There exists only one current in a circuit?

[dalarev]

There exists only one current in a circuit: What I take that to mean is simply that there is a movement of electrons through the circuit wires.

Does that mean that electrons move at different speeds at distinct parts of a circuit like, say, before and after a resistor? No, that would imply a different dQ/dt, therefore different currents. I'm confused as to how we can measure different voltages before and after a resistor in a circuit, but the current be the same along the circuit..

Any simple circuit like the picture attached..

There is a voltage drop across the resistor, but if there's different voltages, why not different currents?



The question came up looking at this thread:

https://www.physicsforums.com/showthread.php?t=316323"

where each of the 6 LED's in series, for instance, each require 15 mA. If the same current occurs through all the LED's, then all you'd need to figure out is the equivalent resistance, yes?

 

[chroot]

Imagine an analogy with water.

A pool of water rests on the ground. A pump is used to move the water from that low altitude to the top of a complex arrangement of pipes, valves, and whatever other trickery you like.

At the top of the arrangement, the drops of water have a large amount of potential energy. As they fall through the arrangement, they lose energy. Perhaps some drops turn a waterwheel or do other useful work, converting their potential energy into some other form.

Make sure to let the pump run for a while, so everything reaches steady-state.

If you then take a horizontal cross-section through the entire arrangement, you will find that the flow of water through it is constant. Most of the water might go through one large pipe, while only a small amount may go through others, but the total flow of water is the same through any horizontal cross-section.

Electrons behave the same way in circuits. A "pump," a battery or other voltage source, gives the electrons energy. They "fall" through the circuit toward lower energies, giving up their potential energy.

- Warren

 

[dalarev]

If you then take a horizontal cross-section through the entire arrangement, you will find that the flow of water through it is constant. Most of the water might go through one large pipe, while only a small amount may go through others, but the total flow of water is the same through any horizontal cross-section.

Different dQ must mean different dt then.. Water flows through larger cross-sections slower than at narrower areas?

 

[chroot]

Different dQ must mean different dt then.. Water flows through larger cross-sections slower than at narrower areas?

Not necessarily. The speed at which water moves through any particular branch has no relationship to the speed at which it moves through any other. All that matters is that water leaves the system at the same rate it enters at the top, and that requires that the flow of water through any cross-section be the same as any other.

- Warren

 

[Chrisas]

Different dQ must mean different dt then.. Water flows through larger cross-sections slower than at narrower areas?

Don't confuse current and voltage concepts. The current is the amount of charge or water that flows through the "pipe". In a single ideal circuit path, that is the same in the circuit at a particular moment in time. The voltage is related to the amount of potential energy the charge has at any given point to transfer to things like resisters.

At one end of the resister a certain amount of charge passes in during a certain amount of time, or I=dQ/dt. There is no where for the current to go except out the other end of the resister. The resister doesn't have charge accumulating someplace inside it. However, the voltage, or potential energy of the charges can be converted into heat, for example. So the charges can lose energy to the resister. This lost energy shows up as a drop in voltage across the resister.

They reason they don't slow down due to lost energy is because the charges in the "pipe" behind them are pushing them along, and so on, back to the battery which is ultimately supplying the energy that the charges lose in the resister.

Now about that water example. If you assume that the water is not compressible and is a closed pluming system (no water leaking out or being added), then the "dQ/dt" doesn't change even though the speed does change. For water dQ is equivalent to the amount of water that passes through a cross section of the pipe in time dt. For a fat pipe, the cross section area is bigger. So the amount of water, or equivalently the volume of water, (call it dQ) that passes by in time dt is
dQ = BigArea * dt * v_slow
where v_slow is the speed of the water.

In a skinny pipe, the cross section area gets smaller, and the speed gets faster in the same proportions so that dQ, the amount of water passing by is the same.
dQ = SmallArea *dt *v_fast

So the amount of water in a time dt is the same in a small pipe or big pipe even though the speeds are different. However, this is not what happens with charge.

I see that chroot beat me to it, but I'll hit submit anyway :)

 

[dalarev]

Not necessarily. The speed at which water moves through any particular branch has no relationship to the speed at which it moves through any other. All that matters is that water leaves the system at the same rate it enters at the top, and that requires that the flow of water through any cross-section be the same as any other.

- Warren

I always thought saying a cross-section implied a cut somewhere in the middle. If not it would just be called the top & bottom. But I do understand what you're saying.

 

[chroot]

You can cut a horizontal cross-section anywhere in the waterworks, add up the flow of all the water through all the branches, and that total will be the same as it would be in any other cross-section.

- Warren

 

[dalarev]

At one end of the resister a certain amount of charge passes in during a certain amount of time, or I=dQ/dt. There is no where for the current to go except out the other end of the resister. The resister doesn't have charge accumulating someplace inside it. However, the voltage, or potential energy of the charges can be converted into heat, for example. So the charges can lose energy to the resister. This lost energy shows up as a drop in voltage across the resister.

