Q. 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 Ω, 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 :
A) 0.25 Ω
B) 0.95 Ω
C) 0.5 Ω
D) 0.75 Ω
Explanation
[Solution] $r = R\left(\frac{l_1}{l_2} - 1\right) = 9.5\left(\frac{3}{2.85} - 1\right) = 0.5$ Ω. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.