Why is there no current flowing across the battery 
Okay so I solved this question which asked me to calculate the value of R such that 
there is no current flowing in the galvanometer branch.
(THE ANSWER IS 100 ohm)
I used the loop rule to write the equations and solve for R.
It was after I solved the question that the trouble started. Upon closer inspection I saw that the potential drop across the 100 ohm resistor was also 2 V (Like the battery on the adjacent branch). So why is no current flowing across the 2 V battery. It can be said that the battery and the 100 ohm resistor are in parallel (Equal potential drops). How is the battery different from the 100 ohm resistor ?
 A: It might be useful to think of some limiting cases to get some intuition.  For example, assume that R starts off being infinite (open circuit), then all the current would flow through the 2V battery, which would be (12-2)/500 = 0.02A.  Now, as you decrease the resistance, some of the current will flow through R as well, but only what is needed to satisfy V=IR, where V=2V.  When R is very large, I will be very small.  Once you get to R=100 ohms, then 100*0.02=2V, so that the 0.02 A when flowing through the resistor R will exactly produce the 2V drop and thus require no current through the 2V battery.  If R goes below 100 ohms, then you will again see current flowing through the 2V battery, but in the other direction.  I hope this helps.
A: Think of a water pipe system. 


*

*The resistor is like a filter. It allows some water through (from any direction) but slows it down.

*The battery is like a pump. It literally forces water from one side to the other, and allows no water to flow the opposite way.


A battery is a fairly complex electrochemical device, but it does indeed act as a one-way charge mover, which via high chemical potentials separate electrons from atoms at one terminal and reunites them at the other. This causes a constant surplus of negative charge at one terminal and a lack at the other, and this charge difference corresponds to an electrical potential energy difference, measured as voltage. Think of voltage as water pressure difference. 
Those surplus electrons want to move, and the only available path they have is through the circuit (and thus they constitute a current - think of current as the water flow) because they aren't able to move through the liquid interior of the battery. This liquid (or maybe in the future non-liquid) electrolyte, as it is called, is an ion conductor, but not an electron conductor. 
Sure, rechargeable batteries (accumulators) are not "one-way streets", so the analogy is not perfect and depends a lot on the exact conditions. But the one-way pump idea works fairly well in most cases. 
A: The 12v battery is supplying current to R through the 500 ohm resistor that is equal to the current R would draw from the 2v battery, so they equal out.
