question_answer

In the given figure equivalent resistance between P and Q will be :

A) \[\frac{14}{9}\Omega \]                               

B)  \[\frac{9}{14}\Omega \]

C) \[\frac{14}{3}\Omega \]                               

D) \[\frac{3}{14}\Omega \]

Correct Answer: C

Solution :

 Key Idea: The circuit forms a balanced Wheatstone's bridge. The ratio of resistances in the opposite arms are \[\frac{3\Omega }{4\Omega }\]and \[\frac{6\Omega }{8\Omega }=\frac{3\Omega }{4\Omega }\]                 Since, ratio of resistances is same, the combination forms a balanced Wheatstone's bridge. Hence, resistance of \[7\Omega \]becomes  ineffective. Equivalent resistance in upper and lower arms are                 \[{{R}_{U}}=3+4=7\Omega \] \[{{R}_{L}}=6+8=14\Omega \] The resistances \[7\Omega \] and \[14\Omega \] are in parallel hence, equivalent resistance is \[\frac{1}{{{R}_{eq}}}=\frac{1}{{{R}_{U}}}\div \frac{1}{{{R}_{L}}}=\frac{1}{7}+\frac{1}{14}=\frac{21}{98}\] \[\Rightarrow \]         \[{{R}_{eq}}=\frac{14}{3}\Omega \] NOTE: In balanced Wheatstone's bridge. If resistance in middle arm is replaced by a galvanometer (G) then there is no effect on  the circuit.