expect a linear relation between voltage and resistance.

Could someone explain what physically happens that causes this?

schematic

simulate this circuit – Schematic created using CircuitLab

Edit

Is my understanding right? According to here

The resistance drops liniar along the path.

In scenario 1 at the start of R1 the resistance the current encounters is 100\$\Omega\$ and at the end of R1 it encounters 0\$\Omega\$.

In scenario 2 at the start of R1 the resistance the current encounters is 200\$\Omega\$ and at the end of R1 it encounters 100\$\Omega\$.

If this is true, why does point A in scenario 2 have more volt left if it has encountered a higher resistance of 200\$\Omega\$ at the beginning opposed to scenario 1 where it encountered 100\$\Omega\$ at the beginning?

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Ronald
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    If you have a piece of pipe that is in an L shape and you put water in at the top if comes out the bottom. If you have u shaped piece of pipe and pour water in at one end you end up with the water staying in the pipe. But the start of the pipe is the same so how does the water you're adding know that it's supposed to stop and not keep flowing downwards? – Andrew Feb 09 '17 at 11:41
  • @Andrew it took a while for me to understand your comment. Well said. – User323693 Feb 09 '17 at 11:44
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    "5V" is not a measure of energy. – Olin Lathrop Feb 09 '17 at 11:52
  • @andrew I know the pressure is needed to keep the current flowing, but is does not explain why the energy at point A is different – Ronald Feb 09 '17 at 12:06
  • @Olin Lathrop I thought the voltage is a measurement of energy because the definition of a volt is Jouls per Coulomb and a Joul is energy right? Or am I getting it wrong? – Ronald Feb 09 '17 at 12:06
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    @Ronald so Coulomb == 1? –  Feb 09 '17 at 12:14
  • @GregoryKornblum 1 Coulomb = equivalent charge of 6.25x10^18 electrons right? – Ronald Feb 09 '17 at 12:59
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    E (Joules) = V(Volts) * Q(charge) and you are correct to say Volts are Joules per Coulomb **but** Joules per Coulomb (Volts) is **not** Joules (energy) – JIm Dearden Feb 09 '17 at 12:59
  • OK, someone wrote it before me. Coulomb is not just 1, it has it's meaning. So joules/coulomb is not equal to joules, that's it. Not equal in any meaning. –  Feb 09 '17 at 13:01
  • Volts are Joules per Coulomb, a single Coulomb is an equivalent charge of `6.25x10^18 electrons`, so the formula is when flatten `Volts = Joules / (charge of 6.25x10^18 electrons)` is this correct? but both are measurements of energy right? – Ronald Feb 09 '17 at 13:07
  • See also http://electronics.stackexchange.com/questions/245610/is-voltage-the-speed-of-electrons/245621#245621 - this is a difficult topic to explain non-mathematically. – pjc50 Feb 09 '17 at 14:24
  • So if I have read correctly coulombs could be an analogy to kg? – Ronald Feb 09 '17 at 14:34
  • 1 Joule of Work (Energy) is needed to move 1 Coulomb of charge across a potential of 1 Volt – user28910 Feb 09 '17 at 16:56
  • The SI units that make up a volt are $$\frac{kgm^{2}}{As^{3}}$$ The SI units that make up a joule are $$\frac{kgm^{2}}{s^{2}}$$ where 'kg' is kilograms, 'm' is meters, 'A' is amperes, and 's' is seconds. That is the difference between voltage and energy. They are not the same units - you're missing a *1/(A*s), which means you must divide energy, in joules, by charge, in coulombs (A*s), to get voltage – DerStrom8 Feb 09 '17 at 19:42
  • So how would the calculation for the voltage in scenario 1 look like in the formula \$V=\frac{kg \times m^{2}}{A \times s^{3}}\$ Could you fill in the variables and explain? – Ronald Feb 09 '17 at 22:03

4 Answers4

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As you may anticipate, the electrons don't actually know anything. Voltage is actually a battery's difference of potentials of its poles to drive electrons. You may think it as its hydraulic counterpart: pressure. Such that pressure drives the water through tight and wide pipes at a certain flow rate, voltage drives electrons through resistance with a certain current.

If you are interested in a more "scientific" explanation, this is what actually happens:

  1. You connect the battery, complete the circuit
  2. The electric field starts to propagate inside the wires, exerting force on electrons while doing so
  3. Some materials show more resistance to electrons and thus slow them down
  4. The slowed down electrons apply their own electric field behind them, taking away some of their potential to move on.
C K
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  • Voltage is not capacity, no matter how do you explain it;) Voltage is difference of potentials. –  Feb 09 '17 at 12:04
  • @GregoryKornblum so, may I say "battery's potential to drive electrons"? :) – C K Feb 09 '17 at 12:05
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    Nope, only difference of potentials :D In engineering words are more important than even in law. Lawyer's mistake will only put someone in jail, engineer's mistake will damage a million dollar equipment or kill someone. –  Feb 09 '17 at 12:09
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How is it possible that the voltage is different at point A going through the same amount of resistance?

In scenario 1 it is 0 volts (as you have labelled it). In scenario 2 is is half the voltage applied i.e. 2.5 volts.

I would expect a linear relation between voltage and resistance.

Not when potential dividers are used unless both resistors change value proportionally.

Ohms law, I = V/R so, in scenario 1 the current is 5/100 = 50 mA. In scenario 2, I = 5/(100 + 100) = 25 mA.

Ohms law again: V = IR so the voltage across R2 is 25 mA x 100 ohms = 2.5 volts.

Andy aka
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The electric fields do the "knowing", do the "exploring", do the "urging".

Electrons explore all possible paths, to get back home to the battery's other terminal. Electrons will go meters off to the side, just because that is another path back from whence they came, even if only 1 electron/second explores that path. Why do electronics do this exploring? because the electric fields urge/push/drive that exploration.

Our circuits often include huge volumes of air, and we think our circuit is just the copper wiring and the resistors and capacitors and ICs and PCBs. But the air is there, extremely high resistance, so we ignore the air.

In your two circuits, the electric fields encounter different length paths, and the volts/meter is different. Most importantly, the volts/resistor is different and we use Ohms Law to handle that situation, while we ignore the air.

analogsystemsrf
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I have found an explanation what I was looking for here, here and here. I think the main thing I needed to realize is as @OlinLathrop and @GregoryKornblum mentioned in a comment, that voltage is not energy itself, voltage is just a difference in "height" so to speak and it tells nothing about the electron itself. It only tells how much energy the electron would have gained if it rolled down the hill.

In the following illustration the voltage is not the ball at the top of the hill, but the voltage is the top of the hill and the ball can be used to transform the energy from potential energy to heat/light or in this example to kinetic energy if it hits the ground.

enter image description here

And as to the diagram here, the 2.5V is there because its not at the ground level yet (0V) and the real energy can be seen in the current because that is the energy carrier and if the current is slowed it loses potential energy. I think the resistance in this situation can be seen as if you drop the ball from the hill and blowing wind agains it upwards (creating drag) it slows down and when it arrives at the ground it has less kenetic energy because its speed was reduced by drag and thus lost gravitational potential energy.

schematic

simulate this circuit – Schematic created using CircuitLab

Correct me if my understanding is still incorrect.

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