Question
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 ohm 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 ohm
-
0.95 ohm
-
0.5 ohm
-
0.75 ohm
Solution
C.
0.5 ohm
Given,
e.m.f., e = 2 Volt
length, l = 4 m
Potential drop per unit length, 
Case 1:
... (i)
where, e' is the emf of the cell
Case 2:
V = 
... (@)
From equations (i) and (ii), we have
e'/V = l1 / l2
e' =l (r+R) and
V = IR
Therefore,
Therefore,