They reason they don't slow down due to lost energy is because the charges in the "pipe" behind them are pushing them along, and so on, back to the battery which is ultimately supplying the energy that the charges lose in the resister.

Any number of charges pass through the resistor. Some are "depleted", or used up by the resistor, while the rest of the charges (in the same cross-section) keep moving along as if nothing happened?

 

[vk6kro]

If you take 10 resistors and place then in series across a power source you can take an ammeter and measure the current by opening up the junctions of the resistors and you will always find it is the same wherever you measure it.
Electrons or charges do not get "used up" as they pass through a resistor.

This is a well known and easy to prove practical result, but you may well choose to believe otherwise if you like.

It is also easy to apply a pulse of voltage to one side of a resistor and measure how long it takes for the pulse to appear at the other side of the resistor. The delay is similar to the delay in a piece of wire with the same dimensions as the resistor. That is, it is a propagation speed delay and nothing else.

 

[dalarev]

Sorry, I guess I made it seem like I was trying to disprove something. I'm just trying to convince myself without a trace of doubt, and since it is a proven law, it should be able to consistently handle any question thrown at it.

In other words, I see it like this: Picture a voltage source in series with a resistor R1, that's it. There will be a certain current through this circuit, I1 (assuming DC). Now if we were to place another resistor R2 in series with this configuration, even if it is placed after R1 (farther from the Vsource), then there will only be one lower current flowing through this new circuit, I2.

So now, the presence of R2 in this circuit affects the current flowing through R1, even though it seems like it should get hit with a more direct impact from Vsource than the farther placed R2 would. In other words, one should think in terms of equivalent resistance in this case.

Now the only confusing aspect about that is if you take the voltage across each R; Then we have voltage dividers, etc. I suppose the source of my confusion is the mix up between voltage/current concepts.

 

[vk6kro]

I suspect you just shifted the goal posts.
Now you have a problem with Ohm's Law?

The resistors and power source are in a loop. Whatever voltage is produced by the power source appears the resistors in the same ratio as their resistance. No resistor is more vulnerable to the voltage than any other because the current flowing through all of them is the same. The current depends on the voltage and the total resistance in the circuit.

The voltage developed across each resistor is proportional to its resistance and these voltages add up to the voltage supplied by the power source.

This seems so elegant and powerful that I just can't see any problem with it.

 

[TurtleMeister]

There is a voltage drop across the resistor, but if there's different voltages, why not different currents?

I think the water analogy has confused you about what current is. It's true that when water flows through a pipe restriction (resistor) it's speed increases. But that's not what current is. Current is the amount that flows past a certain point in a given amount of time. If you measured this in the narrow restricted part of a pipe and compared it to the wider part of the pipe you would find that they are both the same. The different speeds are analogous to voltage. The higher the speed of the water, the higher the voltage drop. So if you have two restrictions in a pipe, one more restrictive than the other, then the one with the most restriction will have the highest speed water going through it (higher voltage drop). But all parts of the pipe will have the same flow rate (current).

The water analogy is a good one for learning the basics of electricity. But after that it can cause confusion, so you should leave it behind. One reason is that an electrical resistor dissipates energy in the form of heat. A restriction in a pipe does not, at least not to the same extent that an electrical resistor does. And the speed of the current carrier (electrons) does not change when it passes through a resistor. The reason for this is that the current carrier in an electrical circuit is moving at near the speed of light, and it's mass is extremely small. So we can think of electricity as a transfer of energy only.

 

[dalarev]

No resistor is more vulnerable to the voltage than any other because the current flowing through all of them is the same. The current depends on the voltage and the total resistance in the circuit.

That's the source of my confusion right there. Even by the way I worded it, R1 being "closer to the power supply", stems wrongfully from the idea that the voltage source supplies charge or something like that. I was picturing charge flow beginning from the Voltage source all the way to ground.

I think the water analogy has confused you about what current is. It's true that when water flows through a pipe restriction (resistor) it's speed increases. But that's not what current is. Current is the amount that flows past a certain point in a given amount of time. If you measured this in the narrow restricted part of a pipe and compared it to the wider part of the pipe you would find that they are both the same. The different speeds are analogous to voltage. The higher the speed of the water, the higher the voltage drop. So if you have two restrictions in a pipe, one more restrictive than the other, then the one with the most restriction will have the highest speed water going through it (higher voltage drop). But all parts of the pipe will have the same flow rate (current).

This makes sense to me. With resistors, V=I*R, so with I held constant, the relation is clear. I suppose then the problem was I didn't intuitively assume that water flows at the same rate along any section of a complex arrangement of pipes, which in a sense, has to be true. Like the poster above said, water particles push the ones in front along, and so on, so they have to travel at the same speed.

Thanks for the help everybody, simple concepts I know, but I deserve to suffer for not paying close attention in my introductory physics courses.