Can someone explain the use of conventional current and voltage in circuits?

[Red_CCF]

1. I understand that conventional current is opposite of electron movement. What I don't get is why we still use this; I've read several websites that just says that it doesn't affect anything without going into more detail.

2. I've never really grasped the concept of voltage. I don't like the analogy that compares voltage to water pressure because I tend to believe pressure as a force whereas voltage is energy related (correct me if I'm wrong). I don't really understand the term potential difference that is used with voltage; what is it a difference of? why is the word potential used when the voltage is related to giving electrons kinetic energy?

3. This question is related to voltage as well. The positive and negative terminals on a DC circuit (a battery for instance) has me confused because if voltage is energy (a scalar quantity) related, how can one side be positive and the other be negative?

4. When electrons are pushed through a resistor, voltage is used up; but if let's say a 9V light bulb is placed on a 9V DC circuit, after the electrons move through the resistor and the voltage is used up, how does the electrons continue moving to the terminals?

5. Another voltage related concept that I don't understand is the negative voltage on AC circuits. How does that work and how come a load can still function with a negative voltage on an AC circuit but not a DC circuit with set positive and negative terminals?

Please correct me if I have any misconceptions in my questions above. I'm going into engineering and I really need to grasp these basic electricity principles.

Thanks for any help that you can provide

 

[chroot]

1. We continue to use this convention for historical reasons (i.e. because millions and millions of books have already been written which use the convention). It does not affect anything mathematically; if we were to use the opposite convention, we'd just have to change plus signs to minus signs and vice versa. There isn't much to gain by re-writing and re-printing all those books. If you want to use the opposite convention in your own calculations, you are welcome to do so.

2. A brick on the floor has no potential energy; if you pick it up and hold it over your head, you have added potential energy. Voltage works essentially the same way, except the term "potential" is used in place of "potential energy." Electrical potential is just potential energy per unit charge, so "potential" is proportional to "potential energy."

To continue the analogy between electricity and gravity, consider an apparatus with a pump. Water is pulled out of a low reservoir by the pump, and pushed up to a higher altitude, where it now has potential energy. It can flow downwards through a water wheel, say, and perform work.

A battery is very much like a pump. It pushes electrons up to a state of higher potential, and they can do work as they flow towards lower potential.

The positive terminal of the battery is the outlet of the pump; the negative terminal is the inlet. (If the charge carriers are electrons, of course).

The "potential difference" created by the battery is the difference in potential between its terminals. This difference in potential is measured in volts.

3. Already answered.

4. It's not really correct to say that voltage is "used up." The voltage is different at each end of the resistor, but that's because electrons have lost energy in their transit through the resistor.

Wires are usually modeled as having zero resistance, but this is not really true. They have a small, but definitely non-zero resistance, on the order or a hundredth or thousandth of an ohm, depending upon the size, composition, etc.

The wire carrying current back to the battery from the 9V bulb actually has a very small difference in potential between its two ends, and that potential difference is what motivates electrons to flow through it back to the battery.

5. Voltage is a "conservative" quantity. There is no such thing as "absolute voltage," just as there is no such thing as "absolute altitude." You must always say what the voltage (and altitude) are measured against. You might measure all voltages with respect to the negative terminal of your battery, and then all your voltages will have positive numerical values. If you choose to measure them with respect to the positive terminal, all your voltages will have negative numerical values.

I hope this helps -- let me know if anything is unclear.

- Warren

 

[Integral]

In additon to chroots execellent response I will add that the Navy tech schools teachs electron flow in all basic electricity courses.

 

[Tac-Tics]

1. I understand that conventional current is opposite of electron movement. What I don't get is why we still use this; I've read several websites that just says that it doesn't affect anything without going into more detail.

When you deal with currents on the scale of small electronics or bigger, the two notions are equivalent. It's all in the math. The current is the amount of positive charge flowing through a wire in a direction. If you flip the signs everywhere, the equality still holds, but now you're talking about negative charge flowing in the opposite direction.

The reason we keep the convention is because it is institutionalized... it's just part of our engineering culture at this point. It would make no more sense to switch it as it would make sense to get the Chinese to give up their writing system in favor of the latin alphabet.

2. I've never really grasped the concept of voltage. I don't like the analogy that compares voltage to water pressure because I tend to believe pressure as a force whereas voltage is energy related (correct me if I'm wrong). I don't really understand the term potential difference that is used with voltage; what is it a difference of? why is the word potential used when the voltage is related to giving electrons kinetic energy?

