TS EAMCET · Physics · Current Electricity
The length of a potentiometer wire is 2.5 m and its resistance is \(8 \Omega\). A cell of negligible internal resistance and emf of 2.5 V is connected in series with a resistance of \(242 \Omega\) in the primary circuit. The potential difference between two points separated by a distance of 20 cm on the potentiometer wire is
- A 1.6 mV
- B 4.8 mV
- C 6.4 mV
- D 3.2 mV
Answer & Solution
Correct Answer
(C) 6.4 mV
Step-by-step Solution
Detailed explanation
\(I = \frac{E_{cell}}{R_{wire} + R_{series}} = \frac{2.5 \text{ V}}{8 \Omega + 242 \Omega} = \frac{2.5}{250} = 0.01 \text{ A}\) \(k = \frac{I \times R_{wire}}{L_{wire}} = \frac{0.01 \text{ A} \times 8 \Omega}{2.5 \text{ m}} = \frac{0.08}{2.5} = 0.032 \text{ V/m}\)…
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