# How do we explain powering a load without converting any current?

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#### Irving

Joined Jan 30, 2016
3,994
I watched the first video, thanks, but haven’t finished the second yet, but it seems you still need a potentially difference for a current, and I’m still trying to understand how current returns to the battery on the negative side of the load.
Maybe this diagram will help...

R1, R2 are the resistances of the wires between battery and bulb, small but real. So the PD of the battery is split across the 3 components R1, Lamp, R2. In many cases R1, R2 could be assumed to be zero as they have relatively little effect, but they exist. Each of R1, R2 converts electrical energy to heat (V * I = 0.05 * 5 = 0.25W) but its tiny compared to the ~60W of the lamp.

#### nsaspook

Joined Aug 27, 2009
13,435
I watched the first video, thanks, but haven’t finished the second yet, but it seems you still need a potentially difference for a current, and I’m still trying to understand how current returns to the battery on the negative side of the load.
It's the same reason per the bike chain analogy that links in the chain return, the charge carriers are linked in a system. The individual links (electrons) didn't posses the energy of the chain (it's just a lump of matter), they move as a group (measured as current) in response to electrical energy to form a confined path for the energy to propagate from point A to point B via the wiring conductors. All of the current conduction electrons are part of a existing sea of charges that slowly (drift speed) move in response to a potential (tension) to equalize (a net neutral conductor) the charge distribution. The current returns to neutralize the distribution of charge all along the conductors in response to the applied potential. This neutralization happens a near light speed in response to changes in potential.

#### Jim@HiTek

Joined Jul 30, 2017
59
Look at it this way...you have a hose. In the middle of the hose is a globe. Inside the glob is a mechanical device designed to delight the kids when it moves. At the bottom on the mechanical device is a set of paddles. You turn on the water, and the stream of water hits each paddle in turn which sets the device spinning and doing all sorts of gyrations. Work is being accomplished. Same as light and heat from a filament. The water is returned to the tank to be pumped out again.

Anyway, the current (water) is doing the work of spinning the device, but it's energy is not lost, the same amount of water comes out of the hose as went in.

#### BobaMosfet

Joined Jul 1, 2009
2,119
Take a simple circuits like a 12v battery and a 5 amp filament bulb.

If we place meters before the bulb and after the bulb we see that current is 5 amps on both the positive and negative sides, while the voltage drops to a few millivolts.

Are we not taught that our electrical power is converted to light and heat in a light bulb? But if the current on the negative side of the circuit is identical to that on the positive side, what exactly was converted to light and heat?

If power = volts x amps and no amps were converted does that mean the bulb was powered completely by voltage alone? If not - what else is powering the bulb?

Thanks.
@Mark Flint No, that is not what we are taught. That is a pedagogical explanation that smooths over the truth- dumbs it down for the masses.

Current flowing through the filament causes a thermochemical reaction resulting in heat and photons being cast off as energy. No electrons are lost in the process- they are simply moved around the circuit.

#### Mark Flint

Joined Jun 11, 2017
145
Maybe this diagram will help...

R1, R2 are the resistances of the wires between battery and bulb, small but real. So the PD of the battery is split across the 3 components R1, Lamp, R2. In many cases R1, R2 could be assumed to be zero as they have relatively little effect, but they exist. Each of R1, R2 converts electrical energy to heat (V * I = 0.05 * 5 = 0.25W) but its tiny compared to the ~60W of the lamp.

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Thanks, yes that makes sense to me. I got confused by a previous comment from someone else who seemed to say there would not be any voltage between the load and the negative terminal. I thought I was getting email notifications on updates to this post but didn’t see any.

#### Mark Flint

Joined Jun 11, 2017
145
Anyway, the current (water) is doing the work of spinning the device, but it's energy is not lost, the same amount of water comes out of the hose as went in.
It is interest you say the current’s energy is not lost. My understanding is it entered the negative terminal and diminished the charge separation in the battery. The voltage drop across the load is not responsible for diminishing the charge separation in the battery, but the current entering the opposite pole is the reason the battery discharges.

#### Mark Flint

Joined Jun 11, 2017
145
@Mark Flint No, that is not what we are taught. That is a pedagogical explanation that smooths over the truth- dumbs it down for the masses.

Current flowing through the filament causes a thermochemical reaction resulting in heat and photons being cast off as energy. No electrons are lost in the process- they are simply moved around the circuit.
If an electron is just a negative charge what does it donate to the reaction, and so what has changed with it (if anything) after that reaction?

#### dcbingaman

Joined Jun 30, 2021
1,065
Take a simple circuits like a 12v battery and a 5 amp filament bulb.

If we place meters before the bulb and after the bulb we see that current is 5 amps on both the positive and negative sides, while the voltage drops to a few millivolts.

