A) \[8\,\Omega \]
B) \[12\,\Omega \]
C) \[\frac{3}{8}\Omega \]
D) \[\frac{8}{3}\Omega \]
Correct Answer: D
Solution :
Two resistances in series are connected parallel with the third. Hence \[\frac{1}{{{R}_{p}}}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\] Þ \[{{R}_{p}}=\frac{8}{3}\Omega \]
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