Electromotive Force, Batteries, Resistors, and Potential Differences

[rtareen]

I have attached a small excerpt from my digital book where they start talking about emf. I am very confused. Let me explain what is confusing to me so that you can clear up what's bothering me.

They start of by saying that an emf device pumps charges by maintaining a potential difference called 𝓔. This potential difference exists between its terminals. So do the electrons move from the negative terminal, go around the circuit due to the potential difference, and then re-enter the battery through the positive terminal? So the battery does work to move it around the circuit? In the previous chapter about resistors we just assumed a potential difference between two points on a resistor. But apparently the potential difference across a resistor and 𝓔 are different quantities.

In the next section they take the charge carriers to be positive. I think they say that 𝓔 is the work the battery does within itself to take the positive charges from the low potential (where it enters) to the high potential (where it leaves so it can go around the circuit). So does this mean the same 𝓔 has two functions? It does work q𝓔 to push the charges around the circuit and the same amount of work q𝓔 to put them back in a position to follow the potential difference it sets up. Please let me know if this is right, or if not, please explain in a way I can understand.

Now let's just focus on what goes on outside of the battery and try to understand Figure 27-1. The battery sets up a potential difference between its terminals so that the positive charge goes around the circuit. Now we can see that there is a resistor halfway between the positive and negative terminals. We know that V = RI. So the voltage difference of this resistor would not be the same as 𝓔 but rather it would be its own resistance times the current that was set up by 𝓔? And is the resistance zero everywhere else besides the resistor? Is that right? If not, please explain in an understandable way. Also, we know that energy is dissipated by a resistor at the rate of $P = \frac{V^2}{R}$. So the charges have less energy going out of the resistor then they did coming in. What energy is this? If the flow is steady then it cannot be Kinetic. Is it electric potential energy? Do the charges have less potential energy because it went from the higher potential to lower potential within the resistor (which has its own potential difference)? Please let me know what's going on.

Now let's look at the more complicated, Figure 27-2. So the positive current moves from the high potential of the positive side of B to the less high potential that is the positive side of A. Is that right? So it is moving basically because of a potential difference given by $V_{B+} - V_{A+}$ ? Next, what happens as it goes through battery A? It says that the chemical energy within batter A is increasing such that battery B is charging battery A? Let me ask you one question. Does the charge simply flow from the high potential (positive side of A) to the low potential (negative side of A) as if nothing else is going on inside the battery? How does this relate to the increasing chemical energy? Next, once the charge gets out of A, what potential is there to keep the charges moving. Does the negative side of A have to have more potential than the negative side of B? Thats the only way I can see the charges still moving in this part of the circuit. And what about the motor and resistor? They say energy is lost to the motor and the resistor. But the potential differences remain the same? So what energy is lost?

 

[Lnewqban]

This article has helped me to translate from electrical to mechanical concepts, with which I am more familiar.

Perhaps the analogies it shows could help you:
https://en.m.wikipedia.org/wiki/Mechanical–electrical_analogies

 

[hutchphd]

For the time being you should not worry about the inner workings of a battery...the details and energy are all chemistry. Interesting but not helpful to understanding circuits. I suggest that you consider them magic devices which consist of a perfect voltage source (magic electron "pump") and a small resistor in series all in a box.

 

[rtareen]

For the time being you should not worry about the inner workings of a battery...the details and energy are all chemistry. Interesting but not helpful to understanding circuits. I suggest that you consider them magic devices which consist of a perfect voltage source (magic electron "pump")

Its good to know I can ignore how the battery takes charges from the lowest to highest potential energy. So youre saying that 𝓔 of the battery only refers to the potential difference it provides for charges to move in the circuit right?

and a small resistor in series all in a box.

So youre saying there can be a resistor even in a battery?

