Help me explain electricity to a friend...

[adamey]

I thought I had a pretty good grasp of electricity, volts, amps, ohms and so on. Until I tried to explain it to a friend and he asked me something I wasn't able to answer.

I was using the usual water example to explain current and voltage. The water flow is representative of current, and a larger pipe (or wire) could carry more water (current) than a smaller one. Voltage is then similar to pressure where a higher pressure could move more water through the same sized pipe.

We then got to disussing watts and that voltage x current gives you watts. We were talking about the idea that you could have a smaller pipe (wire) with a higher pressure (voltage) and that it could carry the same amount of energy (in watts) as a larger pipe (wire) with a lower pressure (voltage).

This came up when we were looking at a 12V booster in the shop that could deliver 12V at over 100A to help start a car. My friend asked how this was possible as the AC outlet has a 15A breaker. I explained we could have 120 volts and 10 amps going into the booster and 12 volts at 100 amps coming out. The wattage was the same (1200 watts).

This is when he said something that had me stumped:

"Since the pressure is higher on the side going into the booster, does that mean the electrons are travelling faster? Just like water would travel faster through a pipe at higher pressure?"

I then realized how the water example isn't always the best at explaining electricity. I said no as electricity moves at the same speed.

"So are there more electrons moving?" This is where I had to think and couldn't come up with an answer and realized I don't know as much as I thought.



So to simplify:

If I had a wire carrying 1V at 1A then I have 1W flowing.

If I increase the voltage to 10V but kept the current at 1A I now have 10W (or 10 times) the energy flowing.

So what is actually changing inside the wire that allows it to carry 10 times the energy by increasing the voltage and leaving current the same? If the electrons aren't moving faster, and we still have the same quantity of electrons (1 amp or 1 coulomb per second), then where is the 10 watts (vs 1 watt) coming from?

 

[Ian Rogers]

Your forgetting one thing... You can't just increase the voltage to 10V as you won't be able to use the same piece of wire... I f you need to keep the current at 1A you will need to resist it more... IE... If 1V, 1A traveled through a wire the wire will have 1 ohms resistance. If you increase the voltage to 10V and still wanted 1A you need to increase the resistance of the wire to 10 ohms.

The water example works fine... Increase the water pressure then increase the flow resistance to suit.

 

[MikeMl]

The other thing you are forgetting is that the "booster" has a lead-acid battery in it, at least mine does. It is charged with a Wall-wart, that can deliver maybe 20W? While starting a V8, the booster is capable of delivering an instantaneous power of 10Vχ700A = 7KW. It may take several hours for the wall-wart to put that back

 

[adamey]

Your forgetting one thing... You can't just increase the voltage to 10V as you won't be able to use the same piece of wire... I f you need to keep the current at 1A you will need to resist it more... IE... If 1V, 1A traveled through a wire the wire will have 1 ohms resistance. If you increase the voltage to 10V and still wanted 1A you need to increase the resistance of the wire to 10 ohms.

The water example works fine... Increase the water pressure then increase the flow resistance to suit.

No, I'm talking about a wire carrying energy to a load. I could have a wire supplied by 1V and carrying 1A to a light bulb that consumes 1W. Or I could use the same wire supplied by 10V and carrying 1A to a light bulb that consumes 10W.

The wire is the same thickness, yet in one example it's carrying 1W and another it's carrying 10W. The current is the same but the voltage has increased. I'm wondering what's happening inside the wire to allow it to carry 10x the energy to my load.

The other thing you are forgetting is that the "booster" has a lead-acid battery in it, at least mine does. It is charged with a Wall-wart, that can deliver maybe 20W? While starting a V8, the booster is capable of delivering an instantaneous power of 10Vχ700A = 7KW. It may take several hours for the wall-wart to put that back

No, this is an industrial booster/charger than can deliver 12V at 100A continuously while it's plugged in. It doesn't have a battery - it's basically a large transformer with some seriously heavy duty rectification diodes.

 

[crutschow]

....................................

This is when he said something that had me stumped:

"Since the pressure is higher on the side going into the booster, does that mean the electrons are travelling faster? Just like water would travel faster through a pipe at higher pressure?"

I then realized how the water example isn't always the best at explaining electricity. I said no as electricity moves at the same speed.

"So are there more electrons moving?" This is where I had to think and couldn't come up with an answer and realized I don't know as much as I thought.

So to simplify:

If I had a wire carrying 1V at 1A then I have 1W flowing.

If I increase the voltage to 10V but kept the current at 1A I now have 10W (or 10 times) the energy flowing.

So what is actually changing inside the wire that allows it to carry 10 times the energy by increasing the voltage and leaving current the same? If the electrons aren't moving faster, and we still have the same quantity of electrons (1 amp or 1 coulomb per second), then where is the 10 watts (vs 1 watt) coming from?

The number of electrons moving (as measured by coulombs) is proportional to the current, independent of voltage. If you increase the current, the electrons will move faster (for normal current flow, the electrons move quite slowly, perhaps a few tenths of a mm per second). A voltage increase has little effect on electron movement for the same current, expect perhaps to squeeze them very slightly closer together. The difference is that now the electrons have more "push" and can thus do more work. Looking at the water analogy, higher pressure means that a given amount of water flow could generate more work, for example if fed to a water turbine.

 

[Ian Rogers]

No, I'm talking about a wire carrying energy to a load. I could have a wire supplied by 1V and carrying 1A to a light bulb that consumes 1W. Or I could use the same wire supplied by 10V and carrying 1A to a light bulb that consumes 10W.

