A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity, (ii) 9.5 , the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is
- 0.25
- 0.95
- 0.5
- 0.75
Answer
(3) 0.5
Solution
With R = (open circuit) the balancing length l1 = 3 m corresponds to the emf; with R = 9.5 the balancing length l2 = 2.85 m corresponds to the terminal voltage.
r = R ( l1l2 - 1 )
= 9.5 ( 32.85 - 1 ) = 9.5 0.152.85
= 0.5
Answer: (3)