 

[rtareen]

Ive crossed out the questions that have been answered so far. Still need help understanding the rest.They start of by saying that an emf device pumps charges by maintaining a potential difference called 𝓔. This potential difference exists between its terminals. So do the electrons move from the negative terminal, go around the circuit due to the potential difference, and then re-enter the battery through the positive terminal? So the battery does work to move it around the circuit? In the previous chapter about resistors we just assumed a potential difference between two points on a resistor. But apparently the potential difference across a resistor and 𝓔 are different quantities.

In the next section they take the charge carriers to be positive. I think they say that 𝓔 is the work the battery does within itself to take the positive charges from the low potential (where it enters) to the high potential (where it leaves so it can go around the circuit). So does this mean the same 𝓔 has two functions? It does work q𝓔 to push the charges around the circuit and the same amount of work q𝓔 to put them back in a position to follow the potential difference it sets up. Please let me know if this is right, or if not, please explain in a way I can understand.

Now let's just focus on what goes on outside of the battery and try to understand Figure 27-1. The battery sets up a potential difference between its terminals so that the positive charge goes around the circuit. Now we can see that there is a resistor halfway between the positive and negative terminals. We know that V = RI. So the voltage difference of this resistor would not be the same as 𝓔 but rather it would be its own resistance times the current that was set up by 𝓔? And is the resistance zero everywhere else besides the resistor? Is that right? If not, please explain in an understandable way. Also, we know that energy is dissipated by a resistor at the rate of $P = \frac{V^2}{R}$. So the charges have less energy going out of the resistor then they did coming in. What energy is this? If the flow is steady then it cannot be Kinetic. Is it electric potential energy? Do the charges have less potential energy because it went from the higher potential to lower potential within the resistor (which has its own potential difference)? Please let me know what's going on.

Now let's look at the more complicated, Figure 27-2. So the positive current moves from the high potential of the positive side of B to the less high potential that is the positive side of A. Is that right? So it is moving basically because of a potential difference given by $V_{B+} - V_{A+}$ ? Next, what happens as it goes through battery A? It says that the chemical energy within batter A is increasing such that battery B is charging battery A? Let me ask you one question. Does the charge simply flow from the high potential (positive side of A) to the low potential (negative side of A) as if nothing else is going on inside the battery? How does this relate to the increasing chemical energy? Next, once the charge gets out of A, what potential is there to keep the charges moving. Does the negative side of A have to have more potential than the negative side of B? Thats the only way I can see the charges still moving in this part of the circuit. And what about the motor and resistor? They say energy is lost to the motor and the resistor. But the potential differences remain the same? So what energy is lost?

 

[rtareen]

Heres another question that might clear up some confusion for me. In the calculus-based introductory sequence, does the value of current remain constant throughout the entire circuit? Or can it change from one part of a circuit to another?

 

[jtbell]

So youre saying there can be a resistor even in a battery?

http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html

A Google search for things like "battery internal resistance" will turn up similar pages.

Many or most introductory physics textbooks also discuss it. My old copy of Serway and Vuille's College Physics lists "internal resistance" in the index (at the end of the book) as a subtopic of "Battery".

 

[jtbell]

In the calculus-based introductory sequence, does the value of current remain constant throughout the entire circuit?

Yes. This follows from conservation of charge, or number of electrons.

 

[DaveE]

Ive crossed out the questions that have been answered so far. Still need help understanding the rest.

They start of by saying that an emf device pumps charges by maintaining a potential difference called 𝓔. This potential difference exists between its terminals. So do the electrons move from the negative terminal, go around the circuit due to the potential difference, and then re-enter the battery through the positive terminal?

Yes

So the battery does work to move it around the circuit?

Yes

In the previous chapter about resistors we just assumed a potential difference between two points on a resistor. But apparently the potential difference across a resistor and 𝓔 are different quantities.

If the resistor is connected across the battery terminals then the battery voltage is the same as the voltage across the resistor. Since the convention in schematics is that the connections (wires) are perfect conductors, that voltage is just the potential between those two circuit nodes. Each node has a connection to a battery terminal and also one side of the resistor.

