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

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$4$ and $12$

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(B) Let the resistances of the two coils 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 $R_p = \frac{R_1 R_2}{R_1 + R_2} = 3 \, \Omega$.Substituting $R_1 + R_2 = 16$ into the parallel formula: $\frac{R_1 R_2}{16} = 3$, which gives $R_1 R_2 = 48$.We have a system of equations: $R_1 + R_2 = 16$ and $R_1 R_2 = 48$.Substituting $R_2 = 16 - R_1$ into the product equation: $R_1(16 - R_1) = 48$.$16 R_1 - R_1^2 = 48 \implies R_1^2 - 16 R_1 + 48 = 0$.Factoring the quadratic equation: $(R_1 - 4)(R_1 - 12) = 0$.Thus, the resistances are $4 \, \Omega$ and $12 \, \Omega$. When used singly, we get $4 \, \Omega$ and $12 \, \Omega$. When in series, we get $4 + 12 = 16 \, \Omega$. When in parallel, we get $\frac{4 	imes 12}{4 + 12} = \frac{48}{16} = 3 \, \Omega$. All given values match.

$A$ technician has only two resistance coils. By using them singly, in series or in parallel, he is able to obtain the resistances $3, 4, 12$ and $16 \, \Omega$. The resistances of the two coils are ........... $\Omega$.

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