# Voltage - Energy drop

I'm having a hard time understanding the nature of voltage and am hoping you guys can help. The main issue is the concept of the voltage drop.

Take the following circuits : In regard to the first, we have a 5v supply, which from my understanding means that if you were able to enclose a coulomb of charge eminating from the negative terminal, you'd find that it has 5 joules of energy.

This means that each electron eminating from the negative terminal has :

• 5 / 6.24x10^18e = 0.8x10^18e joules of energy each

(relative to an electron entering the postive terminal)

Now, as the voltage across the battery is 5v and the voltage drop across the resistor is 5v, this means that an electron entering the resistor will have :

0.8x10^18e joules of energy

and one leaving must have 0 energy.

This leads me to a few questions :

1) Why is less energy lost going through R3 than R1? or rather, why does an electron have more energy after passing through R3 than an electron passing through R1

2) If the electron loses all its energy on exiting R1, why does it still end up moving to the positive terminal? If it's still moving to it, doesn't that mean it's got energy?

3) Why doesn't resistance value have an effect on the total energy loss of an electron passing through the circuit?

I understand how this can be explained through ohms law, and I understand that KVL tells us this must be the case, but I want to understand what is physically going on (what the electrons are doing), as opposed to what the equations say.

I'd also appreciate an answer without using an anolgy if possible, as anologies generally always lead me to more questions.

• Any reason the 5 V resulted in 9 Joules of energy? Aug 6 '15 at 21:08
• woops - I initially had a 9v battery :) (corrected) Aug 6 '15 at 21:12
• Less total resistance in circuit 2, therefore less current going through both resistors. Therefore less energy dissipation in either resistor in circuit 2. Aug 6 '15 at 21:15
• Taking the resistance to an extreme (infinite resistance) there are no electrons moving (through the wires in the loop) but you still have the same drop, the same potential, the same voltage (of course assuming an ideal battery) Aug 6 '15 at 21:17
• Your first energy calculation needs clarification. Either use parentheses or MathJax/LaTeX notation. What quantities are you dividing? Why don't you multiply $q\Delta V$? q = $1.602\times10^{-19} C$ and $\Delta V$ = 5 V. Aug 6 '15 at 22:26

In regard to the first, we have a 5v supply, which from my understanding means that if you were able to enclose a coulomb of charge eminating from the negative terminal, you'd find that it has 5 joules of energy.

This isn't the typical understanding. A 5V (ideal) supply (source) maintains a 5V potential difference across the terminals independent of the current through.

It follows that 5 Joules of work is done by the source on 1 coulomb of (positive) charge in moving through the source from the more negative terminal to the more positive terminal.

Conversely, 5 Joules of work is done on the source by 1 coulomb of charge moving through the source from the more positive terminal to the more negative terminal.

1) Why is less energy lost going through R3 than R1? or rather, why does an electron have more energy after passing through R3 than an electron passing through R1

There is 5V across R1 but only 2.5V across R3. Why? There twice the current through R1 than through R3.

If the electron loses all its energy on exiting R1

It doesn't; the (average) kinetic energy of the electrons entering is the same as the electrons exiting (the current entering is the same as the current exiting).

However, the exiting electrons have less potential energy.

Why doesn't resistance value have an effect on the total energy loss of an electron passing through the circuit?

The resistance has an effect on the power (rate at which work is done); the electrons must lose the same amount of potential energy in travelling through the resistor(s) (all else being equal) but the number of electrons per second is reduced when the resistance is increased.

• Ok - I seem to be getting confused with energy then. I thought that the potential energy of the electron was converted into kinetic energy. In that as the electrons build up at the negative terminal, a repulsion is built up that makes them want to move away. And that how large this replusion is, governs the amount of PE each electron has, which is then converted to KE as the charge is attracted to the other terminal and begins to move. Aug 7 '15 at 6:30

Power in a circuit is defined by P= I x E

P is power in watts or joules per second

I is the flow of electrons in coulombs per second or amps

E is the electormotive force or voltage

the energy of the electrons is lost to heat

the power supply see only the circuit as a whole it dose not see the individual resistors for R1 the total resistance of the circuit is 25 ohms so the current delivered to the circuit would be I=V/R I= .2 amps (I'm not including the resistance of the conductor because in most cases in is very small compared to the load R1 in this case but the power supply does see it ) since R1 is the only resistor in the circuit all the voltage and power is dropped across it minus the power required to move the electrons through the wire. So when the electron leaves R1 it still has enough energy to over come the resistance of the conductor.

in your second circuit R2 and R3

R3 has less voltage dropped across it so if you use P=I X E for R3 voltage drop across the resistor is 2.5 V current in the circuit is I=V/(R2+R3) = .1 amp so the power loss across R3 would be .25 watts the reason R1 is getting more power is it is the only resistor in the circuit so the total power of the circuit in dropped across R1. In the R2 R3 circuit each resistor in dropping half the power since they are equal in resistance

the electron still has energy leaving R1 because not all the voltage of the circuit is dropped across R1 some voltage even though very small is dropped across the wire or conductor it self. the power supply sees the load as RW+R1 . RW=the resistance of the wire

question 3 should be answered above as well

• Thankyou - but as per my question, I was hoping to find out what is "physically going on (what the electrons are doing), as opposed to what the equations say." Aug 7 '15 at 6:18

I'd like to start with a slightly pedantic aside: I think you should try to avoid the phrasing of electrons "having energy"; what they have is potential energy, and potential energy is relative. This is important to remember.

