Alan S
Well-Known Member
How can you get less current out of one end of a wire than you put in to the other?
‘Fessing up. My dubious LiFePo install.
- Thread starter fredrussell
- Start date
lustyd
Well-Known Member
I think this is an important point. While this discussion is interesting, as is Paul's experiment, and I had been thinking similar things, none of the discussion is useful without specifics. It's just old men shouting at clouds until there is some thread of scientific approach. I'm quite sure some will have the specifics so please don't take that as a suggestion otherwise. The whole thing is just voltages and currents so no reason it can't work just fine.
I'm pretty sure that everyone already agreed that modern smart regulated alternators have no problem with lithium, so if it's smart then the alternator wouldn't die and it's not unusual.
The charge profiles are certainly different between lead and lithium, as is the resting voltage. This may very well shorten the life of one or both. If the lithium sits at a higher voltage connected to lead it's as good as leaving a charger on, but with potentially no way to switch it off. That's probably my main concern with permanently linking them, but with a VSR or similar could be a non-issue (assuming the VSR switches off at Lithium voltages while not charging).
Regarding charge profiles - MPPTs reset every morning anyway, so I'm not sure that charge profile is necessarily an argument here, the battery will see charge voltages regularly anyway.
Sorry, that turned into a ramble
lustyd
Well-Known Member
Some is turned to heat in the resistor (aka wire). That's the basics of circuit design - you use various components to put the juice where you want it and account for losses.
MikeBz
Well-Known Member
You can’t. The resistance of the wire results in a voltage drop across the wire (and hence at the battery). A lower voltage at the battery means less current.
MontyMariner
Well-Known Member
Be aware, you reap what you So!
lustyd
Well-Known Member
Usually its the opposite, power is current x voltage
Alan S
Well-Known Member
Your post 15 certainly seemed to imply that the resistance of the long wire reduced the current flowing into the battery but did not reduce the current out of the alternator. That's what puzzled me.
andsarkit
Well-Known Member
Kirchhoff's law in action.
Sea Change
Well-Known Member
Thought provoking.
I've heard of voltage drop, but not current drop.
OTOH resistors (I think) reduce the current flow, and we talk about long skinny wires increasing the resistance in a circuit.
Could somebody better clued up explain what's going on in the real world?
I've heard of voltage drop, but not current drop.
OTOH resistors (I think) reduce the current flow, and we talk about long skinny wires increasing the resistance in a circuit.
Could somebody better clued up explain what's going on in the real world?
Moodysailor
Well-Known Member
Ohm's law - current is directly proportional to voltage (I = V/R)
Electrical resistance is inversely proportional to the cross-sectional area of a conductor. This means that a thinner wire offers more resistance to the flow of current than a thicker wire. When current flows through a wire with resistance, a voltage drop occurs across the wire. This voltage drop reduces the voltage available to the load connected to the wire. A lower voltage reaching the load results in a lower current flow through the load, as current is directly proportional to voltage. A thinner wire has a smaller cross-sectional area, which restricts the flow of electrons, causing a voltage drop across the wire and a reduction in current reaching the load.
Imagine that electricity is like river water - voltage is the speed the water is flowing, and current is the volume of water. In order to get the same amount of water to travel from point a to b it either travels fast or you shift a lot of it. At some point if the resistance is too high, the water (or electricity) can't flow enough. In river terms, this is when floods happen - in electrical terms, this is when the smoke comes out!
Alan S
Well-Known Member
All quite straightforward and understood. It's the bit about the resistance reducing the current to the battery without reducing the current from the alternator that got me.
lustyd
Well-Known Member
Resistors, and wires, will get warmer when the current is too high for their capacity. Just like a fuse does. The heat comes from the current in the wire/fuse/resistor. Current comes from the source, and consumed current * voltage is the power draw on the circuit. If the battery is taking a given number of watts and you drop the voltage, the current may rise to compensate and the alternator work harder to keep up. Generators do similar, and will rev as consumption increases. This is the reason USB PD has a higher Voltage (into the hundreds of Volts now) since the current cannot get bigger in the USB connectors and cables. It's also why electric car chargers use hundreds of volts - it keeps the current down.
billskip
Well-Known Member
That doesn't answer his question.
Sea Change
Well-Known Member
Thanks, I thought I understood all of that but I'm scratching my head a bit.
Intuitively, you'd think that thin wire = higher resistance = reduced power getting through, with the lost power being the wire heating up. But I suspect that's not actually what's happening because a) it doesn't explain the voltage drop, and b) the wire would get very hot to burn off all that excess power.
So the thin wire is affecting both the voltage *and* the current, I think.
It's funny, I've done heaps of electrical work and always just followed the rules, and it's always worked perfectly.
But I've never really asked *why* the voltage drops on long thin runs, and how that affects the rest of the system.
Alan S
Well-Known Member
Can you explain this bit in the context of an alternator charging a LiFePO4 battery via a long wire..
Quote "If the battery is taking a given number of watts and you drop the voltage, the current may rise to compensate"
Quote "If the battery is taking a given number of watts and you drop the voltage, the current may rise to compensate"
lustyd
Well-Known Member
Ohms law is useful for circuit design to match components and voltages, but it doesn't affect power consumption which is P = I * V
billskip
Well-Known Member
If you read ohms law as its written you will understand it.
Anyways any amount of current entering a circuit ..(that circuit could be just a piece of wire)..will be the same amount exiting the circuit..
Ceirwan
Well-Known Member
Modern car alternators actually have surprisingly smart logic.
Depending how new your VW is of course.
But even a lowly Golf has a battery sensor to monitor current in and out and drops the charging voltage when the battery is full. (to try and not fry the expensive AGM batteries that cars with start stop tend to use)
Given its a campervan it makes sense that VW would install some kind of protection on the alternator as owners probably install quite large battery banks.
Depending how new your VW is of course.
But even a lowly Golf has a battery sensor to monitor current in and out and drops the charging voltage when the battery is full. (to try and not fry the expensive AGM batteries that cars with start stop tend to use)
Given its a campervan it makes sense that VW would install some kind of protection on the alternator as owners probably install quite large battery banks.
MikeBz
Well-Known Member
The bigger the voltage drop across the wire the lower voltage applied to the battery terminals. The battery has an internal resistence, so Ohm’s Law applies to the current that the battery draws - hence a lower voltage across the battery terminals means it draws less current.
billskip
Well-Known Member
That's because you are stating a constant..ie "a given number of watts"
As watts is made up in calculations as volts x current if volts go down current must go up to keep your number of watts constant....
However it will not be possible to maintain constant watts if the current goes down and it's not possible to increase the generated voltage.
