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Yes, the output voltage of a battery decreases (roughly) in proportion to the current drawn from it. This behavior is usually characterized as an internal resistance of the battery, although its actual physical origin may be related more to the chemical properties of the battery than to the actual resistance of any conductive part of the battery. For the ...


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It's actually the conductor dielectric interface that matters for ESR. this rises slowly with age and rapidly below 10% SoC while the conductors are constant and conductive particles in the dielectric would discharge it. The impedance or Effective Series resistance , ESR or Rs, internal resistance is inversely related to the cell storage capacity , Ah yet ...


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A battery will need a charging regime to match its chemistry. So, this regime will change over time as the battery state changes. The charging circuit is designed to supply some fraction of the current capacity as some batteries are capable of supplying hundreds, or thousands, of amps for short periods of time - usually seconds. Charging the battery at its ...


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There are only a limited number of battery chemistries and each one has its own voltage so there are only a limited number of voltages. Also, battery voltage drops as they get discharged so you won't find exactly 12V. Search radio-controlled hobby websites. The most likely chemistries available to you, in order of increasing cost, increasing performance, ...


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If the batteries are identical, one battery provides half the current. If they are not identical, e.g. one battery is dead or missing, full 3A. If you connect two batteries together that are unequally charged, e.g. one full and one empty, any arbitrarily large current would flow to equalize the charge levels. Prepare with good margin and fuses.


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If they are identical batteries with identical charge (an ideal assumption and not the case, but its safe to assume so hypothetically) then half the current will be drawn from both each such that the required 3A comes from 1.5A of each of the batteries - they can be seen as mutually exclusive in the way that the current from the 2nd battery doesnt have to go ...


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To make an "apples to apples" comparison, you'd need the capacity rating for both devices in watt hours. What you have is VA for one device and mAh for the other. VA (volt amperes) and watts are a power rating. mAh (milliampere hours) is (sort of) an energy rating. The unit "watt" doesn't have time factored into it. Energy always has ...


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The capacity of the capacitors can be back-calculated. Let's do some math. The capacity of the battery is 2900mAh. Let's say that the phone will run for 6 hours on this. Roughly the phone is then consuming 480mA. Also, a typical Li-ion battery goes from 4.2 ish volts down to 3.4 with a hard cutoff at 3.0v. Armed with these facts suppose we want the phone to ...


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