Assuming a typical lead-acid, 12 V car battery (typically at 13 V or so fully charged), and that it takes roughly 500 A over 3 seconds to start an engine, how long will it take to recharge the battery at any given charge rate?

Here's my attempt from what I remember about physics:

12.8 V * 500 A = 6400 W

Over 3 seconds that's 19,200 joules.

So, in a perfect world where all the current goes right back in to the battery and whatnot, how long does it take to regain all my joules and put them back in my battery?

Given a 2A charge rate:

14 V (output of charger?) * 2 A = 28 watts

Here's where I'm a little shaky. What's next? Divide the joules by the wattage to get time? Seems like it:

19,200 joules / 28 watts = 11.4 minutes.

That's it? 11.4 minutes at 2 A and all 19,200 joules are back? Seems hard to believe. My charger also has a 10A setting. So that means in about 2.5 minutes, it'll be "recharged".

So, are my assumptions correct? Do you really just use the charging voltage to calculate this, it seems like you would need to put the charging voltage in relation to the battery's capacity/voltage/whatever.

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    \$\begingroup\$ 14v*2A = 28 Watts, but that would only be true if your battery was at 0v. If your battery was at 12v, there's only 2v difference, 2V * 2A = 4W = A very long time to recharge it. Hence why automotive alternators typically run to 100A output. \$\endgroup\$ – John U Feb 11 '13 at 11:28
  • \$\begingroup\$ Ah, ok, that was another thing I was unsure about. Thanks. \$\endgroup\$ – Nick Feb 12 '13 at 14:32
  • \$\begingroup\$ It also assumes your charger would manage 2A into a 12v battery, it may in reality be much less as the battery voltage comes up towards 14v. \$\endgroup\$ – John U Feb 12 '13 at 15:13
  • \$\begingroup\$ @JohnU, this is something I too was wondering about since the voltage on the battery will rise just as it would with a charging capacitor - thereby lowering the charge current. However, in the context of automobiles, do altenators act as constant current sources rather than constant voltage sources with high current capacities? \$\endgroup\$ – sherrellbc Jun 30 '14 at 19:20
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    \$\begingroup\$ Using current and voltage introduces LOTS of uncertainties. Your calculations (and the ones in the answers which take your assumptions as correct) are WAY overestimated. You should take the starter motor power (usually 1-1,5 kW, that is still overestimated, if you consider the drop in battery voltage) and the time to start the car. Check here for better calculation and adapt the answers below: physics.stackexchange.com/questions/57794/… Basically it's not 19.2 kJ as you wrote, it's only 4 kJ. \$\endgroup\$ – FarO Apr 16 '15 at 12:56

No, it isn't Joules in = Joules out. To a first approximation, it's Coulombs in = Coulombs out. It's the electrons flowing through the circuit that participate in the chemical reaction inside the battery (but not at 100% efficiency).

Forget about the energy/power/voltage calculation and just make the ampere-seconds for charging equal to the ampere-seconds for discharging, and then multiply by a fudge factor to account for the inefficiencies.

500A × 3s =1500 A-s = 2A × 750s = 10A × 150s

750s = 12.5 minutes

Figure about 90% efficiency, so the 12.5 minutes / 0.90 = about 14 minutes.

  • \$\begingroup\$ Ok that's much simpler. How does the voltage play into this? Specifically, what happens if I charge at 15 V instead of 13V? How does one pick a charge voltage and how does it affect the charging? \$\endgroup\$ – Nick Feb 12 '13 at 14:36
  • \$\begingroup\$ @Nick: The charge voltage simply influences the instantaneous current going into the battery, based on the battery's current internal voltage and internal resistance. Higher voltage implies more current, but the current may need to be limited to a certain value based on the physical construction of the battery. \$\endgroup\$ – Dave Tweed Feb 12 '13 at 14:47
  • \$\begingroup\$ I've never thought much about this. If you connect a discharged battery (say 12V) to a 15V potential, you are essentially shorting 3V across only wire resistance. What is the consequence of this? Assuming 1 Ohm resistance, is this simply the equivalent of charging the battery with 3A continuous current? No further consequence (assuming current can be supplied)? It's also likely that a short enough wire has << 1 Ohm resistance, so the current is bound to be much higher than 3A in this case. \$\endgroup\$ – sherrellbc Jun 30 '14 at 19:14
  • \$\begingroup\$ 500 CCA is the maximum rating of the battery. Normally it only takes 100-200A to start a car so more like 5 minutes to recharge the battery by your calculation. \$\endgroup\$ – SurpriseDog Sep 21 '19 at 16:34

For "Close enough", you can use

Tcharge = Tdischarge * (idischarge / icharge) * k

k is a unitless current efficiency factor and varies with battery chemistry, charge and discharge rates, battery state of charge and phase of the moon (and sometimes whether today is a bank holiday), but for a

  • lead acid battery: about 1.1 to 1.2
  • lithium ion battery: about 1.01
  • nickel-metal hydride (NiMH): about 1.15 to 1.2

This just says that charge and discharge times are inversely proportional to current drain multiplied by a variable constant.