Now let's just focus on what goes on outside of the battery and try to understand Figure 27-1. The battery sets up a potential difference between its terminals so that the positive charge goes around the circuit. Now we can see that there is a resistor halfway between the positive and negative terminals. We know that V = RI. So the voltage difference of this resistor would not be the same as 𝓔 but rather it would be its own resistance times the current that was set up by 𝓔?

The potential difference generated by the battery is the same as the potential across the resistor, which is also V=RI. So, this means that the current must be I=V/R to make this balance. The battery voltage applied across the resistor creates just the right amount of current through the resistor to satisfy this requirement. That is kind of the definition of resistance; for a given applied voltage how much current flows.

And is the resistance zero everywhere else besides the resistor?

Yes, by convention in circuit diagrams. If they want you to worry about the very small resistance of real wires, they must add that in as a resistor (usually the wire resistance is so small that it is insignificant compared to the other components). This is just a convention so that we know what we are all talking about.

Also, we know that energy is dissipated by a resistor at the rate of P=V2R. So the charges have less energy going out of the resistor then they did coming in. What energy is this? If the flow is steady then it cannot be Kinetic. Is it electric potential energy? Do the charges have less potential energy because it went from the higher potential to lower potential within the resistor (which has its own potential difference)? Please let me know what's going on.

All of the energy the electron started with because of the battery potential is dissipated in the resistor. This energy was converted into heat, so the kinetic energy of the atoms/molecules in the resistor and the rest of the environment around it. It is converted by the resistor from electric energy into heat energy. The decrease in energy of the electrons as they flow through the resistor is manifest in the decrease in the potential (voltage) at each point along it's path.

Now let's look at the more complicated, Figure 27-2. So the positive current moves from the high potential of the positive side of B to the less high potential that is the positive side of A. Is that right? So it is moving basically because of a potential difference given by VB+−VA+ ?

Yes. Although if Va > Vb you could have the same drawing and just say the the current is negative (compared to the arrows they drew)

Next, once the charge gets out of A, what potential is there to keep the charges moving.

That would be the sum of the potentials (voltage drops) for each component the current passed through. In this case it would be Vb - I*R - Va = Vm (the voltage drop across the motor, top to bottom)

Does the negative side of A have to have more potential than the negative side of B?

Not necessarily, that would depend on voltages of the batteries. But yes, if you want current to flow through the motor in the direction of the arrows, then you would be correct.

And what about the motor and resistor? They say energy is lost to the motor and the resistor. But the potential differences remain the same?

No the potential drop across each of the components doesn't have to be equal. But, if you follow the current around the whole loop and add up each of the potential changes, they MUST add up to zero (this is called Kirchhoff's voltage law). So Va -I*R -Vb -Vm = 0. This just makes sense if you see that each electron ends up where it started, it can't arrive with more energy than it left with.

So what energy is lost?

All of the energy due to current flow is lost or stored in one of the batteries or as mechanical work. For the current direction shown (so Vb > Va); the resistor dissipates I²*R power as heat (that's the voltage, I*R, times the current I); Battery A is charged at a rate of I*Va (watts); and the motor does mechanical work at the rate of I*Vm = I*(Vb -I*R -Va), assuming that the motor is 100% efficient (which none actually are, they will also make some heat). The energy source, battery B, is discharged at the rate I*Vb.

 

[rtareen]

http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html

A Google search for things like "battery internal resistance" will turn up similar pages.

Many or most introductory physics textbooks also discuss it. My old copy of Serway and Vuille's College Physics lists "internal resistance" in the index (at the end of the book) as a subtopic of "Battery".

This page helped alot! I'll follow your advice as well. Thank you

 

[rtareen]

Everything DaveE said

Thank you so much DaveE! I wish I could give you more than just one like haha

 

[rtareen]

http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html

A Google search for things like "battery internal resistance" will turn up similar pages.

Many or most introductory physics textbooks also discuss it. My old copy of Serway and Vuille's College Physics lists "internal resistance" in the index (at the end of the book) as a subtopic of "Battery".

