Question on How to Measure Current of Battery

Thread Starter

Benjamin0904

Joined Oct 3, 2020
31
Hi everyone! Just recently I stumbled across a problem in one of my electricity projects.
I have a 12V 8Amp Lead Acid Battery. Here is the question:
Imagine that battery was old, but still working, and the label was torn off. How do I figure out that that battery is 8Amps?
I just want some kind of meter to read 8Amps, or some mathematical equation that will equate to 8Amps.
I know how to measure the output voltage with a voltage meter. That's not the problem.
I figured there would be a little ohms law in the equation, but never understood where to begin.
If anybody could simplify this down for me, it would be much appreciated.

Thanks so much,
Benjamin
 

dl324

Joined Mar 30, 2015
12,871
I figured there would be a little ohms law in the equation, but never understood where to begin.
If anybody could simplify this down for me, it would be much appreciated.
It doesn't work that way.

The way you measure battery capacity is you drain it until voltage is at what's considered dead and measure how much current it delivered to that point.

In theory an 8AH battery will provide 8A for 1 hour, or 4A for 2 hours, or 2A for 4 hours, or 16A for half an hour, and so on. At either extreme, other factors need to be considered.

The above doesn't apply to an old battery.
 

WBahn

Joined Mar 31, 2012
26,398
Where are you getting the 8 A rating from? Are you sure it isn't 8 Ah?

Batteries are seldom rated directly in terms of amperage. The exception is usually batteries used where high currents are needed, such as starting engines. Then you will hear ratings like "cold cranking amps" or "reserve amps", the definitions of which are a combination of useful information and marketing hype.
 

MrChips

Joined Oct 2, 2009
23,510
Batteries are spec'd with output voltage, not current.
For example, your lead acid battery is labeled as 12V output.
The 8A you quote is more likely to be 8Ah. In other words, the battery can deliver approx. 8A for one hour or 1A for 8 hours, etc.

Therefore you cannot measure the amperage of a battery the way you are thinking. You can measure how long it can deliver current to a load. For example, you can purchase a 24W automotive bulb and connect it to your 12V battery. The bulb will consume 2A. If it stays bright for 4 hours then the battery is good for 8Ah.
 

Thread Starter

Benjamin0904

Joined Oct 3, 2020
31
Suppose the Lead Acid battery's label was torn off and you knew it was 8Amp Hours, but wanted to measure it just to make sure. How would you do that?
 

boostbuck

Joined Oct 5, 2017
42
The simplest way is to measure the weight of the battery, which will give a rough indication of it's capacity. My 15kWh battery weighs 450kg, and a 90Ah 12V (~=1kWh) car battery is about 25kg. It's a very approximate measure, dependent of battery architecture.
 

WBahn

Joined Mar 31, 2012
26,398
Suppose the Lead Acid battery's label was torn off and you knew it was 8Amp Hours, but wanted to measure it just to make sure. How would you do that?
You can only get a rough confirmation because the capacity of a battery is load, temperature, and history dependent.

If you have an "ideal" battery that has a 8 Ah rating, then in theory if could supply 8 A for 1 hour, 1 A for 8 hours, or 8000 A for 1/1000 hour (i.e., 3.6 seconds) or 1 mA for 8000 hour. But real batteries don't behave that way. First, there's an upper limit to how much current they can deliver, even into a dead short. Second, even if you stay with reasonable current draws, the capacity that you would measure generally decreases with increasing load. So if the 8 Ah that it is rated at was measured at a current draw of 1 A, then if you draw 10 A from it you would probably come up with something less than 8 Ah while if you draw 0.1 A you would probably come up with something more than 8 Ah. The differences might be fairly small or they might be very significant.

Another complication is that there is no magic point at which the battery is dead. So if you draw a constant 1 A of current from the battery, there isn't going to be some sudden change in behavior after 8 hours that signals that the battery capacity has been reached. The battery manufacturer used some kind of specification that described their test criteria and that criteria may or may not be published.

Most battery technologies have their own stereotypical discharge characteristics, but even within a basic family (such as lead-acid) there are subfamilies that have their own very significant characteristics depending on their designed use case. For instance, batteries that are designed for high current and shallow discharge, like most automotive batteries, behave differently than batteries designed for powering modest loads for extended periods of time, resulting is deep discharges but no high currents. So you need to investigate the typical behavior of the specific type of battery that you are dealing with and then devise a test to determine the likely rated capacity accordingly.
 

Thread Starter

Benjamin0904

Joined Oct 3, 2020
31
So I now understand how to measure Amp Hours on a battery. I have one more question that is similar. A DC motor windmill is roughly producing 15v at a constant 400rpm. Is measuring the output Amp Hours the same as with a battery? For example I have a 12v load that draws 1 amp. Can someone clarify? Thanks.
 

WBahn

Joined Mar 31, 2012
26,398
So I now understand how to measure Amp Hours on a battery. I have one more question that is similar. A DC motor windmill is roughly producing 15v at a constant 400rpm. Is measuring the output Amp Hours the same as with a battery? For example I have a 12v load that draws 1 amp. Can someone clarify? Thanks.
Totally unrelated concept.

The Ah rating of a battery is a measure of how much charge that the fully-charged battery can deliver before it is nominally dead. 1 amp-hour is equal to 3600 coulombs, by definition. It is an indirect measure of the total available energy that is stored in the battery when it is fully charged, which is roughly the amp-hour rating multiplied by the terminal voltage (the actual energy is somewhat less than that as you need to use the terminal voltage as the battery is discharging). Neither of these have any meaning whatsoever with a windmill, which can keep delivering charge (and energy) as long as the wind is blowing (until it wears out).
 

