Voltage in Circuits: Explaining Voltage Drops & Flow

[sophiecentaur]

See but then shouodn't voltage drop be uniform because it is only due to change in position.

Why would you expect that? Voltage drop is all about energy lost and there is very little energy lost as a charge flows through a copper wire so 'all' the volts are dropped across the load. Any model that you build in your head must take that sort of thing into account from the very start and it should ring the credibility bell before you move on to the next step.

 

[UMath1]

Voltage drop is about change in potential energy. Potential energy is a function of distance from field source(Kq1q2/r). Therefore, voltage drop must be uniform with respect to displacement towards field source.

Can you explain the PE graph here
https://community.dur.ac.uk/p.m.johnson/electric_circuits/11_bulb_circuit.htm

And can you explain why an electric field cannot act in a open circuit?

 

[davenn]

I also don't understand why the circuit must be closed for the field to be set up. My textbook says that the electrons do not rush to the lightbulb as soon as the circuit is closed, but instead the field acts on the electrons near the lightbulb. But why can't the field act even if the circuit is open?

because the EM field isn't created till there is a charge/electron flow. And for a charge/electron flow to occur, there needs to be a closed circuit.
The EM field then travels along the outside of the wire at close to the speed of light

 

[UMath1]

Why is current needed for the electric field? Electric field is defined as kq1/r^2. If there is a source charge at the terminals can it not create a field acting on the electrons in the lightbulb? Or is that the force is weakened across air and the field is insignificant?

Voltage drop is about change in potential energy. Potential energy is a function of distance from field source(Kq1q2/r). Therefore, voltage drop must be uniform with respect to displacement towards field source.

Can you explain the PE graph here
https://community.dur.ac.uk/p.m.johnson/electric_circuits/11_bulb_circuit.htm

And can you explain why an electric field cannot act in a open circuit?

 

[davenn]

Voltage drop is about change in potential energy. Potential energy is a function of distance from field source(Kq1q2/r). Therefore, voltage drop must be uniform with respect to displacement towards field source.

you didn't read what Sophiecentaur said

again ...

Voltage drop is all about energy lost and there is very little energy lost as a charge flows through a copper wire so 'all' the volts are dropped across the load.

Dave

 

[UMath1]

The issue I am having is that voltage is treated a total energy per unit charge but in electrostatics it was defined as POTENTIAL energy per unit charge. In a circuit, I understand across a wire with zero resistance there should be negligible energy loss. But shouldn't there still be a voltage drop. Wouldn't potential energy be converted to kinetic energy?

 

[sophiecentaur]

Voltage drop is about change in potential energy. Potential energy is a function of distance from field source(Kq1q2/r). Therefore, voltage drop must be uniform with respect to displacement towards field source.

You are missing out something important. The Potential is a function of Position ONLY in an isotropic, uniform medium. Enough said? Your formula relates to free space.

 

[sophiecentaur]

I understand across a wire with zero resistance there should be negligible energy loss. But shouldn't there still be a voltage drop.

There is and you can always measure it if your voltmeter is sensitive enough in any real wire. It is, by definition, zero when R is zero. After all R = V/I.

I think at this point it could be useful if you assume you are not right here and start looking for holes in your own argument. Everyone else can't be wrong, can they?

voltage is treated a total energy per unit charge

That isn't really the definition. Read it again in a textbook and spot the difference. The two definitions are the same (they have to be, don't they). It must be your take on them that's adrift. I'm not sure that anyone else can sort out your problem when all that's needed is to read the definitions and interpret them right.

 

[Tyler Worden]

You are right about the wire. All wire has a resistivity factor eg. Solid copper is around 1.7x (10^-8) ohms per meter, and when using direct current, can be calculated as resistance = (resistivity x length) over cross section area of the wire. So using only wire, “let’s say 1 meter long”, in a circuit you can calculated the potential voltage at any given point. Add a piece of smaller diameter wire at the midpoint of the circuit, and use the same principles, adding the resistance. Knowing the total resistance gives you the current flow which equalizes once the circuit is closed. Current flow gives you field effect around the wire.

Now you can calculate your electrical potential or field potential using k Q / r2 , electron speeds depend on the wire sizes.

 

[UMath1]

How do you use the current flow for the field effect?

