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21.5 Voltage by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.

Electrical circuits can be used for sending signals, storing information, or doing calculations, but their most common purpose by far is to manipulate energy, as in the battery-and-bulb example of the previous section. We know that lightbulbs are rated in units of watts, i.e., how many joules per second of energy they can convert into heat and light, but how would this relate to the flow of charge as measured in amperes? By way of analogy, suppose your friend, who didn't take physics, can't find any job better than pitching bales of hay. The number of calories he burns per hour will certainly depend on how many bales he pitches per minute, but it will also be proportional to how much mechanical work he has to do on each bale. If his job is to toss them up into a hayloft, he will get tired a lot more quickly than someone who merely tips bales off a loading dock into trucks. In metric units,

`"joules"/"second"="haybales"/"second"×"joules"/"haybale".`

Similarly, the rate of energy transformation by a battery will not just depend on how many coulombs per second it pushes through a circuit but also on how much mechanical work it has to do on each coulomb of charge:

`"joules"/"second"="coulombs"/"second"×"joules"/"coulomb"`

or

`"power=current×work per unit charge".`

Units of joules per coulomb are abbreviated as *volts*, 1 V=1 J/C, named after the Italian physicist Alessandro Volta. Everyone knows that batteries are rated in units of volts, but the voltage concept is more general than that; it turns out that voltage is a property of every point in space. To gain more insight, let's think more carefully about what goes on in the battery and bulb circuit.

To do work on a charged particle, the battery apparently must be exerting forces on it. How does it do this? Well, the only thing that can exert an electrical force on a charged particle is another charged particle. It's as though the haybales were pushing and pulling each other into the hayloft! This is potentially a horribly complicated situation. Even if we knew how much excess positive or negative charge there was at every point in the circuit (which realistically we don't) we would have to calculate zillions of forces using Coulomb's law, perform all the vector additions, and finally calculate how much work was being done on the charges as they moved along. To make things even more scary, there is more than one type of charged particle that moves: electrons are what move in the wires and the bulb's filament, but ions are the moving charge carriers inside the battery. Luckily, there are two ways in which we can simplify things:

**The situation is unchanging.** Unlike the imaginary setup in which we attempted to light a bulb using a rubber rod and a piece of fur, this circuit maintains itself in a steady state (after perhaps a microsecond-long period of settling down after the circuit is first assembled). The current is steady, and as charge flows out of any area of the circuit it is replaced by the same amount of charge flowing in. The amount of excess positive or negative charge in any part of the circuit therefore stays constant. Similarly, when we watch a river flowing, the water goes by but the river doesn't disappear.

**Force depends only on position.** Since the charge distribution is not changing, the total electrical force on a charged particle depends only on its own charge and on its location. If another charged particle of the same type visits the same location later on, it will feel exactly the same force.

The second observation tells us that there is nothing all that different about the experience of one charged particle as compared to another's. If we single out one particle to pay attention to, and figure out the amount of work done on it by electrical forces as it goes from point A to point BB along a certain path, then this is the same amount of work that will be done on any other charged particles of the same type as it follows the same path. For the sake of visualization, let's think about the path that starts at one terminal of the battery, goes through the light bulb's filament, and ends at the other terminal. When an object experiences a force that depends only on its position (and when certain other, technical conditions are satisfied), we can define an electrical energy associated with the position of that object. The amount of work done on the particle by electrical forces as it moves from A to B equals the drop in electrical energy between A and B. This electrical energy is what is being converted into other forms of energy such as heat and light. We therefore define voltage in general as electrical energy per unit charge:

The difference in voltage between two points in space is defined as

where `DeltaPE_"elec"` is the change in the electrical energy of a particle with charge `q` as it moves from the initial point to the final point.

The amount of power dissipated (i.e., rate at which energy is transformed by the flow of electricity) is then given by the equation

`=>` The 1.2 V rechargeable battery in figure j is labeled 1800 milliamp-hours. What is the maximum amount of energy the battery can store?

