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  • An experiment is conducted to compare the emf of two cells using a potentiometer. A primary cell and a secondary cell are connected in series through a potentiometer wire of length L = 100 cm. The

An experiment is conducted to compare the emf of two cells using a potentiometer. A primary cell and a secondary cell are connected in series through a potentiometer wire of length L = 100 cm. The balancing length for the primary cell is L1 = 40 cm, and for the secondary cell, it is L2 = 60 cm. If the internal resistance of the potentiometer is negligible, calculate the ratio of the emf (\epsilon_{1} / \epsilon_{2}) of the two cells.

Option: 1

2/3


Option: 2

4/3


Option: 3

2/5


Option: 4

7/3


Answers (1)

best_answer

Given values:

Length of the potentiometer wire (L) = 100 cm
Balancing length for primary cell (L1) = 40 cm
Balancing length for secondary cell (L2) = 60 cm

The ratio of emf of the two cells (\epsilon_{1} / \epsilon_{2}) can be calculated using the formula:

                    \frac{\epsilon_{1}}{\epsilon_{2}} = \frac{L_{1}}{L_{2}}

Step 1: Calculate the ratio of emf of the two cells:

                    \frac{\epsilon_{1}}{\epsilon_{2}} = \frac{40 cm}{60 cm} = \frac{2}{3}

The ratio of emf of the primary cell to the secondary cell is 2/3.

Posted by

SANGALDEEP SINGH

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