 

[Tyler Worden]

The magnetic field around a current-carrying wire is measured perpendicular to the wire and = to the permeability of free space x current (amps) / (2π x distance from the wire). Electric field from any point charge is k Q / r2

You can look up permeability of free space on the web.

 

[UMath1]

So you have to apply magnetic and electric fields?

If I understand correctly, you want me to calculate the net electric and magnetic field at different points on the circuit and then use that to find voltage difference? And why use permeability of free space...isn't it a conducting wire?

Just to clarify, my main goal is to unify my understanding of voltage in circuits and electrostatic conditions.

 

[Tyler Worden]

Ohm’s law E=IR E= voltage I=amperage R= resistance
A simple circuit using 10 volts & 10 (1ohm) resistors

Battery Positive-----R1----R2----R3----R4----R5----R6----R7----R8----R9----R10----Battery Negative
At the Battery positive point in the system the voltage is 10V and the amperage of the circuit is I=E/R 10 volts divided by 10 ohms or 1 amp

The voltage at the point between R1 and R2 is calculated by removing R1 from the equation
E=IR E= 1 x 9 E=9

The voltage at the point between R2 and R3 is calculated by removing R1 & R2 from the equation
E=IR E= 1 x 8 E=8

Etc. Etc. Etc
Wire can thought of in the same manner as the resistors. The further down the wire the greater the resistance.
Now you have your voltage potential at different points.
and that's all from me do your home work.

 

[UMath1]

But that's using ohm's law. I can apply ohm's law and get all the right answers.

My issue is a conceptual one. Voltage seems to be defined differently in circuits and electrostatic conditions. In electrostatics we found difference in voltage using kq/r1- kq/r2. But that doesn't seem to work in circuits.If it did, the voltage drop across wires and resistors would be the same for the same amount of change in r.

 

[UMath1]

For example if a ball falls downwards in vacuum its potential energy gets converted to kinetic energy meaning there is a drop in potential energy.

In the same way, should a charge moving across a wire with zero resistance still have some voltage drop? But according to ohm's law it does not.

 

[sophiecentaur]

For example if a ball falls downwards in vacuum its potential energy gets converted to kinetic energy meaning there is a drop in potential energy.

In the same way, should a charge moving across a wire with zero resistance still have some voltage drop? But according to ohm's law it does not.

The simple particle / KE treatment of what goes on in a superconductor is not relevant but, if no energy is lost in transferring the charge, no PD exists across it and the charges would not flow any faster at the other end. This has been dealt with before on this thread. How many times do you need to be told about this?
The ball falling in a vacuum is not an appropriate analogy here. Why do you want to ignore the accepted theories? Have you studied them formally or have you just read some popular versions?
PS "Ohm's Law" is not a fundamental law of the Universe. It is a description of the relationship between V and I in (essentially) metals at constant temperature. It should really (imo) be described as Ohm's Relationship (but that could be taken as referring to his Girlfriend). Many substances and electronic components do not follow Ohm's Law but people often refer to the ratio of V and I as a 'dynamic resistance' value. If you try to use the 'resistance' found in this way as if it were constant, you will come unstuck.

 

[sophiecentaur]

But that doesn't seem to work in circuits.If it did, the voltage drop across wires and resistors would be the same for the same amount of change in r.

I am not sure what experiment you are referring to here but the PD across points on a (uniform) Potentiometer Wire is directly proportional to the distance between them. Where do you think it doesn't happen?? I ask again, how much of this subject have you learned in a formal way and how much are you cherry picking the bits that interest you? EE (and most of Physics, in fact) is pretty well impossible to understand if you miss out big chunks of the course.

 

[Dale]

Voltage seems to be defined differently in circuits and electrostatic conditions. In electrostatics we found difference in voltage using kq/r1- kq/r2. But that doesn't seem to work in circuits.

You can use Jefimenko's equations in both electrostatics and in circuits.

https://en.m.wikipedia.org/wiki/Jefimenko's_equations#Origin_from_retarded_potentials

It is not a particularly useful approach for most circuits, but certainly it shows that the laws remain the same.

If it did, the voltage drop across wires and resistors would be the same for the same amount of change in r.

That assumes that the charge distribution is the same, which it is not.

 

[UMath1]

I did read the link but I don't understand the notation. Can you explain it simply to me?

 

[Dale]

I did read the link but I don't understand the notation. Can you explain it simply to me?

If this notation is too advanced then you should probably mark future threads as "B" rather than as "I". This is definitely Intermediate level math and very standard notation. Please use the appropriate designation for your threads.

Jefimenko's scalar equation gives the relationship between charge density, $\rho$, and voltage, $\varphi$. For a point charge it reduces to the form you have been using.

The key mistake in your reasoning is that you are not taking into account the fact that $\rho$ is different in the two scenarios. Since $\rho$ is different, so is $\varphi$.

 

[UMath1]

Sorry but I can't change it to basic level at this point and I am still confused. So if I understand correctly, to calculate the total electric potential you must sum the effects of each individual charge in the wire and this would be affected by the charge density. So the charge density in the resistor is different from that in the wire, and so the change in potential is different?

I don't see the clear connection between charge density and voltage though.

 

[Dale]

Sorry but I can't change it to basic level at this point and I am still confused.

I don't think that you will be able to get un-confused yet. I think that you are asking a question that you are not yet ready to learn the answer. That is causing your confusion. In my opinion you need to take a step back and fill in some gaps first. We can answer your questions, and have already done so, but understanding requires asking the right questions too.

So if I understand correctly, to calculate the total electric potential you must sum the effects of each individual charge in the wire and this would be affected by the charge density.

Yes, to calculate the potential from first principles requires knowing the exact charge density throughout the circuit. Alternatively you can just use Kirchoff's voltage and current laws. You get the same answer with a lot less effort.

So the charge density in the resistor is different from that in the wire, and so the change in potential is different?

Yes. The charge density in the entire circuit is different with vs without the resistor.

I don't see the clear connection between charge density and voltage though.

That connection is given by the first equation in the section "origin from retarded potentials" that I pointed to earlier. The one that shows exactly how to calculate $\varphi$ given $\rho$

 

[Dale]

Can I ask what your background and goals are regarding physics and math? Looking at your questions I don't see a clear pattern.

 

[UMath1]

I am currently in AP Calc BC and AP Physics 2. I have completed AP Physics 1 and a self-study of AP Physics C: Mech.

I just want to gain a better understanding of the concepts we are covering in class. Right now I feel as if I can apply a formulas and get the right answers but I don't really feel strong on the concepts. I want to have a unified understanding of voltage in electrostatics and circuits because right now the two seem to be two different ideas.

 

[Dale]

I am currently in AP Calc BC and AP Physics 2. I have completed AP Physics 1 and a self-study of AP Physics C: Mech.

Are the AP physics courses calculus-based or algebra based?

 

[UMath1]

AP Physics 1 & 2 are algebra based. But last year I self studied AP Physics C: Mechanics which was calc-based. This year I am also self-studying AP Physics C: Electricity and Magnetism which is calc-based.

 

[Dale]

OK, so one general point. You are looking to understand unified concepts. That generally requires a mathematical framework which unifies the different concepts. So, when you are studying some physics using algebra you are going to find that a lot of things which you hope would be unified cannot be unified strictly algebraically. Have patience, as you develop the mathematical framework necessary you will revisit those concepts and see how they are unified. Until you do so there is not much that can be done.

For classical EM, the unifying framework is Maxwell's equations:
https://en.wikipedia.org/wiki/Maxwell's_equations#Conventional_formulation_in_SI_units
which can be expressed in terms of potentials as
https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Lorenz_gauge
which is heavily based on vector calculus.

The concepts are unified, but you are probably several semesters of calculus away from being able to follow it. Until then all I can do is to urge patience. You are quite far ahead of where you need to be already, so don't feel upset that you cannot do everything today. Give yourself time to learn in the proper order and time.

 

[AlejandroDes]

I also feel stuck on how voltage is POTENTIAL energy per unit charge. How can friction play a role on potential energy which is based soley only location? If it was losing kinetic energy across the resistor I would understand because friction and resistance slow things down.

I think you're getting confused because you are trying to compare the voltage (electric potencial) with the energy of a mass at certain height (potencial energy). They are not the same thing (the units are not the same). Imagine you have a mass at a certain height the Potencial energy would be given by PE = m*g*h, now divide this PE by the height and you'll get PE/h = m*g [J/m] (Weigth). Now you get a usefull quantity that will offer you information on how the mass will behave under any heigth variation.

