A) \[12\,\Omega \]
B) \[4\,\Omega \]
C) \[6\,\Omega \]
D) \[8\,\Omega \]
Correct Answer: D
Solution :
The last two resistance are out of circuit. Now \[8\Omega \] is in parallel with \[(1+1+4+1+1)\Omega \]. \[\therefore R=8\Omega ||8\Omega =\frac{8}{2}=4\Omega \] Þ \[{{R}_{AB}}=4+2+2=8\Omega \]
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