ExamBro
ExamBro
COMEDK · Physics · 19. Current Electricity

To the potentiometer wire of length \(L\) and \(10 \Omega\) resistance, a battery of emf \(2.5 \mathrm{~V}\) and a resistance \(R\) are connected in series. If a potential difference of \(1 \mathrm{~V}\) is balanced across \(L\) / 2 length, the value of \(R\) in \(\Omega\) will be

  1. A 1
  2. B \(1.5\)
  3. C 2
  4. D \(2.5\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2.5\)

Step-by-step Solution

Detailed explanation

The given situation is shown below. Resistance of potentioneter wire, \(R_{p}=10 \Omega\) BmI, \(E=25 \mathrm{~V}\) Total resistance, \(R^{\prime}=R_{p}+R=(10+R) \Omega\) Current through potentiometer wire, \[ l=\frac{E}{R_{p}}=\frac{25}{10+R} \] Since, potential difference…