Hi everyone,

Let’s say that you have a 12 V battery that delivers 5 Ah, just to make up an example. And let’s also say that you have a 12 V battery that delivers 3.3 Ah.

So, here is my first question, please. If you had a device that drew 1 A, then the 5 Ah battery would supply current at an electromotive force of 12 V for five hours, whereas the 3.3 Ah battery would supply current at an electromotive force of 12 V for 3 hours 18 minutes. Am I correct on this?

And just to be certain that I have that down pat, if you had a device that drew 2 A, then the 5 Ah battery would supply current at a force of 12 V for 150 minutes, whereas the 3.3 Ah battery would supply current at a force of 12 V for 99 minutes. Am I correct on this?

Now, if I am correct so far, then let me finally ask this. Back to the 12 V battery that delivers 5 Ah. What happens after five hours? Does the battery just die, or does it gradually loose force so that it is not longer supplying current for the full five hours? Is the battery perhaps delivering current with an electromotive force less than 12 V as the five-hour period begins to end?

Thank you.

Stanley

 

That would be true for a "perfect" battery. For real batteries the AH rate is normally quoted for a much lower discharge rate. (I think it is for dischrging them ofer a 10 or 20 hour period.) At high discharge rates the AH capacity drops quite a lot. You would need to look at the manufacturers data to find out what the AH rating is for these relativly high currents. A fully charged lead acid battery would be at about 13.8 volts and would drop down to about 11.5 volts at which point you should remove the load to avoid damaging the battery.

Les.

 

When a battery is considered discharged or empty.........it still has current in it. We don't use chargeable batteries until we've sucked all current out of it. Like a gas tank.

We use batteries until they reach a certain voltage. Voltage tells you about your battery....not current.

The amp hour rating is a power density rating. It's how quick you can pull current from it. It's a guide line. We don't keep track of current like we do gasoline.

Think of keeping your gas tank at least 3/4 full at all times. That's how a battery is.

 

Hi everyone,

Let’s say that you have a 12 V battery that delivers 5 Ah, just to make up an example. And let’s also say that you have a 12 V battery that delivers 3.3 Ah.

So, here is my first question, please. If you had a device that drew 1 A, then the 5 Ah battery would supply current at an electromotive force of 12 V for five hours, whereas the 3.3 Ah battery would supply current at an electromotive force of 12 V for 3 hours 18 minutes. Am I correct on this?

And just to be certain that I have that down pat, if you had a device that drew 2 A, then the 5 Ah battery would supply current at a force of 12 V for 150 minutes, whereas the 3.3 Ah battery would supply current at a force of 12 V for 99 minutes. Am I correct on this?

You have the basic concept correct, which is a lot better than most folks you have a very hard time distinguishing between amps and amp-hours.

Now, if I am correct so far, then let me finally ask this. Back to the 12 V battery that delivers 5 Ah. What happens after five hours? Does the battery just die, or does it gradually loose force so that it is not longer supplying current for the full five hours? Is the battery perhaps delivering current with an electromotive force less than 12 V as the five-hour period begins to end?

First note that a current multiplied by time yields an amount of electrical charge. So the amp-hour rating is nominally the total amount of charge that the battery can deliver. But the chemical reactions that produce the charge separations within the battery, and hence the voltage difference that drives the charge through the external circuit, are not like a light switch that turns off at some special moment in time. The reactions get weaker and weaker, which manifests itself as both lowered voltage and as higher internal resistance. So what you will find with most batteries is that the terminal voltage gradually drops as the battery is drained until a point at which it drops off very sharply. The "cutoff" voltage that is used to declare the battery "dead" is usually well above this point. Also, like many things, the harder you push the battery (the more current you draw), the less efficient the chemistry is and so energy is lost to heat and the effective amp-hour capacity drops. Hence, for most batteries, the amp-hour rating is specified at a particular current and you can expect it to be lower at higher current draws and, to some extent, higher at lower current draws.