Battery condition check via voltmeter (with/without Load)

[EEristavi]

Hello,
I read article how to check battery condition appropriately (e.g. car battery).
(Link: https://www.allaboutcircuits.com/textbook/direct-current/chpt-11/battery-ratings/ )
It's written that: in order to get better results, one must check under load.

I'm not EE - I study physics and this fact is not clear for me:
When you see battery, its written 12V (for example), so it is voltage source.
From previous assumption (its fixed voltage source): potential difference shouldn't be dependent on resistance. But as a fact it does - if we see voltage without load it shows different value compared to voltage when we have a load.

So, if its not fixed voltage source, than why giving voltage and amp-hour and not power

 

[anorlunda]

Hello,
I read article how to check battery condition appropriately (e.g. car battery).
(Link: https://www.allaboutcircuits.com/textbook/direct-current/chpt-11/battery-ratings/ )
It's written that: in order to get better results, one must check under load.

I'm not EE - I study physics and this fact is not clear for me:
When you see battery, its written 12V (for example), so it is voltage source.
From previous assumption (its fixed voltage source): potential difference shouldn't be dependent on resistance. But as a fact it does - if we see voltage without load it shows different value compared to voltage when we have a load.

So, if its not fixed voltage source, than why giving voltage and amp-hour and not power

The fixed voltage source is only an approximation.

Batteries produce electricity chemically. Try looking up lead acid battery on Wikipedia.

 

[jim hardy]

From previous assumption (its fixed voltage source): potential difference shouldn't be dependent on resistance. But as a fact it does - if we see voltage without load it shows different value compared to voltage when we have a load.

That shows the 'previous assumption' to be either oversimplified or plain wrong. Like everything else in life, the more you dig the more you learn.

So, if its not fixed voltage source, than why giving voltage and amp-hour and not power

You might start your journey toward "Battery Expert" here
http://batteryuniversity.com/

 

[EEristavi]

Thank you very much for an advice.
I will dig into (this question was bugging me for quite a long time).
Just hope, I don't have to become real battery expert to understand it :D

 

[jim hardy]

Oversimplifying here now,
as the car battery discharges its sulfuric acid leaves the electrolyte making it more like water.
Water being a poor conductor doesn't pass current so easily, resulting in more voltage drop across the electrolyte when current starts to flow..
So a battery that's not fully charged will show reduced voltage when loaded.
A run down car battery can read 12 volts when unloaded but be too weak to illuminate the glovebox light..

 

[EEristavi]

@jim hardy Very nice and easy to understand explanation.

About measured voltage difference when we have a load and when we don't have it:
I'm still reading the source you gave me (BatteryUniversity) and haven't found the answer yet (still reading), but
can you provide easy answer, in a few words?

 

[Rive]

When you see battery, its written 12V (for example), so it is voltage source.
From previous assumption (its fixed voltage source): potential difference shouldn't be dependent on resistance. But as a fact it does - if we see voltage without load it shows different value compared to voltage when we have a load.

So, if its not fixed voltage source, than why giving voltage and amp-hour and not power

Those markings are for the garage personnel, not for engineers. Batteries has internal resistance (it is low for car/lead acid batteries, but still present - the voltage on the battery will drop to 8-10V during starting and it's still OK), their voltage is not '12V' but something between 11 and 13.8V depending on the charge rate and the temperature, without load: their capacity will depend on the discharge surrent and so on...

Batteries are just like any other electrical components. The deeper you dig, the more parameters they has...

 

[Borek]

About measured voltage difference when we have a load and when we don't have it:

Observed voltage is V_{open circuit}-iR - so it all depends on the battery internal resistance. Discharged batteries have a large internal resistance, charged ones much smaller one.

 

[jim hardy]

About measured voltage difference when we have a load and when we don't have it:
I'm still reading the source you gave me (BatteryUniversity) and haven't found the answer yet (still reading), but
can you provide easy answer, in a few words?

Rive and Borek have both given the answer.

Are you familiar with ohm's law ? Voltage drop equals current X resistance ?