Thread Starter

Benjamin0904

Joined Oct 3, 2020
31
What is the measure of output on the windmill than? Is it only 15v and current doesn't matter? I am kind of confused. I am just a newbie in the world of electronics.
 

BobTPH

Joined Jun 5, 2013
3,306
So I now understand how to measure Amp Hours on a battery. I have one more question that is similar. A DC motor windmill is roughly producing 15v at a constant 400rpm. Is measuring the output Amp Hours the same as with a battery? For example I have a 12v load that draws 1 amp. Can someone clarify? Thanks.
Your original question is equivalent to asking how far a car can go on a full tank of gas. The new question is like asking how far the car can go.

Bob
 

BobTPH

Joined Jun 5, 2013
3,306
What is the measure of output on the windmill than? Is it only 15v and current doesn't matter? I am kind of confused. I am just a newbie in the world of electronics.
No, the measure of the windmill’s capacity would be volts and amps at a given wind speed. For example 15V at 2A at 20 MPH.

Bob
 

MrChips

Joined Oct 2, 2009
23,510
What is the measure of output on the windmill than? Is it only 15v and current doesn't matter? I am kind of confused. I am just a newbie in the world of electronics.
The simple answer is to connect a load to the windmill generator and measure both current and voltage.
The power output is rated in watts (W).

power output (W) = current (A) x voltage (V)

The power measured will change depending on the load current (for the same wind speed). There is such a thing called maximum power point (MPP) where you get the maximum power for a given situation.
 

WBahn

Joined Mar 31, 2012
26,398
What is the measure of output on the windmill than? Is it only 15v and current doesn't matter? I am kind of confused. I am just a newbie in the world of electronics.
Current matters, but the relationships is going to depend on the design of the windmill in question. At a given wind speed, the windmill will be able to deliver some maximum current at it's designed output voltage. You can't get more energy out than you can draw from the wind. Up to a point, as the windspeed increases, you can get more power out. How well you can do that depends on the specifics of the mechanical and electrical design of the windmill.
 

WBahn

Joined Mar 31, 2012
26,398
The simple answer is to connect a load to the windmill generator and measure both current and voltage.
The power output is rated in watts (W).

power output (W) = current (A) x voltage (V)
But what load? You need to find the load that results in the voltage falling below some specified threshold.

Also, that load is going to be depending on the speed of the wind (and the density of the air, but that is not a big factor for most ground-mounted windmills).
 

WBahn

Joined Mar 31, 2012
26,398
I know how to measure Voltage. How do I measure the current (A) to estimate Power output?
There are a few ways of doing it, depending on the specifics of the situation.

The "normal" way is to use a small valued resistor in series with the load and measure the voltage across it. Ohm's Law then will tell you the current. Another way is to put an ammeter in series with the load -- which is usually doing the same thing depending on how the ammeter is constructed. Be careful not to just connect the ammeter to the power source -- this usually results in a destroyed ammeter, a possibly damaged source, and potentially other damage such as explosions and fires, depending on the specifics.
 

Sensacell

Joined Jun 19, 2012
2,821
If you take 8 "AA" batteries and connect them in series, you get 12 Volts, but why then can this setup NOT start a car?
The voltage is the same as a normal car battery!??

The answer lies in the battery's ability to deliver power, real useful power.

How many of us tried to "measure the current of a battery" when we first started learning electronics?
This newbie stunt usually results in a damaged meter and /or a damaged battery- basically because a meter on "Amps" range presents almost a dead short across the battery, which causes a huge uncontrolled (destructive) flow of current. (DO NOT TRY THIS)

All voltage sources have internal resistance, this resistance limits the amount of current and power a source can deliver.
A car battery has a very small internal resistance, (look at the huge fat metal output terminals) it can deliver a huge current to start the engine, while a "AA" battery can supply far less.
If you tried to start a car with AA batteries, the voltage would collapse to near nothing.

The safe way we measure this internal resistance is to load the voltage source with something that will draw a known safe level of current, while we monitor the voltage of the source.
We can compute the internal resistance of this source by noting the loaded and unloaded voltage, the more it drops under load, the higher the internal resistance.

When we try to start the car with the AA batteries, almost all the energy is consumed by the internal resistance of the batteries, they just get hot, and nothing happens.
This is because the internal resistance is huge in comparison to the resistance of the starter motor.

When we speak of battery capacity, then we are taking time into account, how many hours can the battery deliver XX Amps- the rating is "Amp-Hours"
 

WBahn

Joined Mar 31, 2012
26,398
The safe way we measure this internal resistance is to load the voltage source with something that will draw a known safe level of current, while we monitor the voltage of the source.
We can compute the internal resistance of this source by noting the loaded and unloaded voltage, the more it drops under load, the higher the internal resistance.
Though it needs to be noted that the resulting internal resistance is the apparent internal resistance at that current draw. You can't just take that resistance and divide it into the open circuit voltage and come up with how much the short circuit current will be. How constant the internal resistance as as the current draw changes (all other factors, such as temperature and state of charge being kept the same) depends on the specifics of the technology. For some battery chemistries it is quite constant over a very wide range of current draws and for others it is heavily current-dependent.
 
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