Charging of a battery (reversing of polarity) I wanted to know that how can we charge a battery using another higher voltage DC supply? I got a question that there is a 12 volt battery with an internal resistance 3ohm Connected with a 100 V DC supply , where the 100 v DC supply is connected to reverse the polarity. I don't quite understand the purpose of reversing the polarity. It will be really nice of you to help.
(P.S. I got bad experiences pls be kind)
 A: You can think of charging the battery as being like charging a leaky capacitor - ie one which has a high (but not infinite) resistance dielectric material between its plates.  
So you connect the + terminal of the DC supply to the + electrode of the battery/capacitor, and likewise for the - terminal/electrode.  The DC supply drags electrons from the + electrode of the capacitor, through the DC power supply unit, to the - electrode of the capacitor.  So the charge on each plate of the capacitor increases.  Meanwhile, some electrons flow as a leakage current through the dielectric, between the plates of the capacitor, undoing the work done by the DC unit. But overall the DC unit wins out.
If instead you connected the + on the DC supply to the - on the battery/capacitor, the DC supply would drag electrons from the - side of the capacitor and dump them at the + side.  This would gradually discharge the capacitor/battery.
A: The battery which if working normally has within it a chemical reaction occurring which produces a potential difference across its terminals which in your case is $12$ volts.  
When an external conducting circuit is connected to the terminals of the battery a current flows from the positive terminal, through the external conducting circuit, to the negative terminal of the battery.
In effect the chemical reaction in the battery is converting chemical energy into electrical energy.
If the external circuit is a $12$ volt power supply with the positive terminal of the power supply connected to the positive terminal of the battery then no current will flow.
The net voltage in the circuit is zero.
With the power supply set at 11 volts a current will flow from the positive terminal of the battery into the positive terminal of the power supply, out of the negative terminal of the power supply and into the negative terminal of the battery.
Chemicals in the battery will be used up to produce the flow of current.  
With a 13 volt power supply the current direction will be reversed with a current flowing from the positive terminal of the power supply into the positive terminal of the battery, out of the negative terminal of the battery and into the negative terminal of the power supply.
This time the chemical reaction in the battery is reversed and the electrical energy supplied by the power supply is converted into chemical energy in the battery.
The battery is being recharged ie the battery's supply of chemicals which can produce electrical energy is increasing and this energy comes from the power supply.
In your example the net voltage in the circuit can be found and knowing the maximum current that can flow will enable you to find the total resistance in the circuit part of which is due to the internal resistance of the battery.
From this you should be able to evaluate the extra series resistance you must add to the circuit to limit the current to 8 amps.