Voltage is confusing inside a wire, but relatively simple to understand in space. Voltage is also called electric potential. If you understand gravitational potential energy, it's almost exactly the same. You can think of potential as being the height of a hill. When you fall off a hill (from a high potential to a lower one), you release energy. To climb a hill (to go from a low potential to a high one), you must consume energy. To walk across a mesa requires no energy.

More technically, the potential has to do with the amount of work per unit of charge it takes to push an object through an electric field. If I have a wad of electrons with a total charge of 5C at one end of a battery and I push it with my fingers to the other end, it will cost me 5 Joules of my hard earned energy. Then, when I move it from that end to where it originally started, I get my 5J back.

3. This question is related to voltage as well. The positive and negative terminals on a DC circuit (a battery for instance) has me confused because if voltage is energy (a scalar quantity) related, how can one side be positive and the other be negative?

The voltage rating on a battery is the total difference between the two end points. I'm not sure on standard conventions, but one is most likely a ground while the other is either above or below ground.

4. When electrons are pushed through a resistor, voltage is used up; but if let's say a 9V light bulb is placed on a 9V DC circuit, after the electrons move through the resistor and the voltage is used up, how does the electrons continue moving to the terminals?

Between any two points A and B, you can talk about the voltage "drop" between A and B. But voltage isn't really a commodity. Imagine the hills on a mountain. A voltage (drop) of zero between two points is like saying that those two points are at the same height. The energy required to move between two points with equal potential energy is zero by definition!

5. Another voltage related concept that I don't understand is the negative voltage on AC circuits. How does that work and how come a load can still function with a negative voltage on an AC circuit but not a DC circuit with set positive and negative terminals?

An AC circuit, by definition, is a non-constant voltage source. It varies (usually sinusoidally) with time. It's usually convenient to call the average voltage in one cycle the "ground".

The "ground" or 0V point is completely arbitrary. If you use the hill analogy, think of 0V like sea level. When you fall off a cliff, it doesn't really matter where sea level is. All that matters is how far you fall before you hit the ground. But we must choose some ground value if we are to have a basis for measuring things.

 

[chroot]

If I have a wad of electrons with a total charge of 5V at one end of a battery and I push it with my fingers to the other end, it will cost me 5 Joules of my hard earned energy.

Eek - be careful here. The volt is not a unit of charge.

The voltage rating on a battery is the total difference between the two end points. I'm not sure on standard conventions, but one is most likely a ground while the other is either above or below ground.

"Ground" is what you make it to be. You are welcome to declare either terminal to be "ground," which means that all other voltages in the circuit will be measured with respect to it. Typically, the most negative terminal is called "ground," so that all other voltages in the circuit are positive.

- Warren

 

[Tac-Tics]

Eek - be careful here. The volt is not a unit of charge.

Corrected. I indeed meant coulomb.

"Ground" is what you make it to be. You are welcome to declare either terminal to be "ground," which means that all other voltages in the circuit will be measured with respect to it. Typically, the most negative terminal is called "ground," so that all other voltages in the circuit are positive.

I implicitly meant the same ground you'd find in an AC outlet. I could have been more clear.

 

[pgardn]

Please correct me if I have any misconceptions in my questions above. I'm going into engineering and I really need to grasp these basic electricity principles.

Thanks for any help that you can provide

If you really want to understand this better it is very important to get a grasp on 4 ideas/concepts...

1. Charge: q
2. Electric fields and strength: E
3. Electric potential energy: U
4. Electric potential and Electric potential diff: V and delta V (often abbreviated as V)

If you can get a grasp on the units and move back and forth between the 4 things above it will make electricity easier. Some general rules for circuits learned along with the above adds a flavor tangible use and it becomes more interesting which makes it easier. I am still learning but I would suggest delving into how each of the above relate to each other and how they are different.

 

[Red_CCF]

Thanks so much for the replies, it cleared a lot of things up for me. However I'm still a bit confused on some of the things I mentioned.

1. ... if we were to use the opposite convention, we'd just have to change plus signs to minus signs and vice versa...

Why do we have to change the signs if we were to assume electron flow is the current?

A battery is very much like a pump. It pushes electrons up to a state of higher potential, and they can do work as they flow towards lower potential.

The positive terminal of the battery is the outlet of the pump; the negative terminal is the inlet. (If the charge carriers are electrons, of course).

The "potential difference" created by the battery is the difference in potential between its terminals. This difference in potential is measured in volts.

