A) \[4\Omega \] and \[16\,\Omega \]
B) \[4\,\Omega \] and \[12\,\Omega \]
C) \[7\,\Omega \] and \[9\,\Omega \]
D) \[6\,\Omega \] and \[10\,\Omega \]
Correct Answer: B
Solution :
Let \[{{\text{R}}_{1}}=\text{4}\,\,\Omega \text{ and }{{\text{R}}_{2}}=\text{12}\,\Omega \] In series, \[\text{R}={{\text{R}}_{1}}+{{\text{R}}_{2}}=\text{4}+\text{12}=\text{16 }\Omega \] In parallel, \[\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}=\frac{1}{4}+\frac{1}{12}\] \[=\frac{3+1}{12}=\frac{4}{12}=\frac{1}{3},\]\[\therefore \]\[3\,\Omega \] We can get the resistances of \[3\,\,\Omega ,4\,\Omega ,12\,\Omega \], and\[16\,\Omega \] only with \[4\,\Omega \] and \[12\,\Omega \]
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