Measuring Current from an AA Cell

Homework alert! Homework alert!

"Does 1.05 amps sound reasonable using a 700mah cell?"

The 'short circuit current' has nothing to do with the Ah capacity of the cell...it's all about the internal resistance of the cell...you don't even need the 1ohm, just put the meter right across the cell (on 10A range).
It's about the cell chemistry....
Don't try the test on a Nicad or a lead acid battery, in fact its a bad thing to do to any cell.
Test it on the load it will be used with.
Del

The 1 ohm resistor is the load. Using an AA 1.2v 700mAh cell and your stated precision resistor:I get a voltage reading of .98 volts, meaning .98 amps. Taking a direct reading as del recommends I get a reading of 3.08 amps. Now, which one is right. I'll ask around.

They are both 'right', youy can work out the internal resistance of the cell if you measure the voltage at the same time...
It's completely irrelevant anyway unless you happen to be driving a 1ohm load...in which case use the .98Amp figure.
But in reality you wouldn't want to draw 1 amp from an AA cell...Unless maybe you were using it to trigger an explosive device or similaar one shot device.
Del

Dear guest #2 If you want to see a mystery, Set up the experiment as guest one does. That is, put the resistor in line with the cell and take the voltage measurement on the two center posts. Then change the meter setting to amps, and put the probes in line with the resistor. The readings will be different. It's a mystery to me!

"And NiCd and NiMH batteries typically have only 1/20th of an ohm internal resistance, so don't short circuit them please. And And... You get the greatest amount of power from a battery when the load resistance equals the internal resistance of the battery."

Surely 0.05 Ohms is practically short circuit?

Exactly...and if you short them you get huge currents and the potential to start a fire very quickly, or to melt metal...

You need a Hall effect current probe to measure the short circuit current. Your college lab should have one, plus a good scope. To get the transient current, use a switch to close the short circuit, and capture the current wave on the scope.

Hmmm, but is shorting out a AA battery really the condition to get the most current through this battery? What about connecting the battery to a switch that then connects to a large capacitor that's reverse biased to 1200 volts. That should deplete that battery in no time. Pop goes the AA cell.

Yes.

I posed my silly yet valid answer, reply #10 , to point out that there really isn't a single peak current that a AA battery can provide. It all depends on what is the circuit attached to the battery. But of course there's a limit to how long a battery can continue to provide that current. One should expect that an initially fresh AA battery rated for 700mAH providing 1A should provide 42 minutes of 1A of current.

So what the OP probably should find out from his data is the series resistance of his voltage source. (As others have pointed out here.) But that is as far as I'll go on this homework problem.

(I just loved Lynlynch's first post.)

Right on. The most amperage that a cell can produce will the be the turn-on transient with a perfect short from anode to cathode. The most power it can provide would be when the external load exactly matches the cell's internal resistance.

What you intend to perform is the "Theorem of maximum power transfer". The explanation of how to do it is at this link:

http://electronicacompleta.com/lecciones/circuito-de-entrada-y-antena/

But unfortunately is in Spanish, using the Google translator can achieve understanding.

In the engine Google You can also search for: "Theorem maximum energy transfer, I suppose there will be some link in English.

I hope this information Help You.

Happy to help you

Study up on A/hr discharge rate this will help you understand.