Is it correct for me to assume that the electrons are already at a higher potential in a battery and flows "down" like water as soon as the circuit is connected?

I understand potential of a charge in space, which is just like gravitational potential energy like Tac Tics suggested, but it's in a wire that's confusing. I have trouble understanding the "ground" that Tac Tics was talking about because each half cell in a voltaic cell (typical battery) produces a potential but in class we were told to add the potential produced in the reactions of each half cell, not find the difference so that is where I am confused since both half cells in a battery produces a positive potential.

Thanks for any help that you guys can provide

 

[pgardn]

Thanks so much for the replies, it cleared a lot of things up for me. However I'm still a bit confused on some of the things I mentioned.



Why do we have to change the signs if we were to assume electron flow is the current?

You don't always have to, it depends on what you are asked.


Is it correct for me to assume that the electrons are already at a higher potential in a battery and flows "down" like water as soon as the circuit is connected?

I guess you could do that, but realize that there are electrons already in the wire ready to flow once a potential diff (uneven heights for you?) is created. The electrons "rolling in" at one terminal have less energy than at the other terminal because you created a hill when you completed the circuit with a battery (pot diff). I think it might be easier to think of one end of the battery having a high electric potential and the other end lower for your hill analogy.

I understand potential of a charge in space, which is just like gravitational potential energy like Tac Tics suggested, but it's in a wire that's confusing. I have trouble understanding the "ground" that Tac Tics was talking about because each half cell in a voltaic cell (typical battery) produces a potential but in class we were told to add the potential produced in the reactions of each half cell, not find the difference so that is where I am confused since both half cells in a battery produces a positive potential.

Not sure what you are getting at here, maybe someone else does.

Thanks for any help that you guys can provide

 

[Red_CCF]

Not sure what you are getting at here, maybe someone else does.

I probably phrased the question really badly. I'm just confused with the positive and negative terminals on a battery (AC/DC circuits in general actually) and how we calculated the potential difference of a battery in the first place. I always thought that each ends of a battery carried a potential (as taught by my chem teacher); but the two potentials are added in calcluations in class not subtracted so I don't really get where the difference comes from.

 

[chroot]

Why do we have to change the signs if we were to assume electron flow is the current?

There are two possible choices. You can label the charge carriers positive, or you can label them negative. The equations will look the same for either choice, except that the plus and minus signs would be reversed. Neither is really "better" than the other in any way at all.

Is it correct for me to assume that the electrons are already at a higher potential in a battery and flows "down" like water as soon as the circuit is connected?

Batteries use chemical reactions, and there is no such thing as one-way chemical reactions. All chemical reactions operate in both directions, until some equilibrium is achieved. In equilibrium, the reaction proceeds at the same rate in both directions.

Inside a battery, the redox reaction that liberates free electrons will operate until an equilibrium is reached, meaning that the electrons headed for the negative terminal experience such a repulsion from the electrons already present that no more can go. It only takes a few, since the repulsion electrons experience from one another is so large. Electrons don't "pile up" at the terminals, ready to "roll down the hill." When the load is connected, the electrons all begin to move more or less at the same time, all through the battery, through its fluid electrolytes and membranes, and the wire.

The best analogy is a closed loop of pipe with an inline pump, filled with water. When the pump starts (the circuit is closed), the water begins moving all through the loop at roughly the same time. A pressure wave propagates around the loop first, though. A wave of disturbance in the electromagnetic field propagates through the circuit in the same fashion, but it is so fast that it is effectively instantaneous by human standards.

I understand potential of a charge in space, which is just like gravitational potential energy like Tac Tics suggested, but it's in a wire that's confusing. I have trouble understanding the "ground" that Tac Tics was talking about because each half cell in a voltaic cell (typical battery) produces a potential but in class we were told to add the potential produced in the reactions of each half cell, not find the difference so that is where I am confused since both half cells in a battery produces a positive potential.

You need two half-cells to make a complete battery, because you need two electrodes. The two half-cells are connected in series, so their voltages add together.

- Warren

 

[Red_CCF]

The equations will look the same for either choice, except that the plus and minus signs would be reversed.

What type of equations are you talking about?

Inside a battery, the redox reaction that liberates free electrons will operate until an equilibrium is reached, meaning that the electrons headed for the negative terminal experience such a repulsion from the electrons already present that no more can go. It only takes a few, since the repulsion electrons experience from one another is so large. Electrons don't "pile up" at the terminals, ready to "roll down the hill." When the load is connected, the electrons all begin to move more or less at the same time, all through the battery, through its fluid electrolytes and membranes, and the wire.

