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(D) To obtain an equivalent resistance of $\frac{46}{3}\, \Omega$,we test the given combinations.Let us evaluate option $D$: $2\, \Omega$ and $4\, \Om

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$2\, \Omega$ and $4\, \Omega$ are in parallel with $6\, \Omega$ and $8\, \Omega$ in series

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(D) To obtain an equivalent resistance of $\frac{46}{3}\, \Omega$,we test the given combinations.Let us evaluate option $D$: $2\, \Omega$ and $4\, \Omega$ are in parallel,and this combination is in series with $6\, \Omega$ and $8\, \Omega$.First,calculate the equivalent resistance of $2\, \Omega$ and $4\, \Omega$ in parallel $(R_p)$:$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} = \frac{2+1}{4} = \frac{3}{4}$$R_p = \frac{4}{3}\, \Omega$Now,add this in series with $6\, \Omega$ and $8\, \Omega$:$R_{eq} = R_p + 6 + 8 = \frac{4}{3} + 14 = \frac{4 + 42}{3} = \frac{46}{3}\, \Omega$This matches the required equivalent resistance.

If you are provided a set of resistances $2\, \Omega, 4\, \Omega, 6\, \Omega$ and $8\, \Omega$. Connect these resistances so as to obtain an equivalent resistance of $\frac{46}{3}\, \Omega$.

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