AP EAMCET · PHYSICS · Current Electricity
In a potentiometer experiment, the balancing with a cell is at length \(250 \mathrm{~cm}\). In shunting the cell with a resistance of \(2 \Omega\), the balancing length becomes \(125 \mathrm{~cm}\). The internal resistance of the cell is
- A \(2 \Omega\)
- B \(4 \Omega\)
- C \(0.5 \Omega\)
- D \(1 \Omega\)
Answer & Solution
Correct Answer
(A) \(2 \Omega\)
Step-by-step Solution
Detailed explanation
Given, balancing length with emf of cell, \(l_1=250 \mathrm{~cm}\) Balancing length with cell and shunt resistance, \(l_2=125 \mathrm{~cm}\) Shunt resistance, \(R=2 \Omega\) Let \(r\) be the internal resistance of cell. Then, using relation, \(r=R\left(\frac{l_1}{l_2}-1\right)\)…
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