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(B) Let the two resistances be $R_1$ and $R_2$.When connected in series,the equivalent resistance is $R_S = R_1 + R_2 = 16\,\Omega$.When connected in

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$4$ and $12\,\Omega$

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(B) Let the two resistances be $R_1$ and $R_2$.When connected in series,the equivalent resistance is $R_S = R_1 + R_2 = 16\,\Omega$.When connected in parallel,the equivalent resistance is $R_P = \frac{R_1 R_2}{R_1 + R_2} = 3\,\Omega$.From the series equation,$R_1 + R_2 = 16$.Substituting this into the parallel equation: $\frac{R_1 R_2}{16} = 3$,which gives $R_1 R_2 = 48$.We need two numbers that add to $16$ and multiply to $48$. These numbers are $4$ and $12$.Checking the options: $1$. Singly: $4\,\Omega$ and $12\,\Omega$ (given in the problem).$2$. Series: $4 + 12 = 16\,\Omega$.$3$. Parallel: $\frac{4 	imes 12}{4 + 12} = \frac{48}{16} = 3\,\Omega$.Thus,the resistances are $4\,\Omega$ and $12\,\Omega$.

By using only two resistance coils—singly,in series,or in parallel—one should be able to obtain resistances of $3$,$4$,$12$,and $16\,\Omega$. The separate resistances of the coils are:

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