A) \[4\Omega \]
B) \[2\Omega \]
C) \[1\,\Omega \]
D) \[8\,\Omega \]
Correct Answer: C
Solution :
Given that the resistance of the total wire is \[4\Omega \] Here, \[\,ABC\,(2\Omega )\]and A\[ADB(2\Omega )\] are in parallel. So, the resistance across any diameter is \[\Rightarrow \]\[\frac{1}{R}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1\]\[\Rightarrow \]\[R=1\Omega \]You need to login to perform this action.
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