Two conductors

Two conductors have resistance Rs=5 ohms when connected in series and Rp=1.2 ohms when connected in parallel. Calculate the value of the resistors.

Correct answer:

R₁ =  3 Ω
R₂ =  2 Ω

Step-by-step explanation:

$$\begin{gathered} S=5\Omega \\ P=1.2\Omega \\ {R}_1+R_2=S \\ 1/R_1+1/R_2=1/P \\ {R}_2+R_1=R_1\cdot R_2/P \\ (S-R_1)+R_1=R_1\cdot (S-R_1)/P \\ (5-R_1)+R_1=R_1\cdot (5-R_1)/1.2 \\ 0.83333333333333R^2-4.167R+5=0 \\ a=0.833333;b=-4.167;c=5 \\ D=b^2-4ac=4.167^2-4\cdot 0.833333\cdot 5=0.6944444444 \\ D>0 \\ {R}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{4.17\pm \sqrt{0.69}}{1.666667} \\ {R}_{1,2}=2.5\pm 0.5 \\ {R}_1=3=3\Omega \\ {R}_2=2 \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} \text{ Verifying Solution: } \\ {S}_2=R_1+R_2=3+2=5\Omega \\ {P}_2=\frac{1}{1/R_1+1/R_2}=\frac{1}{1/3+1/2}=\frac{6}{5}=1.2\Omega \\ {R}_2=2=2\Omega \end{gathered}$$

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