`=>` An ampere-hour is a unit of current multiplied by a unit of time. Current is charge per unit time, so an ampere-hour is in fact a funny unit of *charge*:

`(1 A)(1 hour)=(1 C"/"s)(3600 s)`

`=3600 C`

1800 milliamp-hours is therefore `1800×10^(-3)×3600 C=6.5×10^3 C`. That's a huge number of charged particles, but the total loss of electrical energy will just be their total charge multiplied by the voltage difference across which they move:

`DeltaPE_("elec")=qDeltaV`

`=(6.5×10^3 C)(1.2 V)`

`=7.8 kJ`

`=>` Doorbells are often rated in volt-amps. What does this combination of units mean?

`=>` Current times voltage gives units of power, `P=IDeltaV`, so volt-amps are really just a nonstandard way of writing watts. They are telling you how much power the doorbell requires.

`=>` If a 9.0-volt battery causes 1.0 A to flow through a lightbulb, how much power is dissipated?

`=>` The voltage rating of a battery tells us what voltage difference `DeltaV` it is designed to maintain between its terminals.

`P=IDeltaV`

`=9.0 ADeltaV`

`=9.0 C"/"s*J"/"C`

`=9.0 J"/"s`

`=9.0 W`

The only nontrivial thing in this problem was dealing with the units. One quickly gets used to translating common combinations like `A*V` into simpler terms.

Here are a few questions and answers about the voltage concept.

*Question: *OK, so what *is* voltage, really?

*Answer: *A device like a battery has positive and negative charges inside it that push other charges around the outside circuit. A higher-voltage battery has denser charges in it, which will do more work on each charged particle that moves through the outside circuit.

To use a gravitational analogy, we can put a paddlewheel at the bottom of either a tall waterfall or a short one, but a kg of water that falls through the greater gravitational energy difference will have more energy to give up to the paddlewheel at the bottom.

*Question: *Why do we define voltage as electrical energy divided by charge, instead of just defining it as electrical energy?

*Answer: *One answer is that it's the only definition that makes the equation `P=I*DeltaV` work. A more general answer is that we want to be able to define a voltage difference between any two points in space without having to know in advance how much charge the particles moving between them will have. If you put a nine-volt battery on your tongue, then the charged particles that move across your tongue and give you that tingly sensation are not electrons but ions, which may have charges of `+e`, `-2e`, or practically anything. The manufacturer probably expected the battery to be used mostly in circuits with metal wires, where the charged particles that flowed would be electrons with charges of `-e`. If the ones flowing across your tongue happen to have charges of `-2e`, the electrical energy difference for them will be twice as much, but dividing by their charge of `-2e` in the definition of voltage will still give a result of `9 V`.

*Question: *Are there two separate roles for the charged particles in the circuit, a type that sits still and exerts the forces, and another that moves under the influence of those forces?

*Answer: *No. Every charged particle simultaneously plays both roles. Newton's third law says that any particle that has an electrical force acting on it must also be exerting an electrical force back on the other particle. There are no “designated movers” or “designated force-makers.”

*Question: *Why does the definition of voltage only refer to voltage *differences*?

*Answer: *It's perfectly OK to define voltage as `DeltaV=(DeltaPE_"elec")/q`. But recall that it is only *differences* in interaction energy, `U`, that have direct physical meaning in physics. Similarly, voltage differences are really more useful than absolute voltages. A voltmeter measures voltage differences, not absolute voltages.

**A** A roller coaster is sort of like an electric circuit, but it uses gravitational forces on the cars instead of electric ones. What would a high-voltage roller coaster be like? What would a high-current roller coaster be like?

**B** Criticize the following statements:

- “He touched the wire, and 10000 volts went through him.”
- “That battery has a charge of 9 volts.”
- “You used up the charge of the battery.”

**C** When you touch a 9-volt battery to your tongue, both positive and negative ions move through your saliva. Which ions go which way?

**D** I once touched a piece of physics apparatus that had been wired incorrectly, and got a several-thousand-volt voltage difference across my hand. I was not injured. For what possible reason would the shock have had insufficient power to hurt me?

21.5 Voltage by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.