question_answer

In a Wheatstones network P = 2\[\Omega \], Q = 2\[\Omega \], R = 2\[\Omega \] and S = 3\[\Omega \]. The resistance with which S is to be shunted in order that the bridge may be balanced is

A)  1\[\Omega \]                                   

B)  2\[\Omega \]

C)  4\[\Omega \]                                   

D)  6\[\Omega \]

E)  8\[\Omega \]

Correct Answer: D

Solution :

                Let a resistance r ohm be shunted with resistance S, so that the bridge is balanced. If S is the resultant resistance of S and r then In balanced position \[\frac{P}{Q}=\frac{R}{S}\] \[\frac{2}{2}=\frac{2}{S}\] \[\therefore \]  \[S=2\Omega \] Now      \[\frac{1}{S}=\frac{1}{S}+\frac{1}{r}\] or            \[\frac{1}{r}=\frac{1}{S}-\frac{1}{S}\]    \[=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}\] \[\frac{1}{r}=\frac{1}{6}\] \[r=6\,\Omega \]