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If you have 2 (e.g. 1.5V) batteries in series and one is dead/empty, what is the reason why a device is not working? Lack of voltage (1.5 + 0 is simply not enough for the device to operate)? From my understanding, for batteries in series, the current is the same as for one of them. So, a common sense says to me that it must somehow connected with voltage, not current. Is it true?

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  • $\begingroup$ Maybe your device needs 3V while 1.5V is not sufficient? I'm just brainstorming. $\endgroup$
    – my2cts
    Commented May 20, 2021 at 19:10

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Actually, using a combination of fresh and somewhat depleted batteries was widely used strategy not so lo... well, about 40 years ago (I must remember I am already adult). I remember my grandfather (Electical Engineering PhD) doing so for his portable radio.

The practice was neither that much effective nor absolutely safe. Batteries back then already had a warning on their labels not to mix fresh and depleted batteries.

How it works: The battery-powered devices are built with some tolerance to the power voltage. The portable radio in question used 6 x D-size, 1.5V cells. 6 fresh batteries could give like 10V (they started somewhat above 1.5V each) and the radio worked acceptably down to 5V (like 0.85V per cell). When the radio refuses to work on voltage that low, you change one of the cells and it works some more. Repeat, always changing the oldest battery.

The trick was, those cells (leclanché dry cells, ZnMno2 with ammonium chloride electrolyte) had pretty shallow discharge curve and they could give a meaningful amount of energy below the usual 0.8-0.9V end-of-discharge point.

What could go wrong: Leclanche cells like to leak electrolyte when deeply discharged. Their container is also the negative electrode that is consumed in the discharge process. At some point, the voltage of a particular cell can go below 0, at that point the cell actually consumes energy and (provided it is still gas-tight) builds gas pressure that can rupture the container. Or make the electrolyte leak faster.

Another limitation is that the internal resistance of the cell goes up with the state of charge going down. That's why the trick works only for small discharge currents.

Modern electrochemical cells are way better in regard to their discharge curves. When their voltage falls below their intended end-of-discharge limit, one can extract almost no energy from them. That's why it is pointless to mix fresh and depleted modern batteries. The dead cell just drops some voltage, instead of adding a little.

When we talk about rechargeable cells of any chemistry, over-discharging them usually quickly shortens their useful life (or outright destroys them). That's why it is best to use cells in series that are strictly matched by capacity.

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A battery works by harnessing an electric potential to drive electrons through a conductor. When two batteries are in series, there's a sandwich of different potential regions in the pattern negative-positive-negative-positive. As electrons leave the negative end of this sandwich and through a conductor, electrons are also simultaneously filling and neutralizing the positive end too. Now the positive end is becoming more and more negative, and this causes electrostatic repulsion in the negative region next to the positive end. The repulsion pushes electrons into the next positive region to satisfy the fact that the potential is being depleted and put to use. Now let assume the pattern becomes negative-positive-neutral-neutral, you can see already that it doesn't add up. There isn't a way the electrons leaving the negative end can make it to the positive region needed to be neutralize to have electrostatic neutrality.

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  • $\begingroup$ So, with zero voltage in one of the battery, there will be no loop, so no current flow a all? And if a battery is considered dead (see the reply below), but some voltage still remaining, does it mean that I can have luck (if voltage would be high enough) to run the circuit (but e.g. led will be dimmer)? thanks! $\endgroup$
    – Pavel
    Commented Mar 5, 2019 at 8:06
  • $\begingroup$ If it's truly zero, yes no loop. But if there's still some juice in the weak one, it'll will run the device but you won't be able to harness the full 1.5 V potential in the strong one, at least not for long. $\endgroup$
    – TechDroid
    Commented Mar 5, 2019 at 8:23
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A battery is considered dead when: 1.5 volt battery is .8 volts and below, a 9 volt battery is 5.4 volts and below.

So if you have 1 good battery and 1 dead battery, you will have 2.3 volts instead of 3 volts.

You will lack the voltage required to run a circuit. (The amperage available stays the same).

This is where you have to understand the difference between amperage available in the battery and amperage in the circuit.

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