A) \[2\,\Omega \]
B) \[4\,\Omega \]
C) \[8\,\Omega \]
D) \[16\,\Omega \]
Correct Answer: B
Solution :
Key Idea: It is a balanced Wheatstone bridge. The given circuit can be redrawn as: The ratios of resistances in the arms are \[\frac{P}{Q}=\frac{4}{8}=\frac{1}{2}\] And \[\frac{2}{4}=\frac{1}{2}=\frac{R}{S}\] Hence, \[\frac{P}{Q}=\frac{R}{S}=\frac{1}{2}\] Therefore, bridge is balanced, so potential across\[C\]and \[D\] is same. Also \[4\,\Omega \] and \[8\,\Omega \] are connected in series and \[2\,\Omega \] and \[4\,\Omega \] are also in series. Hence, circuit reduces to as shown. The effective resistance is \[\frac{1}{R'}=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\] \[\Rightarrow \]\[R'=4\Omega \]You need to login to perform this action.
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