Comparison Of Emf Using Potentiometer — the NEET Physics formula with its derivation, variables, validity constraints and worked solver.
Comparison of EMF using Potentiometer Applicable for comparing the EMFs of two primary cells using a potentiometer setup where the current is constant and the wire is uniform. According to the principle of the potentiometer, the potential drop across a length l is directly proportional to the length ( V = kl ) provided the current is constant. When cell 1 is in the circuit, at the null point, no current flows through the galvanometer, so 1 = k l 1 . When cell 2 is in the circuit, at the null point, 2 = k l 2 . Dividing the first equation by the second eliminates the potential gradient k : 1 2 = k l 1 k l 2 = l 1 l 2 . The EMF of the driver cell (auxiliary battery) must be greater than the EMF of the cells being compared. The positive terminals of both cells and the driver cell must be connected to the same end (zero point) of the potentiometer. The potentiometer wire must have a uniform cross-sectional area and composition. Current through the potentiometer wire must remain steady during the experiment. Thinking that the internal resistance of the cells being tested affects the balancing length (since no current flows at the null point, internal resistance does not cause a potential drop). Assuming the balancing length depends on the resistance of the galvanometer. Believing the formula works if the driver cell EMF is lower than the test cell EMF (null point cannot be obtained).