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In a simple circuit which consists of a battery and one resistor, why do electrons lose all their potential energy across this one resistor regardless of the magnitude of the voltage or the resistance? Why is the voltage drop of this one resistor equal the voltage of the battery?

If there is more than just one resistor in series then in the first resistor happens a voltage drop which is not equal to the voltage of the battery. The electrons don't lose all their potential energy across the first voltage. it seems that the resistor knows that there is another resistor after it so it doesn't eat all the electrons' potential and leaves some potential for the next one. Why does this happen?

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    $\begingroup$ Don't forget about what happens to the current $\endgroup$ – BioPhysicist Jul 6 '19 at 15:04
  • $\begingroup$ The motivation for this question isn't clear to me. The change in potential energy of an electron, in moving from the negative terminal to the positive terminal of battery, is determined by the voltage across the battery isn't it? If an electron arrives at the positive terminal from the negative terminal, does the value of the resistance of the path taken change this fact? If the electron is at the junction of two series resistors, it hasn't arrived at the positive terminal, correct? Also, I don't understand what "eating potential" means. $\endgroup$ – Alfred Centauri Jul 6 '19 at 22:23
  • $\begingroup$ @AlfredCentauri The OP is wondering how placing another resistor in the circuit then causes the potential drop across the first one to change. $\endgroup$ – BioPhysicist Jul 7 '19 at 2:53
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It has to do with the definition of electrical potential, or voltage. The potential difference $V$ between two points is defined as the work per unit charge required to move the charge between the two points with units of Joules/coulomb.

As charge Q moves through resistance connected to the terminals of a battery the battery does work of $QV$ to move the charge between the terminals against the electrical resistance of the circuit, and the charge loses potential energy in the form of heat dissipated in the resistors.

For the same voltage across the battery terminals$^1$, the same amount of work will be done and the same loss of potential energy will occur, regardless of the number of resistors connected in series across the terminals. If you have multiple resistors in series, the same total loss in potential energy will simply be divided up among the resistors in proportion to the voltage drop (drop in potential) across each resistor.

The above said, the amount of resistance between the terminals does affect the rate of loss of potential energy for a given voltage (rate of work done per unit charge, or power). The greater the resistance, the lower the rate .

you said "For the same voltage across the battery terminals, the same amount of work will be done and the same loss of potential energy will occur, regardless of the number of resistors connected in series across the terminals" I know that this is true but why how can this be true ?

Maybe it will help if I give you a gravitational potential energy analogy. You know that the difference in gravitational potential between two points depends only on gravitational potential (gravitational potential energy per unit mass),or $gh$ where $h$ is the vertical distance between the points. The difference in electrical potential energy depends only on the electrical potential, $V$, between the two points. Let the atmospheric air consist of air "resistors" between two vertically separated points. Let these air "resistors" be analogous to our electrical resistors.

Now Let's say an object is dropped from some altitude above the surface of the earth greater than a height $h$ and it encounters air resistance. By the time it reaches a height $h$ let's say it has reached its terminal velocity due to air friction and its velocity becomes constant, like the average drift velocity of charge that defines current. Let the acceleration due to gravity be a constant, $g$. At the height $h$ its gravitational potential energy is $mgh$. Think of the air between this height and the ground as now constituting a series of resistors. For each fraction of the total height the object falls it encounters an equal fraction of the total resistance, until it reaches the ground and all of its potential energy is lost (to air friction and kinetic energy(which later dissipates into heating up the ground and sound)).

As an example, call the total air resistance between $h$ and the ground $R_{air}$. Let it consist of two resistors, $\frac{R_{air}}{4}$ representing the resistance falling a height $\frac{h}{4}$ and the second resistance be $\frac{3R_{air}}{4}$ for the final falling height of $\frac{3h}{4}$.

In this example the loss of potential energy in the first air "resistor" is $\frac{mgh}{4}$ and the loss of potential energy in the second "resistor" is $\frac{3mgh}{4}$ for a total potential energy loss of $mgh$. The total loss of potential energy is dissipated as heat due to air friction.

