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Could you please help me develop the intuition on how voltage divider works.

Let me describe the problem with the following example circuit:

enter image description here

As this circuit starts working, the electrons start moving from Vin towards the first resistor having 12 V.

After the electrons have left the first resistor $Z_1$ ($R = 2\ \Omega$) they have lost 4 Volts of energy (according to the voltage divider formula): $$ V_1 = \frac{Z_1}{Z_1+Z_2}\; V_\text{s}.$$

However, if the resistance $Z_2$ had been different, voltage drop at $Z_1$ would be different as well (e.g. if $Z_2$ resistance had been 10 $\Omega$, then voltage drop at $Z_1$ would be 2 V)

So, voltage drop at $Z_1$ depends on resistance of $Z_2$, even while the electrons coming to $Z_1$ do not even ‘know’ there is $Z_2$ ahead (Please let me know if they do know).

So, the question is: how do the electrons ‘know’ how much volts they should drop at $Z_1$, since they have not been at $Z_2$ and cannot know its resistance.

I cannot stop thinking that voltage drop at a certain resistor should depend exclusively on the resistor’s qualities, not other resistors at this circuit.

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  • $\begingroup$ Writing this as a comment, might be wrong, but here's my opinion: for a moment forget that some other device connected b/w Z1 and Z2 by (ideal) wire (having internal resistance 0) is absent. Then we see that as current density j =σE, and E=-grad(V) where -grad(V) refers to the total potential drop per unit length over the wire(from point V_in to ground) so j is affected by σ of the whole wire which IS affected by both Z1 and Z2, so in a sense the electrons drifting ANYWHERE in the wire 'know' that Z2 is ahead as it constitutes the current which is affected by both Z1 and Z2... $\endgroup$ – Aditya Apr 11 at 13:42
  • $\begingroup$ As for the V_out part, by the same equation E=- grad(V) integration of this with limits from V_in to the point where V_out is connected to the main wire yields V_out and so when the (ideal) wire is connected at that point that new wire is almost instantaneously at a single potential of V_out over its entirety. $\endgroup$ – Aditya Apr 11 at 13:48
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    $\begingroup$ Electrons don't "have" volts; the potential difference is a property of the place in the circuit, like altitude is a property of terrain. $\endgroup$ – chrylis -cautiouslyoptimistic- Apr 12 at 1:03
  • $\begingroup$ Point of order: the electrons are going the other direction. $\endgroup$ – imallett Apr 12 at 3:13
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When you connect the voltage (potential difference) across the potential divider, you set up an electric field in and around the resistors and connecting wires. This exerts forces on the charges in the resistors and the wires, in particular on the 'free' electrons, which move so that, within a small fraction of a second, charge is distributed so that the voltages across the two resistors are in accordance with Ohm's law, and hence with your formula.

If this seems both complicated and vague, you are right. I believe that the complexity is in the Physics (which is usually oversimplified); the vagueness is no doubt down to me.

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So, voltage drop at 𝑍1 depends on resistance of 𝑍2, even while the electrons coming to 𝑍1 do not even ‘know’ there is 𝑍2 ahead (Please let me know if they do know).

Well, electrons are not sentient so they don’t know anything. But in the sense that electrons follow the laws of physics and the laws of physics have a mechanism for passing the necessary information about Z2 to all the electrons in the circuit, in that sense the electrons certainly do “know” about Z2. (By the way, it is almost never helpful to think about electrons in circuit theory, the focus should always be voltage and current)

So, the question is: how do the electrons ‘know’ how much volts they should drop at 𝑍1, since they have not been at 𝑍2 and cannot know its resistance.

The information about Z2 is passed backwards to the electrons going through Z1 by the electromagnetic field. After a brief initial transient, there is a surface charge distribution on the components and at the interface between materials of different resistivity.

These surface charges establish an electric field (and the currents establish magnetic fields) which communicates the information about Z2 back to the electrons at Z1. Those electrons merely respond to the local EM field. That field depends on the value of Z2.

The surface charge distribution is the key for understanding questions like this which straddle the boundary between circuit theory and classical electromagnetic theory.

I cannot stop thinking that voltage drop at a certain resistor should depend exclusively on the resistor’s qualities, not other resistors at this circuit.

This is incorrect. You will need to stop thinking it indeed.

