Figure shows a 2.0 V potentiometer used for the determinationof internal resistance of a 1.5 V cell. The balance point of the cell inopen circuit is 76.3 cm. When a resistor of 9.5 W is used in the externalcircuit of the cell, the balance point shifts to 64.8 cm length of thepotentiometer wire. Determine the internal resistance of the cell.
Given: The emf of potentiometer is 2.0 V , the voltage of cell is 1.5 V , the balance point of the cell in open circuit is 76.3 cm and the balance point when a resistor of 9.5 Ω is used in the external circuit is 64.8 cm .
Consider the given circuit.
The internal resistance of the cell is given as,
r = ( l 1 − l 2 l 2 ) R
Where, the internal resistance is r , the balance point of the cell in open circuit is l 1 , the new balance point in the presence of external resistance R is l 2 .
By substituting the given values in the above expression, we get
r = 76.3 − 64.8 64.8 × 9.5 = 1.68 Ω
Thus, the internal resistance of the cell is 1.68 Ω .
Given: The emf of potentiometer is
Consider the given circuit.
The internal resistance of the cell is given as,
Where, the internal resistance is
By substituting the given values in the above expression, we get
Thus, the internal resistance of the cell is
