JEE Mains · Physics · STD 12 - 3. current electricity
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 \(..............\,\Omega\).
- A \(4\)
- B \(5\)
- C \(3\)
- D \(2\)
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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