Charging 12V 150Ah battery I want to charge a 12V battery of 150Ah with a solar panel. The solar panel specs is
12V, 25 Watt.
Can anyone please provide me how to calculate that how much time it will take to charge the battery? Please provide the calculations and formulas.
 A: Watts (electrical power) = Volts $\cdot$ Amps,  so 25W = 12V $\cdot$ 2.1A
150Amp Hour is the total capacity so 150amp $\cdot$ 1hour, 1amp $\cdot$ 150hours, or 2.1amp for 72hours.
That's in an ideal world of course, there are heating losses as you charge the battery, the voltage of the solar panel varies with the load and if you entirely empty a 12V lead acid battery you are likely to damage it. But basically you are looking at 10days of full sunshine 
A: The answers you've got so far from Vladimir and Martin give you a good first-order approximation: power = current x voltage. Energy = power x time. So your 12V battery of 150Ah needs 1800Wh of energy (12 x 150). So a 25W PV panel would need 72 hours at full output (1800Wh/25W).
The equation is:
Peak-hours required = $\frac{V_{batt} \times capacity_{batt}}{Power_{PV}} $
That's an unusual combination of quite a small panel and a very big battery - is the battery designed to be a multi-week store for a low-power system, by any chance?
For a more accurate answer, you need a lot more information.
The output of the panel at any moment will depend on:
(1) the voltage, (which will be determined by the battery, and will change as the battery charges)
(2) the ambient temperature, and 
(3) the amount light hitting the panel. The amount of light hitting the panel will depend on panel tilt, orientation, overshading, weather, altitude, location, time of year, time of day.
For (1), you need the panel IV curve, typically presented on the manufacturer's datasheet. That datasheet will also tell you how output varies with panel temperature. You can then estimate the panel's temperature from the ambient temperature - it will (to the first order) be a fairly steady amount above ambient.
Here are some example power curves for a 12V PV (and it looks like it's approx 125W peak) panel, from altestore.com :

You also need to know about the battery's characteristics as it charges: the heat losses, and how its voltage changes with temperature and % charged. In a more complex system that involves a controller too, you'd need to know how the controller behaves, how it tracks the maximum power point, and what its internal losses are. A controller is very desirable, not only because it will track the PV panel's maximum power point, but also because it will prevent the battery over-charging.
There are various online PV calculators that will combine some of this information to give you estimates of output, such as the US NREL PVWATTS calculator. And there are free online GIS insolation resources to give you the raw source information to do detailed calculations yourself, such as the EU PVGIS database.
A: You need to know the implication of using a 25W  panel, for me that is pretty small.
The idea is 150Ah at 12 V (DC) we need to apply the 10hr charging method, so 10% of 150Ah gives 15A.$$\frac{150\,\text{Ah}}{10\,\text h}=15\,\text A$$
So power of needed solar panel is computed as follows, $12\,\text V\cdot15\,\text A=180\,\text W$. So the power of your panel is 180 W. 
A: 150 Amper*hour is the battery charge capacity. With power of 25 Watt = 25 A*V at V = 12 V you will supply 25/12 Coulomb each second (Amper*s). So the charge time is equal to 150/(25/12) = 72 hours.
A: In the simplest case, and some unmentioned assumptions, the minimum charging time would be given by the formula already mentioned: $$ T = \frac{V \times Ah}{W}$$  this gives 12x150/25 = 72 hours.  
However, this case makes at least two major assumptions, the battery is fully discharged and there are no losses of any kind! 
In the real world, the battery will not be fully discharged and there are energy losses "everywhere." Just to drive the point home, if the battery is already charged, then the time would be zero. 
A typical 12V battery is considered "discharged" when its voltage is reduced to 9V.  This means it has lost only 1/4 of its energy. Everything else remaining the same, it would only need 1/4 of the time (18 hrs) to charge up.  If we assume 50% efficiency, this would bring the time up to 36 hours.  Assuming 8 hrs per day of sunshine, it would take about 4 days.    
