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An experiment is conducted to determine the internal resistance (r) of a cell using a potentiometer. A known emf of E = 1.5 V is connected to the potentiometer. The balancing length for the cell is L = 60 cm, and the total length of the potentiometer wire is Ltotal = 100 cm. If the internal resistance of the potentiometer is negligible, calculate the internal resistance of the cell.

Option: 1

0.6 volt


Option: 2

0.5 volt


Option: 3

1.0 volt


Option: 4

0.7 volt


Answers (1)

best_answer

Given values:

Emf of the cell (E) = 1.5 V
Balancing length (L) = 60 cm
Total length of the potentiometer wire (Ltotal) = 100 cm

The potential difference across the length L of the potentiometer wire is given by:

                    V = \frac{L}{L_{total}} E

The potential difference V is also equal to the product of the current (I) and the total resistance (R) in the circuit:

                    V = I.R

The total resistance R can be written as the sum of the internal resisitance (r) of the cell and the resistance (Rpot) of the potentiometer wire:

                    R = r + Rpot

Substitute the expressions for V and R and solve for r:

                    r = \frac{L}{L_{total}} E - R_{pot}

Step 1: Calculate the resistance of the potentiometer wire:

                    R_{pot} = \frac{L_{total} - L}{L} E

Step 2: Calculate the internal resistance r of the cell:

                    r = E - R_{pot}

Step 3: Substitute the given values and calculate r:

                    r = 1.5 V - \frac{40 cm}{60 cm} 1.5 V = 0.5 V

The internal resistance of the cell is 0.5 V.

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