Question 2 is trivial, so I'll answer that first: there is no such thing as zero resistance, so the wire from R3 to the batter's negative terminal is better thought of as simply a minimal resistor; if we were to rewrite the problem this way, we'd see that some voltage was left to carry the electrons through it back to the battery.

I think questions 1 and 3 can both be boiled down to a single question: what is different, from the electron's point of view, between a resistor with .2 amps of current and a resistor with .1 amps of current. Why does the electron seem to experience them differently?

A completely accurate answer can only be reached through quantum mechanics, but a pre-quantum model, the Drude model, does provide a relatively intuitive, physically meaningful, and surprisingly accurate result. I will not go into the math, because you can find it on Wikipedia, and I would be basically copying that, but I'll try to give you a qualitative answer, which I think the Wikipedia article is lacking somewhat.

We can think of a resistor as a gas of free valence electrons, interspersed with atoms which are, from the electron's point of view, immovable objects. The electrons bounce around between them, but because there is an electric field pushing them in one direction, they tend to drift that way; we call this current. Since the charge of a resistor in these problems is neutral, the number of electrons doesn't change with voltage or current; a higher voltage simply means that the electrons are accelerated more between collisions, and thus their average speed, which is always in the direction of the current, is greater.

However, because the electrons are moving faster, the "mean free time", the average time between collisions, is shorter; they bump into atoms more often. Because a collision is as likely to push the electron backwards as forwards, more collisions means less progress; the about half of the collisions convert work that the field did accelerating the electron forward into backwards motion.

I know you said you didn't like analogies, but let me try a bad one anyway: it's easier to walk through a forest than to run. (This is bad because obviously you can avoid trees, but an electron can't, but whatever. I tried. An almost perfect analogy would be a board with a bunch of nails on it, at a slight angle; the steeper the angle, the more trouble a marble would have getting through; it would still get through faster, but it would come out the other end having lost a greater portion of its potential energy to nails.)

First, a couple of asides:

A. When circuits are drawn, the convention is that straight lines represent perfect conductors. Any real resistance is shown as a resistor, and hence, gets included in any IR drops. You don't go and say "well, there are some other resistances." If they are important, they are shown. Circuit analysis of diagrams is done with this assumption. Same thing for any stray capacitance or inductance. If they are important, you show them.

B. Individual charges don't have potential energy; systems of charges have potential energy.

C. The power consumption of a closed circuit system is zero. The battery supplies power, and the resistors consume power in this circuit. Power comsumption is calculated for each element and then summed. It is calculated with $P_{consumed}= IV$, with $I$ being the conventional current travelling from high potential end to the low potential end ($I$ might be negative).

The current through $R_1$ is 0.20 A and the voltage is 5 V. The power consumed by the resistor is 1 W. The current through the battery is 1 A, from low to high potential, across 5 V. The power "consumed" by the battery is -1 W, meaning it has supplied energy at a rate of 1 J/s. The total power consumed in the closed circuit is 0 W.

For the second circuit the current in all the elements is 0.10 A. The voltage across $R_2$ is 2.5 V and the power consumed is 0.25 W. The same holds for $R_3$ because it is the same resistance. The power consumed by the battery is -0.5 W. The total power consumed by the closed circuit is 0 W.

Now, the real answer for why the electrons continued to move in the circuit is conservation of charge and continuity of flow. Charge is not building up anywhere (like a capacitor), and the battery is continually changing the energy of the electrons.

The total resistance does have an effect. It changes the flow rate of the electrons which changes both the energy supply rate of the battery and the energy consumption rate of the resistors.

• Why is B the case? Aug 7 '15 at 6:19
• Change in potential energy is defined as the negative of the work within a system when on part of the system moves. The zero value (if it's needed) is usually (but not always) referenced to a zero force arrangement. If you have a single point particle, there is no force doing work on it, so no work is done, hence, no change in potential energy. We colloquially speak of the gravitational potential energy ($U_g$) of an object, but actually it's the $U_g$ of the object/planet system. A charge interacting with another charge is a system with potential energy; a charge by itself is not. Aug 7 '15 at 13:11

Your question is right but imagine a ball at some height; so ball has potential energy and when it moves towards the earth surface, it losses some energy due to air but other part of the energy is used to make velocity and this velocity is only zero when it strikes with the earth surface: this means that when an electron strikes with a positive plate, then its energy becomes zero meaning when electrons are leaving from resistance, then they have energy and it will become zero when they strike with positive plate. this further means that when you are measuring voltage across resistance it indicates 5J energy is given to minus one Coulomb charge but it is not indicated 5J energy is used.