Q3.23Written answer
Question 3. 23:
Figure below 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.
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Worked solution
Internal resistance of the cell, r = 1.5 V cell
Balance point of the cell in open circuit, l = 76.3 cm
External resistance, R = 9.5 Ω
New balance point, l1 = 64. 8 cm
The expression for internal resistance is given as
\(\begin{array}{l}r=R\left [ \frac{l}{l_{1}}-1 \right ]\end{array} \)
\(\begin{array}{l}r=9.5\left [ \frac{76.3}{64.8}-1 \right ]=1.69\Omega\end{array} \)
Class 12 Physics NCERT Solutions for Current ElectricityChapter 3 Current Electricity of Class 12 Physics is framed as per the current CBSE Syllabus 2023-24. The important questions in Chapter 3 of Physics