Does that page assume positive current? Because they say that 𝓔 increases the potential, which would let the postive charges flow around the circuit. But what if we were dealing with electrons going around a circuit? Wouldnt 𝓔 decrease the potential since electrons in a circuit flow from low potential (the negative terminal) to high potential (the positive terminal)?

 

[vanhees71]

The current density always has the same direction, no matter whether the conduction charge carriers are negative (as in usual metals, where the charge carriers are electrons) or positive (e.g., in electrolytes, where the charge carriers are both positive ions and negative electrons). The relation between the current density and the velocity of the charge carriers is
$$\vec{j}=q \vec{v},$$
where $q$ is the charge of the charge carriers and $\vec{v}$ their velocity field. Since $\vec{j}$ has always the same direction given the electromotive force $\mathcal{E}$ (and its orientation!) for $q<0$ the velocity of the charge carriers points in the opposite direction.

 

[jtbell]

It might be helpful to distinguish between the electric potential V (in volts) at a point in the circuit, and the electric potential energy E_p = qV (in joules) of a charge located at that point.

The electric potentials at the terminals of the battery depends on the emf of the battery and on where we choose to place the "zero point" (ground) of the circuit. Suppose the emf from the "-" to "+" terminals is +3 volts, and we choose the ground to be at the "-" terminal. Then the "-" terminal has V = 0 and the "+" terminal has V = +3 volts.

If the charge carriers were positive, e.g. positrons, then they would have E_p = 0 at the "-" terminal and E_p = 3(+q) = +3q at the "+" terminal. They would flow through the circuit in the direction of decreasing electric potential energy, i.e. from the "+" terminal to the "-" terminal.

But the charge carriers are actually electrons, with a negative charge. Therefore they have E_p = 0 at the "-" terminal and E_p = 3(-q) = -3q at the "+" terminal. They also flow in the direction of decreasing electric potential energy, but now this is from the "-" terminal to the "+" terminal... the opposite direction that positrons would flow.

The difference in electric potential across the battery is the same in both cases, but the difference in electric potential energy is opposite for the two kinds of particles, therefore they flow in opposite directions.

For an analogy, consider the gravitational potential at the floor and ceiling of a room, the gravitational potential energies of objects with positive and (hypothetically!) negative masses, and the directions that those masses would "fall" freely. The equivalent of a "battery" in this situation would be an elevator that pushes objects from low to high gravitational potential energy.

 

[anorlunda]

Students of circuit analysis should never use the term "electric potential V at a point", they should use only voltage difference between two points. Talking about potential at a point leads to confusion about definitions, as in this thread. It also leads to great confusion if we move the ground symbol around to different points in the circuit.

 

[rtareen]

Thanks everyone for your help! I think I understand enough about single loop circuits to atleast do the problems. Maybe someday when I have some time i'll try to understand them in more depth. But right now time is limited (since were covering 12 chapters in a single semester). But thank you all. I'm going to post a different question now. It will be a lot shorter than this one.

 

[DaveE]

Students of circuit analysis should never use the term "electric potential V at a point", they should use only voltage difference between two points. Talking about potential at a point leads to confusion about definitions, as in this thread. It also leads to great confusion if we move the ground symbol around to different points in the circuit.

Yes, students must learn this first. But, frankly, it isn't that hard.

Then they have to learn to communicate with people out in the world who, more often than not, talk about voltages at a point as shorthand. Everyone that knows electronics knows that there is some implicit ground reference. True it's incomplete, but implicit references are common in all sorts of human communication for the sake of efficiency.

 

[anorlunda]

Then they have to learn to communicate with people out in the world who, more often than not, talk about voltages at a point as shorthand.

You're right about the shorthand, but it also gets confused with the potential from electrostatics which is the physicist's view. IMO, thinking about potentials or about electrons is a huge problem for beginning students. Those concepts impede learning. Here's an example from today.

A while back, I created the following to illustrate the point for students.