V=IR - bad science or am I being thick?

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V=IR - bad science or am I being thick?

Postby tomandlu » Mon Oct 10, 2016 12:28 pm UTC

In this article:

https://www.theguardian.com/technology/2016/oct/10/samsung-how-batteries-work-smartphones-explode

there is the following statement:

The higher the resistance the harder the battery has to work to maintain a usable voltage and so the amount of power it can produce per charge decreases. You might remember this bit from school:

Voltage = Current x Resistance (V=IR)


To me, that sounds wrong. Apart from anything else, the equation, if taken at face value, would imply that voltage rises with resistance. However, my conceptual understanding is that V=IR describes a relationship to voltage rather than a mechanism - i.e. you can't increase the voltage by increasing the resistance; increasing the resistance would just result in lowering the current. So, who's got the wrong end of the stick here?
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Re: V=IR - bad science or am I being thick?

Postby doogly » Mon Oct 10, 2016 12:39 pm UTC

It's an equality, you can change the pieces either way. Or both ways! Let's say I've got some current flowing through a loop, and it satisfied V=IR for some values. I raise the value for R. What happens?

That depends on what is being held fixed. If there's a battery with a constant V that is providing the current, V is fixed and if R goes up, I goes down. If I'm connected to a constant source of I, then if R goes up the V goes up. You can have such a thing,
https://en.wikipedia.org/wiki/Current_source
though constant V's are often more common.
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Re: V=IR - bad science or am I being thick?

Postby Soupspoon » Mon Oct 10, 2016 12:46 pm UTC

((@OP)) You're not wrong in that without a mechanism to raise the voltage (around the element being "V=IR" monitored), making the resistance higher reduces the current to match that available from the constant V (say from a cell or battery of cells) with fixed potential.

If, however, the supply to the component(s) can supply a dynamic voltage whilst monitoring and maintaining the curremt to keep that constant, it must provide greater voltage in proportion to the greater resistance in order to keep the I steady.

What "V=IR" is, is not "ensure this current and this resistance and you will create this voltage" but that for this I and this R to be true, that V must be true. Do not give that V, and you will find the I is not what you stated and/or the R is not what you thought it was, either.

V=IR is the same as R=V/I or I=V/R, and it cares not which values you do not know. It is up to you to work out how you use the equation, in which configuration, to fill in any gaps you have.

And the "harder the battery has to work" is the current, lest that confuse you.

(At the risk of tripping up over something and describing it in a way that a full-on electrical engineer woyld find less than exacting.)
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Re: V=IR - bad science or am I being thick?

Postby tomandlu » Mon Oct 10, 2016 12:59 pm UTC

Okay, so in relation to a (smartphone) battery, is the quote right, wrong or ambiguous?

My sort-of understanding of a battery is that voltage is the potential, current is how quickly that potential is being consumed, and resistance is how you control the flow of current.
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Re: V=IR - bad science or am I being thick?

Postby Soupspoon » Mon Oct 10, 2016 1:11 pm UTC

I say it is right, insofar as that extracted quote goes.

In "water pipe" analogy, resistance is a thin pipe, current is water volume being shifted through the system (faster in the thinner bits, to shove the same volume past the spot) and voltage is the effort available, voltage being the amount of oomph (by gravity feed or pump, your choice) applied.

If you assume the water pressure does not change, all the pipes must now let more water pass through to ensure the constricted part is fed enough to overcome the restriction, and this higher volume of water both depletes the reservoir quicker and potentially creates additional damage to the pipe-components (the restriction may burst, from the high velocity, like blowing overheating a resistor, or certainly do the watery equivalent to the current through an incandescent lightbulb filament, probably producing loud pipe noises if nothing else). But it's a loose analogy, at best.

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Re: V=IR - bad science or am I being thick?

Postby jaap » Mon Oct 10, 2016 2:12 pm UTC

tomandlu wrote:In this article:

https://www.theguardian.com/technology/2016/oct/10/samsung-how-batteries-work-smartphones-explode

there is the following statement:

The higher the resistance the harder the battery has to work to maintain a usable voltage and so the amount of power it can produce per charge decreases. You might remember this bit from school:

Voltage = Current x Resistance (V=IR)


To me, that sounds wrong. Apart from anything else, the equation, if taken at face value, would imply that voltage rises with resistance. However, my conceptual understanding is that V=IR describes a relationship to voltage rather than a mechanism - i.e. you can't increase the voltage by increasing the resistance; increasing the resistance would just result in lowering the current. So, who's got the wrong end of the stick here?


