Question
Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.
$\frac{1}{2}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$
From aboves equations we get that$\frac{1}{2}=\frac{(\text{R}_1+\text{R}_2)}{\text{R}_1\text{R}_2}$
$\frac{1}{2}=\frac{9}{(9-\text{R}_2)\text{R}_2}$
$9\text{R}_2-\text{R}_2{^2}=18$
$\text{R}_2{^2}-9\text{R}_2+18=0$
$(\text{R}_2-6)(\text{R}_2-3)=0$
$\text{R}_2=6,3$
So if R2 6 ohms, then R1 = 9 - 6 = 3 ohms. If R2 = 3 ohms, then R1 = 9 - 3 = 6 ohms.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
