AP EAMCET · PHYSICS · Current Electricity
The wire of potentiometer has resistance \(4 \Omega\) and length \(1 \mathrm{~m}\). It is connected to a cell of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\). The current flowing through the potentiometer wire is
- A \(0.1 \mathrm{~A}\)
- B \(0.2 \mathrm{~A}\)
- C \(0.4 \mathrm{~A}\)
- D \(0.8 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(0.4 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Resistance of potentiometer wire, \(R=4 \Omega\) Emf of cell, \(E=2 \mathrm{~V}\) Internal resistance, \(r=1 \Omega\) Current flowing through the potentiometer wire, \(I=\frac{E}{R+r}=\frac{2}{4+1}=\frac{2}{5}=0.4 \mathrm{~A}\)
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