Question

The resistance of $10m$ potentiometer wire is $1\Omega /m$. Battery accumulator of $2.2V$ and negligible internal resistance and high resistance connecting in series with the wire. For achieving $2.2mV/m$ potential gradient how much high resistance will have to take?

Answer 1

Length of potentiometer wire
$L_{AB}=1m$
Density of resistance of wire $=1\Omega /m$
Resistance of potentiometer
$R_{AB}=10\times \frac{1\Omega }{m}\times m=10\Omega$
emf of cell $\left(E\right)=2.2Volt$
Internal resistance $\left(r\right)=0$
Potential gradient $\left(k\right)=\frac{2.2mV}{10m}=2.2\times {10}^{-3}Vc{m}^{-1}$
$k=\frac{V_{AB}}{L_{AB}}=\frac{I_{AB}\times R_{AB}}{L_{AB}}$
$2.2\times {10}^{-3}=\frac{2.2}{R_{equi}}\times \frac{10}{10}[\because I_{AB}=\frac{2.2}{R_{equi}}]$
$R_{equi}=\frac{2.2}{2.2\times {10}^{-3}}=1000\Omega$
$\underset{resistance}{R_{equi}}=\underset{1000=R+10\Rightarrow R=990\Omega }{unknownresistance+potentiometer}$.