Class 12 Physics - Chapter Current Electricity NCERT Solutions | Figure shows a 2.0 V potentiometer used

Welcome to the NCERT Solutions for Class 12th Physics - Chapter Current Electricity. This page offers a step-by-step solution to the specific question from Exercise 1, Question 24: figure shows a 2 0 v potentiometer used for the de....
Question 24

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.

Answer

Internal resistance of the cell = r

Balance point of the cell in open circuit, l1 = 76.3 cm

An external resistance (R) is connected to the circuit with R = 9.5 Ω 

New balance point of the circuit, l2 = 64.8 cm

Current flowing through the circuit = I

The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω .

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