A) 3 and 4
B) 4 and 12
C) 12 and 16
D) 16 and 3
Correct Answer: D
Solution :
If we take \[{{R}_{1}}=4\,\,\Omega \,{{R}_{2}}=12\,\,\Omega \], then in series resistance \[R={{R}_{1}}+{{R}_{2}}\] \[=4+12\] \[=16\,\,\Omega \] In parallel, resistance \[R=\frac{4\times 12}{4+12}=3\,\,\Omega \] So, \[{{R}_{1}}=4\,\Omega \] and \[{{R}_{2}}=12\,\,\Omega \]
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