Are we not taught that our electrical power is converted to light and heat in a light bulb? But if the current on the negative side of the circuit is identical to that on the positive side, what exactly was converted to light and heat?

If power = volts x amps and no amps were converted does that mean the bulb was powered completely by voltage alone? If not - what else is powering the bulb?

Thanks.
First of all 'Voltage' is measured in Joules per Coulomb. If it takes 10 Joules of work to push 1 C of charge 'up hill' like in a battery charger, then there is 10V of potential difference. The current is simply how fast C is passing through some device in seconds. And power is just how fast work is being done Joules per second. So if I have 12 Joules per C and multiply it by C per second I get Joules per Second which is watts. A resistor/bulb has no ability to store energy thus it must be dissipated as heat and/or light. If you have an energy storage device such as a capacitor or inductor then you can have a current through the device with no loss of energy if it happens for example to be a sinusoidal waveform. In this case energy is getting stored in the capacitor or inductor and returned to the source from the capacitor or inductor thus no energy loss in an 'ideal' capacitor or inductor. So it is possible to have certain devices using no energy and yet having voltages and currents across them. But the energy is still there, it is simply being stored and released by capacitors and inductors. Of course even these devices consume some power as there is no perfect capacitor or inductor. Long winded explanation sorry.

#### dcbingaman

Joined Jun 30, 2021
1,065
If an electron is just a negative charge what does it donate to the reaction, and so what has changed with it (if anything) after that reaction?
Another example from real life:
Imagine you have a hill so many feet high (say ten feet), that is the same as voltage. Now imagine you have 1000 1 pound rocks at the bottom of the hill. If you move all of the rocks to the top of the hill you will have performed 10000 foot pounds of work. The number of rocks that went from the bottom of the hill to the top is the same, just as the number of electrons passing from one side of the resistor to the other is the same. Power is how quickly work is done. So if it takes you say 10 hours then the power is 1000 foot-pounds per hour. Now I come along with my front end loader, and scoop up all 1000 of those rocks and move them to the top of the hill in say 1/10 of an hour (6 minutes) we both have performed the same amount of work 10000 foot-pounds but with the tractor I get 100,000 foot-pounds per hour of POWER. Now when you did it by hand, your muscles heated up some to perform that work, because it took you 10 hours the energy had more time to dissipate into the air, thus smaller temperature increase. But my tractor will have significantly more heat because the same amount of energy must now be dissipated in much less time (via the combustion engine) to do that work. It is important to note both my front end loader and you did the same amount of work, but because I did the work in significantly less time the power that is work per unit time was much higher and much more heat gets generated in the engine. The light bulb is no different. This is Fun! Hope you enjoyed that.

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#### crutschow

Joined Mar 14, 2008
34,697
It is interest you say the current’s energy is not lost.
The current does not have energy by itself, it transports the potential energy from the source to the load.
If an electron is just a negative charge what does it donate to the reaction, and so what has changed with it (if anything) after that reaction?
The electron is not changed in any way, shape, or form.
It simply acts as a passive way to transport the energy.

As a loose analogy, think of the electrons in the wire as similar to water molecules in a closed loop pipe with a pump on one end generating pressure (voltage) and a small restriction in the pipe (resistance) at the other end of the loop.
Water flowing through restriction will cause a drop in pressure with a energy loss through the restriction proportion to the flow rate (current) and pressure drop (voltage drop).
The water, of course, does not change in any manner through this process.

You seem to be trying to read more into it than there is.
There's no mystical process to the transfer of the energy, it's simply the current times the voltage drop at the load.

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#### dcbingaman

Joined Jun 30, 2021
1,065
So - just to be clear - nothing is getting converted, ie, from one form of energy to another - such as from chemical (battery) to light/heat. I'm thinking about the laws of thermodynamics - doesn't something need to be converted to end up being heat?
Yes, the electrons are 'colliding' with the atoms of the material increasing their kinetic energy, that is increasing the temperature of the bulb. When the bulb temperature gets above a certain point the heat energy is high enough to emit shorter and shorter wavelength photons. Shorter wavelength photons carry more energy. Look up black body radiation for more information.

#### AnalogKid

Joined Aug 1, 2013
11,148
Don't mean to be pedantic, but isn't that Watt's law?
Unless I am mis-remembering something ...

Joule's Law: P = I^2 x R

Watt's Law: P = E^2 / R

ak

#### dcbingaman

Joined Jun 30, 2021
1,065
Unless I am mis-remembering something ...

