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  1. How can we say a battery,say Li-ion battery, is dead when it is 2.8V itself instead of 0V? How can the 2.8V not give current? What is happening? This question, is applicable to all other batteries also. Why do we say a battery is dead when it is not completely 0V, but when it is hanging some very around 2.8V for Lithium ion and 1.1V for the AA battery whose nominal voltage is 1.5V and others?

  2. I have a Li-ion battery which I tried to charge. The maximum voltage mentioned in the battery is 4.4V max. And its nominal voltage is 3.85V. But, when I connect the battery to the USB Charger to charge the dead battery which was at 2.8V, the voltage line reads 4.6V. How is the 4.6V obtained and won't the battery get damaged as we are feeding 4.6V which is more than its maximum 4.4V voltage which is mentioned?

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    \$\begingroup\$ 1) different batteries have different chemistry and you seem to ask this without even trying to do own research, there is plenty material about this subject 2) In you other question you already have an answer to that, there is a charger chip between supply input and battery. Please don't ask same question in multiple places before getting an answer. \$\endgroup\$ – Justme Jun 15 at 8:56
  • \$\begingroup\$ Does this answer your question? Battery Discharged - But Potential difference exists without any current flow \$\endgroup\$ – winny Jun 15 at 9:03
  • \$\begingroup\$ No. I am not clear on why you call a battery dead when it has some voltage left in it? And is that voltage not capable to provide current? Can someone provide a real example of a battery that will not work because it has less voltage (the dead condition, even if it has some voltage left) \$\endgroup\$ – Newbie Jun 15 at 9:56
  • \$\begingroup\$ A dead battery is one that can't be successfully recharged to provide a useful battery with broadly the same energy capacity as a new battery. 2.8 volts is not a dead battery exactly but you shouldn't use it as a power source because as soon as the voltage drops to something closer to 2.5 volts, it will be unusable and non-re-chargeable. Then it can be regarded as dead. \$\endgroup\$ – Andy aka Jun 15 at 11:03
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Short answer: Think of your dead battery as the same battery but with very very reduced capacity. If you leave it alone, it shows some voltage. But as soon as you plug it into a device to extract current from it, it gets empty within milliseconds. When you try to charge it, it appears to fill up quickly.

Discussion: There are many failure modes for each type of battery, some leave them at 0V, some cause them to have high internal resistance, some leave them with reduced charge capacity, some make them leaky. These failure modes can occur alone or simultaneously. The two indications that you mentioned, both seem to be caused by a battery having too little capacity.

But even for a good lithium ion battery, the amount of charge available in the battery is not proportional to it's voltage. An Li ion battery delivers most of its charge between 3.4 V and 4.2 V. Below this voltage, it does have some charge, and it will definitely supply current, but its voltage will fall towards zero very quickly if you draw current from it.

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  • \$\begingroup\$ Thank you. Take a 12V lead acid battery as an example. Why do we say the battery is dead when it drops to around 11.9V? It has voltage right? So, it must provide current even if it is 11.9V, right? \$\endgroup\$ – Newbie Jun 15 at 10:03
  • \$\begingroup\$ To be honest, I wouldn't call it dead when it is at 11.9 V. :-) There is a lot of difference between a dead battery and an empty battery. You can charge up an empty battery (if it is chargeable) and it will be just like a new one. But a dead battery is actually defective. It's broken from inside. Also, the voltage of every battery falls as you use it. So a "12V" lead acid battery has about 13.2 V when it is fully charged, about 11.8 V when it has 50% juice left and most people would say its empty when it has about 10.5 V. \$\endgroup\$ – Abdullah Baig Jun 15 at 10:33
  • \$\begingroup\$ So, at 10.5 V, it can still supply a little current, but its voltage will fall to zero from here very quickly. \$\endgroup\$ – Abdullah Baig Jun 15 at 10:33

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