Do Inverters change the lithium-ion batterie's mA capacity and wattage?

let me say if I have a pack of lithium batteries connected in series and parallels, and i want to connect it to an Inverter, BUT...BEFORE I connect it to inverter, lets say it gives out 12 volts "hypothetically" some informations like 3.7 volt batteries, and 2500 mAh,
lets say then the pack gives 12 volts in DC and hypothetically again I calculate it ((still in DC)) is ..... W = V* Amp ... then it means wattage = 11.1 V * 5 Ah which is 55.5 W. then we got 55.5 watts, THEN....after we want to change the dc into ac it gives you 120 volts "please hypothetically" just answer me this, ARE the Wattage And Ampere are remains the same as DC? what would it be the result after AC?

Answer

I think this question needs some serious sorting out. It has several parts:

1. 
   

   Battery cell. A "primary battery cell" stores ENERGY, which is a certain amount of Watts over certain time. A cell with 3.7V nominal and C=2500 mAh holds about 9.25 W-hours of energy. You can dump all 9 Watts in one hour, or you can sip 1 watt for 9 hours. All would depend on connected load. A 370 Ohm resistive load will work from this cell for 250 hours. However, every cell has certain limit on discharge rate. Normally the cells are made for 1C - 2C discharge rate (C is rated capacity of one cell), but some special battery packs used in Radio Control area can have discharge rates of 15C-25C-50C.

   
   


2. 
   

   A Battery of cells. In general, the energy stored in any battery is a sum of individual cells, regardless of how they are connected, in series, in parallel, or properly mixed. So if you have a 6-cell battery, 2 in parallel, and three in series, you have a 55.5 W-Hour battery. Not "55W", but "55 Watt-Hours". Again, depending on load, you can get 100W in half hour, or 9 W in six hours, or 1 W for 55 hours. If your battery is 2P3S battery with typical 18650 cells, it likely can be discharged at 2C rate , or up to 10 A in your case. So, "hypothetically" you can get 120 W from 2P3S battery (with 2C discharge rate) for about 20 minutes. Or, more hypothetically, you can get 1200W for 2 minutes, if your cells can discharge at 20C rate, hypothetically.

   
   


3. 
   

   Inverter. Inverters (AC-DC converters), or any other switching mode DC-DC devices of this sort, they CONVERT a input into output. They are limited by amount of POWER they can deal with, instant POWER. Ideal converters convert power at 100%, but real converters have some losses. A safe assumption is to have 20% losses. Say you have a 120-W capable converter that handles hypothetically any voltage level, and the 120 W is output rating. This means that if your output is 12 V, the converter will deliver up to 10 A of current to a 1.2-Ohms load. Or 1 A to 12-Ohms load, whichever is connected to its output. The input, however, will use 1.2 A at 120 V, or 20% more power than it delivers to output. But if you use, say mains AC and the inverter forever, you can transfer infinite amount of energy, watt-by-watt, hour by hour, not limited. So, P(in) = P(out)/0.8, and V(out)I(out) = V(in)I(in) *0.8. So no, "amperage" won't be the same on both ends of converter, it will be inversely proportional to voltage ratio.

   
   



4. 
   

   If you plan to attach a 12 V DC to 120 V AC converter/inverter to your battery, it has very little to do with battery energy storage capacity. But same efficiency considerations and power limits will apply. So the 55.5 Whr battery with 12DC-120AC inverter will be able to deliver only 44.4 Whr of energy on 120 V side. 11 Whr will be dissipated as heat, lost in electronic components of the inverter. Again, if you attempt to get, say, 120 W out of this design (120 V at 1 A), the inverter must be rated as "120 W", and your battery will last only 22 minutes. So to get 1 A at 120 V output, your 12-V battery must be delivering about 13.5 A of current, sorry.

   
   

I hope this answers your question, "what would it be the result after AC?"

EDIT: considering the OP question literally, the answer is "no", nothing can change capacity of Li-Ion cells/battery. However, considering the [battery+inverter] as a "black box" with 120V output, the inverter will decrease the effective box capacity from 55 Wh to about 44 Wh first, due to internal losses (assumed at 20%). Second, the effective mAh rating of the box will be inversely proportional to output voltage. So the "box" capacity will be 44Wh/120V=366 mAh.