Joule's Law: P = I^2 x R

Watt's Law: P = E^2 / R

ak
Yes, but they come about from base unit
1 - P=I*V
2 - I=V/R
3 - V=IR

Thus by replacement of 2 into 1
P = I*V = V/R*V=V^2/R

Thus by replacement of 3 into 1
P = I * IR = I^2*R

And it all works out if you remember V = Joules per C, I=C per second and R=V/I

#### KeepItSimpleStupid

Joined Mar 4, 2014
5,088
The wires and the measuring gizmo drops voltage. The "measuring gizmo"is usually a voltmeter across a resistpr.
That voltmeter might look like a 10 Meg resistor as well, so it's siphoning a tiny amount of current.

#### AnalogKid

Joined Aug 1, 2013
11,148
Yes, but they come about from base unit
My pseudo-question was about the correct assignment of names to the equations.

ak

#### Ya’akov

Joined Jan 27, 2019
9,237
If I create a circuit where I have a pump puliing water from a tank and returning it there, and I use a paddle wheel in between to power something with the water current, there will be a pressure drop analogous to the voltage drop as the energy in the water current is dissipated into the work being done by the wheel, but necessarily no water is consumed and it is all returned to the tank.

Now, adjust this setup so that there are two tanks, one up high and the other low. We “charge” the device by lifting the water, then run it until the upper tank is empty. All the water, stripped of the potential energy of being lifted against gravity, is returned to the lower tank. No water is lost, only the energy. The current (total water volume) doesn’t change, the pressure does.

In a battery, the potential energy and kinetic energy are complementary electrochemical processes, the analogy is very close, though.

#### nsaspook

Joined Aug 27, 2009
13,435
If I create a circuit where I have a pump puliing water from a tank and returning it there, and I use a paddle wheel in between to power something with the water current, there will be a pressure drop analogous to the voltage drop as the energy in the water current is dissipated into the work being done by the wheel, but necessarily no water is consumed and it is all returned to the tank.

Now, adjust this setup so that there are two tanks, one up high and the other low. We “charge” the device by lifting the water, then run it until the upper tank is empty. All the water, stripped of the potential energy of being lifted against gravity, is returned to the lower tank. No water is lost, only the energy. The current (total water volume) doesn’t change, the pressure does.

In a battery, the potential energy and kinetic energy are complementary electrochemical processes, the analogy is very close, though.
The conceptual problem with this analog (and with most analogies) is that current never carries the electrical energy in a electrical circuit. Here the water (as the direct analogy to current) seems to gain energy, travel in a pipe circuit, is latter stripped of energy at a load and returns without energy in the transport system to be energized. That doesn't really happen with circuit current and leads to misconceptions of energy flow in circuits when the water analogy is completely unusable like AC circuits.

It's not "flawed" as in wrong but like all analogies it's a limited metaphor for explaining rudimentary electrical properties of the real thing. Some are a lot better than others on how far you can stretch the analog. Electricity & water don’t mix very well IMO.

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#### Ya’akov

Joined Jan 27, 2019
9,237
The conceptual problem with this analog (and with most analogies) is that current never carries the electrical energy in a electrical circuit. Here the water (as the direct analogy to current) seems to gain energy, travel in a pipe circuit, is latter stripped of energy at a load and returns without energy in the transport system to be energized. That doesn't really happen with circuit current and leads to misconceptions of energy flow in circuits when the water analogy is completely unusable like AC circuits.
The intention of the analogy was very narrow. I realize it is very one dimensional in application but the idea of the storage of potential energy and the creation of kinetic energy is narrowly applicable.

Tjere are certainly dimensions of the electrical circuit not accounted for.

#### nsaspook

Joined Aug 27, 2009
13,435
The intention of the analogy was very narrow. I realize it is very one dimensional in application but the idea of the storage of potential energy and the creation of kinetic energy is narrowly applicable.

Tjere are certainly dimensions of the electrical circuit not accounted for.
Sure, I understand but most people first exposed to the analogy don't have the breath of knowledge to keep it narrow. They usually don't understand the principles of closed-loop hydraulics either.
Teachers assume you know how water flows through a pipe so they make an analogy with electrons in a wire. The endgame is when you don't understand hydraulics you get two things wrong.

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#### Ya’akov

Joined Jan 27, 2019
9,237
Sure, I understand but most people first exposed to the analogy don't have the breath of knowledge to keep it narrow. They usually don't understand the principles of closed-loop hydraulics either.
I suppose the one thing I want to take away from it is that a pressure and voltage drop are cogently analogous when trying to work out this problem. We can see how water can’t change in volume in such a system, there has to be as much on each side, with a difference in pressure.

It’s very hard to deconstruct it, though, since the elements are inherently interactive and the picture isn’t complete without the other parts simultaneously accounted for.

I agree that water analogies for electrical circuits can be very misleading.

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