A) 3 and 4
B) 4 and 12
C) 12 and 16
D) 16 and 3
Correct Answer: B
Solution :
When resistances \[4\Omega \] and \[12\Omega \] are connected in series \[=4+12=16\Omega \] When these resistance are connected in parallel. \[\frac{1}{{{R}_{P}}}=\frac{1}{4}+\frac{1}{12}\,\,\,\Rightarrow \,\,\,{{R}_{P}}=\frac{4\times 12}{4+12}=\frac{4\times 12}{16}=3\Omega \]
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