Notice also that in that quote, resistance refers to the internal resistance of the battery. The phone itself will have a particular required values for its input voltage and current for it to work properly (which are related to the phone's own resistance by V=I*R). If the battery's internal resistance rises, it would take more power to run at the required values.

Electric power has the formula P=V*I or P=R*I^2. You can use the former to get the power needed by the phone itself. If you knew the battery's internal resistance, you could use the second formula to get the amount of power wasted as heat generated inside the battery itself.

As others have said, what actually happens when the internal resistance increases depends on the battery type. It might produce a lower voltage, a lower current, or both, or it might still produce close to the right current and voltage but then run out of charge quicker and run hotter.

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Re: V=IR - bad science or am I being thick?

Postby Tub » Mon Oct 10, 2016 2:14 pm UTC

The higher the resistance the harder the battery has to work to maintain a usable voltage[...]

True. If you need a constant voltage (because your transistors require a certain voltage to work), then higher resistance means you need higher current.
A battery contains charge, which is I*t (current * time), usually specified in (milli-)Ampere-hours. If you need a higher current, you get less time before the charge is used up.
[..]and so the amount of power it can produce per charge decreases.

Nonsensical to a physicist who equates "power" with "work over time". But they mean to say that your battery doesn't last as long, which is true.

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Re: V=IR - bad science or am I being thick?

Postby Minerva » Sat Nov 12, 2016 4:42 am UTC

V=IR is very "good" science, at least if every bit of electrical engineering people have ever built is anything to go by.

"The higher the resistance the harder the battery has to work to maintain a usable voltage and so the amount of power it can produce per charge decreases."

This explanation is terrible.

Consider an ideal voltage source (ignoring battery discharge, or internal resistance, or current limiting, or things catching fire) connected in a circuit with a single resistor.

"The higher the resistance the harder the battery has to work to maintain a usable voltage"

Not true.
The voltage across the voltage source is equal to the voltage across the resistor, and it remains the same, regardless of the resistance value.
In real-world batteries this won't quite be true, and there will be some voltage sag under high load - this can be understood when you think about the nonzero internal resistance of the battery. This resistance has some voltage drop proportional to the current drawn through it. (V=IR).

Power dissipation in the resistor equals power "drain" from the battery, and this is P=V2/R.
So if the resistance is higher then the battery literally does a lesser rate of work - the opposite of what it says.

(P=V2/R is just a mathematical mashup of V=IR and P=VI. You will also note that P = I2R).

"and so the amount of power it can produce per charge decreases"

This is a poor wording/explanation because power is a time derivative - a "flow rate" unit - and charge is not.

One volt is the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. This is the definition in "rate units - current and power. A watt is the power required to move one coulomb per second (one amp) through a potential difference of a volt.

To say the same thing in energy/charge units, without the time derivative, a joule is the work required to move an electric charge of one coulomb through a potential difference of one volt. (Analogous to the electron-Volt, one joule is one coulomb-Volt.)
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Re: V=IR - bad science or am I being thick?

Postby KnightExemplar » Sat Nov 12, 2016 6:14 pm UTC

Everyone step back for a bit, and remember one thing: Mathematical models are made, and then we find components that match those models. Or... we find a bunch of cheap stuff, then we make math that describes them. That's it.

A diode for example has I = x^V (yes, exponential amounts of current are let through the higher the voltage is), effectively turning any voltage-source into a current source (in practice, batteries are current-limited and voltage-limited. If you hit the current-limit first, then voltage will drop to meet the maximum current. So we usually pretend that diodes are one-way with a "voltage drop", because that's a lot easier than exponents). Or things catch on fire (aka: fuses, which catch on fire at 3A safely, preventing other things from catching on fire). We also tend to not use things that catch on fire (exception: Lithium Ion, because consumers want even more energy storage in smaller and lighter form factors. So we put up with the whole Lithium-fire issue)

There are plenty of electrical components that don't follow Ohm's law: Capacitors, Inductors, Transistors... in fact, that's why they are used. Because we can manipulate electricity when these different things affect electricity differently.

-------------

The three "ideal" electrical components: capacitors (dv = IC), inductors (di = LV), and resistors (V=IR) don't even match up to all passives! Ferrite Beads are utterly ridiculous components: acting like an inductor at less than 10MHz, and acting like a resistor at 100MHz, and then they act like a capacitor at 1GHz (I'm grossly simplifying here, but that's kinda how people think of Ferrite Beads). We use Ferrite Beads because their unique frequency-dependent nature allows for filtering applications. Also, they're cheap as all fuck and very effective.