The wire is the same thickness, yet in one example it's carrying 1W and another it's carrying 10W. The current is the same but the voltage has increased. I'm wondering what's happening inside the wire to allow it to carry 10x the energy to my load

I know you are... The wire will also have a current rating. Current flow is the same at every instance of the circuit. If the resistance (restriction) is the light bulb (10W) as long as the wire can cope with 1A, all will be fine. Just because a wire can take currents up to 16A doesn't mean it can't carry less.

If 600 cars are travelling down the M4 in a jam and roadworks restricts flow to 100 cars an hour, then only 100 cars an hour moves all the way along, even if the road was designed for 2 to 3 thousand cars an hour.

 

[nsaspook]

One problem with going too far with the "water" analogy is that "electrons" don't do the work in a electrical circuit as the actual kinetic energy from the electron flow in a DC circuit is a tiny fraction of the energy flowing from source to load via the wire. The heat generated from the movement of electrons (current flow) in the wire is an effect of the fields that travel outside the wire.

https://www.st-andrews.ac.uk/~jcgl/Scots_Guide/audio/part6/page2.html

We therefore have to conclude that the real burden of the signal and energy transfer is being carried elsewhere, and not simply by the electrons in the wires.

So your question is a good one.

So what is actually changing inside the wire that allows it to carry 10 times the energy by increasing the voltage and leaving current the same? If the electrons aren't moving faster, and we still have the same quantity of electrons (1 amp or 1 coulomb per second), then where is the 10 watts (vs 1 watt) coming from?

 

[crutschow]

One problem with going too far with the "water" analogy is that "electrons" don't do the work in a electrical circuit as the actual kinetic energy from the electron flow in a DC circuit is a tiny fraction of the energy flowing from source to load via the wire. The heat generated from the movement of electrons (current flow) in the wire is an effect of the fields that travel outside the wire.

The water molecules don't do the work either. It's not the kinetic energy in the water that carries the energy, it's the pressure times the flow volume, analogous to volts times current.

And the heat generated in the wire is due to the resistance of the wire, not from any fields outside the wire. If it were the fields, then you wouldn't have superconducting wire that can carry current with no resistance or heat loss.

 

[nsaspook]

The water molecules don't do the work either. It's not the kinetic energy in the water that carries the energy, it's the pressure times the flow volume, analogous to volts times current.

And the heat generated in the wire is due to the resistance of the wire, not from any fields outside the wire. If it were the fields, then you wouldn't have superconducting wire that can carry current with no resistance or heat loss.

Water:
Pressure is potential energy stored in a dam. When water moves down a pipe to spin a turbine that is expressed as kinetic energy ie "energy in motion"
Electrical energy: battery
The battery voltage is separated charge (potential energy) , that charge is like the water in the dam. When that charge moves (because of an electric field created from that charge separation) via the electrons (charge carriers) in the wire the energy of the electricity is moved in fields (mainly magnetic with low resistance wire) that surround the wire moving at the speed of light that travel down (energy in motion) both sides of a completed circuit to the load. The resistance of the wire to the movement of charge caused by that electric field at the source (battery) creates a voltage field from the difference in charge density along the wire (voltage drop). This combined EM energy field heats the atoms/electrons in the wire. (just like a induction heater)

But with a superconductor you have all the energy contained in one field (magnetic) and because there is no resistance there is no voltage field so there is no energy movement transported by the current flow in the superconductive wire as all the energy used to produce the current in the coil goes into creating the magnetic field.

 

[Ratchit]

adamey,

It is never a good idea to use an analogy to explain another physical system. Analogies are good for making a point, not modeling another system. The often used method of using hydraulic principles to explain electrical principles just doesn't work very well. Do hydraulic engineers use electrical principles to explain hydraulics?

We then got to disussing watts and that voltage x current gives you watts. We were talking about the idea that you could have a smaller pipe (wire) with a higher pressure (voltage) and that it could carry the same amount of energy (in watts) as a larger pipe (wire) with a lower pressure (voltage).

Energy is measured in joules or kilowatt-hours, not watts. Power is measured in watts. One joule per second is one watt.

I think you know that current represents the number of charge carriers moving past a point per unit of time. What you probably don't know is that volts is the energy density of the charges, or joules/coulomb. If there is a higher voltage at one end of a load, the charge carriers are going to travel through the load to where the voltage or energy density is less. During their travel through the load, they are going to lose energy in the form of heat, and arrive at the other end with less energy and thereby less energy density per unit of charge.

If I had a wire carrying 1V at 1A then I have 1W flowing.

If I increase the voltage to 10V but kept the current at 1A I now have 10W (or 10 times) the energy flowing.

So what is actually changing inside the wire that allows it to carry 10 times the energy by increasing the voltage and leaving current the same? If the electrons aren't moving faster, and we still have the same quantity of electrons (1 amp or 1 coulomb per second), then where is the 10 watts (vs 1 watt) coming from?

Assuming the resistance of the wire is negligible, the energy density of the charge delivered to the load at 10 volts is 10 times what it was at 1 volt. You will have to increase the resistance of the load to 10 ohms to keep the current at 1 amp. So the energy density of the charge drops to zero after passing through the load and 10 watts of power is dissipated as heat.

So by defining voltage what is really is (energy density of the charge in joules/coulomb) instead of what it is not (pressure in pascals), all of the confusion is eliminated.

Ratch