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I have an old 9 volt battery that I've measured to 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 to a multimeter, and it just read 0.35 mA. How come it's so low?

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If that's .35ma and not 35ma your battery is really dead. Which is a good thing, because shorting out a power supply with an ammeter is a good way to destroy the meter (or blow the fuse, if present) –  Chris Stratton Dec 26 '12 at 2:07

3 Answers 3

up vote 13 down vote accepted

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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Two reasons: One, the battery is old (read: nearly dead) and two, batteries have internal resistance which prevents them from delivering high amounts of current. (Compare to a capacitor, which can deliver very high currents.)

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Capacitors have internal resistance too. As with batteries, it's larger in some than others. It's not normally smart to short circuit either. –  Chris Stratton Dec 25 '12 at 21:07
    
The dead battery may also drop the voltage when connected to such a load. –  Chetan Bhargava Dec 25 '12 at 21:15

9v Batteries tend to have a nominal 500mah of capacity (Alkaline) while the newer Lithium ones have nominal 1200mah. Even ignoring the internal current resistance of the battery, it couldn't provide 4 amps of current if it wanted to.

Similarly, that internal current resistance is the same reason that you can hook up a 3v coin cell to a led without a current limiting resistor. The battery already limits the current due to it's design and material.

And at 7v, the battery is pretty much dead.

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Sure it could, for a whole 7.5 minutes, even. –  Ilmari Karonen Dec 26 '12 at 1:03
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If you ignore the internal resistance, then it could provide 4 amps. It is precisely because of the high internal resistance that it can't. –  Olin Lathrop Dec 26 '12 at 1:15

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