JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2023
- Medium
- Numerical
A null point is found at \(200\,cm\) in potentiometer when cell in secondary circuit is shunted by \(5\,\Omega\). When a resistance of \(15\,\Omega\) is used for shunting null point moves to \(300\,cm\). The internal resistance of the cell is \(..............\,\Omega\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\(\frac{\varepsilon}{ r +5} \times 5=200\,x\)
\(\frac{\varepsilon \times 15}{ r +15}=300\,x\)
\(\Rightarrow r =5\)
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