A potentiometer used to measure the emf of a cell, Ԑ is shown in the figure. AC is the potentiometer wire of length 100 cm and resistance 5 ohm. The emf of a standard cell is 6volt and rheostat R is adjusted to 5 ohm. If the null point is obtained at B with AB = 60 cm, what is the emf of the cell?
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Since galvanometer is used to find the null point when the battery of unknown EMF is connected to the potentiometer wire at a particular point(null position), no current flows through the unknown battery. Hence, there is no drop across the internal resistance of the battery.
Therefore, the potential across the given segment of wire is the absolute EMF of the unknown battery =
100
30
E
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