When the two known resistance ' R ' and ' S ' are connected in the left and right gaps of a meter bridge respectively, the null point is found at a distance ' l 1 ' fro

Question

When the two known resistance ' R ' and ' S ' are connected in the left and right gaps of a meter bridge respectively, the null point is found at a distance ' l 1 ​ ' from the zero end of a meter bridge wire. An unknown resistance ' X ' is now connected in parallel with ' S ' and null point is found at a distance ' l 2 ​ ' form zero end of meter bridge wire. The unknown resistance ' X ' is

TestHub · Accepted Answer

For the initial condition of the meter bridge:

$\frac{R}{S} = \frac{l _{1}}{100 - l _{1}}$

When unknown resistance $X$ is connected in parallel with $S$ , the equivalent resistance in the right gap becomes $S^{'} = \frac{SX}{S + X}$ .

For the new null point $l_{2}$ :

$\frac{R}{S ^{'}} = \frac{l _{2}}{100 - l _{2}}$

Substituting $S^{'}$ :

$\frac{R ( S + X )}{SX} = \frac{l _{2}}{100 - l _{2}}$

Dividing the second equation by the first equation:

$\frac{S + X}{X} = \frac{l _{2} ( 100 - l _{1} )}{l _{1} ( 100 - l _{2} )}$

$\frac{S}{X} + 1 = \frac{l _{2} ( 100 - l _{1} )}{l _{1} ( 100 - l _{2} )}$

$\frac{S}{X} = \frac{l _{2} ( 100 - l _{1} ) - l _{1} ( 100 - l _{2} )}{l _{1} ( 100 - l _{2} )}$

$\frac{S}{X} = \frac{100 l _{2} - l _{1} l _{2} - 100 l _{1} + l _{1} l _{2}}{l _{1} ( 100 - l _{2} )}$

$\frac{S}{X} = \frac{100 ( l _{2} - l _{1} )}{l _{1} ( 100 - l _{2} )}$

$X = \frac{S l _{1} ( 100 - l _{2} )}{100 ( l _{2} - l _{1} )}$

Answer: $\frac{S l _{1} ( 100 - l _{2} )}{100 ( l _{2} - l _{1} )}$

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