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
$\frac{1}{2}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$
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 $R_2= 6$ ohms, then $R_1 = 9 - 6 = 3 $ ohms.
If $R_2 = $3 ohms, then $R_1 = 9 - 3 = 6$ 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
$\frac{1}{2}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$
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 $R_2= 6$ ohms, then $R_1 = 9 - 6 = 3 $ ohms.
If $R_2 = $3 ohms, then $R_1 = 9 - 3 = 6$ ohms.
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