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$Z_2$ incfluences the electron flow - the current - in the entire system. And the voltage drop over $Z_1$ depends on the current through it.

  • Imagine having a hole instead of $Z_2$ (corresponding to a very, very large $R_2\to \infty$). All current will flow between $V_{in}$ and $V_{out}$.

  • Now imagine having a wire with no resistor $Z_2$ (corresponding to a very, very small $R_2\approx 0$). The current now divides, and some flow to ground (through $Z_2$), while some flows to $V_{out}$.

Roughly speaking, in the latter case, you'll allow for more current to flow to more "outtakes". The current $I$ that flows from $V_{in}$ and through $Z_1$ will thus increase. Ohm's law then tells us that a different current through a resistor means a different voltage drop across it:

$$V=RI$$

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So, the question is: how do the electrons ‘know’ how much volts they should drop at $Z_1$, since they have not been at $Z_2$ and cannot know its resistance.

Electrons do not "know how much volts they should drop at $Z_1$". Basically, they behave based on the electrical force that drives them and the electrical resistance that retards their movement.

The total voltage drop across the series combination of $Z_1$ and $Z_2$ is fixed at $V_{in}$. Since $Z_1$ and $Z_2$ are in series, the current in both is the same and is, by Ohms law

$$I=\frac{V_{in}}{(Z_{1}+Z_{2})}$$

The voltage drop across each resistor is then, again by Ohms law, the current times each resistance. Or, $V_{1}=IZ_{1}$ and $V_{2}=IZ_{2}$. Finally

$$V_{out}=V_{2}=IZ_{2}=V_{in}\frac{Z_{2}}{(Z_{1}+Z_{2})}$$

The fraction of the total voltage $V_{in}$ that appears at $V_{out}$ depends on ratio of $Z_{2}$ to the total resistance. That essentially is the voltage divider rule.

I cannot stop thinking that Voltage drop at a certain resistor should depend exclusively on the resistor’s qualities, not other resistors at this circuit.

The voltage drop across each resistor cannot depend exclusively on the value of that resistance, since the current in that resistor depends on total resistance in the series circuit, and not just that resistor.

Hope this helps.

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how do the electrons ‘know’ how much volts they should drop at Z1, since they have not been at Z2 and cannot know its resistance.

Physicists don't like analogy, but here, in this case, there is one that can clearly show the answer to your question.

In electrical circuits, resistors can be analogous to mechanical springs in various situations.

<enter image description here

How do spirals "know" how many meters to stretch?

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  • $\begingroup$ Well, $this$ physicist $does$ like analogies, provided their limitations are recognised. Perhaps your spring analogy could be made more cogent by pointing out that if the ends of the spring combination are suddenly moved outwards, the disturbances travel non-instantaneously (as waves), so the spring lengths take a finite time to become what the textbooks say they do. $\endgroup$ – Philip Wood Apr 12 at 10:25
  • $\begingroup$ @PhilipWood - I agree. $\endgroup$ – João Bosco Apr 12 at 13:36
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The electrons take their time to converge to stable currents described by Ohm’s law. During that time the currents can vary significantly. The study of what happens in the first moments after energizing a circuit is of great practical importance, large industrial machines have specialized control systems to start the machine correctly (typically by gradually increasing the voltage), avoiding undesired side effects such as it catching fire.

In the first nanoseconds after connecting a battery there’s an excess of electrons on the battery’s negative terminal. These electrons, being repulsed by the Coulomb force, eagerly rush into the circuit. Upon encountering resistance, they slow down and get rear-ended by electrons that follow them. This causes electrons to get too close to each other and the repulsive force between them increases, forcing the ones in the rear to slow down and to spill into parallel routes, if any.

After some time the system stabilizes and the density of electrons at each point is just right so everything is in a dynamic equilibrium. At the battery’s negative terminal the electrons are packed more tightly, which causes them to possess energy like a compressed spring does (and the amount of that energy per one coulomb worth of electrons is the definition of electric potential), while this density drops gradually towards the positive terminal. Should you add another resistor between any two points of the circuit, the electrons will likewise rush into it (from both ends), causing the entire circuit to rearrange its potential levels until the system becomes steady again.

You could imagine connecting a series of pipes to a water supply, the maths are very similar.

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