Question

In a meter bridge, the balance point is found at a distance $l_1$ with resistances R and S as shown in the figure.
An unknown resistance X is now connected in parallel to the resistance S and the balance point is found at a distance $l_2$. Obtain a formula for X in terms of $l_1,l_2$ and S.

Answer 1

Let the length of wire $AB$ be $L$

now,

When $x$ is not connected

$\frac{R}{l_1}=\frac{S}{L-l}$ [balancing condition]

$\Rightarrow L-l_1=\frac{l_{1}S}{R}$

$\Rightarrow L=\frac{l_{1}S}{R},l_1=\frac{l_{1}S+l_{1}R}{R}$

$\Rightarrow L=\frac{l_1(S+R)}{R}.....\left(1\right)$

Now

When the resistance $x$ is connected in parallel to $S$

then effective resistance of $S$ and $X$ will be given by:

$\frac{1}{Reff}=\frac{1}{S}+\frac{1}{X}.....\left(ii\right)$

where,

$Reff=$ effective resistance of $S$ and $X$

Now balancing condition for meter bridge

$\frac{R}{l_2}=\frac{Reff}{L-l_2}$

$\frac{l_2}{R}=\frac{L-l_2}{Reff}$

Putting value of $\left(\frac{1}{Reff}\right)$ from eqn $\left(ii\right)$

$\frac{l_2}{R}=(L-l_2)[\frac{1}{5}+\frac{1}{X}]$

$\Rightarrow \frac{1}{5}+\frac{1}{X}=\frac{l_2}{R[\frac{l_1(S+R)}{R}-l_2]}$

$\frac{1}{S}+\frac{1}{X}=\frac{l_2}{l_1(S+R)-l_{2}R}$

$\frac{1}{S}+\frac{1}{X}=\frac{l_2}{l_{1}S+l_{1}R-l_{2}R}$

$\Rightarrow \frac{1}{X}=\frac{l_2}{l_{1}S+R(l_1-l_2)}-\frac{1}{S}$

$\frac{1}{X}=\frac{S{l}_2-[l_{1}S+R(l_1-l_2\left)\right]}{S[l_{1}S+R(l_1-l_2\left)\right]}$

$\frac{1}{X}=\frac{S{l}_2-S{l}_1-R(l_1-l_2)}{l_1{s}^2+RS(l_1-l_2)}$

$\frac{1}{X}=\frac{(l_2-l_1)(S+R)}{l_1{s}^2+RS(l_1-l_2)}$

$\Rightarrow X=\frac{l_1{S}^2+RS(l_1-l_2)}{(l_2-l_1)(S+R)}$