The "constant" varies because of many factors. Lithium chemistries have no secondary reactions that "eat up" current input. NimH (and NiCd) have secondary chemical reactions that make gases, heat and other fun stuff and consume some of the supplied energy.

Note: Current ratios are not the same as Energy charge ratios.
When charging, the current flow through the internal resistance will cause a drop in voltage between input and battery_proper, so Vin must be greater than Vbattery_proper as the current drop across the internal resistance is lost.

When discharging, the internal resistance again drops the voltage, but Vout will now be lower than Vbattery_proper due to internal drops. So you lose both ways. Overall,

(energy efficiency) = k * (Vout,mean / Vin,mean)

At high currents (such as from a car cranking a starter-motor), up to about half the total voltage may be dropped across the internal resistance. That means that a less than fully charged, less than good condition 12 V car battery may measure 6 V at the terminals during cranking. The same battery will require up to 13.6&nbap;V when charging.

So, voltage efficiency, if discharged by cranking and charged when the battery is almost fully charged, is equal to 6 / 13.6 = ~44%. This is after the 90% efficiency mentioned above for lead acid.
So, for example, a near fully charged lead acid battery that is a "bit tired" may manage 0.9&nbsp:* 0.44 = ~40% energy efficiency for discharged energy over charge energy.

  • \$\begingroup\$ Wait, are you saying that LiIon and NiMH are "over-unity"?!? \$\endgroup\$ – Dave Tweed Feb 11 '13 at 0:23
  • \$\begingroup\$ @DaveTweed - No - I'm saying that I had charge and discharge threads in mind and mixed them. (I usually think in terms of "how much current do I get out for 1 A in" - he is asking in terms of "how much must I put in to get xxx out". I spotted this during editing and wondered if anyone would pick up on it before I corrected it in the final version. You did :-). \$\endgroup\$ – Russell McMahon Feb 11 '13 at 0:31
  • \$\begingroup\$ +1 For bringing in phase of the moon and bank holidays. I always make sure to bring those in when talking about carburetors with people. \$\endgroup\$ – Nick Mar 25 '14 at 3:03

Even if your battery delivers 500A, that's the PEAK current, when the motor is stopped and there is no Back EMF, so basically the motor is a small resistance and inductance. After the motor stars spinning, the back EMF lowers the current drained for the battery. I guess these huge currents are drained for ms only.

  • 3
    \$\begingroup\$ Don't guess. In fact, the starter motor for an automobile engine is operating close to stall current for the duration of the cranking, until the engine catches and removes the mechanical load from the starter. It really does require a few hp (= a few kW) to crank a modern high-compression engine. However, keep in mind that the automobile's charging system, with a capacity of 50 to 100A or more, can restore the cranking charge within a minute or two. \$\endgroup\$ – Dave Tweed Feb 12 '13 at 14:50
  • \$\begingroup\$ @DaveTweed, where did the assumption in the OP come from then for 500A cranking current? \$\endgroup\$ – sherrellbc Jun 30 '14 at 19:17
  • \$\begingroup\$ @sherrellbc: I don't know; you'll have to ask him. That's just the value we're all using here for concreteness in the discussion. Substitute whatever value you like into the equations. \$\endgroup\$ – Dave Tweed Jun 30 '14 at 19:31

If it takes 500 A for 3 seconds to start the engine, 1500 ampere-seconds are used. If the battery is recharged at 1 A, then it takes 1500 seconds (25 minutes) to get 1500 ampere-seconds.


The resting voltage of the full battery is 13.8 Volts at normal temperature. The alternator's regulator puts out 14.4 to 14.7 Volts. This overpotential is necessary or else the battery would never get full, taking infinite time to charge. Between the resting and charging voltage is a difference of less than 1 Volt, which is just low enough that it doesn't cause electrolysis of the water (i.e. "boiling" the battery dry) under normal circumstances.

The internal resistance of the battery is so low that this 1 Volt difference would cause about 50 amps to flow into the battery even when it's full. However, the battery plates and the electrolyte also form a crude electrolytic capacitor, which charges up to this difference and closes the gap. This capacitor charge is what you see when you measure a recently charged battery, and it takes about a minute to dissipate by leaking back through itself.

In effect, you have a battery with a leaky capacitor connected in series. When you start the car, some of the charge in the plates is used, and when the alternator starts putting charge back in, it charges up this capacitor first and whatever leaks through is what actually charges the plates. This is why a lead-acid battery needs the overpotential to charge - charging at exactly 13.8 Volts would never get it full.

So, it doesn't much matter how large your alternator is - the battery will take whatever it wants to take, and so it actually depends on the battery how long it takes to charge back after cranking the car. As the battery ages, develops sulfation and the plates corrode off, the capacitor effect becomes stronger and it takes longer and longer to actually top up the battery.

Eventually when it gets really bad, it will seem like the battery holds charge just as long as you keep the jumper cables on, because all the charge is in the capacitor and none in the actual plates.


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