In the determination of internal resistance of a cell of e.in.f. 1.5 V with potentiometer, the balancing length is 75 cm in the open circuit and when a resistance of 10`Omega` is used the balancing length is found to be 65 cm. Calculate the internal resistance of the cell.

Explanation

Data supplied,
` l_(1)=75cm, " "l_(2)=65cm," "R=10Omega`
Internal resistance `r=((l_(1)-l_(2))/(l_(2)))R=((75-65)xx10)/(65)=1.54Omega`