question_answer

In the given figure the equivalent resistance between P and Q is :

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 \[\frac{1}{R}=\frac{1}{7}+\frac{1}{14}=\frac{14+7}{14\times 7}\]