The wire of potentiometer has resistance \(4 \Omega\) and length \(1 \mathrm{~m}\). It is connected to a…
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
\(0.1 \mathrm{~A}\)
\(0.2 \mathrm{~A}\)
\(0.4 \mathrm{~A}\)
\(0.8 \mathrm{~A}\)
Solution
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,
\(I=\frac{E}{R+r}=\frac{2}{4+1}=\frac{2}{5}=0.4 \mathrm{~A}\)