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 \[\frac{1}{R}=\frac{1}{R'}+\frac{1}{R'\,'}\] \[=\frac{1}{6}+\frac{1}{12}=\frac{3}{12}\] \[\Rightarrow \] \[R=4\,\Omega \].You need to login to perform this action.
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