Do the lead plates inside the battery have resistance ?
Does the sulfuric acid solution between the plates have resistance?

When load current flows through those resistances , voltage between the battery terminals drops. That effect is less pronounced when the battery is fully charged.

Try it - connect a good digital voltmeter across your car battery and note reading.
Turn the key to ON but do NOT start the engine. Note reading.
Now turn on accessories - heater fan, headlights, windshield wipers, rear window defroster, seat warmers, emergency flashers, whatever else it has. Note reading.

Continue your studies at Battery University. Become an expert in automobile battery charging systems. . Your friends will seek out your help.

 

[EEristavi]

Rive and Borek have both given the answer.

Well, maybe, but it's not clear to me :/

Are you familiar with ohm's law ?

Yes, I'm very well familiar with ohms law and I still can't get it.

Voltage drop equals current X resistance ?

True, but if we have a current source (we have fixed current).However, the amperage varies and it depends on the load.

So , we have neither a fixed current or fixed voltage.
Am I wrong?

P.S. If I'm not wrong, then I haven't studied this case yet and this is why I can't understand it yet.

 

[phinds]

The battery creates a current through the load but that same current also has to go through the battery's internal resistance, so the voltage created on the load is less than the cell value. These values are just made up and do not pretend to represent those of an actual battery except that one always wants a very small resistance in the battery relative to the load, else the load voltage would REALLY be low.

This is just an example of how a 12v cell with a 100ohm internal resistance can only deliver 10.9 volts to a 1000 ohm load. For a 100 ohm load, it would deliver only 6 volts. Get it?

 

[davenn]

True, but if we have a current source (we have fixed current).

no we don't

Observed voltage is Vopen circuit-iR - so it all depends on the battery internal resistance. Discharged batteries have a large internal resistance, charged ones much smaller one.

Another way to look at what Borek said, is , when the internal resistance of the battery increases as it discharges, then the battery is less capable of supplying current to an external load and that is why it isn't a constant current supply

The current capability of the battery is totally ( pretty much) dependent on the internal resistance

 

[EEristavi]

Ok, Summing all the responses I finally got it!

Thank you all for such a great help and the effort - I learned more than I asked )

 

[jim hardy]

I learned more than I asked

Best thing we can learn is how much we don't know. It's the beginning of wisdom . Carry On, young man !

 

[Averagesupernova]

Best thing we can learn is how much we don't know. It's the beginning of wisdom . Carry On, young man !

I know some teenagers I sure wish I could get to believe that.

 

[sophiecentaur]

Just hope, I don't have to become real battery expert to understand it :D

Batteries are tricky things. You can have the feeling of understanding them without becoming an expert but imo you need to be pretty much an expert before you can safely predict what's going to happen and get the best out of a battery.

The current capability of the battery is totally ( pretty much) dependent on the internal resistance

Perhaps that should be re-stated as "can be characterised in terms of 'an internal resistance'". It's all too easy to think that there's a real resistance in there somewhere but it's only a short hand for describing how Energy is Lost when Energy is Delivered by the battery. Of course, we all use the term Internal Resistance all the time because it suits us for most charge storage devices.

 

[anorlunda]

Perhaps that should be re-stated as "can be characterised in terms of 'an internal resistance'". It's all too easy to think that there's a real resistance in there somewhere but it's only a short hand for describing how Energy is Lost when Energy is Delivered by the battery. Of course, we all use the term Internal Resistance all the time because it suits us for most charge storage devices.

I would like to second that point. I prefer to think of these things as a nonlinear V versus I curve. At any point on such a curve, we can draw a straight line tangent which can be described as a fixed voltage and a fixed resistance. However, the resistance changes as a function of the operating point. My only point is that a voltage plus internal resistance is one way to describe it, but not the only way.

I tried to find a simple V-I curve for a lead acid battery to illustrate, but those are hard to find. Time comes into the picture because as we draw current, the state of charge is changing. Therefore, most curves show state of charge on the horizontal axis (see the first picture below). Nonlinear V-I curves are better illustrated by a solar cell (see the second picture below) than a battery.