Q. 234.6( 14 Votes )

Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.


Answer :

Balance point for open circuit, l1 = 78.3 cm
Resistance of the external resistor, R = 9.5 Ω

Balance point for this resistor, l2 = 64.8 cm

The relation connecting internal resistance r and balance point is,


r=(78.3-64.8)/64.8× 9.5
 =0.177 × 9.5
 =1.685  Ω

Therefore, the value of the unknown resistance, X, is 1.685  Ω.


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