A) \[\frac{14}{9}\Omega \]
B) \[\frac{9}{14}\Omega \]
C) \[\frac{3}{14}\Omega \]
D) \[\frac{14}{3}\Omega \]
Correct Answer: D
Solution :
Key Idea: The given Wheatstone bridge is balanced. The ratio of resistances in opposite arms of the bridge is \[\frac{P}{Q}=\frac{3}{4}\] \[\frac{R}{S}=\frac{6}{8}=\frac{3}{4}\] \[\frac{P}{Q}=\frac{R}{S}=\frac{3}{4}\] Hence,\[7\,\,\Omega \]resistance is ineffective. The circuit now has \[3\,\,\Omega \] and \[4\,\,\Omega \]. resistances in series, and \[6\,\,\Omega \], and \[8\,\,\Omega \] in series, \[R=3\,\,\Omega +4\,\,\Omega =7\,\,\Omega \] The effective resistance \[7\,\,\Omega \] and \[14\,\,\Omega \] are now connected in parallel, hence \[\frac{1}{R}=\frac{1}{7}+\frac{1}{14}=\frac{14+7}{14\times 7}\]You need to login to perform this action.
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