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(B) Let the two resistors be $R_1$ and $R_2$.For series connection: $R_1 + R_2 = 9 \ \Omega$  ---$(1)$For parallel connection: $\frac{R_1 R_2}{R_1 + R

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$3 \ \Omega$ and $6 \ \Omega$

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(B) Let the two resistors be $R_1$ and $R_2$.For series connection: $R_1 + R_2 = 9 \ \Omega$  ---$(1)$For parallel connection: $\frac{R_1 R_2}{R_1 + R_2} = 2 \ \Omega$  ---$(2)$Substituting $(1)$ into $(2)$: $\frac{R_1 R_2}{9} = 2 \implies R_1 R_2 = 18 \ \Omega^2$.We know that $(R_1 - R_2)^2 = (R_1 + R_2)^2 - 4R_1 R_2$.$(R_1 - R_2)^2 = (9)^2 - 4(18) = 81 - 72 = 9$.Therefore,$R_1 - R_2 = 3 \ \Omega$  ---$(3)$.Adding $(1)$ and $(3)$: $2R_1 = 12 \implies R_1 = 6 \ \Omega$.Subtracting $(3)$ from $(1)$: $2R_2 = 6 \implies R_2 = 3 \ \Omega$.Thus,the resistances are $6 \ \Omega$ and $3 \ \Omega$.

Two resistors are connected $(a)$ in series $(b)$ in parallel. The equivalent resistance in the two cases are $9 \ \Omega$ and $2 \ \Omega$ respectively. Then the resistances of the component resistors are

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