Can any one tell how much energy can be stored in the a single battery (12v) ?

If I want to store 10 kWh of energy then how to calculate the number of batteries required.

Editing notes:

power -> energy
kW -> kWh

  • 7
    \$\begingroup\$ You can't store power, but energy. It depends on the size of the battery. But you can know the energy stored multiplying the charge capacity (Ah) times the voltage. \$\endgroup\$ – clabacchio Mar 28 '12 at 12:39
  • 3
    \$\begingroup\$ Please close this question. Quite apart from the problem with the confusion between energy and power, it is like asking "how much water is there in a lake?" There is only one answer - it depends on how big it is. \$\endgroup\$ – uɐɪ Mar 28 '12 at 13:12
  • 2
    \$\begingroup\$ Once the basic units are sorted out here is enough value in the question to make it worth answering. \$\endgroup\$ – Russell McMahon Mar 28 '12 at 13:25
  • \$\begingroup\$ @RussellMcMahon it depends; asking how much energy a battery can store is quite pointless, while how can derive the energy stored in a battery from the capacity, while beeing a mere conversion, seems more focused. \$\endgroup\$ – clabacchio Mar 28 '12 at 14:18
  • \$\begingroup\$ @clabacchio, we need to know how long he needs 10kW for before you can edit that honestly. \$\endgroup\$ – Kortuk Mar 28 '12 at 16:35

No, since batteries store energy, not power. Your question makes no sense since the units are wrong. It is like asking what percent interest your wallet holds.

Batteries store energy. Power is energy per time. This also means that energy can be expressed as power times time, like the kiloWatt-hours used to express the electric energy your house consumes during a billing period. Another common measure of energy is the Joule. A Watt (a unit of power) is one Joule per second. A kiloWatt-hour is therefore 3.6 MJ.

Batteries are usually rated in units of current times time. This does not directly tell you how much energy the battery can store, but can be a more useful value in deciding how long a circuit will run from a battery. For example, a car battery might be rated for 50 Ah. That means in theory it could source 50 A continously for 1 hour and then go dead. In practise it's never that simple, and there are various factors that effect battery capacity (the current*time rating), like temperature rate of discharge, previous history, etc.


10 kW from 12V -> 833Amps

10kWh from 12V batteries -> 833Ah capacity

Or seventeen 50Ah car batteries in parallel

  • \$\begingroup\$ You forgot the time aspect: your answer assumes the 10kW must be delivered for one hour. A single car battery can deliver 100..200A, so for a short time period 4 batteries might be enough. \$\endgroup\$ – Wouter van Ooijen Mar 28 '12 at 18:12
  • \$\begingroup\$ The question as framed does not have a time element. The discharge rate could be at 1mA meaning that the batteries would take 833000 hours or nearly 100 years to discharge (ignoring self-discharge effects) \$\endgroup\$ – uɐɪ Mar 29 '12 at 6:46
  • \$\begingroup\$ Someone has edited the question, replacing kW with kWh.... \$\endgroup\$ – Wouter van Ooijen Mar 29 '12 at 14:44

Olin's answer is pretty good, but it's worth noting that batteries are rated in amp hours because many factors which affect the amount of voltage a battery can deliver in any particular situation have much less effect on the total amount of charge it will be able to deliver. A battery which would be 90% depleted after delivering 3600 Coulombs (1AH) at 12.0 volts under one set of circumstances would probably be 90% depleted after delivering 3600 Coulombs (1AH) at 10.2 volts, even though in the latter scenario it would have delivered 15% less usable energy.

It's also worth noting that trying to get charge from a battery quickly (i.e. trying to draw large amounts of current) will generally cause the output voltage to sag, reducing the amount of energy delivered per unit of charge consumed, and consequently reducing the total amount of energy one will be able to extract before the battery is depleted. Some batteries exhibit this behavior to a much larger extent than others. A simplified way of thinking about this is to think of a battery as being like a parking lot, with some combination of parking places and travel lanes. A parking lot in which every parking space had its own dedicated lane to the outside could be loaded and unloaded very quickly, but wouldn't be able to accommodate very many cars. A parking lot which crammed cars within a few inches of each other and had no dedicated travel lanes except right at the exit might hold ten times as many cars, but getting cars in and out would be very slow. Most parking lots are, of course, somewhere between those two extremes, but there is still considerable variation.

