Hi, I have a question which goes like this: There is a potentiometer consisting of a 2.0V driver cell, slide wire and galvanometer with a 2.0kΩ series resistor used to find the emf of a Daniel cell. The emf of the Daniel cell is found to be 1.08 V. Now, approximately what current flows through the galvanometer if the protective resistor is in use and the sliding contact is moved to (i) one end (ii) the other end of the slide wire? (The resistances of the galvanometer and of the cells may be neglected) The answers my teacher gave me are (i) I=V/R =1.08/2000 = 0.54 A (ii) I=(E-V)/R =(2.0-1.08)/2000 =0.46x10^3 A My problem is that I don't understand why in (i), current=(voltage of Daniel cell)/(resistance of the protective resistor) and in (ii) the similar doubt. Can anyone explain to me? Thanks in advance!