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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
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\(0.4 \mathrm{~A}\)
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,
\(I=\frac{E}{R+r}=\frac{2}{4+1}=\frac{2}{5}=0.4 \mathrm{~A}\)
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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