I have an old 9 volt battery that I've measured at 7.2 volts. I've measured the resistance of the wires in the circuit I'm using, and it came out to 1.5 ohms. Therefore, according to Ohm's law I'd expect over 4 amps of current.
However, when I connect it up to a 0.2A fuse, it didn't blow. I tried connecting it in series with a multimeter, and it just read 0.35 mA. How come it's so low?
Answer
You did the math right but overlooked one key issue, which is that all batteries have some internal resistance by themselves. As a rough first approximation, you can think of a battery as a ideal voltage source with some resistance in series. The short circuit current of the battery is therefore its voltage divided by that internal resistance.
9V batteries are particularly bad at internal resistance. You have a nearly dead one, which will be even worse. As a battery is used up, its voltage goes down and its internal resistance goes up.
Since this battery is pretty much dead anyway, you can have some fun with it without much loss. Let's characterize it according to the simplified model of it being a voltage source with series resistance, otherwise known as a "Thevenin source". First, measure the battery voltage without any load. Even a crappy voltmeter will have at least 1 MΩ resistance, which draws so little current that there is no effective voltage drop accross the internal resistor. Measuring the battery with a voltmeter therefore tells you its voltage source value directly.
To measure its internal resistance, you use two measurements of its voltage at different currents. If we assume for now that the internal voltage source puts out the same in both cases, the difference in external voltage is the difference in current times the internal resistance. You already have one measurement, which is essentially at 0 current. You could put a known resistance accross the battery to draw a few mA and see how much its external voltage drops as a result. Try it. In the extreme case you can short the battery by putting a ammeter accross it. At that point the current will be the internal voltage divided by the internal resistance. That's right in theory, but in reality the internal voltage source will drop due to some chemical effects at high current. It will also drop rapidly over time, so if you do this, you have to take the measurement quickly.
Other things you could do is to put a fixed resistance accross the battery and record the voltage over time as it goes completely dead. From that you can calculate and plot power output, total energy output, or just see the flatness or lack thereof of the voltage. Or measure the voltage shortly after connecting different resistances accross the battery. That lets you calculate internal resistance as a function of current. See how constant or not it is.
Batteries are not simple. Getting some intuition about them can be valuable.
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