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JEE Mains · Physics · STD 12 - 3. current electricity

An ideal battery of \(4\, V\) and resistance \(R\) are connected in series in the primary circuit of a potentiometer of length \(1\, m\) and resistance \(5\,\Omega \) . The value of \(R\), to give a difference of \(5\, mV\) across \(10\, cm\) of potentiometer wire, is: ................ \(\Omega\)

  1. A \(490\)
  2. B \(480\)
  3. C \(395\)
  4. D \(495\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(395\)

Step-by-step Solution

Detailed explanation

\(i=\frac{4}{5+R}\) \(V_{A B}=i(5)=\frac{20}{5+R}\) \(V_{A P}=\frac{V_{A B}}{L}(0.1)=\frac{20}{5+R}\left(\frac{0.1}{1}\right)=\frac{2}{5+R}\) Now, \(\frac{2}{5+R}=5 \times 10^{-3}\) \(\Rightarrow \quad R=395\, \Omega\)
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