# AMPS

Discussion in 'Electronic Basics' started by Average Shmo, Mar 14, 2006.

1. ### Average ShmoGuest

Hi,

Is there any way to know how many AMPs are comming out of a battery? Most
batteries purchased at the shops dont tell you.

I have a voltmeter here but that only (you guessed it) shows volts!

2. ### Tom BiasiGuest

Amps is a rate of current flow. It depends on the load for a particular
voltage. The battery will have a maximum rating which is determined by the
physical makeup of the battery. You need the specs on your battery.
Tom

3. ### Greg NeillGuest

That's because it depends upon the load connected to the battery.
If you know something about the load, like its power consumption or
its resistance (assuming a resistive load), then you can work out
the current being supplied at a given voltage. Otherwise, purchase
an ammeter and measure the current!

4. ### John FieldsGuest

---
The amount of current a battery can deliver into a load depends on
the battery voltage, the internal resistance of the battery and the

If you had an ammeter you could connect it in series with the load
and measure the current.

If the load resistance was fixed and you knew what it was, or if you
had an ohmmeter and could measure it, you could divide the load
resistance into the battery voltage and get amps as the quotient.

If you have neither an ammeter nor an ohmmeter you might be pretty
much shit out of luck.

Except...

You could get a very small valued resistor (a 'shunt') and put it in
series with the load and the battery, then measure the voltage drop
across the shunt.

Your circuit would look like this: (View in Courier)

+--------+
| |
|+ |
[BATTERY] +<--------+
| | |+
| [SHUNT] [VOLTMETER
| | |
+--------+<--------+

and you could then determine the current in the shunt by using Ohm's
law:

E
I = ---
R

Let's say that the resistor has a resistance of 0.1 ohm and your

Then, the current in the resistor will be:

E 0.02V
I = --- = ------- = 0.2 amperes
R 0.1R

since the current in a series circuit is everywhere the same, that
200mA would also be flowing through the load and the battery, so
that's how much current would be coming from the battery.

There's a small error in there due to the resistance of the resistor
causing less current to flow than if it weren't there, but we can
get rid of it like this:

First, measure the battery voltage when it's under load:

+--------+<--------+
| | |
|+ | |
[BATTERY] + |
| | |+
| [SHUNT] [VOLTMETER
| | |
+--------+<--------+

Just for grins, let's say it's 11.8V. Record that voltage as E1.

Second, measure the voltage across the shunt (as shown earlier) and
record that as E2.

Third, plug those values into this equation:

R2(E1-E2)
R1 = ----------
E2

Where R1 will be the resistance of the load and R2 is the resistance
of the shunt, like so:

R2(E1-E2) 0.1R * (11.8V - 0.02V)
R1 = ---------- = ----------------------- = 58.9 ohms
E2 0.02V

So, now that you know the resistance of the load, you can find the
current the battery's pumping through it by removing the shunt,
measuring the battery voltage with the load connected to it, and
using Ohm's law:

E
I = ---
R

To find out what current is coming out of the battery.

Or perhaps what you _really_ want to know is how to determine the
capacity of a battery (that is, how many amps can you get out of it
for how long)?

^^^^^^^
charge

6. ### DecaturTxCowboyGuest

I'm going to take a different take on the question. Perhaps he's asking
how many amps the battery can supply at maximum load (lowest circuit
resistance).

Batteries are "current limited" by the size of the internal electrodes
affected by the chemical process.

A "AA" battery could source up to .5 Amps, "C" battery 1 Amp, and "D"
cell up to 1.5 Amps. Two "D" cells in series is still 1.5 Amps. but two
in parallel would be 3 Amps.

A five pound lead-acid battery might source up to 20 Amps and a forty
pound one perhaps 100 Amps as the larger battery has bigger lead
electrodes.

OK John.

9. ### DecaturTxCowboyGuest

Well, OK..I was primarily considering the internal components of the
battery. But even then, you're referring to the internal resistance vs
current sourcing capacity. A small nuance of difference, but its all
part of the big picture.