I thought that electrons headed for the positive terminal on a battery?

So basically once a wire is connected to the ends of the batteries and a circuit is complete, the redox reaction from the half cella produce a voltage that pushes the electrons. So the electrons are not piled up before the wire is connected but they move after the wire is connected due to the redox reaction of the battery? So there isn't an excess of electrons at one terminal (each half cell is balanced) before a circuit is made?

You need two half-cells to make a complete battery, because you need two electrodes. The two half-cells are connected in series, so their voltages add together.

- Warren

This is what is confusing me about potential difference. If we add the potentials produced by each half cell together, where does the "difference" come in? What is it a difference of?


Thanks for the help

 

[mikelepore]

What type of equations are you talking about?

For example, the direction of an electric field is defined to be the direction of a force that would be experienced by a positive charge if it were placed into a space where the electric field exists. In that electric field, a positive charge would spontaneously move from an initial position of high potential and high potential energy to a new location of low potential and low potential energy. In that same electric field, a negative charge would spontaneously move from an initial point of low potential and high potential energy to a new location of high potential and low potential energy. In both cases, the change in the particle's potential energy is negative, and the work done by the electric field on the charge is positive. In the case of a negative charge, the reason that work done by the electric field is positive is because of the multiplication of two negative signs: a negative charge having a displacement in the direction opposite to the electric field. Because of these sign conventions, we define current as the flow of net charge, as though we had positive charge carriers.

 

[Red_CCF]

For example, the direction of an electric field is defined to be the direction of a force that would be experienced by a positive charge if it were placed into a space where the electric field exists. In that electric field, a positive charge would spontaneously move from an initial position of high potential and high potential energy to a new location of low potential and low potential energy. In that same electric field, a negative charge would spontaneously move from an initial point of low potential and high potential energy to a new location of high potential and low potential energy. In both cases, the change in the particle's potential energy is negative, and the work done by the electric field on the charge is positive. In the case of a negative charge, the reason that work done by the electric field is positive is because of the multiplication of two negative signs: a negative charge having a displacement in the direction opposite to the electric field. Because of these sign conventions, we define current as the flow of net charge, as though we had positive charge carriers.

But I thought that electric potential is proportional to electric potential energy. Then how come a negative charge can have a low potential and higher potential energy?

 

[jtbell]

Because of the negative sign on the charge.

$$PE = QV$$

When Q is negative, increasing V decreases the PE.

 

[mikelepore]

But I thought that electric potential is proportional to electric potential energy. Then how come a negative charge can have a low potential and higher potential energy?

In the water analogy, I visualize the water in the mountaintop reservoir, etc., as having high potential energy because it didn't fall down yet - it's in a position where it CAN fall down. So what if you have a negative charge very near the negative electrode, that is, a point in space where there is a relatively low potential. It didn't yet "fall" toward the positive electrode. When it does so, it will lose potential energy as it gains kinetic energy.

 

[Red_CCF]

Because of the negative sign on the charge.

$$PE = QV$$

When Q is negative, increasing V decreases the PE.

Oh, my textbook taught us to use the magnitude of the charges, so I think that's where my confusion arises from. Is my textbook wrong in doing that then because it's known to have a lot of errors

 

[Red_CCF]

In the water analogy, I visualize the water in the mountaintop reservoir, etc., as having high potential energy because it didn't fall down yet - it's in a position where it CAN fall down. So what if you have a negative charge very near the negative electrode, that is, a point in space where there is a relatively low potential. It didn't yet "fall" toward the positive electrode. When it does so, it will lose potential energy as it gains kinetic energy.

But if I were to consider potential using magnitude only, would the electron at a negative electrode have a high potential in terms of magnitude? Other than this confusion I understand everything else you explained. Thanks for helping me=)

 

[pgardn]

Oh, my textbook taught us to use the magnitude of the charges, so I think that's where my confusion arises from. Is my textbook wrong in doing that then because it's known to have a lot of errors

This might have been in the section where the text was explaining the force exerted by two charged objects on each other (Coulomb's Law)? The sign you might get when you calculate the electric force between two objects does not tell one about the direction of the Force. Some texts just use absolute value signs and put q (amount of charge) inside. They don't want a + or - sign to confuse you about the direction of the Force (and of course the Force has to be the same magnitude, but direction... that's for you to determine based on like charges repel yada yada). And if you are given a coordinate system you might have to look a the force vectors and decide what direction the net force is directed based on that coordinate system for more than two charged objects.

