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  • In a laboratory experiment, a potentiometer is used to determine the internal resistance (r) of a cell. The potentiometer has a length of L and a total resistance R. A standard cell of known emf E

In a laboratory experiment, a potentiometer is used to determine the internal resistance (r) of a cell. The potentiometer has a length of L and a total resistance R. A standard cell of known emf E is used as the comparison. The null point is obtained when the potential difference across the potentiometer wire is balanced by the emf of the standard cell.

Given:

Length of potentiometer wire, L=1.5 \mathrm{~m}

Total resistance of potentiometer, R=20 \Omega

Emf of standard cell, E=1.2 \mathrm{~V}

Current at null point, I=0.05 \mathrm{~A}

The internal resistance r of the cell is:

Option: 1

1 \Omega


Option: 2

4 \Omega


Option: 3

3 \Omega


Option: 4

5 \Omega


Answers (1)

At the null point, the potential difference across the potentiometer wire is balanced by the emf of the standard cell: 
E=I \cdot R

The total potential drop across the potentiometer wire is given by:
 

Equating the two expressions for E, we get:

I \cdot R=I \cdot(r+R)

Solving for r, the internal resistance of the cell:

r=\frac{E}{I}-R

Substitute the given values:

r=\frac{1.2 \mathrm{~V}}{0.05 \mathrm{~A}}-10 \Omega=24 \Omega-20 \Omega=4 \Omega

The internal resistance r of the cell is =4 \Omega.

Posted by

Kshitij

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