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
- A 1
- B \(1.5\)
- C 2
- D \(2.5\)
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…
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