Question 24: NCERT Solutions for 12th Class Physics: Chapter 3-Current Electricity
Figure 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 9.5 Ω 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.
In the open circuit, the balance point is obtained for the emf of 1.5 V.
When the external circuit is connected, a current is drawn from the cell of 1.5 V in external resistance of 9.5 E1. Now the balance point is obtained for terminal potential
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NCERT Solutions for 12th Class Physics: Chapter 3-Current Electricity