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Two resistors when connected in parallel give the resultant value of $2\,\Omega$, when connected in series the value becomes $9\,\Omega$.

Calculate the value of each resistance.

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closed as off-topic by John Rennie, DavePhD, JamalS, Kyle Kanos, Olin Lathrop May 21 at 13:41

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4  
Why should we? ${}$ –  Awesome May 21 at 11:59

1 Answer 1

If the resistors $R_1$ and $R_2$ are connected in parallel, the net resistance over the entire configuration is

$$\left( \frac{1}{R_1}+\frac{1}{R_2} \right)^{-1} = 2 \, \Omega$$

If they are connected in series, the resistance is simply $R_1 + R_2 = 9 \, \Omega$. By making a substition,

$$\frac{1}{R_1} + \frac{1}{9-R_1} = \frac{1}{2}$$

Combine the fractions, to yield,

$$\frac{(9-R_1)+R_1}{R_1(9-R_1)} = \frac{9}{9R_1 -R^2_1}= \frac{1}{2}$$

Multiply through by the denominator to obtain a quadratic equation,

$$R_1^2 - 9R_1 +18=0$$

By applying the quadratic formula, we obtain, $R_1 =3$ or $R_1=6$. If we employ the second relation originally provided in the question, we see the cases reduce to,

$$\boxed{R_1 = 3, \, \, R_2=6}$$

or vice versa.

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-1 for doing the lazy student's homework for him. –  Olin Lathrop May 21 at 13:42
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@OlinLathrop: Many come to the physics S.E. expecting their homework to be completed for them, and may hand it in as their own. By providing the answer, I am facilitating that, but as the post provided an explanation to the student, it may also mean he/she will now understand the required concepts and methods, which I think outweighs the harm done by allowing him/her to obtain a puny extra mark on their homework. In other words, sometimes people need to be spoonfed answers, before they can continue on their own. –  JamalS May 21 at 13:49

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