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I have two brand new AA batteries. I measured the voltage and current of each battery and they are both 1.5v and 5A respectively. I connected these same batteries in series and I measured the voltage. I expected the voltage and current to increase to 3v and 10A respectively. The voltage doubled to 3v as expected, but the current only increased to 7A. Why didn’t the current double? I didn’t add any resistance I just doubled the voltage.

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  • \$\begingroup\$ What is the resistance of a battery? (hypothetical question, don't measure it, it might damage your multimeter) \$\endgroup\$ – immibis Jul 12 '18 at 5:01
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    \$\begingroup\$ Did you just connect your meter to the battery to measure the current? \$\endgroup\$ – Solar Mike Jul 12 '18 at 5:02
  • \$\begingroup\$ Sorry I’m not an electronic engineer or expert at all, my major is CS >_<. I’m not sure about the resistance of battery, but yes I just connected the meter to battery. \$\endgroup\$ – AznBoyStride Jul 12 '18 at 5:17
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    \$\begingroup\$ I just connected the meter to battery For voltage measurements that's OK and the way to do it. For current measurements it is not, the meter is a short circuit during current measurement so you made a short circuit. Alkaline AA batteries can survive this and the current they can supply isn't that large (however, don't do this with rechargeable AA batteries). But do the same with a car battery or a mains outlet and you will regret that! Smoke, sparks, fire and melting wires will be what you'll measure! \$\endgroup\$ – Bimpelrekkie Jul 12 '18 at 7:06
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You put two batteries in series so that doubled the resistance.

Each battery has an internal resistance, so you need to include that effect.

If you measured the current directly across the battery, then you were lucky you did not damage your meter - you should have a load in the circuit - even just a bulb will suffice.

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I measured the voltage and current of each battery and they are both 1.5v and 5A respectively

Batteries do not have an inherent current in the way they have an inherent voltage. Instead, batteries will have an advertised capacity in the form of Amp-hours (Ah) or milliamp-hours (mAh).

It appears you attempted to measure the "current" of the battery by putting an ammeter directly across the battery terminals.

schematic

simulate this circuit – Schematic created using CircuitLab

Ammeters are designed to have as low a resistance as possible. Your ammeter may have a resistance of m\$\Omega\$ or even \$\mu\Omega\$, which is on the same order as a length of copper wire. So you've essentially shorted the battery to itself.

schematic

simulate this circuit

This is almost never a good idea. If you were using a bigger battery, you would have likely burnt out the fuse in your ammeter (assuming it has a fuse). In an ideal world, Ohm's Law predicts the current you'd measure would be infinite! The only thing that limited the current that you measured (5A and 7A) was the internal resistances of the batteries as they were trying to source as much current as they possibly could. There are very few applications where shorting a battery across itself is useful or informative. With certain kinds of batteries (read: Lithium), you could cause an explosion doing this.

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