How does the first air "resistor" know that there is a second air resistor so that it does not "eat up" all the gravitational potential energy? It doesn't, nor does it need to. The loss of gravitational potential energy depends only on the difference in gravitational potential between two points and it doesn't matter what is between the two points. If there were no air, the loss in gravitational potential energy would be the same. The difference is that loss will equal the increase in kinetic energy of our object rather than be dissipated as heat due to air resistance.

Hope this helps.

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    $\begingroup$ I think the OP wants to understand why this happens. The OP seems to understand that the voltage will be divided between the resistors. $\endgroup$ – BioPhysicist Jul 6 '19 at 16:13
  • $\begingroup$ @Aaron Stevens Thing is Aaron that’s what I thought I was doing. If the total loss of potential energy per unit charge depends only on potential difference between the battery terminals, then it doesn’t matter what the resistance is between the terminals. What does matter is the rate of loss in potential energy (power). I think I’ll point that out $\endgroup$ – Bob D Jul 6 '19 at 16:26
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    $\begingroup$ I guess my interpretation of the question is that the OP wants to know the mechanism behind why it charges between the cases rather than just an argument of why it must be true. The OP seems to know it's truth, so I don't think they need more convincing of that fact. $\endgroup$ – BioPhysicist Jul 6 '19 at 16:56
  • $\begingroup$ @Bod D you said "For the same voltage across the battery terminals, the same amount of work will be done and the same loss of potential energy will occur, regardless of the number of resistors connected in series across the terminals" I know that this is true but why how can this be true ? $\endgroup$ – Mahmoud Amin Jul 6 '19 at 17:36
  • $\begingroup$ @MahmoudAmin I am preparing a revision to may answer that may help. Please stand by. $\endgroup$ – Bob D Jul 6 '19 at 18:08
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Formally, this is called Kirchoff’s current and voltage laws.

But you asked why this happens. What mechanism is it that make sure that the current behaves this way?

Imagine that current goes through the first resistor of two, and then stops before going through the second. This means that charge is building up at that intermediate point. That means the voltage there, hence the voltage across the second resistor, will increase. That voltage will drive more and more current (Ohm’s law) through that second resistor.

This process automatically stabilizes, with the charge on that point not changing, when the current in equals the current out: the current in both resistors is the same.

At that point, the intermediate connection has exactly the right voltage for everything to balance: charges lose the right amount of energy in the first and second resistor for everything to balance.

For most circuits, this happens so incredibly fast that we just consider the adjustment to be instantaneous. It’s only in very special cases (fast signals, very high resistance or large wiring) that it has to be considered at all. Usually Kirchoff’s Laws work so automatically well that we don’t think about it.

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The voltage drop across an ohmic resistor is given by Ohm's law:

$$V=IR$$

Therefore, for a set resistance the potential drop is determined by the current. So really, your question about adding a resistor becomes Why does adding a resistor to the circuit decrease the current flowing through the circuit?

For resistors in series, the total resistance of the circuit increases as you add more resistors. This makes sense if you thinking of adding resistors in series as just adding a longer section of the circuit that resists the flow of current$^*$ More resistance means less current, which means less potential drop across each resistor.


$^*$ Note that for resistors in parallel the argument is not the same. It's not true in general that adding a resistor to a circuit increases the overall resistance in the circuit. This answer is only concerned with your specific series example.

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It's simply due to energy conservation. What does a battery? It uses its own internal chemical stored energy to create potential difference across its terminals and when it's terminals connected to the circuit it potential energy goes to electrons and if there is a resistor attached to the circuit. then first what resistors do in the circuit? They dissipate the energy of electron which comes from a potential difference of battery terminals. Hence if there only loss of energy can happen by a resistor is that circuit is isolated then energy provided battery is equal to the energy dissipated by resistor for which potential drop across the total resistance is equal to the potential difference across battery terminals. It's nothing but KIRCHOFF'S LAW.

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    $\begingroup$ The OP wants to understand why this happens. Why does the energy dissipated changes from one case to the next. $\endgroup$ – BioPhysicist Jul 6 '19 at 15:38

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