How at balancing point in a potentiometer does the galvanometer record zero deflection? Why is no current flowing at this point of time in the secondary circuit? Can someone explain what is going on in the the potentiometer wire and secondary circuit giving reference to electric potentials at the nodes.
If you notice, the cell in the primary circuit and the secondary circuit are oppositely aligned. This means that if we take the two circuits to be independent, the current will flow in opposite directions for the wire in common.
At a certain point on the wire the circuit becomes equivalent to a Wheatstone Bridge and there is no current flowing through the potentiometer wire. Hence the galvanometer shows zero deflection.
A key element of potentiometer is a uniform wire, the potentiometer wire, across which a known potential difference (voltage) is placed. Going from the positive end of the potentiometer wire to the negative is analogous to going downhill. It is, of course, electrical potential rather than gravitational potential that is decreasing.
When using a potentiometer you link a point, A, in the secondary circuit to one end (let's say the positive end) of the potentiometer wire. Suppose that there is a fall in potential (a voltage drop) in the secondary circuit between A and another point, B. And suppose the fall in potential to be less than that across the potentiometer wire (from the end connected to A to the other end).
In that case there will be a point P along the potentiometer wire that is at the same potential as B; it's as far 'downhill' from A as B is. P and B are, as it were, at the same level. There is no pd between them. Current will therefore not flow through a galvanometer connected between them.