This might be why you are focusing on the magnitude of a charged object?
Sorry if I misread your thinking.

 

[user111_23]

Voltage is called a difference because it is a difference of electric potential. Electric potential is the potential energy per unit of charge or V=PE/q.

Here's how I understood voltage:

Let's say you build a classic zinc and copper battery with dilute sulfuric acid as the electrolyte. When a zinc bar is dipped inside the acid, it loses electrons and becomes negatively charged. If I move an electron towards the zinc bar, it is repelled by the electric field surrounding the zinc bar. It also gains potential energy since I would have to move the electron towards the negative terminal. But at the copper terminal, there is less electric potential. This difference in electric potential is called the electric potential difference or voltage.

Voltage is the change in potential energy per unit of charge or $$\Delta$$V=$$\Delta$$PE/q. Since the movement of electrons by the chemical reactions push is pushing the electrons from high potential to low potential, voltage is really just the work done per unit of charge.

Hope I didn't confuse you. :tongue2:

 

[mikelepore]

As soon as you say "potential" you have no choice but to use positives and negatives. The positive side of something is more positive than its negative side, and the negative side of something is more negative than its positive side. If you didn't want to talk about that relative sense, then you wouldn't discuss "potential."

The occasion to neglect the negative sign of a charge is you want to talk about the magnitude of the force on it. If you're wondering how hard it got pushed, but if you don't care which direction it got pushed. But you don't have much occasion to talk about force magnitudes when you're solving electric circuits. Instead, you have your sense of direction in every pairs of +- signs that represents a voltage and every arrow that represents a current.

 

[Red_CCF]

As soon as you say "potential" you have no choice but to use positives and negatives. The positive side of something is more positive than its negative side, and the negative side of something is more negative than its positive side. If you didn't want to talk about that relative sense, then you wouldn't discuss "potential."

The occasion to neglect the negative sign of a charge is you want to talk about the magnitude of the force on it. If you're wondering how hard it got pushed, but if you don't care which direction it got pushed. But you don't have much occasion to talk about force magnitudes when you're solving electric circuits. Instead, you have your sense of direction in every pairs of +- signs that represents a voltage and every arrow that represents a current.

My concern is that, since voltage is energy related, and energy is a scalar quantity, how can it have direction?

 

[mikelepore]

A scalar doesn't have a direction, but you can point in the direction where the scalar becomes greater or the direction where the scalar becomes less. This kind of direction is the "gradient" of the scalar. You've probably seen topographical maps where squiggley lines indicate the various elevations on a hill -- the elevation at each point is a scalar, but you still have direction for traveling uphill or downhill. In an electric circuit, the electric field, which is a vector, is the gradient of the potential, which is a scalar. (More accurately, it the negative of the gradient, since a positive charge spontaneously moves to a lower potential.)

 

[jtbell]

This might have been in the section where the text was explaining the force exerted by two charged objects on each other (Coulomb's Law)? The sign you might get when you calculate the electric force between two objects does not tell one about the direction of the Force.

That's because many introductory textbooks use the version of Coulomb's Law that gives only the magnitude of the force. This version really should be written using absolute-value signs on the charges to make this clear:

$$F = \frac {1}{4 \pi \epsilon_0} \frac {|Q_1| | Q_2|}{r^2}$$

In a version of Coulomb's law that gives the force as a vector (magnitude and direction), you do have to use positive or negative values for the charges as appropriate in order to get the direction right:

$$\vec F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1 Q_2} {r^2} \hat r$$

or

$$\vec F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1 Q_2} {r^3} \vec r$$

 

[pgardn]

That's because many introductory textbooks use the version of Coulomb's Law that gives only the magnitude of the force. This version really should be written using absolute-value signs on the charges to make this clear:

$$F = \frac {1}{4 \pi \epsilon_0} \frac {|Q_1| | Q_2|}{r^2}$$

In a version of Coulomb's law that gives the force as a vector (magnitude and direction), you do have to use positive or negative values for the charges as appropriate in order to get the direction right:

$$\vec F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1 Q_2} {r^2} \hat r$$

or

$$\vec F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1 Q_2} {r^3} \vec r$$

The high school texts most often do not. I have looked at quite a few.
Most of the more contempary college first year texts do not use the last two equations. They go absolute value (at least they use it). The older introductory versions will use the last two.

 

[mikelepore]

Same thing with F=qE, in many introductory courses they use absolute values.

Also in E=V/d, they drop the negative sign [reference table for New York Regents physics].