Potentiometer : how that balancing point is reached?

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When potentiometer is used to measure and compare emf of cells, how that balncing point is reached?...
this is the circuit diagram what i mentioned..
p23pot1.jpg


in the description it is said that at that balancing point (current through that galvanometer shows zero), potential drop across AC is equal to E emf of the cell.

But my doubt is that " by Kirchoff's voltage rule , consider that loop A-G-C-A, potential drop across AC is equal to E without any conditions like balanced or not. Then we have to say that each and every point is a balanced point, right?"
 
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In general, if the arrangement is not balanced, there will be a pd across the galvanometer, G. This will only be zero when the pd between A and C is equal to the emf of the cell
 
i can't understand.
even if it is balanced, there should be a pd across the galvanometer.(kirchoff's voltage rule).
the voltage in that closed loop should be zero...
technician said:
In general, if the arrangement is not balanced, there will be a pd across the galvanometer, G.