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.You need to login to perform this action.
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