I did some math on it. I’m not sure if it’s correct. Perhaps a Doper math whiz could double check it for me.
We first need to figure out how much energy a car battery can store. If the battery is 640 CCA, then it must be capable of sourcing 640 amps at 7.2 V for 30 seconds. This is a power of 4608 W, and the energy produced over these 30 seconds is 138240 J. But the battery doesn’t go dead after 30 seconds, so the total energy in the battery must be higher. I found out the “reserve energy” is a better number to use. According to this page, a typical car battery has around 2,000,000 J when you calculate the energy based on reserve energy. So let’s use that figure.
Now let’s assume your car battery is completely dead. This means it contains no energy, and the voltage is 0 V. How long will it take a charger to pump 2,000,000 J into the battery using a current of 1.2 A?
Let’s assume that, while we’re charging the battery at 1.2 A, the battery voltage linearly increases over time. Let’s also assume the battery is fully charged (and thus the energy in the battery is 2,000,000 J) when the voltage reaches 12.6 V. The time it takes to fully charge the battery is T seconds. We need to solve for T.
We know that
P = I*V
and thus
p(t) = I*v(t) where I = 1.2 amps.
When t = 0, v(t) = 0 V. And when t = T, v(t) = 12.6 V. Assuming a linear increase in voltage over time,
v(t) = (12.6/T)*t, where t is in seconds.
Therefore
p(t) = I*(12.6/T)*t
Energy is the integral of power:
E = 2000000 = ∫ p(t) dt, where the limits of the integral are t = 0 to t = T.
E = ∫ I*(12.6/T)*t dt
E = I*(12.6/T) ∫ t dt
E = I*(12.6/T)*((T^2)/2 - (0^2)/2)
E = I*(12.6/T)(T^2)/2 = I12.6*T/2
Solving for T:
T = 2E/(I12.6) = 22000000/(1.212.6)
T = 264550 seconds ≈ 3 days
This assumes, of course, the battery will take a charge.