  • \$\begingroup\$ While the analogy is helpful to understand the effect, why does that effect happen in batteries? I know that different batteries have different chemical makeups, does that affect it at all? \$\endgroup\$ – NickHalden Mar 28 '12 at 16:46
  • \$\begingroup\$ Batteries have resistance, which loses energy in heat loss due to I2R dissipation. But supercat's answer sort of touches on two other effects: (1) higher current use causes the battery voltage to reach its "end-of-discharge" voltage more quickly (you think it's empty sooner than it actually is) due to IR drop, and (2) higher current use actually makes the available amp-hour capacity slightly smaller. I think this 2nd effect is due to chemical diffusion: the battery electrodes are at the surface of the battery, but energy is stored volumetrically, so it takes a small amt of time to diffuse out. \$\endgroup\$ – Jason S Mar 28 '12 at 17:08
  • \$\begingroup\$ Good reminder that Ah are not energy, but Wh are. \$\endgroup\$ – user42875 Apr 16 '15 at 23:55

As people have noted, what is stored is, effectively, energy

Energy is a measure of how much work has been done or can be done.
Power is the rate and which work is done or the amount of work needs to be done or can be done.

A car may operate at 40 HP while driving under certain conditions.
That's power - as the name horsepower implies.
Power is the instantaneous measure of work being done.

If the same car operated for one hour at 40 HP then the energy used is 40 horsepower.hours. We do not usually hear HP.hours mentioned BUT we do hear of kilowatt hours or kWh when measuring electrical energy.

Batteries are often rated in Ampere.hours or Amp.hours or Ah.
A.h are actually NOT a measure of energy but they imply energy if we know the voltage as well.

The proper units of power (= instantaneous work rate) for a battery is Watts. The proper units of energy (= work done or doable) for a battery is Watt.seconds or Joules.

If we work for one second at a power of one Watt we do 1 Watt second of work or 1 Joule of work and use 1 Joule of energy.

For interest, we do about one Joule of work by lifting 0.1 kg a height of one metre against sea level gravity.

Now - all the above is true [E&OE] but liable to confuse. There is a vast amount about this on the internet. For now

Power = rate of doing work Watts or kiloWatts (1000 Watts = 1 kW.)
Energy or work done is measured in Joules.

1000 Joules = 1 kiloJoule = 1 kJ.

In one hour at one Watt we use 1 W x 3600 s = 3600 Joule = 3.6 kJ

Battery energy = Volts_average x Amp hours capacity = Watt hour capacity.

Battery energy density:

Energy density can be measured in two ways.

Volumetric energy density tells use how many Watt hours can be fitted into 1 litre
Mass energy density tells use how many Watt hours can be fitted into 1 kilogram.

How much battery capacity / mass / volume is needed to provide a certain number of Joule can be determined from data for the battery chemistry used.

Table below is somewhat out of date but gives some idea.

                        Wh/kg   Wh/l  

Carbon Zinc             9       60-120  
Alkaline                162     398     
Lithium                 140-340 410-710  
Lithium Ion             105-130 270-325   
Lithium Polymer         120     250        
NiCd                    40-60        
NimH                    60-80        
SLA                     30           
Silver oxide            130    500   

More later maybe ...


For a standard car battery


I=2.25 amps

Therefore power = 12*2.25= 27watts

Standard depletion time = 20hours = 72000seconds

Therefore energy = 27*72000= 2000kJ

  • 3
    \$\begingroup\$ "I=2.25 amps". Where did this universal specification come from? -1 \$\endgroup\$ – Transistor May 18 '16 at 21:48

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