Finding EMF of a battery using potentiometers

objective to find out the emf of an unknown battery using potentiometer let us look at how a potentiometer is used to measure the emf of a battery we know that balancing equation for a potentiometer is b equals to v naught l by l naught where v is the potential difference across the terminals of the potentiometer l is the balancing length v naught is the potential drop across a b and l naught is the length of the wire a b

suppose the terminals of the potentiometer are connected across a battery with emf e at null deflection point the current through the battery is zero and the voltage difference across lm is equal to emf of the battery by identifying the corresponding length and substituting it in the balancing equation the emf of the cell can be measured accurately

one easy way of finding the emf of an unknown battery is to compare it with a standard battery of known emf two batteries one of known emf e1 and another of unknown emf e2 are connected to potentiometer using a two-way key as shown the driving circuit should have a strong battery so that the potential difference across a b is larger than the emf of the two batteries when the key is plugged in s1 the first battery gets connected to the driving circuit when the key is taken out and pressed into the other plug s2 the first battery gets disconnected and the second battery gets connected when the key is plugged into s1 the jockey is slid to find the null deflection point p1 on the wire a b let the balancing length ap1 be l1 e1 is equal to v naught into l1 by l naught where l naught is the length of a b and v naught is the potential difference across it key is taken out and gets inserted into s2 let null deflection occur at a point p2 this time let the balancing length ap2 be l2 e2 is equal to v naught into l2 by l naught upon dividing the two equations we get e2 by e1 equals to l2 by l1 the unknown emf e2 is equal to e1 into l2 by l1

for example in a potentiometer arrangement a battery of emf e1 equal to 5 volts gives a balance point at l1 equal to 240 centimeters now when an unknown battery is connected the balance point shifts to l2 equal to 360 centimeters then the unknown emf e2 is given by the relation e2 is equal to e1 into l2 by l1 thus e2 equals to 5 into 360 by 240 which comes out to be 7.5 volts

summary emf of an unknown battery e2 is equal to emf of a standard battery e1 into balancing length of unknown battery l2 by balancing length of the known battery