We can do the same for the electric problem. Imagine we have a bunch of charged particles (negative and positive) and we take the time and work to separate all the positive particles from the negative particles (you'll get escentially a battery). Now, that we invested some work (energy) into the separation from it's natural state (like putting work into lifting something) we have stored potencial energy in this "battery". Let's say that this Energy is proportional to the amount of particles we had: PE = k*Q. If we divide this PE by the amount of charge we'll get PE/Q = k [J/Co] (Voltage), so k = V.

Now for the resistor part, resistors do not cause a voltage drop by themselves. What causes the voltage drop is given by a current flowing through a resistor. Also don't take into account kinetic energy. Electrons in a wire do not move at the speed of ligth that would be impossible. In electricity we are transfering energy from one place to another but it doesn't mean that the electron(charge) has to cover the distance. Is way easier to just pass "energy" to the next electron and so on. Now say that in a wire you have a cross section with a finithe amount of electrons (I would use 100). So you have 100 electrons passing energy to the next 100 which will then pass it to the next 100 and so on. Suddenly, you get to a part where the system isn't closed anymore (electrons can "lose" some of the energy while passing it) so each time an electron passes the energy to the next one, some of it gets out and you end up with a lesser amount of energy. Less energy is equal to less voltage. So the voltage drop is given by the amount of energy that the electrons lost.

I hope this helped.

 

[Jesra]

I feel that I don't understand how voltage works in a circuit. I understand voltage to be electric potential energy per unit charge (kq/r). In the case of a circuit, electrons flow from low potential to high potential. But I don't understand how resistors cause a voltage drop. Isn't voltage based on position? How can the resistors cause a drop in potential energy? I can understand the resistor causing a drop in kinetic energy slowing the electrons down, but how does it lower the voltage? And if the total voltage is equal to the sum of the voltages of the resistors, then the voltage difference in wire after the last resistor and the positive terminal would be zero, right? Then how would the electrons be able to flow back to the positive terminal? Wouldn't they just stop?

The best analogy I can think of is a river flowing downhill that turns a turbine. But in this case the turbine doesn't cause a drop in the water's potential energy. It only takes some of the water's kinetic energy.

My understanding is voltage does not change with resistance, but amperes do. Voltage is an assist to predict what distance an amplitude will travel through x ohm resistance.

''
Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance,[1] one arrives at the usual mathematical equation that describes this relationship:[2]

where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current.[3]''

https://en.wikipedia.org/wiki/Ohm's_law

So, doing variable relation conversions:

amperes=voltage/resistance, resistance (amperes)=volts, resistance=volts/amperes

Electrons flow from negative potential to positive potential, but we need to keep in mind that the interpretation of electron flow is temporally dependent (e.g. how can an alternating current have direction when the 'current' is both positive and negative in an oscillating relation (but that relation is with the return line to its generator). The direction of an electron flow is as indicative to voltage polarity as is the ampere difference at two points in the flow is to resistance. Different voltage generating systems may have different rules, such as battery voltage variations and power line voltage variations. e.g. I have measured 'dead' batteries with voltages only a volt less than live batteries and I am reasonably certain two dead batteries in series add up to one dead battery at 24 volts. :-)

 

[Chandra Prayaga]

See but then shouodn't voltage drop be uniform because it is only due to change in position. I dom't see why it is more across resistors and almost zero in the wire. Can you show me how this would work with kq/r?

If you can explain the potential energy graph on this page :
https://community.dur.ac.uk/p.m.johnson/electric_circuits/11_bulb_circuit.htm that would be very helpful.

I also don't understand why the circuit must be closed for the field to be set up. My textbook says that the electrons do not rush to the lightbulb as soon as the circuit is closed, but instead the field acts on the electrons near the lightbulb. But why can't the field act even if the circuit is open?

There are two questions here. I will address the first one first. Each resistor in your circuit is the equivalent of several miles of a conducting wire. You can easily calculate how long a wire you will need to replace a 1 ohm resistor. If you did that, you will see that the potential drop is indeed uniform across that very long wire. You are effectively taking a very very long wire, packaging it into a tiny wound bundle, and calling it a resistor. This is a model. Real resistors are made differently, but the model is pretty good.

Now the second one:

If you don't close the circuit, the electron no longer has a low resistance path to go through. The air gap has a very high resistance. If you increase the potential difference to very high values, you can even cause a breakdown through the air gap.