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.

Answer

Two resistance when connected in series, resultant value is 9 ohms. Two resistance when connected in parallel, resultant values is 2 ohms. Let the two resistance be $R_1$ and $R_2$ If connected in series, then $9=R_1+R_2 R_1=9-R_2$ If connected in parallel, then $\frac{1}{2}=\frac{1}{R_1}+\frac{1}{R_2}$
From aboves equations we get that $\frac{1}{2}=\frac{\left( R _1+ R _2\right)}{ R _1 R _2}$
$\frac{1}{2}=\frac{9}{\left(9-R_2\right) R_2}$
$9 R_2-R_2^2=18$
$R_2^2-9 R_2+18=0$
$\left(R_2-6\right)\left(R_2-3\right)=0$
$R_2=6,3$
So if $R_2 6$ ohms, then $R_1=9-6=3$ ohms. If $R_2=3$ ohms, then $R_1=9-3=6$ ohms.

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