Get Answers to all your Questions

header-bg qa

A potentiometer experiment is conducted to determine the internal resis- tance of a cell. A standard cell of known EMF 1.2V is connected to the potentiometer. The null point is obtained at a length of 60 cm. When the cell being tested is connected instead of the standard cell, the null point is obtained at a length of 48 cm. Calculate the internal resistance of the cell being tested.

Option: 1

8.8Ω


Option: 2

5.0Ω


Option: 3

4.8 Ω


Option: 4

5.10 Ω


Answers (1)

best_answer

The internal resistance (r) of a cell can be calculated using the formula:

r = Length of Potentiometer Wire × Known EMF/ Change in Potentiometer Length

Substitute the given values:

Length of Potentiometer Wire = 60 cm
                           Known EMF = 1.2 V

Change in Potentiometer Length = 60 cm − 48 cm = 12 cm
Calculate the internal resistance:
r=\frac{60\times 1.2}{12} = 6.0\Omega
Hence, the internal resistance of the tested cell is 6.0 Ω.

Posted by

manish painkra

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE