question_answer

The equivalent resistance between points \[0.98\text{ gauss},\text{ }0.9\sqrt{2}\text{ gauss}\]and\[B\] is :                                                     [BHU PMT-2002]                

A)                  \[1.07\text{ gauss},\text{ 0}\text{.11}\sqrt{2}\text{ gauss}\]                                     

B)                  \[\text{4}000\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]

C)                  \[\text{2 eV}\]                                

D)                  \[\text{0}\text{.5}\,\text{eV}\]

Correct Answer: A

Solution :

                 Key Idea: The given Wheatstone bridge is balanced.                 The ratio of resistances in the arms of the bridge are                                                 \[\frac{P}{Q}=\frac{2}{4}=\frac{1}{2}\]                                   ?(1)                                 \[\frac{R}{S}=\frac{4}{8}=\frac{1}{2}\]                                    ?(2)                 From Esq. (1) and (2), we observe that since, ratio of resistances is equal, hence bridge is balanced and no current flows through arm\[CD\]. The circuit now reduces to as shown. The \[2\,\Omega \] and \[4\,\Omega \] resistances are in series and \[4\,\Omega \] and \[8\,\Omega \] are also in series.                                 \[\therefore \]                  \[R'=2+4=6\,\Omega \]                                                 \[R''=4+8=12\,\Omega \] The resistances of \[6\,\Omega \] and \[12\,\Omega \] are in parallel, therefore equivalent resistance is                                 \[\frac{1}{R}=\frac{1}{R'}+\frac{1}{R'\,'}\]                 \[=\frac{1}{6}+\frac{1}{12}=\frac{3}{12}\]                 \[\Rightarrow \]                               \[R=4\,\Omega \].