Two resistances R1 and R2 give combined resistance of 4.5 ohms when in series and 1 ohm when in para

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Two resistances R1 and R2 give combined resistance of 4.5 ohms when in series and 1 ohm when in parallel. The resistances are

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1.5 ohms and 3 ohms Explanation: Given: Series resistance R_s = 4.5 ohms, Parallel resistance R_p = 1 ohm. Formula Used: $$R_s = R_1 + R_2$$, $$R_p = \frac{R_1 imes R_2}{R_1 + R_2}$$ Steps: Define the system of equations: From the definition of series and parallel resistance, we have $R_1 + R_2 = 4.5$ and $\frac{R_1 R_2}{R_1 + R_2} = 1$.. Simplify to find the product: Substitute $R_1 + R_2 = 4.5$ into the parallel equation: $R_1 R_2 = 1 imes 4.5 = 4.5$.. Form the quadratic equation: Using the properties of roots for a quadratic equation $x^2 - (sum)x + (product) = 0$, we get $x^2 - 4.5x + 4.5 = 0$. Multiply by 2 to clear decimals: $2x^2 - 9x + 9 = 0$.. Solve for resistances: Factor the quadratic equation $2x^2 - 6x - 3x + 9 = 0 \implies 2x(x - 3) - 3(x - 3) = 0 \implies (2x - 3)(x - 3) = 0$. This gives $x = 1.5$ and $x = 3$. Answer: C is correct because the values 1.5 ohms and 3 ohms satisfy both the series sum of 4.5 ohms and the parallel equivalent of 1 ohm. Exam Tip: This problem is a specific application of Vieta's formulas in quadratic equations; it is frequently used to verify stability conditions in filter circuits.

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