I have Duracell Ultra Power AAA batteries, current draw is around 2.5A when shorting for single unit.

I have just tried connecting them in series and parallel then measure discharge current.


  • Parallel: A bit increased current sourcing

  • Series: doubled current sourcing

It doesn't make sense if we think in terms of V=IR since series connection also increase internal resistance.

So my question is why? I doubt battery chemistry is the limiting parameter not the internal resistance ( maybe it is low enough?)

(BTW my end goal is to use 2 AA batteries to step up to 3.6V while supplying 2A for short bursts)

  • 3
    \$\begingroup\$ The wire you're using to short the batteries has its own resistance. The device you're using to measure the current uses a shunt resistor which also has its resistance. If you take all of these into account, I'm sure you'll find it all follows Ohm's law. \$\endgroup\$ Sep 2, 2012 at 11:45
  • \$\begingroup\$ To measure the discharge current more accurately, I suggest you perform a 4-wire measurement. For that, short the batteries using something with a known resistance(e.g. 10x 5% 10Ω resistors in paralell). This will be your shunt resistor. Then, measure the voltage on your shunt and the batteries. This way you should be able to calculate the internal resistance of your AAA batteries. \$\endgroup\$ Sep 2, 2012 at 12:03
  • 1
    \$\begingroup\$ @Jonny: A 1 Ohm current sense resistor probably won't work since that is significantly more than the battery internal resistance. This is in fact exactly the problem he is seeing. I think his ammeter has a relatively high current sense resistance. \$\endgroup\$ Sep 2, 2012 at 12:46

3 Answers 3


I think this is due to how you are measuring the current. Your results all make sense if the resistance of your current meter is higher than the internal resistance of the battery at the measured current.

Think of the limiting case where the batteries are pure voltage source (no resistance) and you put a fixed resistance accross them (your current meter in this case). Under these conditions, the current would be proportional to the total voltage, and you'd get pretty much what you measured.

The other limiting case is where each battery has a fixed finite internal resistance. This can be thought of as a resistance in series with a ideal battery. In this example, the "short" is truly a short of 0 Ohms. Now the short circuit current of any one battery is fixed, which is the battery voltage divided by its internal resistance. In this case adding any number of batteries in series won't change the short circuit current, but putting then in parallel multiplies the current by the number of batteries.

From your measurements, I think what is going on is close to case 1, not case 2.

  • \$\begingroup\$ I have tried another ammeter and case 1 was my problem indeed. \$\endgroup\$
    – Ozgur
    Sep 4, 2012 at 7:26

You can estimate the internal resistance (hence the shourt-circuit current), without using an ammeter. All you need is a single external resistor and a voltmeter.

First, a battery can be modeled as a series of an ideal voltage source of \$V_{BATT}\$ volts and a resistor of \$R_{BATT}\$ ohms. \$R_{BATT}\$ is the battery's internal resistance. This model is not perfect, because \$R_{BATT}\$ actually varies with charge, load, temperature, etc. But it's close enough, so a typical rechargeable AA / AAA battery may look like this:

Battery ESR model

\$V_{BATT}\$ is easily measured with a voltmeter - it's the open-circuit voltage of the battery. Now, to calculate \$R_{BATT}\$, we need that extra resistor of \$R_{LOAD}\$ ohms:

Battery ERS Measurements

When the \$R_{LOAD}\$ resistor closes the circuit, it forms a voltage divider with \$R_{BATT}\$. If the voltage across \$R_{LOAD}\$ is \$V_{LOAD}\$, then:

$$ V_{LOAD} = V_{BATT} * \frac {R_{LOAD} }{ R_{BATT} + R_{LOAD}} $$ or, by rearranging: $$ R_{BATT} = R_{LOAD} * ( \frac{V_{BATT}}{V_{LOAD}} - 1) $$

All values on the right-hand side can be measured directly.

Note: On the schematic above, when SW1 is open (or \$R_{LOAD}\$ is absent), VM measures \$V_{BATT}\$; when SW1 is closed (i.e. \$R_{LOAD}\$ completes the circuit), VM measures \$V_{LOAD}\$

Note: There are a couple of things to consider, when choosing the value of the load resistor:

  1. It has to be low enough, so that the \$V_{LOAD}\$ measurement yields a useful result. With an internal resistance typically in the range of 0.05 ~ 1 ohm, a 100 ohm load resistor would drop 99+% of the voltage and would introduce a lot of error in calculation.
  2. It has to be high enough, so that the discharge current is not too high - batteries don't like that and may be permanently damaged if (near-)short-circuited for too long. I can't provide exact values, but anything under 1A should be OK.
  3. It has to be high enough, for the specific resistor package / wattage, so that the resistor itself does not catch fire in the process. The power dissipated in the resistor is approximately \$V_{BATT}^2/R_{LOAD}\$

A common 10 ohm 1/4W resistor should be fine for a quick test of AA/AAA batteries: Discharge current is 0.12 ~ 0.15A, dissipated power is 0.144 ~ 0.225W. To be on the safe side, a 1/2W resistor would be a better choice.


Maybe a shorter answer would be more to the point:

  • paralleled, => 1.5V
  • series => 3V

Suppose the resistance is 1 Ohm =>

  • 1.5/1 = 1.5A
  • 3/1=3A

Extend this to your case: the finite multimeter resistance (+wires) plus internal ones.

  • 1
    \$\begingroup\$ Not quite. If each battery has an internal resistance of 1 ohm, then in parallel the total resistance is 0.5 ohm and in series the total resistance is 2 ohm. \$\endgroup\$
    – DimKo
    Sep 2, 2012 at 16:40
  • \$\begingroup\$ True, but this simple numeric example was the sake of calculation; in practice, either them or the ampermeter's are small enough. \$\endgroup\$
    – Vlad
    Sep 2, 2012 at 19:43

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