Question

Potentiometer wire is in 2 segments where diameter of wire BC is double of that of AB, while material is same as well as length of each segment is equal to 2m. When switch S is open, distance of jockey in meter from end A is 2.4 m. Distance of jockey in meters from end A when switch S is closed.

Answer 1

Let current through potentiometer wire be I when 5 is open $\Rightarrow$ $V_{AD}=4=I[R_{AB}+R_{BD}]$

Where $R_{AB}=4R_{BC}$ & $R_{BD}=\frac{R_{BC}}{2}\times 0.4=0.2R_{BC}$

$4=I[R_{AB}+\frac{0.2}{4}{R}_{AB}]\Rightarrow I=\frac{16}{4.2R_{AB}}$

When S is closed $\Rightarrow$ P.d across $5\Omega$ is $\Delta v=5i=\frac{4}{6}+5=\frac{10}{3}$v

Let new position of jockey be 'D' from A

$\Delta v=V_{A{D}^{\prime }}\Rightarrow \frac{10}{3}=I{R}_{A{D}^{\prime }}$

$\frac{10}{3}=\frac{16}{4.2R_{AB}}{R}_{A{D}^{\prime }}\Rightarrow R_{A{D}^{\prime }}=0.875R_{AB}$

$L_{A{D}^{\prime }}=1.75$m