Question

Length of potentiometer wire is $L$. In primary circuit battery of $2.5V$ and $10ft$ resistance is combined in series. Balance length is $\frac{L}{2}$ for emf $1.0\Omega$. If the resistance of primary circuit becomes double then what is the new balance length?

Answer 1

Length of wire $=LPotential\left(V\right)=2.5\left(V\right)$
Let a resistance of $R_1$ ohm is placed in series with battery, then current
$i=\frac{E}{R+R_1}$
Given : $E=2.5V,R=10\omega$
Let due to current $i$ the potential gradient
$k=ip....\left(2\right)$
but $\rho =\frac{R}{L}=\frac{10}{L}....\left(3\right)$
$k=\frac{2.55}{10+R_1}\times \frac{10}{L}....\left(4\right)$
Cell of $1$ volt is balanced at $L/2$ length.
therefore $V=k{i}^{\prime }$
$1=k\frac{L}{2}$
$\therefore k=\frac{2}{L}....\left(5\right)$
From equation (4) and (5)
$\frac{2}{L}=\left(\frac{2.5}{10+R_1}\right)\times \frac{10}{L}$
$2=\frac{25}{10+R_1}$
$R_1=2.5\Omega$
When we double the resistance $R_1$, connected in series with potentiometer then total resistance $\left(R^{\prime }\right)=R+2R_1$
$R^{\prime }=10+5=15\Omega$
In this condition if current in potentiometer is $i$, then
$i=\frac{E}{R^{\prime }}=\frac{2.5}{15}=\frac{1}{6}amp$
If new gradient is $k^{\prime }$
$k^{\prime }=i\times \rho =\frac{1}{6}\times \frac{10}{L}=\frac{5}{3L}$
If cell of $1$ volt give null point at distance $L$
$V=L{k}^{\prime }$
$L=\frac{V}{k^{\prime }}=\frac{1\times 3L}{5}=0.6Lm$.