So why do we use V=IR? Because its easy to remember, easy to calculate, and there's a bunch of super-cheap components called resistors that follow the math. It turns out that you can sorta-kinda-maybe pretend that everything has some combination of resistance / capacitance / inductance. (except when you can't: like transistors or diodes) We can pretend that batteries are a Voltage source with internal resistance. But lets be honest here: a lead-acid battery is about Pb + HSO4- -> PbSO4 + H+ + 2e-, and any discussion about batteries without discussing the chemical reaction is utter shit.

But it turns out that Voltage Source + Resistor is actually kinda close to how batteries "feel like" when you're working with them. Its easy math, but it tells us jack shit about how batteries really work. You really need to discuss chemistry.

In fact, we can pretend that batteries are a current source with an internal parallel resistance. And that of course, can be converted into a voltage source + resistor. Its mathematically equivalent. Pretend it to be whatever you want it to be, it doesn't matter as long as the math comes out right at the end.

It also means that these models are utterly worthless for describing things chemically. Because these are ideal mathematical constructs that were designed to work on anything. Just like how all of physics turns out to be a Taylor series polynomial, all of electronics turn out to be a Thevenin-equivalent circuit because Thevenin-equivalent circuits can represent a lot of shit!

---------

So why do Lithium-based batteries explode? Well...

Image

The further you go to the top-left, the more reactive the elements are. We started with Lead-acid batteries (using the metal Pb, #82 on the periodic table). As a metal, Lead is heavy... but we can use more "powerful" elements if we want to get lighter batteries. Latter batteries (NiCD or NiMH) use Nickle #28 on the periodic table. Notice that it's "up" and "left" from Pb, which means it is more reactive.

You know what? Lets just cut to the chase. What's the most reactive metal on the Periodic Table? Well... that's easy. The one on the top-left, a little substance known as Lithium. #3 on the periodic table, a single valance electron in the 2s shell. Its literally explosive. Yeah, lets pack a whole punch of that into a metal can and see what happens.

So yeah, Lithium-batteries explode because Lithium is a crazy metal.

Now yes, technically... Lithium-ion batteries just use Lithium-ions. They don't (typically) have solid bunches of Lithium lying around, ready to explode. But when Lithium-ion batteries are used incorrectly, Lithium builds up and the batteries can explode.

A typical Lithium-ion battery actually uses Lithium-Metal-Oxides (like LiCoO2) and LiC (Lithiated Carbon), and they transfer the lithium-atoms back and forth to store energy, or to release its stored energy. But if those Li atoms get out of whack, the Lithium can break free from its bonds and start reacting with the outside world (which is explosive). Or thermal runaway just... makes the whole battery get overheated to the point of fire. Lots of different reasons why Lithium-ion batteries can explode.

Sooo... no one really knows why Samsung's batteries explode. There are too many legitimate reasons and possibilities. Undervoltage cause lithium plating. Overcharging can cause lithium plating. Thermal runaway causes fires. Whatever the issue is... its Samsung's problem to figure out.
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Re: V=IR - bad science or am I being thick?

Postby Jorpho » Tue Nov 15, 2016 7:10 pm UTC

KnightExemplar wrote:The further you go to the top-left, the more reactive the elements are.
Clarification: metals in the lower left are more reactive than those in the upper left. Lithium in water will fizz around enthusiastically; sodium will ignite, and it gets more exciting from potassium on down. (Francium is too radioactive to be isolated in meaningful quantities, but that's just because of the size of the nucleus.) The further away from the nucleus the outermost electron gets, the more easily it is released.

Metals in the upper left are, however, more "electronegative" and will result in greater voltage being produced when used in a battery. Because reasons.

A typical Lithium-ion battery actually uses Lithium-Metal-Oxides (like LiCoO2) and LiC (Lithiated Carbon)
I think LiPol (lithium polymer) is the preferred term.

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Re: V=IR - bad science or am I being thick?

Postby KnightExemplar » Tue Nov 15, 2016 11:46 pm UTC

Jorpho wrote:
A typical Lithium-ion battery actually uses Lithium-Metal-Oxides (like LiCoO2) and LiC (Lithiated Carbon)
I think LiPol (lithium polymer) is the preferred term.


LiPol IIRC is a less explosive version of the chemistry. Its still a Lithium-Ion battery but they're changing the chemistry to the point that it does deserve a new name.

Jorpho wrote:
KnightExemplar wrote:The further you go to the top-left, the more reactive the elements are.
Clarification: metals in the lower left are more reactive than those in the upper left. Lithium in water will fizz around enthusiastically; sodium will ignite, and it gets more exciting from potassium on down. (Francium is too radioactive to be isolated in meaningful quantities, but that's just because of the size of the nucleus.) The further away from the nucleus the outermost electron gets, the more easily it is released.

Metals in the upper left are, however, more "electronegative" and will result in greater voltage being produced when used in a battery. Because reasons.


I stand corrected. It seems like I got it backwards.
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Re: V=IR - bad science or am I being thick?

Postby jaap » Wed Nov 16, 2016 7:01 am UTC

Jorpho wrote:
KnightExemplar wrote:The further you go to the top-left, the more reactive the elements are.
Clarification: metals in the lower left are more reactive than those in the upper left. Lithium in water will fizz around enthusiastically; sodium will ignite, and it gets more exciting from potassium on down. (Francium is too radioactive to be isolated in meaningful quantities, but that's just because of the size of the nucleus.) The further away from the nucleus the outermost electron gets, the more easily it is released.

Although each atom is more reactive further down the alkali metal column of the table, they also have more mass, so per kg the metals do necessarily not get more reactive. The tv show Brainiac faked their "experiment" because nothing much happened when they dropped caesium in a bathtub of water:
https://en.wikipedia.org/wiki/Brainiac: ... ed_results

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Re: V=IR - bad science or am I being thick?

Postby Zamfir » Wed Nov 16, 2016 12:26 pm UTC


LiPol IIRC is a less explosive version of the chemistry. Its still a Lithium-Ion battery but they're changing the chemistry to the point that it does deserve a new name.

Not even that, it's mostly a packaging format. Lithium polymer is lithium ion in a soft bag, instead of a hard cylinder. I think there's some difference in the solvents, they have to use special solvents that are gelly instead of fully liquid. But AFAIK, most or all li-ion chemistries can be packaged as li-poly.

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Re: V=IR - bad science or am I being thick?

Postby Zamfir » Wed Nov 16, 2016 12:29 pm UTC


LiPol IIRC is a less explosive version of the chemistry. Its still a Lithium-Ion battery but they're changing the chemistry to the point that it does deserve a new name.

Not even that, it's mostly a packaging format. Lithium polymer is lithium ion in a soft bag, instead of a hard cylinder. I think there's some difference in the solvents, they have to use special solvents that are gelly instead of fully liquid. But AFAIK, most or all li-ion chemistries can be packaged as li-poly.

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Re: V=IR - bad science or am I being thick?

Postby KnightExemplar » Wed Nov 16, 2016 3:44 pm UTC

Zamfir wrote:

LiPol IIRC is a less explosive version of the chemistry. Its still a Lithium-Ion battery but they're changing the chemistry to the point that it does deserve a new name.

Not even that, it's mostly a packaging format. Lithium polymer is lithium ion in a soft bag, instead of a hard cylinder. I think there's some difference in the solvents, they have to use special solvents that are gelly instead of fully liquid. But AFAIK, most or all li-ion chemistries can be packaged as li-poly.


Interesting.

https://en.wikipedia.org/wiki/Lithium_polymer_battery

This article concerns the second, more extended meaning (among the general public), while the first meaning (understood in research and academia) is discussed only in the last section.


Seems like your definition is more common (soft pouch). As in, anything labeled "Lithium Polymer" just means a new packaging. I've assumed that it meant the "first meaning" from Wikipedia, which actually describes a different chemistry.
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Re: V=IR - bad science or am I being thick?

Postby Zamfir » Wed Nov 16, 2016 5:33 pm UTC

Their "first meaning" is still a kind of packaging, really. They refer to batteries with a solid elecrrolyte, that is, a solid that can transport ions through it but not electrons. That's special chemistry indeed, but not in the basic reactions of the battery. The goal is to make a truly solid battery. Current lipo batteries have an electrolyte that's rather viscous (or it would flow around in the pouch), but in the end it's still a liquid solvent with ions flowing through it.

So they took the idea of the solid battery, and made something vaguely like it without the need for the weird solid electrolytes.

Here's a fun visit to a Chinese battery factory. https://learn.sparkfun.com/tutorials/ho ... s-are-made

I am always amused how anything, no matter how high tech at some point, eventually ends up as a bunch of dubious machines in a warehouse.


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